2024 Cambridge IAS-Level Physics (9702) 模擬試題連答案詳解

Let \(E\) be the Young modulus of the material. The cross-sectional area of wire X is \(A_X = \frac{\pi d^2}{4}\). The cross-sectional area of wire Y is \(A_Y = \frac{\pi (2d)^2}{4} = 4 A_X\). The extension \(x\) of a wire under load \(F\) is given by \(x = \frac{F L}{A E}\). For wire X: \(x_X = \frac{F L}{A_X E}\). The elastic potential energy stored in wire X is \(U_X = \frac{1}{2} F x_X = \frac{F^2 L}{2 A_X E}\). For wire Y: \(x_Y = \frac{F (2L)}{A_Y E} = \frac{2 F L}{4 A_X E} = \frac{F L}{2 A_X E}\). The elastic potential energy stored in wire Y is \(U_Y = \frac{1}{2} F x_Y = \frac{F^2 L}{4 A_X E}\). The ratio of the stored energies is \(\frac{U_X}{U_Y} = \frac{\frac{F^2 L}{2 A_X E}}{\frac{F^2 L}{4 A_X E}} = 2.0\).

1 mark for the correct option C. 1 mark for calculating the ratio of stored elastic potential energies.

In a stationary wave, the distance between any two adjacent nodes is equal to half a wavelength, \(\frac{\lambda}{2}\). The distance from the second node to the fifth node contains three of these half-wavelength intervals: \(5 - 2 = 3\) intervals. Therefore, the distance is \(3 \times \frac{\lambda}{2} = 36\text{ cm}\), which gives \(\frac{3}{2}\lambda = 36\text{ cm}\). Solving for the wavelength, we get \(\lambda = \frac{36 \times 2}{3} = 24\text{ cm}\).

1 mark for the correct option C. 1 mark for finding the wavelength from node spacing.

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

1 mark for the correct option B. 1 mark for writing the change in momentum as m(u+v) and applying Newton's second law.

Let the original resistance of the wire be \(R\). The power dissipated is originally \(P = \frac{V^2}{R}\). When cut into three pieces of equal length, each piece has a resistance of \(R' = \frac{R}{3}\). When connected in parallel, the equivalent resistance \(R_{\text{eq}}\) is given by \(\frac{1}{R_{\text{eq}}} = \frac{3}{R'} = \frac{9}{R}\), which means \(R_{\text{eq}} = \frac{R}{9}\). The total power dissipated in this arrangement is \(P_{\text{new}} = \frac{V^2}{R_{\text{eq}}} = \frac{V^2}{R/9} = 9 \left(\frac{V^2}{R}\right) = 9P\).

1 mark for the correct option D. 1 mark for calculating the new parallel resistance and corresponding power.

Since the two cells are connected in series opposing each other, their net electromotive force is \(E_1 - E_2\). The total resistance in the closed loop is the sum of the external and internal resistances: \(R_{\text{total}} = R_1 + R_2 + r_1 + r_2\). Applying Kirchhoff's second law, the current in the loop is \(I = \frac{\text{Net e.m.f.}}{R_{\text{total}}} = \frac{E_1 - E_2}{R_1 + R_2 + r_1 + r_2}\). Kirchhoff's second law is a statement of the conservation of energy.

1 mark for the correct option B. 1 mark for applying Kirchhoff's loop rule and connecting it to energy conservation.

The Young modulus \(E\) of steel is constant: \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A x}\), which gives \(x = \frac{F L}{A E}\). For the second wire, let the required force be \(F'\): \(x = \frac{F' (2L)}{(A/2) E} = \frac{4 F' L}{A E}\). Setting the extensions equal: \(\frac{F L}{A E} = \frac{4 F' L}{A E}\), which simplifies to \(F = 4F'\). Thus, \(F' = \frac{F}{4}\).

1 mark for the correct option A. 1 mark for utilizing the Young modulus formula to relate the forces.

For a closed-open pipe of length \(L_c\) vibrating in its fundamental mode, the wavelength is \(\lambda = 4 L_c\). For an open-open pipe of length \(L_o\) vibrating in its first overtone (second harmonic), the wavelength of the sound is \(\lambda = L_o\). Since the wavelength is the same in both cases: \(4 L_c = L_o\), which gives the ratio \(\frac{L_c}{L_o} = \frac{1}{4}\).

1 mark for the correct option A. 1 mark for expressing wavelength in terms of pipe lengths.

Let the direction to the right be positive. The total initial momentum of the system is \(p_i = m_1 u_1 + m_2 u_2 = (2.0 \times 6.0) + (4.0 \times (-3.0)) = 12.0 - 12.0 = 0\text{ kg m s}^{-1}\). According to the conservation of momentum, the total final momentum is also zero: \(p_f = (m_1 + m_2) v = 0\), which gives \(v = 0\text{ m s}^{-1}\). Since the kinetic energy after the collision is zero, whereas it was non-zero beforehand (\(54\text{ J}\)), kinetic energy is not conserved. A collision where kinetic energy is not conserved and the objects stick together is inelastic.

1 mark for the correct option B. 1 mark for calculating final velocity using conservation of momentum and identifying the collision type.

The strain energy \(W\) stored in an elastically stretched wire of length \(L\), cross-sectional area \(A\) and Young modulus \(E\) under load \(F\) is given by: \(W = \frac{1}{2} F x\). Since Young modulus is given by \(E = \frac{F L}{A x}\), the extension is \(x = \frac{F L}{A E}\). Substituting this expression for \(x\) gives: \(W = \frac{F^2 L}{2 A E}\). For the second wire: the length is \(L_2 = 2L\), the diameter is half of the first wire, so the cross-sectional area is \(A_2 = \frac{1}{4} A\), the material is the same, so the Young modulus is \(E\), and the load \(F\) is the same. The strain energy stored in the second wire is: \(W_2 = \frac{F^2 L_2}{2 A_2 E} = \frac{F^2 (2L)}{2 (\frac{1}{4} A) E} = 8 \left( \frac{F^2 L}{2 A E} \right) = 8W\).

1 mark for identifying that strain energy is proportional to length and inversely proportional to cross-sectional area, and correctly calculating the ratio of 8.

A stationary wave in a pipe closed at one end has a node (N) at the closed end and an antinode (A) at the open end. If there is exactly one other node inside the pipe, the sequence of nodes and antinodes from the closed end to the open end is: N (closed end) - A - N (inside) - A (open end). The distance between adjacent nodes and antinodes is \(\frac{\lambda}{4}\), and the distance between adjacent nodes is \(\frac{\lambda}{2}\). Thus, the total length \(L\) of the pipe is: \(L = \frac{\lambda}{2} + \frac{\lambda}{4} = \frac{3\lambda}{4}\). This gives the wavelength: \(\lambda = \frac{4L}{3}\). Using the wave equation \(v = f \lambda\), the frequency \(f\) is: \(f = \frac{v}{\lambda} = \frac{3v}{4L}\).

1 mark for relating the pipe length to three-quarters of a wavelength and solving for frequency to obtain \(\frac{3v}{4L}\).

Since the blocks are initially at rest, the total initial momentum of the system is zero. According to the principle of conservation of momentum, the total momentum after release must also be zero: \(3m (-v) + m (v_Q) = 0\), which gives the speed of block Q as \(v_Q = 3v\). The kinetic energy of block P is: \(E_k = \frac{1}{2} (3m) v^2 = \frac{3}{2} m v^2\). The kinetic energy of block Q is: \(E_Q = \frac{1}{2} m v_Q^2 = \frac{1}{2} m (3v)^2 = \frac{9}{2} m v^2\). In terms of \(E_k\), \(E_Q = 3 \left( \frac{3}{2} m v^2 \right) = 3 E_k\). The total kinetic energy of both blocks is: \(E_{\text{total}} = E_k + E_Q = E_k + 3 E_k = 4 E_k\).

1 mark for using momentum conservation to find Q's speed, calculating its kinetic energy, and summing both energies to find \(4 E_k\).

Let the initial resistance of the wire of length \(L\) be \(R\). The power dissipated is: \(P = \frac{V^2}{R}\). When the wire is cut into two equal halves, the resistance of each half is \(R_{\text{half}} = \frac{R}{2}\). When connected in parallel across the same battery of e.m.f. \(V\), each half experiences the full potential difference \(V\). The power dissipated in each half is: \(P_{\text{half}} = \frac{V^2}{R_{\text{half}}} = \frac{V^2}{R/2} = 2 \left(\frac{V^2}{R}\right) = 2P\). The total power dissipated by both halves is: \(P_{\text{total}} = P_{\text{half}} + P_{\text{half}} = 2P + 2P = 4P\).

1 mark for identifying that cutting the wire halves the resistance, and connecting them in parallel increases the total power dissipation to \(4P\).

By Kirchhoff's second law, the sum of e.m.f.s around a closed loop is equal to the sum of the potential drops across the resistors: \(\sum E = \sum I R\). Since the batteries oppose each other and the current is in the direction of \(E_1\), the net e.m.f. is \(E_1 - E_2\). The potential drops are the potential difference across \(R_1\) (which is \(I R_1\)) and the potential difference across \(R_2\) (which is \(V_2 = 4.5\text{ V}\)): \(E_1 - E_2 = I R_1 + V_2\). Substituting the given values: \(12 - E_2 = (1.5 \times 3.0) + 4.5 \implies 12 - E_2 = 4.5 + 4.5 = 9.0\text{ V}\). This yields \(E_2 = 12 - 9.0 = 3.0\text{ V}\).

1 mark for applying Kirchhoff's second law with opposing e.m.f.s and calculating \(E_2 = 3.0\text{ V}\).

During stretching, the force required at each extension is greater than the force exerted by the rubber band at the same extension during unloading. This is represented by the loading curve lying above the unloading curve. Therefore, the work done on the rubber band during stretching, \(W_1\) (represented by the area under the loading curve), is greater than the work done by the rubber band during unloading, \(W_2\) (represented by the area under the unloading curve), so \(W_1 > W_2\). The difference \((W_1 - W_2)\) is the area of the hysteresis loop, which represents the net energy lost as heat (thermal energy released) during the cycle.

1 mark for correctly identifying that \(W_1 > W_2\) and the difference is the thermal energy released (represented by the area of the loop).

First, calculate the wavelength \(\lambda\) of the microwaves using the wave equation: \(c = f \lambda \implies \lambda = \frac{c}{f} = \frac{3.0 \times 10^8\text{ m s}^{-1}}{10 \times 10^9\text{ Hz}} = 0.030\text{ m} = 30\text{ mm}\). In a stationary wave pattern, intensity maxima correspond to antinodes and intensity minima correspond to nodes. The distance between an adjacent node and antinode is \(d = \frac{\lambda}{4} = \frac{30\text{ mm}}{4} = 7.5\text{ mm}\).

1 mark for calculating the wavelength \(\lambda = 30\text{ mm}\) and dividing by 4 to get the distance between a node and adjacent antinode as \(7.5\text{ mm}\).

In a short time interval \(\Delta t\), the spacecraft travels a distance \(V \Delta t\). The volume of the dust cloud swept out by the front surface of the spacecraft is \(\text{Volume} = A \times (V \Delta t)\). The mass \(\Delta m\) of dust collected in this time is: \(\Delta m = \rho \times \text{Volume} = \rho A V \Delta t\). The dust particles are initially at rest, and after colliding they travel at the speed \(V\) of the spacecraft. The change in momentum \(\Delta p\) of this collected dust is: \(\Delta p = \Delta m \times V = (\rho A V \Delta t) V = \rho A V^2 \Delta t\). By Newton's third law, the magnitude of the force on the spacecraft is equal to the rate of change of momentum of the dust: \(F = \frac{\Delta p}{\Delta t} = \rho A V^2\).

1 mark for deriving the swept mass per unit time \(\frac{dm}{dt} = \rho A V\) and multiplying by \(V\) to obtain the rate of change of momentum \(\rho A V^2\).

The extension of a wire is given by:
\(x = \frac{FL}{AE}\)

The work done \(W\) in stretching a wire elastically is given by:
\(W = \frac{1}{2} F x = \frac{F^2 L}{2AE}\)

For the second wire, let the work done be \(W_2\):
\(W_2 = \frac{F_2^2 L_2}{2A_2 E_2}\)

Given that the material is the same, \(E_2 = E\). Substituting the new dimensions and force:
\(W_2 = \frac{(2F)^2 (2L)}{2(2A)E} = \frac{4F^2 \times 2L}{4AE} = 4 \left( \frac{F^2 L}{2AE} \right) = 4W\)

1 mark for the correct option C.
Method: Deduce the relationship between work done and the parameters of the wire, substitute the changed variables, and correctly solve for the ratio.

1. **Maximum Elastic Strain Energy**: At the maximum extension, the total work done to stretch the band is represented by the entire area under the loading curve, which is \(A_1\). This is the maximum strain energy stored in the material at that point.
2. **Thermal Energy Generated**: During unloading, only \(A_2\) of that energy is recovered as mechanical work. The remaining energy represented by the area of the hysteresis loop, \(A_1 - A_2\), is dissipated as thermal energy (internal energy increase of the rubber molecules).

1 mark for the correct option A.
Method: Identify that the area under the loading curve is the maximum strain energy stored, and the difference between loading and unloading areas represents the thermal energy dissipated.

For a closed tube (with water acting as the boundary), resonance occurs at odd quarter-wavelengths:
First resonance: \(L_1 + c = \frac{\lambda}{4}\)
Second resonance: \(L_2 + c = \frac{3\lambda}{4}\)
where \(c\) is the end correction.

Subtracting the first equation from the second eliminates the end correction \(c\):
\(L_2 - L_1 = \frac{\lambda}{2}\)
\(73.5\text{ cm} - 24.5\text{ cm} = 49.0\text{ cm} = 0.490\text{ m}\)
\(\lambda = 2 \times 0.490\text{ m} = 0.980\text{ m}\)

Now, use the wave equation \(v = f\lambda\):
\(v = 340\text{ Hz} \times 0.980\text{ m} = 333.2\text{ m s}^{-1} \approx 333\text{ m s}^{-1}\)

1 mark for the correct option B.
Method: Recognize that the difference between adjacent resonance lengths is half a wavelength, calculate the wavelength, and use the wave equation to find the speed of sound.

In a stationary wave, all particles within a single loop (between two adjacent nodes) oscillate in phase (phase difference is \(0\text{ rad}\)).

Particles in adjacent loops (separated by exactly one node) oscillate completely out of phase with each other. Therefore, the phase difference between any particle in one loop and any particle in the next loop is always \(180^\circ\) or \(\pi\text{ rad}\), regardless of their physical distance.

Since there is exactly one node between \(X\) and \(Y\), they reside in adjacent loops. Thus, their phase difference is \(\pi\text{ rad}\).

1 mark for the correct option C.
Method: Recall the phase characteristics of stationary waves, specifically that particles in adjacent loops separated by a single node are in antiphase.

Take the direction to the right as positive.

**1. Conservation of Linear Momentum:**
Initial momentum \(p_{\text{initial}} = m_A u_A + m_B u_B\)
\(p_{\text{initial}} = (2.0 \times 6.0) + (4.0 \times -3.0) = 12.0 - 12.0 = 0\text{ kg m s}^{-1}\)

Final momentum \(p_{\text{final}} = m_A v_A + m_B v_B\)
Since momentum is conserved:
\(0 = (2.0 \times -2.0) + (4.0 \times v_B)\)
\(0 = -4.0 + 4.0 v_B \implies v_B = +1.0\text{ m s}^{-1}\) (to the right).

**2. Kinetic Energy Analysis:**
Initial kinetic energy \(E_{k\text{,initial}} = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2\)
\(E_{k\text{,initial}} = \frac{1}{2}(2.0)(6.0)^2 + \frac{1}{2}(4.0)(-3.0)^2 = 36.0 + 18.0 = 54.0\text{ J}\)

Final kinetic energy \(E_{k\text{,final}} = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2\)
\(E_{k\text{,final}} = \frac{1}{2}(2.0)(-2.0)^2 + \frac{1}{2}(4.0)(1.0)^2 = 4.0 + 2.0 = 6.0\text{ J}\)

Since \(E_{k\text{,final}} < E_{k\text{,initial}}\), kinetic energy is not conserved, so the collision is inelastic.

1 mark for the correct option B.
Method: Apply conservation of linear momentum to calculate the final velocity of trolley B, then calculate and compare total kinetic energy before and after the collision to identify that it is inelastic.

Let the initial direction of motion of the ball be positive.
Initial velocity \(u = +20\text{ m s}^{-1}\)
Final velocity \(v = -30\text{ m s}^{-1}\)

Change in momentum \(\Delta p = m(v - u) = 0.15\text{ kg} \times (-30 - 20)\text{ m s}^{-1} = -7.5\text{ N s}\)
The magnitude of the impulse is \(7.5\text{ N s}\).

The impulse is equal to the area under the force-time graph. For a triangular force profile:
\(\text{Impulse} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \Delta t \times F_{\text{max}}\)

Substitute the known values:
\(7.5 = \frac{1}{2} \times (8.0 \times 10^{-3}\text{ s}) \times F_{\text{max}}\)
\(7.5 = 4.0 \times 10^{-3} \times F_{\text{max}}\)
\(F_{\text{max}} = \frac{7.5}{4.0 \times 10^{-3}} = 1875\text{ N} \approx 1.9\text{ kN}\)

1 mark for the correct option C.
Method: Calculate the change in momentum (taking direction into account), equate it to the area of the triangular force-time graph, and solve for the peak force.

The initial power dissipated by the single wire is:
\(P = \frac{V^2}{R}\)

When the wire is cut in half, the resistance of each half is \(R_1 = R_2 = \frac{R}{2}\).

When connected in parallel across the same voltage supply \(V\), each half experiences the full potential difference \(V\). The power dissipated by each half is:
\(P_{\text{half}} = \frac{V^2}{R/2} = 2 \left(\frac{V^2}{R}\right) = 2P\)

The total power dissipated by both halves in parallel is:
\(P_{\text{total}} = P_{\text{half}} + P_{\text{half}} = 2P + 2P = 4P\)

Alternatively, the equivalent resistance of the parallel combination is \(R_{\text{eq}} = \frac{R/2}{2} = \frac{R}{4}\).
Hence, \(P_{\text{total}} = \frac{V^2}{R_{\text{eq}}} = \frac{V^2}{R/4} = 4 \frac{V^2}{R} = 4P\).

1 mark for the correct option D.
Method: Relate resistance to length, find the new equivalent resistance of the parallel combination, and use \(P = V^2 / R\) to determine the new total power.

From Kirchhoff's second law applied to the loop:
\(E = I(R + r)\)

Create two simultaneous equations for the two states:
State 1: \(E = 1.0\text{ A} \times (5.0\ \Omega + r) = 5.0 + r\)
State 2: \(E = 0.50\text{ A} \times (12.0\ \Omega + r) = 6.0 + 0.5r\)

Since \(E\) is constant, equate the two expressions:
\(5.0 + r = 6.0 + 0.5r\)
\(0.5r = 1.0\)
\(r = 2.0\ \Omega\)

1 mark for the correct option B.
Method: Construct two loop equations using \(E = I(R+r)\), set them equal to eliminate \(E\), and solve for the internal resistance \(r\).

The elastic potential energy stored in a wire is given by \(E = \frac{1}{2} F x\), where \(x\) is the extension. Using the definition of Young's modulus, the extension is \(x = \frac{F L}{A E_y}\), where \(E_y\) is the Young's modulus of the material. Substituting this into the energy formula gives \(E = \frac{F^2 L}{2 A E_y}\). For the first wire, \(E_1 = \frac{F^2 L}{2 A E_y}\). For the second wire, \(E_2 = \frac{F^2 (L/2)}{2 (2A) E_y} = \frac{F^2 L}{8 A E_y} = \frac{1}{4} E_1\).

1 mark for identifying the relationship between energy, force, area, and length, and calculating the correct ratio of 1/4.

The work recovered during unloading (elastic strain energy released) is represented by the area under the unloading line on a force-extension graph. The unloading line goes from \((4.0\text{ mm}, 120\text{ N})\) to \((1.0\text{ mm}, 0\text{ N})\). This is a right-angled triangle with base equal to the change in extension, \(\Delta x = 4.0\text{ mm} - 1.0\text{ mm} = 3.0\text{ mm} = 3.0 \times 10^{-3}\text{ m}\), and height equal to the maximum force of \(120\text{ N}\). Elastic energy recovered \(= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (3.0 \times 10^{-3}\text{ m}) \times 120\text{ N} = 0.18\text{ J}\). The work done to produce plastic deformation is the energy retained by the wire: \(\text{Plastic work done} = \text{Total loading work} - \text{Elastic energy recovered} = 0.34\text{ J} - 0.18\text{ J} = 0.16\text{ J}\).

1 mark for calculating the area under the unloading line (0.18 J) and subtracting it from the total loading work to get 0.16 J.

The wavelength of the sound wave is \(\lambda = \frac{v}{f} = \frac{340}{512} \approx 0.664\text{ m} = 66.4\text{ cm}\). For the first resonance in a closed pipe: \(L_1 + e = \frac{\lambda}{4}\), where \(e\) is the end correction. Here, \(\frac{\lambda}{4} = 16.6\text{ cm}\), so \(e = 16.6 - 16.0 = 0.6\text{ cm}\). For the second resonance: \(L_2 + e = \frac{3\lambda}{4}\). Hence, \(L_2 + 0.6 = 3 \times 16.6 = 49.8\text{ cm}\), which yields \(L_2 = 49.2\text{ cm}\).

1 mark for determining the wavelength, calculating the end correction as 0.6 cm, and finding the second resonant length as 49.2 cm.

For a string of length \(L = 1.20\text{ m}\) vibrating in its third harmonic, there are 3 loops. The length of each loop is \(\frac{L}{3} = 0.40\text{ m}\). Nodes are located at \(0\text{ m}\), \(0.40\text{ m}\), \(0.80\text{ m}\), and \(1.20\text{ m}\). Point \(P\) (at \(0.10\text{ m}\)) lies in the first loop (between \(0\text{ m}\) and \(0.40\text{ m}\)). Point \(Q\) (at \(0.50\text{ m}\)) lies in the second loop (between \(0.40\text{ m}\) and \(0.80\text{ m}\)). Since they lie in adjacent loops, they vibrate completely out of phase with each other. The phase difference is therefore \(180^\circ\) (or \(\pi\text{ rad}\)).

1 mark for identifying the positions of nodes, determining that points P and Q are in adjacent loops, and finding the phase difference of 180 degrees.

By conservation of linear momentum: \((3m)v = (3m + m)V\), where \(V\) is the common velocity after the collision. This gives \(V = \frac{3}{4}v\). The initial kinetic energy is \(E_k = \frac{1}{2} (3m) v^2 = \frac{3}{2} m v^2\). The final kinetic energy is \(E_{kf} = \frac{1}{2} (4m) V^2 = 2 m \left(\frac{3}{4}v\right)^2 = \frac{9}{8} m v^2\). The loss in kinetic energy is \(\Delta E_k = E_k - E_{kf} = \frac{3}{2} m v^2 - \frac{9}{8} m v^2 = \frac{3}{8} m v^2\). The fraction of initial kinetic energy lost is \(\frac{\Delta E_k}{E_k} = \frac{3/8}{3/2} = \frac{1}{4}\).

1 mark for using conservation of momentum to find the final velocity, computing the initial and final kinetic energies, and calculating the correct fraction of 1/4.

According to Newton's second law, force is the rate of change of momentum: \(F = \frac{dp}{dt} = v \frac{dm}{dt}\) because the belt speed \(v\) is kept constant. Given \(\frac{dm}{dt} = 12\text{ kg s}^{-1}\) and \(v = 2.5\text{ m s}^{-1}\), the horizontal force is \(F = 2.5 \times 12 = 30\text{ N}\).

1 mark for identifying the force as the rate of change of momentum and performing the correct calculation to obtain 30 N.

The resistance of a wire is given by \(R = \rho \frac{L}{A} = \rho \frac{4L}{\pi d^2}\). Let the resistance of wire \(X\) be \(R_X = R_0 = \rho \frac{4L}{\pi d^2}\). The resistance of wire \(Y\) is \(R_Y = (3\rho) \frac{4(2L)}{\pi (2d)^2} = 3\rho \frac{8L}{4 \pi d^2} = 1.5 R_0\). Since the wires are in parallel, they experience the same potential difference \(V\). The power dissipated is \(P = \frac{V^2}{R}\). Therefore, the ratio of power is \(\frac{P_X}{P_Y} = \frac{R_Y}{R_X} = \frac{1.5 R_0}{R_0} = 1.5 = \frac{3}{2}\).

1 mark for relating resistance to resistivity and dimensions, finding the resistance ratio, and using the parallel power relation to calculate the ratio 3/2.

First, calculate the equivalent resistance of the parallel combination: \(R_p = \frac{6.0 \times 12.0}{6.0 + 12.0} = 4.0\ \Omega\). The total resistance of the circuit is \(R_{\text{total}} = R_p + r = 4.0 + 2.0 = 6.0\ \Omega\). The total current in the circuit is \(I_{\text{total}} = \frac{E}{R_{\text{total}}} = \frac{12.0}{6.0} = 2.0\text{ A}\point\) The terminal potential difference across the parallel combination is \(V = I_{\text{total}} \times R_p = 2.0 \times 4.0 = 8.0\text{ V}\point\) The current through the \(12.0\ \Omega\) resistor is therefore \(I_{12} = \frac{V}{12.0} = \frac{8.0}{12.0} \approx 0.67\text{ A}\point\)

1 mark for calculating the parallel equivalent resistance, finding the total circuit current, obtaining the terminal potential difference, and using it to find the branch current of 0.67 A.

The strain energy stored in a wire under tension is given by \(E_s = \frac{1}{2} F x\), where \(x\) is the extension.
Using Young's modulus, \(Y = \frac{FL}{Ax}\), the extension can be expressed as \(x = \frac{FL}{AY}\).
Substituting this into the strain energy formula gives:
\(E_s = \frac{F^2 L}{2AY}\)
For a constant force \(F\) and material of the same Young's modulus \(Y\), the strain energy is proportional to the ratio of length to cross-sectional area: \(E_s \propto \frac{L}{A}\).
For the second wire, \(L_2 = 2L\) and \(A_2 = 0.5A\), so:
\(\frac{L_2}{A_2} = \frac{2L}{0.5A} = 4 \left(\frac{L}{A}\right)\)
Therefore, the strain energy stored in the second wire is \(4E\).

1 mark for identifying that strain energy is proportional to L/A for a constant force, and calculating the new strain energy as 4E.

For an air column closed at one end, the fundamental mode of vibration (first harmonic) has a wavelength \(\lambda_1\) given by \(L = \frac{\lambda_1}{4}\), which gives \(f_1 = \frac{v}{4L}\).
The next possible mode of vibration, known as the first overtone or third harmonic, has a wavelength \(\lambda_3\) given by \(L = \frac{3\lambda_3}{4}\).
Therefore, the frequency of the first overtone is:
\(f_3 = \frac{3v}{4L} = \frac{3 \times 340}{4 \times 0.60} = 425\text{ Hz}\).

1 mark for using the correct relation for the third harmonic frequency of a closed pipe and obtaining 425 Hz.

Let the direction to the right be positive.
Initial momentum of the system:
\(p_i = m_X u_X + m_Y u_Y = (2.0 \times 6.0) + (4.0 \times (-3.0)) = 12.0 - 12.0 = 0\text{ kg m s}^{-1}\).
Since the total initial momentum is zero, by the conservation of momentum, the final momentum must also be zero, meaning the two combined blocks come to a complete stop: \(v_f = 0\text{ m s}^{-1}\).
Initial kinetic energy of the system:
\(E_{k,i} = \frac{1}{2} m_X u_X^2 + \frac{1}{2} m_Y u_Y^2 = \frac{1}{2}(2.0)(6.0)^2 + \frac{1}{2}(4.0)(-3.0)^2 = 36.0 + 18.0 = 54.0\text{ J}\).
Final kinetic energy of the system:
\(E_{k,f} = 0\text{ J}\).
Therefore, the loss in kinetic energy is:
\(\Delta E_k = 54.0\text{ J} - 0\text{ J} = 54\text{ J}\).

1 mark for calculating that the final velocity is 0 m/s and finding the total initial kinetic energy of 54 J, which is completely lost.

In the series connection:
Total resistance is \(R_s = R + 2R = 3R\).
The current is \(I = \frac{V}{3R}\).
The power \(P\) dissipated in the resistor of resistance \(R\) is:
\(P = I^2 R = \left(\frac{V}{3R}\right)^2 R = \frac{V^2}{9R}\).
In the parallel connection:
The potential difference across the resistor of resistance \(R\) is equal to the e.m.f. of the cell, \(V\).
The power \(P_{\text{parallel}}\) dissipated in the resistor of resistance \(R\) is:
\(P_{\text{parallel}} = \frac{V^2}{R}\).
By comparing the two expressions, we find:
\(P_{\text{parallel}} = 9P\).

1 mark for correctly comparing the power in series and parallel configurations and finding that the parallel power is 9 times larger.

The total potential difference across the external circuit is:
\(V = 3.2\text{ V} + 4.8\text{ V} = 8.0\text{ V}\).
The power dissipated internally in the battery is \(P_{\text{int}} = I V_{\text{int}} = 0.32\text{ W}\).
Using this, the potential difference across the internal resistance is:
\(V_{\text{int}} = \frac{P_{\text{int}}}{I} = \frac{0.32\text{ W}}{0.40\text{ A}} = 0.80\text{ V}\).
According to Kirchhoff's second law, the sum of e.m.f.s in a closed loop equals the sum of potential drops:
\(E = V + V_{\text{int}} = 8.0\text{ V} + 0.80\text{ V} = 8.8\text{ V}\).

1 mark for using the internal power dissipation to find internal voltage drop, and applying Kirchhoff's second law to find E = 8.8 V.

Since the wires are connected in series, they experience the same tensile force \(F\).
The extension is given by \(x = \frac{FL}{AY}\), where \(A = \frac{\pi d^2}{4}\).
Therefore, the extension \(x \propto \frac{1}{d^2 Y}\).
For wire \(P\): \(x_P \propto \frac{1}{d^2 E_Y}\).
For wire \(Q\): \(x_Q \propto \frac{1}{(2d)^2 (2E_Y)} = \frac{1}{8d^2 E_Y}\).
The ratio is:
\(\frac{x_P}{x_Q} = \frac{1}{1/8} = 8\).

1 mark for establishing the relationship between extension, diameter, and Young modulus, and determining the ratio of 8.

In a stationary wave, all particles in a single loop (between two adjacent nodes) oscillate in phase. Particles in adjacent loops oscillate completely out of phase with each other (phase difference of \(\pi\text{ rad}\)).
The distance between adjacent nodes is \(\frac{1}{2}\lambda\).
Since \(P\) is at an antinode, it is at the midpoint of a loop, which is \(\frac{1}{4}\lambda\) away from the nodes on either side.
Because the distance from \(P\) to \(Q\) is \(\frac{1}{3}\lambda\) (which is greater than \(\frac{1}{4}\lambda\)), \(Q\) must lie in an adjacent loop.
Since \(Q\) is in the adjacent loop, the phase difference between \(P\) and \(Q\) is exactly \(\pi\text{ rad}\).

1 mark for identifying that Q is in an adjacent loop to P, hence the phase difference is pi rad.

Taking the upward direction as positive:
Initial velocity \(u = -8.0\text{ m s}^{-1}\)
Final velocity \(v = +6.0\text{ m s}^{-1}\)
Change in momentum \(\Delta p = m(v - u) = 0.20 \times (6.0 - (-8.0)) = 0.20 \times 14.0 = 2.8\text{ N s}\).
The magnitude of the average resultant force \(F\) is:
\(F = \frac{\Delta p}{\Delta t} = \frac{2.8\text{ N s}}{0.14\text{ s}} = 20\text{ N}\).

1 mark for calculating the correct change in momentum (accounting for direction) and using Newton's second law to find a force of 20 N.

(a) Ultimate tensile stress is the maximum tensile stress that a material can withstand before it breaks.

(b) In elastic deformation, the material returns to its original length/shape when the deforming force is removed. In plastic deformation, there is permanent deformation and the material does not return to its original length/shape when the force is removed.

(c)(i) Force on wire: \(F = m g = 4.5 \times 9.81 = 44.15\text{ N}\).
Young modulus: \(E = \frac{\text{stress}}{\text{strain}} = \frac{F L}{A x}\).
Rearranging for extension \(x\):
\(x = \frac{F L}{A E} = \frac{44.15 \times 2.4}{1.5 \times 10^{-7} \times 1.1 \times 10^{11}} = \frac{105.96}{16500} = 6.42 \times 10^{-3}\text{ m} = 6.4\text{ mm}\).

(ii) Elastic potential energy stored: \(E_p = \frac{1}{2} F x = 0.5 \times 44.15 \times 6.42 \times 10^{-3} = 0.142\text{ J} \approx 0.14\text{ J}\).

(iii) The final length will be greater than the original length. When loaded beyond the elastic limit into the plastic region, some layers of atoms slide past each other, leading to permanent deformation. When the load is removed, the wire recovers only its elastic strain, leaving a permanent extension.

(a) Maximum force divided by original cross-sectional area before breaking [1]

(b) Elastic: returns to original shape when load is removed [1]
Plastic: permanent extension/does not return to original shape when load is removed [1]

(c)(i) Use of \(F = m g\) to get \(44.15\text{ N}\) [1]
Formula \(x = \frac{F L}{A E}\) used [1]
Calculation to yield \(6.4 \times 10^{-3}\text{ m}\) (accept \(6.4\text{ mm}\) or \(6.42\text{ mm}\)) [1]

(ii) Formula \(E_p = \frac{1}{2} F x\) or \(E_p = \frac{1}{2} k x^2\) [1]
Calculation to yield \(0.14\text{ J}\) (accept \(0.142\text{ J}\)) [1]

(iii) Final length is longer / permanent extension remains [1]
Explanation: atoms have slid past each other / permanent deformation occurred because plastic limit/elastic limit was exceeded [1]

(a) Two differences:
1. Progressive waves transfer energy through the medium, whereas stationary waves store energy (there is no net transfer of energy).
2. In progressive waves, all particles have the same amplitude (assuming no damping), whereas in stationary waves, the amplitude varies from zero at nodes to a maximum at antinodes.
(Other acceptable differences: phase difference variation, shape propagation).

(b)(i) Sound waves travel down the tube and reflect at the water surface. The incident and reflected waves, which have the same frequency and amplitude, travel in opposite directions and superpose to form a stationary wave.

(ii) A displacement node (zero amplitude) is formed at the closed end (water surface) and a displacement antinode (maximum amplitude) is formed at the open end of the tube.

(iii) For the first resonance of a tube closed at one end:
\(L = \frac{\lambda}{4} \implies \lambda = 4 L = 4 \times 0.160\text{ m} = 0.640\text{ m}\).
Speed of sound: \(v = f \lambda = 512 \times 0.640 = 327.68\text{ m s}^{-1} \approx 328\text{ m s}^{-1}\).

(iv) The new frequency is \(f' = 1536\text{ Hz}\).
\(\frac{f'}{f} = \frac{1536}{512} = 3\).
Since the second resonance (third harmonic) for a tube closed at one end occurs at odd harmonics (\(3 f\), \(5 f\), etc.), the tube will resonate at \(1536\text{ Hz}\) with three quarter-wavelengths fitting exactly in the tube of length \(16.0\text{ cm}\). Thus, resonance is heard.

(a) Difference 1: Stationary waves do not transfer energy, progressive waves do. [1]
Difference 2: Stationary waves have nodes/antinodes (varying amplitude), progressive waves have constant amplitude. [1]

(b)(i) Incident wave reflects at the closed water end [1]
Incident and reflected waves superpose/interfere (having same frequency and amplitude, traveling in opposite directions) [1]

(ii) Displacement node at the water surface/closed end [1]
Displacement antinode at/near the open end [1]

(iii) Correct expression \(\lambda = 4 L = 0.640\text{ m}\) [1]
Correct calculation: \(v = f \lambda = 328\text{ m s}^{-1}\) (accept \(327.7\text{ m s}^{-1}\)) [1]

(iv) Ratio \(f' / f = 3\) or \(\lambda' = \lambda / 3\) calculated [1]
Conclusion: Yes, resonance occurs because it is an odd harmonic (third harmonic / \(3 \lambda / 4\)) [1]

(a) The total momentum of a closed system remains constant, provided there are no external forces acting on it.

(b)(i) Let direction to the right be positive.
Initial momentum: \(p_i = m_A u_A + m_B u_B = 0.350 \times 2.40 + 0.150 \times (-1.20) = 0.840 - 0.180 = 0.660\text{ kg m s}^{-1}\).
Final momentum: \(p_f = m_A v_A + m_B v_B = 0.350 \times 0.84 + 0.150 \times v_B = 0.294 + 0.150 v_B\).
By conservation of momentum:
\(0.294 + 0.150 v_B = 0.660 \implies 0.150 v_B = 0.366 \implies v_B = 2.44\text{ m s}^{-1}\).
Since the sign is positive, the direction is to the right.

(ii) Initial Kinetic Energy:
\(E_{k,i} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2 = 0.5 \times 0.350 \times (2.40)^2 + 0.5 \times 0.150 \times (1.20)^2 = 1.008 + 0.108 = 1.116\text{ J}\).
Final Kinetic Energy:
\(E_{k,f} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = 0.5 \times 0.350 \times (0.84)^2 + 0.5 \times 0.150 \times (2.44)^2 = 0.1235 + 0.4465 = 0.570\text{ J}\).
Since \(E_{k,f} < E_{k,i}\), kinetic energy is not conserved, meaning the collision is inelastic.

(iii) Change in momentum of glider B:
\(\Delta p_B = m_B (v_B - u_B) = 0.150 \times (2.44 - (-1.20)) = 0.150 \times 3.64 = 0.546\text{ N s}\).
Using \(F = \frac{\Delta p}{\Delta t}\):
\(14.0 = \frac{0.546}{\Delta t} \implies \Delta t = \frac{0.546}{14.0} = 0.0390\text{ s}\).

(a) Total momentum remains constant [1]
For a closed system / no external force acting [1]

(b)(i) Correct signs for velocities (e.g., \(-1.20\text{ m s}^{-1}\) for B) [1]
Correct application of conservation of momentum: \(0.350 \times 2.40 + 0.150 \times (-1.20) = 0.350 \times 0.84 + 0.150 v_B\) [1]
Velocity of B: \(2.44\text{ m s}^{-1}\) to the right (must specify direction) [1]

(ii) Calculation of initial kinetic energy: \(1.12\text{ J}\) [1]
Calculation of final kinetic energy: \(0.57\text{ J}\) [1]
Conclusion: Inelastic because KE is not conserved (energy lost) [1]

(iii) Correct change in momentum of glider B: \(0.546\text{ N s}\) (or glider A) [1]
Time of contact: \(0.0390\text{ s}\) (accept \(0.039\text{ s}\)) [1]

(a) Potential difference is the electrical energy converted into other forms of energy per unit charge passing through a component: \(V = W/Q\).

(b)(i) Total resistance of circuit: \(R_{total} = R + r\).
Current in circuit: \(I = \frac{E}{R+r}\).
Power dissipated in \(R\): \(P = I^2 R = \frac{E^2 R}{(R+r)^2}\).

(ii) Power \(P = \frac{\text{Energy}}{\text{time}}\).
Base unit of Energy (Joule): \(\text{N m} = \text{kg m s}^{-2} \times \text{m} = \text{kg m}^2\text{ s}^{-2}\).
Therefore, base unit of Power: \(\frac{\text{kg m}^2\text{ s}^{-2}}{\text{s}} = \text{kg m}^2\text{ s}^{-3}\).

(c)(i) Using \(E = I(R + r)\):
Case 1: \(E = 2.0(4.0 + r) = 8.0 + 2.0r\)
Case 2: \(E = 1.0(10.0 + r) = 10.0 + r\)
Equating Case 1 and Case 2:
\(8.0 + 2.0r = 10.0 + r \implies r = 2.0\ \Omega\).
Substituting \(r\) back into Case 2:
\(E = 10.0 + 2.0 = 12.0\text{ V}\).

(ii) Useful power output when \(R = 10.0\ \Omega\):
\(P_{out} = I^2 R = (1.0)^2 \times 10.0 = 10.0\text{ W}\).
Total power produced by the cell:
\(P_{total} = E I = 12.0 \times 1.0 = 12.0\text{ W}\).
Efficiency: \(\eta = \frac{P_{out}}{P_{total}} = \frac{10.0}{12.0} \approx 0.833\text{ or } 83.3\%\).

(a) Energy transferred per unit charge [1]

(b)(i) Expression \(I = \frac{E}{R+r}\) used or implied [1]
Correct final formula: \(P = \frac{E^2 R}{(R+r)^2}\) [1]

(ii) Power defined as Work done/time [1]
Substitution of base units for work/time to yield \(\text{kg m}^2\text{ s}^{-3}\) [1]

(c)(i) Set up of two correct simultaneous equations [1]
Solving for \(r = 2.0\ \Omega\) [1]
Solving for \(E = 12.0\text{ V}\) [1]

(ii) Useful power calculated (\(10.0\text{ W}\)) or formula for efficiency \(\frac{R}{R+r}\) identified [1]
Correct calculation of efficiency as \(83.3\%\) (or \(0.833\)) [1]

(a) Kirchhoff's first law states that the sum of the currents entering a junction is equal to the sum of the currents leaving the junction. This is a consequence of the conservation of charge.

(b) Kirchhoff's second law states that the sum of the e.m.f.s in any closed loop is equal to the sum of the p.d.s in that loop. It expresses the conservation of energy because the total electrical potential energy given to the charges by the source equals the total electrical energy transformed in the circuit components per unit charge.

(c)(i) At junction X: \(I_1 + I_2 = I_3\).

(ii) In the left loop (moving clockwise from Y through left branch, then down central branch):
\(+E_1 - I_1 R_1 - I_3 R_3 = 0 \implies 9.0 - 2.0 I_1 - 5.0 I_3 = 0\) (or equivalent).

(iii) In the right loop (moving clockwise from Y through central branch, then down right branch):
\(+I_3 R_3 + I_2 R_2 - E_2 = 0 \implies 5.0 I_3 + 3.0 I_2 - 2.0 = 0\), which rearranges to:
\(2.0 - 3.0 I_2 - 5.0 I_3 = 0\) (moving anticlockwise through the right branch and down central branch).

(iv) We have the simultaneous equations:
1) \(2 I_1 + 5 I_3 = 9\)
2) \(3 I_2 + 5 I_3 = 2\)
Substitute \(I_3 = I_1 + I_2\) into both equations:
\(2 I_1 + 5(I_1 + I_2) = 9 \implies 7 I_1 + 5 I_2 = 9\)
\(3 I_2 + 5(I_1 + I_2) = 2 \implies 5 I_1 + 8 I_2 = 2\)

Multiply first by 8 and second by 5:
\(56 I_1 + 40 I_2 = 72\)
\(25 I_1 + 40 I_2 = 10\)
Subtracting the two equations:
\(31 I_1 = 62 \implies I_1 = 2.0\text{ A}\).

Substitute \(I_1 = 2.0\text{ A}\) back:
\(7(2.0) + 5 I_2 = 9 \implies 14 + 5 I_2 = 9 \implies 5 I_2 = -5 \implies I_2 = -1.0\text{ A}\).

Then, \(I_3 = I_1 + I_2 = 2.0 + (-1.0) = 1.0\text{ A}\).

(a) Sum of currents entering a junction equals sum of currents leaving [1]
Conservation of charge [1]

(b) Sum of e.m.f.s = sum of p.d.s around a closed loop [1]
Energy is conserved as charge moves around a closed loop [1]

(c)(i) \(I_1 + I_2 = I_3\) [1]

(ii) Correct equation: \(9.0 = 2.0 I_1 + 5.0 I_3\) [1]

(iii) Correct equation: \(2.0 = 3.0 I_2 + 5.0 I_3\) [1]

(iv) Method of solving simultaneous equations shown [1]
Correct values for \(I_1 = 2.0\text{ A}\) and \(I_2 = -1.0\text{ A}\) [1]
Correct value for \(I_3 = 1.0\text{ A}\) (consistent with others) [1]

(a)(i) Moment is the product of the force and the perpendicular distance from the pivot to the line of action of the force.
(ii) A couple is a pair of equal and opposite parallel forces that act on an object but not along the same line, producing a rotational force (torque) without any net translation.

(b) For an object to be in equilibrium:
1. The net (resultant) force acting on the object must be zero.
2. The net (resultant) moment (torque) acting about any point must be zero.

(c)(i) Uniform shelf has weight \(W = 24.0\text{ N}\) acting at its center of gravity (distance \(0.60\text{ m}\) from end A).
Tension \(T\) acts at end B (distance \(1.20\text{ m}\) from A) at an angle of \(30.0^\circ\).
Taking moments about the pivot A:
\(\Sigma M_A = 0 \implies (W \times 0.60) - (T \sin(30.0^\circ) \times 1.20) = 0\)
\(24.0 \times 0.60 = T \times 0.500 \times 1.20\)
\(14.4 = 0.600 T \implies T = 24.0\text{ N}\).

(ii) For horizontal equilibrium, the horizontal force from the pivot \(H_A\) must balance the horizontal component of the tension:
\(H_A = T \cos(30.0^\circ) = 24.0 \times \cos(30.0^\circ) = 20.78\text{ N} \approx 20.8\text{ N}\).

(iii) For vertical equilibrium, the sum of upward forces must equal the sum of downward forces:
\(V_A + T \sin(30.0^\circ) = W\)
\(V_A + 24.0 \times \sin(30.0^\circ) = 24.0\)
\(V_A + 12.0 = 24.0 \implies V_A = 12.0\text{ N}\).
Since \(V_A\) is positive, it must act upwards.

(a)(i) Force multiplied by perpendicular distance from pivot to line of action of force [1]
(ii) Pair of equal and opposite parallel forces [1]
Separated by a distance/producing rotational torque only [1]

(b) Resultant force is zero [1]
Resultant moment about any point is zero [1]

(c)(i) Taking moments about A with weight acting at center (0.60 m) [1]
Correct moment equation: \(24.0 \times 0.60 = T \sin(30.0^\circ) \times 1.20\) [1]
Correct calculation for \(T = 24.0\text{ N}\) [1]

(ii) Equating horizontal component \(H_A = T \cos(30.0^\circ)\) [1]
Calculation: \(20.8\text{ N}\) (accept \(21\text{ N}\)) [1]

(iii) Vertical force balance: \(V_A + T \sin(30.0^\circ) = 24.0\) [1]
Calculation: \(12.0\text{ N}\) acting upwards (must state direction) [1]