2025 Cambridge IAS-Level Physics (9702) 模擬試題連答案詳解

When the wire is fully loaded at 120 N, the elastic strain energy stored in the wire is equal to the energy that can be recovered during unloading, which is the area under the unloading curve: \(E_{\text{strain}} = 0.15\text{ J}\). The total work done during loading is \(0.32\text{ J}\). The energy dissipated as heat is the difference between the total work done during loading and the recovered work during unloading: \(E_{\text{dissipated}} = 0.32\text{ J} - 0.15\text{ J} = 0.17\text{ J}\).

1 mark: Correctly identifies that the elastic strain energy is the recovered work (0.15 J) and the dissipated thermal energy is the net work done during the cycle (0.17 J).

Stress is given by \(\sigma = F/A\). Since wire Y has twice the diameter of wire X, its cross-sectional area is four times larger (\(A_Y = 4 A_X\)). Under the same load \(F\), the stress in Y is \(\sigma_Y = F / (4 A_X) = 0.25 \sigma_X\). Both wires are made of the same metal, so they share the same Young modulus \(E\). Strain is given by \(\varepsilon = \sigma / E\). Therefore, the strain in Y is \(\varepsilon_Y = \sigma_Y / E = 0.25 \sigma_X / E = 0.25 \varepsilon_X\).

1 mark: Deduces stress is 0.25 of X because area is 4 times larger, and strain is 0.25 of X because Young's modulus is constant.

Comparing the given equation to \(x = x_0 \cos(\omega t)\), the amplitude is \(x_0 = 0.040\text{ m}\) and the angular frequency is \(\omega = 5.0 \pi\text{ rad s}^{-1}\). The maximum velocity of the particle is \(v_0 = \omega x_0 = (5.0 \pi) \times 0.040 = 0.20 \pi\text{ m s}^{-1}\). The maximum kinetic energy is \(E_{\text{k,max}} = \frac{1}{2} m v_0^2 = \frac{1}{2} \times 0.15 \times (0.20 \pi)^2 = 0.003 \pi^2 \approx 0.0296\text{ J} \approx 0.030\text{ J}\).

1 mark: Correctly identifies amplitude and angular frequency, calculates maximum velocity, and computes the maximum kinetic energy to 2 significant figures.

Using the potential divider formula, \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\). In bright light: \(V_{\text{out, bright}} = 9.0 \times \frac{300}{1200 + 300} = 1.8\text{ V}\). In darkness: \(V_{\text{out, dark}} = 9.0 \times \frac{6000}{1200 + 6000} = 7.5\text{ V}\). The change in the output voltage is \(\Delta V_{\text{out}} = 7.5\text{ V} - 1.8\text{ V} = 5.7\text{ V}\).

1 mark: Calculates the output voltages under both conditions and takes the difference to find the correct change in voltage.

The total useful work done per second (power output) is the sum of the rate of potential energy increase and the rate of kinetic energy increase: \(P_{\text{out}} = \frac{\Delta E_{\text{p}}}{\Delta t} + \frac{\Delta E_{\text{k}}}{\Delta t} = \left(\frac{m}{t}\right) g h + \frac{1}{2} \left(\frac{m}{t}\right) v^2 = (8.0 \times 9.81 \times 15.0) + (0.5 \times 8.0 \times 2.0^2) = 1177.2 + 16.0 = 1193.2\text{ W}\). The power input is \(1.6\text{ kW} = 1600\text{ W}\). Therefore, efficiency \(\eta = \frac{1193.2}{1600} \times 100\% \approx 74.6\% \approx 75\%\).

1 mark: Recognizes that total useful power output includes both potential and kinetic energy rate terms, then calculates efficiency.

Using the grating equation: \(d \sin \theta = n \lambda\). For \(n = 2\), \(d \sin(24.0^\circ) = 2 \times (540 \times 10^{-9})\), which gives grating spacing \(d \approx 2.655 \times 10^{-6}\text{ m}\). For the maximum possible order, \(\sin \theta \le 1\), which requires \(n_{\text{max}} \le d/\lambda = 2.655 \times 10^{-6} / 540 \times 10^{-9} \approx 4.92\). Since \(n\) must be an integer, \(n_{\text{max}} = 4\). The total number of maxima is \(2 n_{\text{max}} + 1 = 2(4) + 1 = 9\) (including the central maximum and the orders on both sides).

1 mark: Computes the grating spacing, determines the maximum possible integer order, and finds the total number of maxima as twice the maximum order plus one.

At terminal velocity \(v_{\text{t}}\), the upward drag force equals the downward weight: \(k v_{\text{t}}^2 = m g\). When \(v = 0.5 v_{\text{t}}\), the drag force is \(F = k (0.5 v_{\text{t}})^2 = 0.25 k v_{\text{t}}^2 = 0.25 m g\). The net downward force on the sky-diver is \(F_{\text{net}} = m g - 0.25 m g = 0.75 m g\). From Newton's second law, \(F_{\text{net}} = m a\), so the acceleration is \(a = 0.75 g\).

1 mark: Equates weight and drag at terminal speed, determines drag at half speed is 0.25mg, and uses Newton's second law to find a = 0.75g.

When damping is increased: (1) The amplitude of the oscillations at all frequencies decreases, causing the peak of the resonance curve to become broader and flatter. (2) The peak frequency (the frequency at which resonance occurs) shifts to a slightly lower frequency. Therefore, the peak becomes broader and shifts to a lower frequency.

1 mark: Correctly identifies that increased damping decreases the peak amplitude (broader peak) and decreases the resonant frequency.

The Young modulus \(E\) is given by:

\[E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{\Delta L / L} = \frac{F L}{A \Delta L}\]

Since the cross-sectional area \(A = \frac{\pi d^2}{4}\), we can write:

\[E = \frac{4 F L}{\pi d^2 \Delta L}\]

Thus, the ratio of the Young moduli is:

\[\frac{E_X}{E_Y} = \left(\frac{F_X}{F_Y}\right) \left(\frac{L_X}{L_Y}\right) \left(\frac{d_Y}{d_X}\right)^2 \left(\frac{\Delta L_Y}{\Delta L_X}\right)\]

We are given:
- \(F_X = F_Y\)
- \(L_X = 3 L_Y\)
- \(d_X = 0.5 d_Y \implies \frac{d_Y}{d_X} = 2\)
- \(\Delta L_X = 4 \Delta L_Y\)

Substituting these values in:

\[\frac{E_X}{E_Y} = 1 \times 3 \times (2)^2 \times \frac{1}{4} = 3.0\]

Award 1 mark for the correct option (D).
- Method: Recall the formula for the Young modulus and express it in terms of diameter and length.
- Accuracy: Calculate the ratio correctly to obtain 3.0.

Within the limit of proportionality, the force-extension relationship is linear: \(F = k x\).

The work done to extend the wire from \(x_1\) to \(x_2\) is equal to the area under the force-extension graph from \(x_1\) to \(x_2\), which forms a trapezium:

\[\text{Work Done} = \frac{1}{2} (F_1 + F_2)(x_2 - x_1)\]

Expanding this expression:

\[\text{Work Done} = \frac{1}{2}(F_1 x_2 - F_1 x_1 + F_2 x_2 - F_2 x_1)\]

Since the material behaves elastically, \(F_1 = k x_1\) and \(F_2 = k x_2\), meaning:

\[F_1 x_2 = (k x_1) x_2 = k x_1 x_2\]
\[F_2 x_1 = (k x_2) x_1 = k x_1 x_2\]

Therefore, \(F_1 x_2 = F_2 x_1\).

Substituting this back into the expanded expression simplifies the middle terms:

\[\text{Work Done} = \frac{1}{2}(F_2 x_2 - F_1 x_1)\]

Award 1 mark for the correct option (B).
- Method: Use the area of a trapezium under a linear F-x graph, then apply Hooke's law to simplify the expression.

Since the displacement is at its positive maximum at \(t = 0\), we use the cosine function to describe the displacement:

\[x = A \cos(\omega t) = A \cos\left(\frac{2\pi}{T} t\right)\]

Substituting \(t = \frac{5}{6}T\):

\[x = A \cos\left(\frac{2\pi}{T} \times \frac{5}{6}T\right) = A \cos\left(\frac{5\pi}{3}\right)\]

Evaluating the cosine term in radians (or degrees where \(\frac{5\pi}{3}\text{ rad} = 300^\circ\)):

\[\cos\left(\frac{5\pi}{3}\right) = +0.50\]

Thus, the displacement is:

\[x = +0.50A\]

Award 1 mark for the correct option (C).
- Method: Express SHM displacement as a cosine function and substitute the correct time fraction.
- Accuracy: Correctly evaluate the trigonometric value including its sign.

As damping is increased:
1. The peak amplitude of the forced oscillation decreases (becomes flatter/broader).
2. The frequency at which resonance occurs (the resonant frequency) shifts slightly to a lower frequency.

Therefore, in water, the peak amplitude is lower and occurs at a slightly lower frequency.

Award 1 mark for the correct option (D).
- Method: Recall the effects of damping on both the peak amplitude and the resonant frequency in forced oscillations.

1. **Terminal p.d. (\(V\))**:
The current in the circuit is given by \(I = \frac{E}{R + r}\). When \(R\) decreases, the total resistance of the circuit decreases, so current \(I\) increases.
The terminal potential difference is \(V = E - Ir\). Since \(I\) has increased, the internal 'lost volts' \(Ir\) increases, which means \(V\) decreases.

2. **Power dissipated in internal resistance (\(P_r\))**:
The power dissipated internally is \(P_r = I^2 r\). Since the current \(I\) has increased and \(r\) is constant, \(P_r\) must increase.

Award 1 mark for the correct option (B).
- Method: Analyze circuit current change due to resistance change, then determine the effect on both terminal voltage and internal power dissipation.

Because the car travels at a constant speed, the net force along the slope is zero.
The forward force \(F\) required of the motor must equal the sum of the resistive force and the component of the gravitational force acting down the slope:

\[F = F_{\text{resistive}} + m g \sin\theta\]
\[F = 400 + \left(1200 \times 9.81 \times \sin(5.0^\circ)\right)\]
\[F \approx 400 + 1026 = 1426\text{ N}\]

The power output of the motor is:

\[P = F v = 1426\text{ N} \times 15\text{ m s}^{-1} \approx 21390\text{ W} = 21\text{ kW}\]

Award 1 mark for the correct option (C).
- Method: Resolve gravitational force parallel to the slope and sum with resistive force. Use \(P = Fv\) to calculate power.
- Accuracy: Round to two significant figures to get \(21\text{ kW}\).

Using the diffraction grating equation \(d \sin\theta = n\lambda\):
For \(n = 3\):

\[d \sin(36^\circ) = 3 \lambda \implies \frac{\lambda}{d} = \frac{\sin(36^\circ)}{3} \approx 0.196\]

To find the maximum possible order of diffraction \(n_{\max}\), we set \(\sin\theta \le 1\):

\[n_{\max} \le \frac{d}{\lambda} = \frac{1}{0.196} \approx 5.1\]

Since \(n\) must be an integer, the highest observable order is \(n_{\max} = 5\).

The total number of maxima observed includes the central maximum (zero-order) and both the positive and negative orders:

\[\text{Total maxima} = 2 \times n_{\max} + 1 = 2(5) + 1 = 11\]

Award 1 mark for the correct option (C).
- Method: Find the ratio of \(\lambda / d\) using the given values, then calculate the highest possible order under the physical limit \(\sin\theta \le 1\). Determine the total maxima as \(2n+1\).

When the ball is released, it is at rest, so the drag force is zero. The initial acceleration is at its maximum, driven by gravity net of upthrust.

As the ball's speed increases, the viscous drag force increases. This drag acts in the opposite direction to motion, reducing the net downward force and therefore reducing the acceleration.

As acceleration drops, velocity increases at a slower rate, which means the drag increases more slowly as well. Thus, the acceleration decreases asymptotically towards zero as the ball approaches its terminal velocity.

Award 1 mark for the correct option (C).
- Method: Qualitatively relate speed, drag force, net force, and acceleration for an object falling in a viscous medium.

Young modulus \(E = \frac{\text{stress}}{\text{strain}}\), so \(\text{strain} = \frac{\text{stress}}{E} = \frac{F}{A E}\). Since both wires are made of the same metal, they have the same Young modulus \(E\). Since they support equal loads, \(F\) is the same. Thus, the strain is inversely proportional to the cross-sectional area \(A\). Since \(A = \frac{\pi d^2}{4}\), strain is inversely proportional to the square of the diameter \(d\), and is independent of length \(L\). Therefore, \(\frac{\text{strain}_X}{\text{strain}_Y} = \frac{d_Y^2}{d_X^2} = \frac{(2d)^2}{d^2} = 4\).

1 mark: Correctly identifies that strain depends on cross-sectional area and is independent of original length, then calculates the ratio of the strains to be 4.

The Young modulus is given by \(E = \frac{F L_0}{A \Delta L}\), which gives the extension \(\Delta L = \frac{F L_0}{A E}\). The elastic potential energy stored in the wire is \(E_p = \frac{1}{2} F \Delta L = \frac{F^2 L_0}{2 A E}\). Substituting the given values: \(E_p = \frac{(120)^2 \times 2.0}{2 \times (1.5 \times 10^{-6}) \times (2.0 \times 10^{11})} = \frac{28800}{6.0 \times 10^5} = 0.048\text{ J}\).

1 mark: Correctly calculates extension or uses the combined formula to find the stored elastic potential energy as 0.048 J.

The equation of motion for the falling skydiver is \(m g - F = m a\), where \(F = k v^2\). Rearranging for acceleration gives \(a = g - \frac{k v^2}{m}\). Substituting the values: \(a = 9.81 - \frac{0.32 \times 25^2}{80} = 9.81 - \frac{200}{80} = 9.81 - 2.50 = 7.31\text{ m s}^{-2}\).

1 mark: Calculates the resistive force at the given speed, applies Newton's second law to find net force, and determines acceleration to be 7.31 m s^-2.

Using the grating equation \(d \sin \theta = n \lambda\), we have \(d = \frac{n \lambda}{\sin \theta}\). For \(n = 3\), \(\lambda = 633 \times 10^{-9}\text{ m}\), and \(\theta = 48.0^\circ\): \(d = \frac{3 \times 633 \times 10^{-9}}{\sin 48.0^\circ} \approx 2.555 \times 10^{-6}\text{ m}\). The number of lines per millimetre is \(N = \frac{10^{-3}\text{ m}}{d} = \frac{10^{-3}}{2.555 \times 10^{-6}} \approx 391\text{ lines mm}^{-1}\).

1 mark: Correctly applies the diffraction grating formula to find slit spacing, then calculates the number of lines per millimetre to be 391.

Using the equation \(V = E - I r = E \frac{R}{R + r}\): for \(R = 6.0\ \Omega\), \(4.5 = E \frac{6.0}{6.0 + r} \implies E = 0.75(6.0 + r)\). For \(R = 12.0\ \Omega\), \(5.4 = E \frac{12.0}{12.0 + r} \implies E = 0.45(12.0 + r)\). Equating the expressions: \(0.75(6.0 + r) = 0.45(12.0 + r) \implies 4.5 + 0.75 r = 5.4 + 0.45 r \implies 0.30 r = 0.9 \implies r = 3.0\ \Omega\). Substituting back gives \(E = 0.75(6.0 + 3.0) = 6.75\text{ V}\).

1 mark: Formulates two equations using the potential divider relation with internal resistance, solves them simultaneously to find E = 6.75 V and r = 3.0 \Omega.

The useful power output is \(P_{\text{out}} = F v = m g v = 120 \times 9.81 \times 0.50 = 588.6\text{ W}\). Since the efficiency is \(60\%\), \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{588.6}{0.60} = 981\text{ W}\).

1 mark: Calculates the useful output power and divides by the efficiency to find the input electrical power of 981 W.

The total energy is \(E_{\text{total}} = E_k + E_p = 3 E_p + E_p = 4 E_p\). The total energy is given by \(E_{\text{total}} = \frac{1}{2} k x_0^2\) and the potential energy by \(E_p = \frac{1}{2} k x^2\), where \(x_0 = 4.0\text{ cm}\) is the amplitude and \(x\) is the displacement. Therefore, \(\frac{1}{2} k x_0^2 = 4 \left(\frac{1}{2} k x^2\right) \implies x_0^2 = 4 x^2 \implies x = \frac{x_0}{2} = \frac{4.0\text{ cm}}{2} = 2.0\text{ cm}\).

1 mark: Correctly relates total energy to potential energy and uses the energy formulas to find the displacement of 2.0 cm.

For a system undergoing forced oscillations, increasing the damping decreases the maximum amplitude at resonance (the peak of the resonance curve becomes lower and broader). It also slightly decreases the resonant frequency (the peak shifts to a slightly lower frequency).

1 mark: Recognises that higher damping lowers both the peak amplitude and the resonant frequency.

Tensile strain is defined as \(\epsilon = \frac{\text{stress}}{E} = \frac{F}{A E}\). Since both wires are made of the same metal, their Young’s modulus \(E\) is identical. The force \(F\) applied to both is also the same. Therefore, the strain is inversely proportional to the cross-sectional area: \(\epsilon \propto \frac{1}{A} \propto \frac{1}{d^2}\), where \(d\) is the wire diameter. Since wire \(P\) has half the diameter of wire \(Q\) (\(d_P = 0.5 d_Q\)), its cross-sectional area is a quarter of wire \(Q\)'s area (\(A_P = 0.25 A_Q\)). Hence, \(\frac{\epsilon_P}{\epsilon_Q} = \frac{A_Q}{A_P} = \frac{1}{0.25} = 4\). The original length of the wires does not affect their strain under these conditions.

1 mark for selecting the correct ratio C, derived from the inverse-square relationship between strain and diameter for a constant force and Young's modulus.

The work done in stretching a wire is given by \(W = \frac{1}{2} F x\). Using Hooke's law, \(F = \frac{E A x}{L}\), where \(E\) is the Young's modulus. Substituting \(F\) into the work done formula yields \(W = \frac{E A x^2}{2 L}\). For the second wire, the area is \(2A\), the original length is \(3L\), and the extension is \(x\). Thus, the work done on the second wire is: \(W_2 = \frac{E (2A) x^2}{2 (3L)} = \frac{2}{3} \left(\frac{E A x^2}{2 L}\right) = \frac{2}{3} W\).

1 mark for the correct option A, determined by expressing the work done in terms of Young's modulus, area, length, and extension.

The speed \(v\) of a particle in simple harmonic motion is given by \(v = \omega \sqrt{x_0^2 - x^2}\). The maximum speed is \(v_{\text{max}} = \omega x_0\). We are given that \(v = 0.5 v_{\text{max}}\), which means: \(\omega \sqrt{x_0^2 - x^2} = 0.5 \omega x_0\). Dividing by \(\omega\) and squaring both sides gives: \(x_0^2 - x^2 = 0.25 x_0^2 \implies x^2 = 0.75 x_0^2 \implies x = \sqrt{0.75} x_0 \approx 0.866 x_0\). Rounding to two significant figures, this is \(0.87 x_0\).

1 mark for selecting option D, using the SHM velocity-displacement relation.

The total energy of a simple harmonic oscillator is equal to its maximum kinetic energy, which occurs at the equilibrium position where speed is maximum. The maximum speed is given by \(v_{\text{max}} = \omega A\). Since the angular frequency \(\omega = 2 \pi f\), the maximum speed is \(v_{\text{max}} = 2 \pi f A\). The total energy \(E\) is: \(E = \frac{1}{2} m v_{\text{max}}^2 = \frac{1}{2} m (2 \pi f A)^2 = \frac{1}{2} m (4 \pi^2 f^2 A^2) = 2 \pi^2 m f^2 A^2\).

1 mark for option C, derived by linking total energy to maximum kinetic energy and substituting angular frequency.

For a negative temperature coefficient (NTC) thermistor, a decrease in temperature results in an increase in its resistance. Since the thermistor and the fixed resistor are connected in series, the total resistance of the circuit increases. With a constant e.m.f., this increase in total resistance causes the current in the circuit to decrease. According to Ohm's law, the potential difference across the fixed resistor is given by \(V = I R\). Since the current \(I\) decreases and \(R\) is constant, the potential difference across the fixed resistor also decreases.

1 mark for the correct row A, identifying the increase in thermistor resistance, and the subsequent decrease in both total current and potential difference across the fixed resistor.

Since the car moves up the slope at a constant speed, the net force acting on it is zero. The forward driving force \(F\) of the engine must exactly balance the components of the forces acting down the slope. These forces are the resistive force \(F_{\text{res}}\) and the component of the weight down the slope \(W_x = m g \sin\theta\). Thus, \(F = F_{\text{res}} + m g \sin\theta = 400 + 1200 \times 9.81 \times \sin(5.0^\circ) \approx 400 + 1026 = 1426\text{ N}\). The useful power output is \(P = F v = 1426\text{ N} \times 15\text{ m s}^{-1} \approx 21390\text{ W} \approx 21\text{ kW}\).

1 mark for selecting C, based on summing the gravity component and resistive force, then multiplying by velocity.

The diffraction grating equation is \(d \sin \theta = n \lambda\). The grating spacing \(d\) is the distance between adjacent lines, which is equal to \(d = \frac{1}{N}\). We are given that the order \(n = 3\) and the angle \(\theta = 30^\circ\). Substituting these values into the equation yields: \(\frac{1}{N} \sin(30^\circ) = 3 \lambda \implies \frac{1}{N} (0.5) = 3 \lambda \implies \lambda = \frac{0.5}{3N} = \frac{1}{6N}\).

1 mark for the correct answer A, derived from the grating equation using the relationship between spacing and lines per metre.

The forces acting on the skydiver are the weight \(m g\) acting downwards and the air resistance \(R\) acting upwards. The net downward force is given by \(F_{\text{net}} = m g - R\). According to Newton's second law, \(F_{\text{net}} = m a\). Given that \(a = \frac{1}{3} g\), we can substitute this in: \(m g - R = m \left(\frac{1}{3} g\right) \implies R = m g - \frac{1}{3} m g = \frac{2}{3} m g\).

1 mark for option C, applying Newton's second law to the downward motion of the skydiver.

The wires are made of the same material, so they have the same Young modulus \(E = \frac{\text{stress}}{\text{strain}}\). This means \(\text{strain} = \frac{\text{stress}}{E}\). The stress \(\sigma\) is given by \(\sigma = \frac{F}{A}\), where the force \(F = Mg\) is the same for both wires. The cross-sectional area \(A = \frac{\pi d^2}{4}\) is proportional to the square of the diameter \(d^2\). Therefore, stress is inversely proportional to the square of the diameter, \(\sigma \propto \frac{1}{d^2}\). Since \(E\) is constant, the strain \(\varepsilon\) is also proportional to \(\frac{1}{d^2}\). The diameter of wire Y is twice that of wire X, so the strain in wire Y is \(\frac{1}{2^2} = \frac{1}{4}\) of the strain in wire X. Thus, the ratio of the strain in wire X to the strain in wire Y is 4.

1 mark for the correct option C. [1] for realizing that strain is independent of length for a given stress and is inversely proportional to the cross-sectional area, leading to the ratio of 4.

The net accelerating force \(F\) acting on the object is the difference between its constant weight \(W = mg\) acting downwards and the upward drag force \(F_D = kv\). Thus, \(F = mg - kv\). This equation is in the form \(y = mx + c\), where \(F\) is on the vertical axis and \(v\) is on the horizontal axis. The vertical intercept is \(mg\) (a positive non-zero value) and the gradient is \(-k\) (a constant negative value). Therefore, the graph of \(F\) against \(v\) is a straight line with a negative gradient.

1 mark for the correct option C. [1] for formulating the net force equation \(F = mg - kv\) and identifying it as a linear relationship with a negative gradient.

Using the diffraction grating equation \(d \sin \theta = n \lambda\), for \(n = 3\) and \(\theta = 42^\circ\), we have \(d \sin(42^\circ) = 3\lambda\). This gives \(\frac{d}{\lambda} = \frac{3}{\sin 42^\circ} \approx \frac{3}{0.6691} \approx 4.48\). The highest possible order of maximum \(n_{\text{max}}\) occurs when \(\sin \theta \le 1\), which means \(n \le \frac{d}{\lambda}\). Therefore, \(n \le 4.48\). Since the order \(n\) must be an integer, the highest observable order of maximum is 4.

1 mark for the correct option B. [1] for calculating \(d/\lambda\) as approximately 4.48 and rounding down to the nearest integer 4.

Using the potential divider formula, the output voltage across the fixed resistor \(R\) at temperature \(T_1\) is given by \(V_{\text{out}} = V_{\text{in}} \left( \frac{R}{R + R_{\text{th}}} \right)\), where \(R = 3.0\text{ k}\Omega\), \(V_{\text{in}} = 12\text{ V}\), and \(V_{\text{out}} = 8.0\text{ V}\). Substituting these values: \(8.0 = 12 \left( \frac{3.0}{3.0 + R_{\text{th}}} \right) \Rightarrow \frac{2}{3} = \frac{3.0}{3.0 + R_{\text{th}}} \Rightarrow 3.0 + R_{\text{th}} = 4.5 \Rightarrow R_{\text{th}} = 1.5\text{ k}\Omega\). At the higher temperature \(T_2\), the resistance of the thermistor is halved, so \(R'_{\text{th}} = 0.75\text{ k}\Omega\). The new voltmeter reading is \(V'_{\text{out}} = 12 \left( \frac{3.0}{3.0 + 0.75} \right) = 12 \left( \frac{3.0}{3.75} \right) = 12 \times 0.8 = 9.6\text{ V}\).

1 mark for the correct option B. [1] for calculating the thermistor resistance at \(T_1\) as \(1.5\text{ k}\Omega\), halving it to \(0.75\text{ k}\Omega\) at \(T_2\), and calculating the new voltmeter reading as \(9.6\text{ V}\).

Since Hooke's law is obeyed, \(F = kx\), and the elastic potential energy stored is \(E_p = \frac{1}{2} F x = \frac{1}{2} k x^2\). When the force is doubled to \(2F\), the extension also doubles to \(2x\) because extension is directly proportional to force. The new elastic potential energy stored is \(E'_p = \frac{1}{2} (2F) (2x) = 4 \left( \frac{1}{2} F x \right) = 4 E_p\). The increase in stored energy is \(\Delta E_p = E'_p - E_p = 4 E_p - E_p = 3 E_p\).

1 mark for the correct option C. [1] for finding that doubling the force quadruples the stored energy, and subtracting the initial energy to find the increase.

For simple harmonic motion, the speed \(v\) of a particle at displacement \(x\) is given by \(v = \omega \sqrt{A^2 - x^2}\), where the angular frequency is \(\omega = \frac{2\pi}{T}\). Substituting \(x = \frac{1}{2}A\): \(v = \left(\frac{2\pi}{T}\right) \sqrt{A^2 - \left(\frac{1}{2}A\right)^2} = \left(\frac{2\pi}{T}\right) \sqrt{\frac{3}{4}A^2} = \left(\frac{2\pi}{T}\right) \frac{A\sqrt{3}}{2} = \frac{\pi A \sqrt{3}}{T}\).

1 mark for the correct option B. [1] for using the SHM velocity formula and substituting \(x = 0.5A\) and \(\omega = 2\pi/T\) correctly.

The useful work output is \(W_{\text{out}} = mgh = 80 \times 9.81 \times 15 = 11772\text{ J}\). The useful power output is \(P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{11772}{20} = 588.6\text{ W}\). Since the efficiency \(\eta\) is \(60\%\) (or 0.60), the input electrical power is \(P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{588.6}{0.60} = 981\text{ W} \approx 980\text{ W}\).

1 mark for the correct option C. [1] for calculating the useful output power and dividing by the efficiency of 0.60 to obtain the input power.

Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with the upward direction taken as positive: initial velocity \(u = +12\text{ m s}^{-1\)}, acceleration \(a = -g = -9.81\text{ m s}^{-2\)}, and time \(t = 4.0\text{ s}\). The vertical displacement is \(s = -h\) (since the final position is a distance \(h\) below the starting point). Substituting these values: \(-h = (12 \times 4.0) + \frac{1}{2}(-9.81)(4.0)^2 = 48 - 78.48 = -30.48\text{ m}\). Therefore, the height of the cliff is \(h \approx 30\text{ m}\).

1 mark for the correct option A. [1] for using the correct kinematic equation with consistent signs for velocity and acceleration to find the displacement.

(a)(i) Tensile stress is defined as force divided by cross-sectional area: \(\sigma = F/A\).
(ii) Tensile strain is defined as extension divided by original length: \(\varepsilon = x/L\).

(b)(i) Diameter \(d = 0.85\text{ mm} = 0.85 \times 10^{-3}\text{ m}\).
Cross-sectional area:
\[A = \frac{\pi d^2}{4} = \frac{\pi \times (0.85 \times 10^{-3})^2}{4} = 5.67 \times 10^{-7}\text{ m}^2\]
Force:
\[F = mg = 6.5 \times 9.81 = 63.8\text{ N}\]
Tensile stress:
\[\sigma = \frac{F}{A} = \frac{63.8}{5.67 \times 10^{-7}} = 1.12 \times 10^8\text{ Pa} \approx 1.1 \times 10^8\text{ Pa}\]

(b)(ii) From Young modulus definition:
\[E = \frac{\sigma}{\varepsilon} \implies \varepsilon = \frac{\sigma}{E} = \frac{1.124 \times 10^8}{1.1 \times 10^{11}} = 1.02 \times 10^{-3}\]
Extension:
\[x = \varepsilon \times L = 1.022 \times 10^{-3} \times 2.4 = 2.45 \times 10^{-3}\text{ m} \approx 2.5\text{ mm} \text{ (or } 2.5 \times 10^{-3}\text{ m)}\]

(c) The applied stress of \(1.1 \times 10^8\text{ Pa}\) is less than the elastic limit of \(2.8 \times 10^8\text{ Pa}\). Therefore, the wire undergoes elastic deformation only, and will return to its original length when the load is removed.

(a)(i) B1: force per unit cross-sectional area (accept force / area where area is cross-sectional)
(a)(ii) B1: extension per unit original length (accept change in length / original length)
(b)(i) C1: calculation of cross-sectional area (5.67 x 10^-7 m^2)
C1: calculation of force (63.8 N or 65 N)
A1: stress = 1.1 x 10^8 Pa (allow 1.12 x 10^8 Pa)
(b)(ii) C1: use of E = stress / strain
C1: use of strain = extension / original length
A1: extension = 2.5 mm or 2.5 x 10^-3 m (allow 2.45 mm)
(c) B1: wire returns to original length / yes
B1: (because) stress is less than the elastic limit

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

(b)(i) Extension \(x = 15.6\text{ cm} - 12.0\text{ cm} = 3.6\text{ cm} = 0.036\text{ m}\).
From Hooke's law:
\[k = \frac{F}{x} = \frac{15.0}{0.036} = 416.7\text{ N m}^{-1} \approx 4.2 \times 10^2\text{ N m}^{-1}\]

(b)(ii) Work done \(W\) in stretching the spring is equal to the elastic potential energy stored:
\[W = \frac{1}{2} F x = \frac{1}{2} \times 15.0 \times 0.036 = 0.27\text{ J}\]

(c)(i) For two identical springs in parallel, the effective spring constant \(k_{\text{p}}\) is:
\[k_{\text{p}} = k_1 + k_2 = 416.7 + 416.7 = 833.4\text{ N m}^{-1} \approx 8.3 \times 10^2\text{ N m}^{-1}\text{ (or } 840\text{ N m}^{-1}\text{ using the show-that value)}\]

(c)(ii) Extension of the combination:
\[x_{\text{p}} = \frac{F}{k_{\text{p}}} = \frac{25.0}{833.4} = 0.030\text{ m}\]
Elastic potential energy stored:
\[E_{\text{p}} = \frac{1}{2} F x_{\text{p}} = \frac{1}{2} \times 25.0 \times 0.030 = 0.375\text{ J} \approx 0.38\text{ J}\text{ (or } 0.37\text{ J using } 840\text{ N m}^{-1}\text{)}\]

(a) B1: elastic: returns to original shape/length when load removed
B1: plastic: permanent/irreversible extension/deformation when load removed
(b)(i) C1: extension = 3.6 cm or 0.036 m
A1: k = 15.0 / 0.036 = 417 N m^-1 (must show at least 3 s.f. to justify 4.2 x 10^2)
(b)(ii) C1: use of W = 0.5 * F * x or W = 0.5 * k * x^2
A1: W = 0.27 J
(c)(i) C1: recognition of parallel rule kp = k1 + k2
A1: kp = 833 N m^-1 (or 840 N m^-1 if using 420)
(c)(ii) C1: use of Ep = 0.5 * F * x or Ep = F^2 / (2 * kp)
A1: Ep = 0.38 J (or 0.37 J if using 840)

(a) Simple harmonic motion requires:
1. Acceleration is directly proportional to displacement from the equilibrium position.
2. Acceleration is always directed towards the equilibrium position (or opposite to displacement).

(b)(i) Angular frequency:
\[\omega = 2\pi f = 2\pi \times 4.5 = 9.0\pi \approx 28.3\text{ rad s}^{-1}\text{ (or } 28\text{ rad s}^{-1}\text{)}\]

(b)(ii) The block remains in contact with the platform as long as the downward acceleration of the platform does not exceed the acceleration due to gravity \(g = 9.81\text{ m s}^{-2}\).
At the highest point, the downward acceleration is:
\[a_{\text{max}} = \omega^2 A\]
For contact to be maintained:
\[\omega^2 A \le g \implies A_{\text{max}} = \frac{g}{\omega^2} = \frac{9.81}{(28.27)^2} = \frac{9.81}{799} = 0.0123\text{ m} \approx 1.2 \times 10^{-2}\text{ m}\text{ (or } 1.2\text{ cm)}\]

(c) Maximum velocity:
\[v_{\text{max}} = \omega A_{\text{max}} = 28.27 \times 0.01227 = 0.347\text{ m s}^{-1}\]
Maximum kinetic energy:
\[E_{\text{k,max}} = \frac{1}{2} m v_{\text{max}}^2 = \frac{1}{2} \times 0.15 \times (0.347)^2 = 0.00903\text{ J} \approx 9.0 \times 10^{-3}\text{ J}\text{ (or } 9.0\text{ mJ)}\]

(a) B1: acceleration is (directly) proportional to displacement
B1: acceleration is directed towards a fixed point / equilibrium position / opposite to displacement
(b)(i) C1: use of omega = 2 * pi * f
A1: omega = 28 rad s^-1 (allow 28.3 rad s^-1)
(b)(ii) C1: realization that max downward acceleration = g at limit
C1: use of a = omega^2 * A
A1: Amax = 1.2 cm or 1.2 x 10^-2 m (allow 1.23 cm)
(c) C1: use of vmax = omega * A (vmax = 0.35 m s^-1)
C1: use of Ek = 0.5 * m * v^2
A1: Ek,max = 9.0 mJ or 9.0 x 10^-3 J (allow 9.03 mJ)

(a) Electromotive force (e.m.f.) is defined as the energy converted from chemical (or other) forms to electrical energy per unit charge that passes through the source.

(b)(i) The circuit diagram should show:
- A loop with a DC source (battery), a fixed resistor, and an LDR connected in series.
- Correct symbols for battery, fixed resistor, and LDR.
- A voltmeter connected in parallel specifically across the LDR.

(b)(ii) Total resistance of the circuit:
\[R_{\text{total}} = 1800\ \Omega + 450\ \Omega = 2250\ \Omega\]
Using the potential divider equation, the potential difference across the LDR (voltmeter reading) is:
\[V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R_{\text{total}}} = 6.0 \times \frac{450}{2250} = 1.2\text{ V}\]

(b)(iii) When light intensity decreases, the resistance of the LDR increases. Since the circuit is a potential divider, the output voltage is proportional to the fraction of the total resistance held by the LDR. An increase in the LDR's resistance means it takes a larger share of the total resistance, and thus a larger share of the 6.0 V e.m.f. Consequently, the potential difference across the LDR, and therefore the voltmeter reading, increases.

(a) B1: energy transferred from chemical/other forms to electrical energy
B1: per unit charge (or equation E = W/Q with terms defined)
(b)(i) B1: battery, resistor, and LDR in series with correct symbols
B1: voltmeter connected in parallel across the LDR only
(b)(ii) C1: calculation of total resistance (2250 ohms)
C1: use of potential divider formula or current (I = 6 / 2250 = 2.67 mA)
A1: V = 1.2 V
(b)(iii) M1: resistance of LDR increases
M1: total resistance of circuit increases (reducing total current) OR LDR occupies a larger proportion of total resistance
A1: potential difference across the LDR / voltmeter reading increases

(a)(i)
- Initially, at \(t = 0\), the only force acting is weight, so her acceleration is equal to \(g\) (approx \(9.81\text{ m s}^{-2}\)).
- As velocity increases, the upward air resistance (drag force) increases.
- The net downward force \(F = mg - D\) decreases.
- Since \(a = F/m\), the acceleration decreases progressively over time.
- Eventually, drag equals weight, the net force is zero, and acceleration becomes zero (terminal velocity).

(a)(ii) The velocity-time graph starts at the origin (0,0) with a steep positive gradient equal to \(9.81\text{ m s}^{-2}\). The gradient decreases continuously (curves downwards) and eventually approaches a horizontal line (zero gradient) at the terminal velocity value \(v_{\text{T}} = 54\text{ m s}^{-1}\).

(b)(i) The rate of loss of gravitational potential energy is:
\[\frac{\Delta E_{\text{p}}}{\Delta t} = \frac{mg \Delta h}{\Delta t} = mgv\]
Substituting the values:
\[\text{Rate} = 75\text{ kg} \times 9.81\text{ m s}^{-2} \times 54\text{ m s}^{-1} = 39730\text{ W} = 39.73\text{ kW} \approx 40\text{ kW}\]

(b)(ii) At terminal velocity, the velocity is constant, so there is no change in kinetic energy (\(\Delta E_{\text{k}} = 0\)). By conservation of energy, all the lost gravitational potential energy must be converted to internal (thermal) energy of the air and skydiver through work done against air resistance. Therefore, the rate of work done against air resistance is equal to the rate of loss of GPE, which is \(40\text{ kW}\) (or \(39.7\text{ kW}\)).

(a)(i) B1: acceleration is initially g / maximum and decreases over time
B1: air resistance / drag increases with speed
B1: net/resultant force decreases
B1: acceleration becomes zero when drag equals weight
(a)(ii) B1: curve starting at origin with decreasing gradient
B1: gradient becomes zero / levels off at terminal velocity
(b)(i) C1: use of rate of GPE loss = m * g * v
A1: calculation shows 39730 W (or 39.7 kW), which rounds to 40 kW
(b)(ii) B1: rate of work done is 40 kW / 39.7 kW
B1: explanation that kinetic energy is constant, so all GPE lost goes to work done against drag (thermal energy)

(a) Diffraction is the spreading of a wave as it passes through a slit, aperture, or past the edge of an obstacle.

(b)(i) Using the diffraction grating equation:
\[d \sin\theta = n\lambda\]
Given:
- \(n = 2\)
- \(\lambda = 633\text{ nm} = 633 \times 10^{-9}\text{ m}\)
- \(\theta = 38.0^\circ\)

Calculate grating spacing \(d\):
\[d = \frac{n\lambda}{\sin\theta} = \frac{2 \times 633 \times 10^{-9}}{\sin(38.0^\circ)} = \frac{1.266 \times 10^{-6}}{0.61566} = 2.056 \times 10^{-6}\text{ m}\]
The number of lines per metre \(N_{\text{m}}\) is:
\[N_{\text{m}} = \frac{1}{d} = \frac{1}{2.056 \times 10^{-6}} = 4.863 \times 10^5\text{ lines m}^{-1}\]
Convert to lines per millimetre (\(1\text{ mm} = 10^{-3}\text{ m}\)):
\[N = 4.863 \times 10^5 \times 10^{-3} = 486.3\text{ lines mm}^{-1} \approx 486\text{ lines mm}^{-1}\text{ (or } 490\text{ lines mm}^{-1}\text{ to 2 s.f.)}\]

(b)(ii) The maximum possible angle of diffraction is \(\theta = 90.0^\circ\), where \(\sin\theta = 1\).
Using \(d \sin\theta = n\lambda\):
\[n_{\text{max}} = \frac{d \sin(90.0^\circ)}{\lambda} = \frac{2.056 \times 10^{-6}}{633 \times 10^{-9}} = 3.25\]
Since the order \(n\) must be an integer, the highest observable order is \(n = 3\).

(c) The number of visible maxima increases. According to \(\sin\theta = n\lambda/d\), if \(\lambda\) decreases (from 633 nm to 450 nm), the angle of diffraction \(\theta\) for any given order decreases. This allows more orders (up to \(n = 4\)) to fit within the visible range of \(\theta < 90^\circ\).

(a) B1: spreading of a wave
B1: when passing through a gap / slit / aperture OR around an obstacle
(b)(i) C1: use of d * sin(theta) = n * lambda
C1: calculation of d = 2.06 x 10^-6 m
C1: use of N = 1/d
A1: N = 486 lines mm^-1 (allow 490 lines mm^-1)
(b)(ii) C1: use of sin(theta) <= 1 or theta = 90 degrees
A1: highest order is 3 (must be an integer, do not accept 3.25)
(c) B1: number of visible maxima increases
B1: smaller wavelength results in smaller diffraction angles (so more orders fit within 90 degrees)

In this experiment, the moment of the forces about the pivot must balance when the rule is horizontal. The clockwise moment due to the suspended load \( M \) and the weight of the rule \( W \) is balanced by the anticlockwise moment due to the tension in the spring. Tension \( T = k e \), where \( k \) is the spring constant and \( e \) is the extension. Taking moments about the pivot: \( T \cdot x = M g \cdot 0.95 + W \cdot 0.50 \implies k e x = M g \cdot 0.95 + W \cdot 0.50 \implies e = \frac{(0.95 M g + 0.50 W)}{k} \frac{1}{x} \). This is in the form \( e = \frac{a}{x} + b \), where \( a = \frac{0.95 M g + 0.50 W}{k} \) and \( b \) is the intercept (ideally zero, but non-zero due to small systematic levelling errors or spring end-effects). By plotting \( e \) against \( 1/x \), we obtain a straight line where the gradient is equal to \( a \) and the y-intercept is equal to \( b \).

Table of results (8 marks): 1 mark: 6 sets of readings recorded with correct trend (as \( x \) increases, \( e \) decreases). 1 mark: Range of \( x \) used includes \( 20.0\text{ cm} \) and \( 70.0\text{ cm} \) (or within 2.0 cm of these). 1 mark: Column headings include appropriate units for raw and processed variables: \( x / \text{cm} \), \( l / \text{cm} \), \( e / \text{cm} \), and \( 1/x / \text{cm}^{-1} \). 1 mark: Raw data for \( x \) and \( l \) recorded to the nearest millimetre. 1 mark: \( 1/x \) calculated to the same or one more significant figures as \( x \). 1 mark: Extension \( e \) calculated correctly from raw data. 2 marks: Quality of data - points on the graph show low scatter about the line of best fit. Graph (5 marks): 1 mark: Axes chosen such that the plotted points occupy more than half the grid in both directions, scales are linear and easy to read. 1 mark: All points plotted accurately within half a small square. 1 mark: Straight line of best fit drawn with a balanced distribution of points. 1 mark: Gradient calculated from a triangle where the hypotenuse is at least half the length of the line of best fit. 1 mark: y-intercept determined by direct read-off (if x-axis starts at zero) or via substitution of a point on the line into \( y = mx + c \). Analysis and Constants (7 marks): 1 mark: Correct calculation of gradient with unit. 1 mark: Correct calculation of y-intercept with unit. 1 mark: Equating constant \( a \) to gradient and constant \( b \) to y-intercept. 1 mark: Value of \( a \) given to 2 or 3 s.f. with units (e.g., \( \text{m}^2 \) or \( \text{cm}^2 \)). 1 mark: Value of \( b \) given to 2 or 3 s.f. with units (e.g., \( \text{m} \) or \( \text{cm} \)).

For vertical oscillations of a floating cylinder, the restoring force is proportional to the displacement from equilibrium: \( F = -\rho A g x \). The equation of motion is \( (m + m_0) \frac{d^2x}{dt^2} = -\rho A g x \). This gives a simple harmonic motion with period \( T = 2\pi \sqrt{\frac{m + m_0}{\rho A g}} \). Squaring both sides gives \( T^2 = \frac{4\pi^2}{\rho A g} (m + m_0) \), which confirms the suggested relationship \( T^2 = k(m + m_0) \) where \( k = \frac{4\pi^2}{\rho A g} \). In the experiment, the student calculates \( k_1 = \frac{T_1^2}{m_1 + m_0} \) and \( k_2 = \frac{T_2^2}{m_2 + m_0} \) and compares their fractional/percentage difference to the experimental uncertainty (typically around 10% due to reaction time in stopwatch measurements).

Measurements and period (5 marks): 1 mark: Value of \( t \) for 10 oscillations recorded to 0.01 s with repetitions. 1 mark: Value of \( T \) calculated correctly as \( t/10 \) with unit (s). 1 mark: Raw values of \( m \) recorded correctly. 1 mark: Second set of readings for \( m = 100\text{ g} \) showing larger \( t \) and \( T \). 1 mark: Correct conversion of mass units if applicable. Uncertainty and analysis (5 marks): 1 mark: Absolute uncertainty in \( t \) estimated in the range 0.1 s to 0.3 s to account for human reaction time. 1 mark: Correct calculation of percentage uncertainty in \( T \) (equal to percentage uncertainty in \( t \)). 1 mark: Two values of \( k \) calculated correctly with consistent units. 1 mark: Statement comparing the percentage difference between the two \( k \) values to the experimental tolerance (e.g., 10%). 1 mark: Logical conclusion drawn: if difference is within tolerance, relationship is supported; otherwise, it is not. Limitations and Improvements (8 marks - 1 mark for each limitation and 1 mark for each corresponding improvement, max 4 pairs): 1. Limitation: Oscillations decay quickly due to damping / hard to count 10 oscillations. Improvement: Use a video camera with a playback timer / slow-motion capture to count frames. 2. Limitation: Cup tilts or hits the sides of the container during oscillation. Improvement: Use a wider container / guide the cup vertically using a low-friction vertical guide rod through the center of the cup. 3. Limitation: Difficult to release the cup consistently without introducing lateral motion. Improvement: Use an electromagnet and a small iron washer at the bottom to release the cup vertically from a fixed position. 4. Limitation: Mass inside the cup shifts, causing the cup to list to one side. Improvement: Secure the masses to the bottom of the cup symmetrically using Blu-Tack or adhesive.