An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V2) Cambridge International AS Level Physics (9702) paper. Not affiliated with or reproduced from Cambridge.
Paper 1 (Multiple Choice)
Answer all forty multiple choice questions. Each question carries one mark. Choose the single best option (A, B, C, or D) and mark on the separate answer sheet.
40 Question · 40 marks
Question 1 · Multiple Choice
1 marks
Which physical quantity can be expressed in terms of SI base units as \(\text{kg m}^2\text{ s}^{-3}\text{ A}^{-1}\)?
A.electric current
B.electromotive force
C.electrical resistance
D.electric charge
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Worked solution
Electromotive force (e.m.f.) and potential difference are defined as energy transferred per unit charge: \(V = \frac{W}{Q}\). - Work \(W = F \times s\), with base units \((\text{kg m s}^{-2}) \times \text{m} = \text{kg m}^2\text{ s}^{-2}\). - Charge \(Q = I \times t\), with base units \(\text{A s}\).
Dividing the two gives: \[\text{Units of } V = \frac{\text{kg m}^2\text{ s}^{-2}}{\text{A s}} = \text{kg m}^2\text{ s}^{-3}\text{ A}^{-1}\] Thus, the correct quantity is electromotive force.
Marking scheme
B [1 mark] - A is incorrect (base unit \(\text{A}\)) - C is incorrect (resistance is \(V/I = \text{kg m}^2\text{ s}^{-3}\text{ A}^{-2}\)) - D is incorrect (charge is \(\text{A s}\))
Question 2 · Multiple Choice
1 marks
A small sphere is projected horizontally with an initial speed \(u\) from the edge of a cliff of height \(h\). The sphere hits the level ground below at an angle of \(45^\circ\) to the horizontal.
Air resistance is negligible. The acceleration of free fall is \(g\).
What is the height \(h\) of the cliff in terms of \(u\) and \(g\)?
A.\(\frac{u^2}{4g}\)
B.\(\frac{u^2}{2g}\)
C.\(\frac{u^2}{g}\)
D.\(\frac{2u^2}{g}\)
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Worked solution
Horizontal motion has constant velocity, so just before hitting the ground: \[v_x = u\] Since the velocity vector makes an angle of \(45^\circ\) with the horizontal: \[\tan 45^\circ = \frac{v_y}{v_x} = 1 \implies v_y = v_x = u\] For the vertical motion, starting from rest (\(u_y = 0\)): \[v_y^2 = u_y^2 + 2gh = 2gh\] Substituting \(v_y = u\): \[u^2 = 2gh \implies h = \frac{u^2}{2g}\]
Marking scheme
B [1 mark] - A corresponds to an incorrect kinematic relation \(4gh\). - C corresponds to forgetting the factor of \(1/2\) in \(v^2 = 2as\). - D corresponds to an inverted fraction factor.
Question 3 · Multiple Choice
1 marks
A trolley of mass \(2.0\text{ kg}\) travels at a speed of \(6.0\text{ m s}^{-1}\) along a frictionless track and collides with a stationary trolley of mass \(4.0\text{ kg}\). After the collision, the two trolleys stick together and move with a common speed.
What is the loss of kinetic energy during the collision?
A.\(12\text{ J}\)
B.\(24\text{ J}\)
C.\(36\text{ J}\)
D.\(48\text{ J}\)
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Worked solution
By conservation of linear momentum: \[m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\] \[(2.0)(6.0) + (4.0)(0) = (2.0 + 4.0) v\] \[12.0 = 6.0 v \implies v = 2.0\text{ m s}^{-1}\]
Loss of kinetic energy: \[\Delta E_{\text{k}} = 36\text{ J} - 12\text{ J} = 24\text{ J}\]
Marking scheme
B [1 mark] - A is the remaining kinetic energy, not the loss. - C is the initial kinetic energy. - D is an arithmetic error from incorrect velocity calculation.
Question 4 · Multiple Choice
1 marks
A uniform solid block of wood has a side length of \(0.10\text{ m}\) and floats in equilibrium in liquid water of density \(1000\text{ kg m}^{-3}\). The base of the block is horizontal, and it is submerged to a depth of \(0.060\text{ m}\).
What is the density of the wood?
A.\(170\text{ kg m}^{-3}\)
B.\(400\text{ kg m}^{-3}\)
C.\(600\text{ kg m}^{-3}\)
D.\(1670\text{ kg m}^{-3}\)
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Worked solution
For an object floating in equilibrium: \[\text{Weight of block} = \text{Upthrust}\] \[\rho_{\text{wood}} V_{\text{total}} g = \rho_{\text{water}} V_{\text{submerged}} g\] Since the cross-sectional area \(A\) is uniform: \[\rho_{\text{wood}} (A \times h_{\text{total}}) = \rho_{\text{water}} (A \times h_{\text{submerged}})\] \[\rho_{\text{wood}} = \rho_{\text{water}} \times \frac{h_{\text{submerged}}}{h_{\text{total}}} = 1000 \times \frac{0.060}{0.10} = 600\text{ kg m}^{-3}\]
Marking scheme
C [1 mark] - A corresponds to \(1000 \times (0.10 - 0.060) / 0.24\). - B corresponds to subtracting the submerged ratio from 1 (unsubmerged fraction). - D corresponds to inverted ratio \(1000 / 0.60\).
Question 5 · Multiple Choice
1 marks
A vehicle of mass \(1200\text{ kg}\) travels up a slope inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = 0.050\), at a constant speed of \(20\text{ m s}^{-1}\).
The total resistive force opposing the motion of the vehicle is \(400\text{ N}\). The acceleration of free fall \(g\) is \(9.81\text{ m s}^{-2}\).
What is the useful power output delivered by the engine of the vehicle?
A.\(8.0\text{ kW}\)
B.\(12\text{ kW}\)
C.\(20\text{ kW}\)
D.\(28\text{ kW}\)
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Worked solution
Since the speed is constant, the forward driving force \(F\) equals the sum of the resistive force and the component of weight acting down the slope: \[F = mg\sin\theta + F_{\text{resistive}}\] \[F = (1200 \times 9.81 \times 0.050) + 400 = 588.6 + 400 = 988.6\text{ N}\]
The power output is: \[P = Fv = 988.6\text{ N} \times 20\text{ m s}^{-1} = 19772\text{ W} \approx 20\text{ kW}\]
Marking scheme
C [1 mark] - A considers only the resistive force (\(400 \times 20 = 8.0\text{ kW}\)). - B considers only the gravitational component (\(588.6 \times 20 \approx 12\text{ kW}\)). - D results from using \(\cos\theta\) instead of \(\sin\theta\) or arithmetic error.
Question 6 · Multiple Choice
1 marks
A stationary observer detects sound of frequency \(540\text{ Hz}\) from a siren on an emergency vehicle approaching at a constant speed \(v_s\). When the vehicle moves directly away from the observer at the same speed \(v_s\), the observed frequency is \(460\text{ Hz}\).
The speed of sound in air is \(340\text{ m s}^{-1}\).
What is the speed \(v_s\) of the vehicle?
A.\(14\text{ m s}^{-1}\)
B.\(27\text{ m s}^{-1}\)
C.\(30\text{ m s}^{-1}\)
D.\(54\text{ m s}^{-1}\)
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Worked solution
Using the Doppler effect formula for a moving source: When approaching: \[f_1 = f_s \left( \frac{v}{v - v_s} \right) = 540\text{ Hz}\] When receding: \[f_2 = f_s \left( \frac{v}{v + v_s} \right) = 460\text{ Hz}\]
B [1 mark] - A corresponds to halving the frequency shift. - C is an approximation from assuming linear frequency shift. - D corresponds to \(2 v_s\).
Question 7 · Multiple Choice
1 marks
A potential divider circuit consists of a fixed resistor of resistance \(6.0\text{ k}\Omega\) connected in series with a negative temperature coefficient (NTC) thermistor across a power supply of constant electromotive force (e.m.f.) and negligible internal resistance.
The output potential difference \(V_{\text{out}}\) is measured across the fixed \(6.0\text{ k}\Omega\) resistor.
The temperature of the thermistor is increased.
Which row correctly describes the changes that occur?
A.Resistance of thermistor: decreases | Potential difference \(V_{\text{out}}\): decreases
B.Resistance of thermistor: decreases | Potential difference \(V_{\text{out}}\): increases
C.Resistance of thermistor: increases | Potential difference \(V_{\text{out}}\): decreases
D.Resistance of thermistor: increases | Potential difference \(V_{\text{out}}\): increases
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Worked solution
1. For an NTC thermistor, an increase in temperature causes the release of more charge carriers, so its resistance decreases. 2. In a series circuit, total resistance \(R_{\text{total}} = R_{\text{fixed}} + R_{\text{thermistor}}\) decreases. 3. The circuit current \(I = \frac{E}{R_{\text{total}}}\) increases. 4. The output potential difference across the fixed resistor is \(V_{\text{out}} = I R_{\text{fixed}}\), which increases as current \(I\) increases.
Marking scheme
B [1 mark] - A fails to recognize that \(V_{\text{out}}\) is across the fixed resistor. - C and D assume a positive temperature coefficient behavior.
Question 8 · Multiple Choice
1 marks
A nucleus undergoes \(\beta^+\) (beta-plus) decay. During this process, a proton in the nucleus changes into a neutron.
Which row gives the change in quark flavour and the leptons emitted in this decay?
A.Quark change: \(u \to d\) | Leptons emitted: positron and electron neutrino
B.Quark change: \(u \to d\) | Leptons emitted: positron and electron antineutrino
C.Quark change: \(d \to u\) | Leptons emitted: electron and electron neutrino
D.Quark change: \(d \to u\) | Leptons emitted: electron and electron antineutrino
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Worked solution
- A proton has quark composition \(uud\) and a neutron has quark composition \(udd\). - In \(\beta^+\) decay, a proton changes into a neutron, so an up quark changes to a down quark (\(u \to d\)). - To conserve charge, lepton number, and other quantum numbers, a positron (\(\text{e}^+\)) and an electron neutrino (\(\nu_e\)) are emitted.
Marking scheme
A [1 mark] - B emits an antineutrino, which violates lepton number conservation in \(\beta^+\) decay. - C and D correspond to \(\beta^-\) decay where a down quark changes to an up quark (\(d \to u\)).
Question 9 · multiple_choice
1 marks
What are the SI base units of electrical resistivity?
A.\(\text{kg m}^2 \text{s}^{-3} \text{A}^{-1}\)
B.\(\text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\)
C.\(\text{kg m}^3 \text{s}^{-2} \text{A}^{-2}\)
D.\(\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\)
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Worked solution
Resistivity \(\rho\) is given by \(\rho = \frac{R A}{L}\). Resistance \(R = \frac{V}{I} = \frac{W}{Q I} = \frac{W}{I^2 t}\). In SI base units: Work done \(W\) has base units \(\text{kg m}^2 \text{s}^{-2}\). Therefore, \(R\) has base units \(\frac{\text{kg m}^2 \text{s}^{-2}}{\text{A}^2 \text{s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\). Multiplying by cross-sectional area \((\text{m}^2)\) and dividing by length \((\text{m})\): \([\rho] = (\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}) \times \text{m} = \text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\).
Marking scheme
B [1 mark]: Correct identification of base units of resistivity. A, C, D are incorrect combinations of powers of m, s, and A.
Question 10 · multiple_choice
1 marks
A small ball is projected horizontally from the edge of a vertical cliff of height \(h\). The ball hits the horizontal ground below at a horizontal distance \(x\) from the base of the cliff.
At the instant before impact, the velocity of the ball is directed at an angle of \(60^\circ\) below the horizontal. Air resistance is negligible.
Which expression gives \(h\) in terms of \(x\)?
A.\(\frac{x}{2\sqrt{3}}\)
B.\(\frac{\sqrt{3}}{2} x\)
C.\(\sqrt{3} x\)
D.\(\frac{3}{2} x\)
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Worked solution
Let \(u\) be the initial horizontal velocity and \(t\) be the time of flight. Horizontal motion: \(x = u t \implies t = \frac{x}{u}\). Vertical motion: \(v_y = g t\) and \(h = \frac{1}{2} g t^2 = \frac{v_y^2}{2g}\). At impact, \(\tan 60^\circ = \frac{v_y}{u} = \sqrt{3} \implies v_y = u\sqrt{3}\). Using \(x = u t = u \left(\frac{v_y}{g}\right)\), we have: \(\frac{h}{x} = \frac{\frac{v_y^2}{2g}}{\frac{u v_y}{g}} = \frac{v_y}{2u} = \frac{\tan 60^\circ}{2} = \frac{\sqrt{3}}{2}\). Therefore, \(h = \frac{\sqrt{3}}{2} x\).
Marking scheme
B [1 mark]: Correct kinematic derivation yielding \(h = \frac{\sqrt{3}}{2} x\). A: Inverted trigonometric ratio. C: Omitted factor of \(\frac{1}{2}\) in distance formula. D: Miscalculated \(\tan 60^\circ\).
Question 11 · multiple_choice
1 marks
Two gliders, X and Y, move towards each other on a horizontal frictionless track.
Glider X has mass \(0.40\text{ kg}\) and moves to the right with a speed of \(1.5\text{ m s}^{-1}\). Glider Y has mass \(0.20\text{ kg}\) and moves to the left with a speed of \(3.0\text{ m s}^{-1}\).
The gliders undergo a perfectly elastic head-on collision.
What are the velocities of X and Y after the collision?
A.\(1.5\text{ m s}^{-1}\) to the left for X, and \(3.0\text{ m s}^{-1}\) to the right for Y
B.\(3.0\text{ m s}^{-1}\) to the left for X, and \(1.5\text{ m s}^{-1}\) to the right for Y
C.\(0.5\text{ m s}^{-1}\) to the left for X, and \(1.0\text{ m s}^{-1}\) to the right for Y
D.\(0\text{ m s}^{-1}\) for X, and \(0\text{ m s}^{-1}\) for Y
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Worked solution
Initial total momentum \(p_i\): \(p_i = m_X u_X + m_Y u_Y = (0.40)(+1.5) + (0.20)(-3.0) = 0.60 - 0.60 = 0\text{ kg m s}^{-1}\). Since the collision is perfectly elastic: 1. Total momentum is conserved: \(0.40 v_X + 0.20 v_Y = 0 \implies v_Y = -2 v_X\). 2. Relative speed of separation equals relative speed of approach: Relative speed of approach \(= 1.5 - (-3.0) = 4.5\text{ m s}^{-1}\). Relative speed of separation \(= v_Y - v_X = (-2v_X) - v_X = -3v_X = 4.5\text{ m s}^{-1}\). Thus, \(v_X = -1.5\text{ m s}^{-1}\) (i.e. \(1.5\text{ m s}^{-1}\) to the left) and \(v_Y = -2(-1.5) = +3.0\text{ m s}^{-1}\) (i.e. \(3.0\text{ m s}^{-1}\) to the right).
Marking scheme
A [1 mark]: Correct application of conservation of momentum and relative speed in an elastic collision. B: Swapped velocities. C: Inelastic outcome. D: Completely inelastic outcome.
Question 12 · multiple_choice
1 marks
A uniform solid cylinder of cross-sectional area \(A\) and length \(L\) floats upright in a liquid of density \(\rho\). A fraction \(f\) of the length of the cylinder is submerged beneath the surface of the liquid.
The atmospheric pressure is \(p_0\) and the acceleration of free fall is \(g\).
What is the difference between the downward force exerted by the atmosphere on the top flat face of the cylinder and the upward hydrostatic force exerted by the liquid on the bottom flat face of the cylinder?
A.\(\rho g L A\)
B.\(f \rho g L A\)
C.\(p_0 A + f \rho g L A\)
D.\((1 - f) \rho g L A\)
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Worked solution
The downward force exerted by the atmosphere on the top face is: \(F_{\text{top}} = p_0 A\). The depth of the bottom face below the liquid surface is \(h = f L\). The pressure at the bottom face is \(p_{\text{bottom}} = p_0 + \rho g (f L)\). The upward force on the bottom face is: \(F_{\text{bottom}} = p_{\text{bottom}} A = p_0 A + f \rho g L A\). The magnitude of the difference between these two forces is: \(F_{\text{bottom}} - F_{\text{top}} = (p_0 A + f \rho g L A) - p_0 A = f \rho g L A\).
Marking scheme
B [1 mark]: Correctly relates hydrostatic force difference to upthrust \(f \rho g L A\). A: Misses the submerged fraction \(f\). C: Includes atmospheric pressure term twice. D: Uses the unsubmerged fraction \((1-f)\).
Question 13 · multiple_choice
1 marks
An electric motor pulls a crate of mass \(450\text{ kg}\) at a constant speed of \(0.80\text{ m s}^{-1}\) up a ramp inclined at \(30^\circ\) to the horizontal.
A constant frictional force of \(600\text{ N}\) opposes the motion of the crate. The overall efficiency of the motor system is \(75\%\).
What is the electrical input power to the motor?
A.\(1.7\text{ kW}\)
B.\(2.2\text{ kW}\)
C.\(2.4\text{ kW}\)
D.\(3.0\text{ kW}\)
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Worked solution
The force \(F\) required to pull the crate at constant speed must balance both the component of weight down the slope and the frictional force: \(F = m g \sin 30^\circ + F_{\text{friction}}\) \(F = (450 \times 9.81 \times \sin 30^\circ) + 600 = 2207.25 + 600 = 2807.25\text{ N}\).
The useful mechanical output power is: \(P_{\text{out}} = F v = 2807.25 \times 0.80 = 2245.8\text{ W}\).
D [1 mark]: Correctly determines total force, output power, and divides by efficiency. A: Multiplied by efficiency instead of dividing. B: Useful output power \(P_{\text{out}}\) without accounting for efficiency. C: Omitted friction force.
Question 14 · multiple_choice
1 marks
Two wires, P and Q, are made of the same metal and have circular cross-sections.
Wire P has length \(L\) and diameter \(d\). Wire Q has length \(2L\) and diameter \(2d\).
A tensile force \(F\) is applied to wire P, causing an extension \(e\) and storing elastic potential energy \(E\). Both wires obey Hooke's law.
A tensile force \(2F\) is applied to wire Q.
What are the extension and the elastic potential energy stored in wire Q?
A.extension \(= e\), energy \(= 2E\)
B.extension \(= 2e\), energy \(= 2E\)
C.extension \(= e\), energy \(= 4E\)
D.extension \(= \frac{1}{2}e\), energy \(= E\)
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Worked solution
Extension \(e = \frac{F L}{A E_Y}\) where \(A = \frac{\pi d^2}{4}\). For wire P: \(e_P = e \propto \frac{F L}{d^2}\). For wire Q: \(e_Q \propto \frac{(2F)(2L)}{(2d)^2} = \frac{4FL}{4d^2} = \frac{FL}{d^2} = e\).
Elastic potential energy is \(E_p = \frac{1}{2} F_{\text{applied}} e\). For wire P: \(E = \frac{1}{2} F e\). For wire Q: \(E_Q = \frac{1}{2} (2F) e_Q = \frac{1}{2} (2F) e = 2 \left(\frac{1}{2} F e\right) = 2E\).
Marking scheme
A [1 mark]: Correctly identifies that extension is \(e\) and stored energy is \(2E\). B: Incorrect scaling of area with diameter. C: Incorrect squared dependence on force in energy formula without matching extension. D: Miscalculated ratios.
Question 15 · multiple_choice
1 marks
Light of wavelength \(580\text{ nm}\) is incident normally on a diffraction grating. The third-order maximum is formed at an angle of \(54^\circ\) to the central zero-order maximum.
What is the total number of maxima visible in the interference pattern produced on a semicircular screen centered on the grating?
A.3
B.6
C.7
D.8
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Worked solution
Using the grating equation \(d \sin \theta = n \lambda\): For \(n = 3\) and \(\theta = 54^\circ\): \(d = \frac{3 \times 580 \times 10^{-9}}{\sin 54^\circ} = \frac{1.74 \times 10^{-6}}{0.8090} \approx 2.151 \times 10^{-6}\text{ m}\).
For the highest observable order \(n_{\text{max}}\), \(\sin \theta \le 1\): \(n_{\text{max}} \le \frac{d}{\lambda} = \frac{2.151 \times 10^{-6}}{580 \times 10^{-9}} \approx 3.71\). Since \(n\) must be an integer, \(n_{\text{max}} = 3\).
The total number of maxima is \(2 n_{\text{max}} + 1 = 2(3) + 1 = 7\) (orders \(n = -3, -2, -1, 0, 1, 2, 3\)).
Marking scheme
C [1 mark]: Correctly identifies \(n_{\text{max}} = 3\) and computes total maxima as \(2n+1 = 7\). A: Gave \(n_{\text{max}}\) only. B: Omitted central zero-order maximum \((2n)\). D: Incorrectly rounded up \(n_{\text{max}}\) to 4.
Question 16 · multiple_choice
1 marks
A cell of electromotive force (e.m.f.) \(E\) and internal resistance \(r\) is connected in series with a fixed resistor of resistance \(R\) and a variable resistor of resistance \(R_v\).
When \(R_v\) is set to \(4.0\ \Omega\), the current in the circuit is \(1.2\text{ A}\). When \(R_v\) is set to \(10.0\ \Omega\), the current decreases to \(0.80\text{ A}\).
What is the value of \((R + r)\)?
A.\(6.0\ \Omega\)
B.\(8.0\ \Omega\)
C.\(12\ \Omega\)
D.\(14\ \Omega\)
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Worked solution
The circuit equation is \(E = I(R + r + R_v)\). Let \(S = R + r\).
For the first setting: \(E = 1.2 (S + 4.0) = 1.2 S + 4.8\)
For the second setting: \(E = 0.80 (S + 10.0) = 0.80 S + 8.0\)
Equating both expressions for \(E\): \(1.2 S + 4.8 = 0.80 S + 8.0\) \(0.40 S = 3.2\) \(S = \frac{3.2}{0.40} = 8.0\ \Omega\).
Thus, \(R + r = 8.0\ \Omega\).
Marking scheme
B [1 mark]: Correct setup and solution of simultaneous equations yielding \(8.0\ \Omega\). A: Calculation error in subtraction. C: Substituted into one equation incorrectly. D: Added total resistance of one state instead of \((R+r)\).
Question 17 · Multiple Choice
1 marks
Which row gives the correct derived SI base units for electrical resistivity?
A.\(\text{kg m}^2 \text{s}^{-3} \text{A}^{-1}\)
B.\(\text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\)
C.\(\text{kg m}^3 \text{s}^{-2} \text{A}^{-2}\)
D.\(\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\)
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Worked solution
Resistivity \(\rho\) is defined by \(R = \frac{\rho L}{A}\), so \(\rho = \frac{R A}{L}\). Resistance is \(R = \frac{V}{I}\), where potential difference is work done per unit charge: \(V = \frac{W}{Q} = \frac{W}{I t}\). In SI base units: \[ [W] = \text{kg m}^2 \text{s}^{-2} \] \[ [V] = \frac{\text{kg m}^2 \text{s}^{-2}}{\text{A s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-1} \] \[ [R] = \frac{\text{kg m}^2 \text{s}^{-3} \text{A}^{-1}}{\text{A}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2} \] Therefore, the base units of resistivity are: \[ [\rho] = [R] \times \frac{[A]}{[L]} = (\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}) \times \frac{\text{m}^2}{\text{m}} = \text{kg m}^3 \text{s}^{-3} \text{A}^{-2} \]
Marking scheme
B is correct [1 mark]. A incorrectly omits the length dimension factor for resistivity. C has incorrect powers for time \(\text{s}\). D gives the base units of resistance, not resistivity.
Question 18 · Multiple Choice
1 marks
A stone is thrown vertically upwards from the top edge of a cliff of height \(35\text{ m}\) with an initial velocity of \(12\text{ m s}^{-1}\). The stone falls freely under gravity and lands at the base of the cliff.
Air resistance is negligible.
What is the total time taken for the stone to reach the base of the cliff?
A.\(1.2\text{ s}\)
B.\(2.4\text{ s}\)
C.\(4.2\text{ s}\)
D.\(5.6\text{ s}\)
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Worked solution
Taking the upwards direction as positive: Initial displacement \(s = -35\text{ m}\) Initial velocity \(u = +12\text{ m s}^{-1}\) Acceleration \(a = -g = -9.81\text{ m s}^{-2}\)
Using the kinematic equation: \[ s = ut + \frac{1}{2}at^2 \] \[ -35 = 12t - 4.905t^2 \] \[ 4.905t^2 - 12t - 35 = 0 \]
Solving the quadratic equation for positive time \(t\): \[ t = \frac{12 + \sqrt{(-12)^2 - 4(4.905)(-35)}}{2(4.905)} = \frac{12 + \sqrt{144 + 686.7}}{9.81} = \frac{12 + 28.82}{9.81} \approx 4.2\text{ s} \]
Marking scheme
C is correct [1 mark]. A corresponds to the time to reach maximum height (\(12/9.81 \approx 1.2\text{ s}\)). B corresponds to the time to return to the release level (\(2.4\text{ s}\)). D arises from using incorrect signs in the quadratic equation.
Question 19 · Multiple Choice
1 marks
A uniform solid cylinder of cross-sectional area \(A\) and height \(h\) is suspended vertically from a light string and is completely submerged in a stationary liquid of density \(\rho\).
The cylinder is made of a material of density \(\sigma\), where \(\sigma > \rho\). The acceleration of free fall is \(g\).
What is the tension in the string?
A.\(\rho A h g\)
B.\(\sigma A h g\)
C.\((\sigma - \rho) A h g\)
D.\((\sigma + \rho) A h g\)
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Worked solution
The forces acting on the cylinder in vertical equilibrium are: 1. Tension \(T\) acting upwards. 2. Upthrust \(U\) acting upwards, equal to the weight of displaced liquid: \(U = \rho V g = \rho A h g\). 3. Weight \(W\) acting downwards: \(W = \sigma V g = \sigma A h g\).
Equating forces: \[ T + U = W \] \[ T = W - U = \sigma A h g - \rho A h g = (\sigma - \rho)Ahg \]
Marking scheme
C is correct [1 mark]. A is the upthrust on the cylinder. B is the total weight of the cylinder. D incorrectly adds the upthrust and weight.
Question 20 · Multiple Choice
1 marks
A car of mass \(1200\text{ kg}\) travels along a level horizontal road. Initially, it moves at a constant speed of \(25\text{ m s}^{-1}\) against a total resistive force of \(640\text{ N}\).
The car then accelerates uniformly from \(25\text{ m s}^{-1}\) to \(30\text{ m s}^{-1}\) in a time of \(8.0\text{ s}\). The resistive force remains constant at \(640\text{ N}\).
What is the average useful power output of the car engine during this \(8.0\text{ s}\) interval?
A.\(18\text{ kW}\)
B.\(21\text{ kW}\)
C.\(38\text{ kW}\)
D.\(42\text{ kW}\)
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Worked solution
Acceleration of the car: \[ a = \frac{v - u}{t} = \frac{30 - 25}{8.0} = 0.625\text{ m s}^{-2} \]
Resultant force required for acceleration: \[ F_{\text{net}} = ma = 1200 \times 0.625 = 750\text{ N} \]
Total forward driving force provided by the engine: \[ F_D = F_{\text{net}} + F_{\text{resistive}} = 750 + 640 = 1390\text{ N} \]
Average speed during the uniform acceleration: \[ v_{\text{avg}} = \frac{25 + 30}{2} = 27.5\text{ m s}^{-1} \]
Average power output: \[ P_{\text{avg}} = F_D \times v_{\text{avg}} = 1390 \times 27.5 = 38225\text{ W} \approx 38\text{ kW} \]
Marking scheme
C is correct [1 mark]. A considers only the power against resistive forces (\(640 \times 27.5 \approx 17.6\text{ kW}\)). B considers only the accelerating power (\(750 \times 27.5 \approx 20.6\text{ kW}\)). D uses the maximum velocity \(30\text{ m s}^{-1}\) rather than the average velocity (\(1390 \times 30 = 41.7\text{ kW}\)).
Question 21 · Multiple Choice
1 marks
A tensile force \(F\) is applied to a metal wire of original length \(L\) and circular cross-section of radius \(r\), producing an extension \(x\) within Hooke's law limit.
A second wire is made of the same metal and has an original length \(2L\) and radius \(2r\).
What tensile force must be applied to the second wire to produce the same extension \(x\)?
A.\(0.5 F\)
B.\(F\)
C.\(2F\)
D.\(4F\)
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Worked solution
From the definition of Young modulus \(E\): \[ E = \frac{\sigma}{\varepsilon} = \frac{F/A}{x/L} = \frac{F L}{\pi r^2 x} \implies F = \frac{E \pi r^2 x}{L} \]
For the second wire with radius \(2r\) and length \(2L\): \[ F_2 = \frac{E \pi (2r)^2 x}{2L} = \frac{E \pi (4r^2) x}{2L} = 2 \left(\frac{E \pi r^2 x}{L}\right) = 2F \]
Marking scheme
C is correct [1 mark]. A incorrectly squares the length ratio. B incorrectly assumes the doubled radius and doubled length cancel linearly. D neglects the doubling of the length.
Question 22 · Multiple Choice
1 marks
A train travels directly away from a stationary observer at a constant speed of \(28\text{ m s}^{-1}\). The horn on the train emits sound of constant frequency \(560\text{ Hz}\).
The speed of sound in air is \(336\text{ m s}^{-1}\).
What is the frequency of the sound observed by the stationary observer?
A.\(517\text{ Hz}\)
B.\(560\text{ Hz}\)
C.\(607\text{ Hz}\)
D.\(611\text{ Hz}\)
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Worked solution
Using the Doppler effect equation for a source moving away from a stationary observer: \[ f_o = f_s \left(\frac{v}{v + v_s}\right) \]
Substituting the given values: \[ f_o = 560 \times \left(\frac{336}{336 + 28}\right) = 560 \times \left(\frac{336}{364}\right) \]
A is correct [1 mark]. B is the unshifted source frequency. C arises from an arithmetic mistake in the denominator. D is obtained by using \(v - v_s\) (source moving towards the observer: \(560 \times 336 / 308 \approx 611\text{ Hz}\)).
Question 23 · Multiple Choice
1 marks
A parallel beam of monochromatic light of wavelength \(600\text{ nm}\) is incident normally on a diffraction grating with \(400\text{ lines per mm}\).
What is the total number of principal intensity maxima (including the central maximum) that can be observed?
A.4
B.5
C.8
D.9
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Using the diffraction grating equation \(d \sin\theta = n\lambda\): For the maximum order \(n_{\text{max}}\), \(\sin\theta \le 1\): \[ n_{\text{max}} \le \frac{d}{\lambda} = \frac{2500\text{ nm}}{600\text{ nm}} = 4.17 \]
Since \(n\) must be an integer, \(n_{\text{max}} = 4\).
The total number of observed maxima is: \[ N = 2n_{\text{max}} + 1 = 2(4) + 1 = 9 \]
Marking scheme
D is correct [1 mark]. A gives only the maximum order \(n = 4\). B considers orders on one side plus the central maximum (\(4 + 1 = 5\)). C accounts for both sides but forgets the central zero-order maximum (\(2 \times 4 = 8\)).
Question 24 · Multiple Choice
1 marks
A potential divider circuit consists of a battery of electromotive force (e.m.f.) \(12.0\text{ V}\) and negligible internal resistance connected in series with a fixed resistor of resistance \(4.0\text{ k}\Omega\) and a negative temperature coefficient (NTC) thermistor.
A high-resistance voltmeter is connected across the \(4.0\text{ k}\Omega\) fixed resistor.
At temperature \(T_1\), the resistance of the thermistor is \(8.0\text{ k}\Omega\). When the temperature changes to \(T_2\), the reading on the voltmeter becomes \(8.0\text{ V}\).
What is the decrease in the resistance of the thermistor?
A.\(2.0\text{ k}\Omega\)
B.\(4.0\text{ k}\Omega\)
C.\(6.0\text{ k}\Omega\)
D.\(8.0\text{ k}\Omega\)
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Worked solution
Initial state at temperature \(T_1\): Thermistor resistance \(R_{\text{th}, 1} = 8.0\text{ k}\Omega\).
Final state at temperature \(T_2\): Voltmeter reading across the fixed resistor \(R = 4.0\text{ k}\Omega\) is \(V_R = 8.0\text{ V}\).
Since the total e.m.f. is \(12.0\text{ V}\), the potential difference across the thermistor is: \[ V_{\text{th}} = 12.0\text{ V} - 8.0\text{ V} = 4.0\text{ V} \]
Circuit current at \(T_2\): \[ I = \frac{V_R}{R} = \frac{8.0\text{ V}}{4.0\text{ k}\Omega} = 2.0\text{ mA} \]
Resistance of the thermistor at \(T_2\): \[ R_{\text{th}, 2} = \frac{V_{\text{th}}}{I} = \frac{4.0\text{ V}}{2.0\text{ mA}} = 2.0\text{ k}\Omega \]
C is correct [1 mark]. A is the final resistance of the thermistor \(R_{\text{th}, 2} = 2.0\text{ k}\Omega\). B is the potential difference across the thermistor or an incorrect division. D is the initial resistance of the thermistor.
Question 25 · Multiple Choice
1 marks
What are the SI base units of electrical resistivity \(\rho\)?
A.\(\text{kg m}^{3} \text{s}^{-3} \text{A}^{-2}\)
B.\(\text{kg m}^{2} \text{s}^{-3} \text{A}^{-2}\)
C.\(\text{kg m}^{3} \text{s}^{-2} \text{A}^{-1}\)
D.\(\text{kg m} \text{s}^{-3} \text{A}^{-2}\)
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Worked solution
Resistivity is given by \(\rho = \frac{R A}{L}\). Resistance \(R = \frac{V}{I} = \frac{P}{I^2} = \frac{\text{work done}}{I^2 \times t}\). In SI base units: Work \(= \text{kg m}^2 \text{s}^{-2}\), so \([R] = \frac{\text{kg m}^2 \text{s}^{-2}}{\text{A}^2 \text{s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\). Therefore, \([\rho] = [R] \times \frac{[A]}{[L]} = (\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}) \times \text{m} = \text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\).
Marking scheme
A is correct [1 mark]. B confuses resistance with resistivity. C omits time power correctly. D has incorrect power of metre.
Question 26 · Multiple Choice
1 marks
A ball is projected horizontally at a speed of \(18\text{ m s}^{-1}\) from the top of a vertical cliff of height \(45\text{ m}\). Air resistance is negligible.
What is the speed of the ball just before it strikes the horizontal ground below?
A.\(22\text{ m s}^{-1}\)
B.\(30\text{ m s}^{-1}\)
C.\(35\text{ m s}^{-1}\)
D.\(48\text{ m s}^{-1}\)
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Worked solution
The horizontal component of velocity remains constant: \(v_x = 18\text{ m s}^{-1}\). For the vertical component of motion: \(v_y^2 = u_y^2 + 2 g h = 0 + 2 \times 9.81 \times 45 = 882.9\text{ m}^2 \text{s}^{-2}\). The resultant speed \(v\) is given by: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{18^2 + 882.9} = \sqrt{324 + 882.9} = \sqrt{1206.9} \approx 34.7\text{ m s}^{-1} \approx 35\text{ m s}^{-1}\).
Marking scheme
C is correct [1 mark]. A incorrectly calculates average speed. B gives only the vertical velocity component (\(v_y = 29.7\text{ m s}^{-1}\)). D adds horizontal and vertical speeds directly without Pythagoras.
Question 27 · Multiple Choice
1 marks
A vehicle of mass \(4000\text{ kg}\) travelling at \(12\text{ m s}^{-1}\) collides head-on with a car of mass \(1000\text{ kg}\) travelling in the opposite direction at \(18\text{ m s}^{-1}\). The two vehicles lock together after the collision.
What is the total loss in kinetic energy during the collision?
A.\(90\text{ kJ}\)
B.\(270\text{ kJ}\)
C.\(360\text{ kJ}\)
D.\(450\text{ kJ}\)
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Worked solution
Taking the initial direction of the truck as positive: Total initial momentum \(p_{\text{initial}} = (4000 \times 12) + (1000 \times -18) = 48000 - 18000 = 30000\text{ kg m s}^{-1}\). Common final velocity \(v = \frac{30000}{4000 + 1000} = 6.0\text{ m s}^{-1}\). Initial kinetic energy \(E_{k1} = \frac{1}{2}(4000)(12^2) + \frac{1}{2}(1000)(18^2) = 288000 + 162000 = 450000\text{ J} = 450\text{ kJ}\). Final kinetic energy \(E_{k2} = \frac{1}{2}(5000)(6.0^2) = 90000\text{ J} = 90\text{ kJ}\). Loss in kinetic energy \(\Delta E_k = 450 - 90 = 360\text{ kJ}\).
Marking scheme
C is correct [1 mark]. A is the final kinetic energy. B calculates loss by incorrectly treating velocities as having the same initial sign. D is the total initial kinetic energy.
Question 28 · Multiple Choice
1 marks
A solid cylinder of cross-sectional area \(2.5 \times 10^{-3}\text{ m}^2\) and height \(0.40\text{ m}\) is fully submerged vertically in a liquid of uniform density \(1200\text{ kg m}^{-3}\). The upper face of the cylinder is at a depth of \(0.15\text{ m}\) below the surface.
What is the difference in hydrostatic pressure between the bottom face and top face of the cylinder, and what is the upthrust acting on the cylinder?
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Worked solution
Pressure difference between bottom and top: \(\Delta p = \rho g \Delta h = 1200 \times 9.81 \times 0.40 = 4708.8\text{ Pa} \approx 4.7 \times 10^3\text{ Pa}\). Upthrust \(F = \Delta p \times A = 4708.8 \times (2.5 \times 10^{-3}) \approx 11.8\text{ N} \approx 12\text{ N}\).
Marking scheme
B is correct [1 mark]. A uses the depth of the top face (\(0.15\text{ m}\)) instead of the height. C uses total depth to bottom face for upthrust. D uses total depth \(0.55\text{ m}\) for the pressure difference.
Question 29 · Multiple Choice
1 marks
A metal wire of original length \(2.0\text{ m}\) and cross-sectional area \(0.80\text{ mm}^2\) is suspended vertically and extended within its limit of proportionality by a tensile force of \(120\text{ N}\). The Young modulus of the metal is \(2.0 \times 10^{11}\text{ Pa}\).
What is the elastic potential energy stored in the stretched wire?
A.\(4.5 \times 10^{-2}\text{ J}\)
B.\(9.0 \times 10^{-2}\text{ J}\)
C.\(1.8 \times 10^{-1}\text{ J}\)
D.\(3.6 \times 10^{-1}\text{ J}\)
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B is correct [1 mark]. A uses \(\frac{1}{4}Fx\). C forgets the factor of \(1/2\) and evaluates \(Fx\). D evaluates \(2Fx\).
Question 30 · Multiple Choice
1 marks
An emergency vehicle sounds a siren of constant frequency \(680\text{ Hz}\) while travelling at a constant speed of \(30.0\text{ m s}^{-1}\) directly towards a stationary observer. The speed of sound in air is \(340\text{ m s}^{-1}\).
What is the frequency of the siren detected by the stationary observer?
A.\(625\text{ Hz}\)
B.\(680\text{ Hz}\)
C.\(710\text{ Hz}\)
D.\(746\text{ Hz}\)
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Worked solution
Using the Doppler effect formula for a moving source approaching a stationary observer: \(f_o = f_s \left(\frac{v}{v - v_s}\right)\). \(f_o = 680 \times \left(\frac{340}{340 - 30.0}\right) = 680 \times \frac{340}{310} \approx 745.8\text{ Hz} \approx 746\text{ Hz}\).
Marking scheme
D is correct [1 mark]. A uses the receding formula (\(v + v_s\) in denominator). B assumes no Doppler shift. C is an arithmetic rounding error.
Question 31 · Multiple Choice
1 marks
A cell has electromotive force (e.m.f.) \(E\) and internal resistance \(r\). When a resistor of resistance \(4.0\,\Omega\) is connected across its terminals, the terminal potential difference is \(8.0\text{ V}\). When a resistor of resistance \(9.0\,\Omega\) is connected across its terminals, the terminal potential difference is \(9.0\text{ V}\).
What are the values of \(E\) and \(r\)?
A.\(E = 10\text{ V},\; r = 1.0\,\Omega\)
B.\(E = 10\text{ V},\; r = 2.0\,\Omega\)
C.\(E = 12\text{ V},\; r = 1.0\,\Omega\)
D.\(E = 12\text{ V},\; r = 2.0\,\Omega\)
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Worked solution
Terminal p.d. \(V = E - I r = E - \left(\frac{V}{R}\right)r \implies E = V\left(1 + \frac{r}{R}\right)\). For \(R = 4.0\,\Omega\): \(E = 8.0\left(1 + \frac{r}{4.0}\right) = 8.0 + 2.0 r\). For \(R = 9.0\,\Omega\): \(E = 9.0\left(1 + \frac{r}{9.0}\right) = 9.0 + 1.0 r\). Equating the two expressions for \(E\): \(8.0 + 2.0 r = 9.0 + 1.0 r \implies 1.0 r = 1.0 \implies r = 1.0\,\Omega\). Substituting \(r\) back: \(E = 8.0 + 2.0(1.0) = 10\text{ V}\).
Marking scheme
A is correct [1 mark]. B has correct e.m.f. but incorrect internal resistance. C and D result from incorrect formulation of the potential divider/terminal p.d. equation.
Question 32 · Multiple Choice
1 marks
In a \(\beta^+\) decay process occurring inside a nucleus, a proton decays into a neutron.
Which row correctly identifies the change in quark flavour and the other fundamental particles emitted during this decay?
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Worked solution
A proton has quark composition \(uud\) and a neutron has quark composition \(udd\). In \(\beta^+\) decay, a proton changes into a neutron, so an up quark changes into a down quark (\(u \to d\)). To conserve charge and lepton number, a positron (\(\beta^+\)) and an electron neutrino (\(\nu_e\)) are emitted.
Marking scheme
A is correct [1 mark]. B pairs \(\beta^+\) with an antineutrino (violating lepton number conservation). C has the reverse quark transition. D describes \(\beta^-\) decay.
Question 33 · Multiple Choice
1 marks
Which quantity has the same SI base units as the Young modulus?
A.energy \(\times\) density
B.energy per unit volume
C.momentum per unit area
D.power per unit volume
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Worked solution
The Young modulus is stress divided by strain. Stress has units of \(\text{N m}^{-2}\) and strain is dimensionless. In SI base units, \(1\text{ N} = 1\text{ kg m s}^{-2}\), so the Young modulus has base units of \(\text{kg m s}^{-2} \times \text{m}^{-2} = \text{kg m}^{-1}\text{s}^{-2}\). Energy per unit volume has units of \(\text{J m}^{-3} = (\text{kg m}^2\text{s}^{-2}) \text{m}^{-3} = \text{kg m}^{-1}\text{s}^{-2}\), which is identical.
Marking scheme
B [1 mark] for identifying energy per unit volume as having base units \(\text{kg m}^{-1}\text{s}^{-2}\).
Question 34 · Multiple Choice
1 marks
A stone is thrown horizontally from the top of a vertical cliff of height \(h\) with an initial horizontal speed \(v\). The acceleration of free fall is \(g\). Air resistance is negligible.
Just before the stone hits the horizontal ground below, its velocity makes an angle \(\theta\) with the horizontal.
Which expression gives \(\tan \theta\)?
A.\(\frac{\sqrt{2gh}}{v}\)
B.\(\frac{v}{\sqrt{2gh}}\)
C.\(\frac{2gh}{v^2}\)
D.\(\frac{\sqrt{gh}}{v}\)
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Worked solution
The horizontal component of velocity remains constant: \(v_x = v\). For the vertical component, using \(v_y^2 = u_y^2 + 2gs\) with \(u_y = 0\) and \(s = h\), we get \(v_y = \sqrt{2gh}\). The tangent of the angle with the horizontal is given by \(\tan \theta = \frac{v_y}{v_x} = \frac{\sqrt{2gh}}{v}\).
Marking scheme
A [1 mark] for correctly resolving perpendicular velocity components to find \(\tan \theta = \frac{\sqrt{2gh}}{v}\).
Question 35 · Multiple Choice
1 marks
A toy rocket of total mass \(0.40\text{ kg}\) travels horizontally at a velocity of \(15\text{ m s}^{-1}\). An internal explosion splits the rocket into two sections, \(X\) and \(Y\).
Section \(X\) has a mass of \(0.10\text{ kg}\) and continues in the original direction at a speed of \(36\text{ m s}^{-1}\).
What is the speed of section \(Y\) immediately after the explosion?
A.\(4.0\text{ m s}^{-1}\)
B.\(8.0\text{ m s}^{-1}\)
C.\(12\text{ m s}^{-1}\)
D.\(21\text{ m s}^{-1}\)
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Worked solution
By conservation of linear momentum: \(p_{\text{initial}} = p_{\text{final}}\). Initial momentum = \(0.40\text{ kg} \times 15\text{ m s}^{-1} = 6.0\text{ N s}\). Mass of section \(Y = 0.40 - 0.10 = 0.30\text{ kg}\). Momentum of section \(X = 0.10\text{ kg} \times 36\text{ m s}^{-1} = 3.6\text{ N s}\). Momentum of section \(Y = 6.0 - 3.6 = 2.4\text{ N s}\). Speed of section \(Y = \frac{2.4\text{ N s}}{0.30\text{ kg}} = 8.0\text{ m s}^{-1}\).
Marking scheme
B [1 mark] for applying conservation of momentum to determine \(v_Y = 8.0\text{ m s}^{-1}\).
Question 36 · Multiple Choice
1 marks
A uniform solid cylinder of cross-sectional area \(A\) and length \(L\) floats upright in a liquid of density \(\rho\). The depth of the cylinder submerged in the liquid is \(h\).
The cylinder is then pushed vertically downwards so that it is submerged by an additional depth \(x\), where \((h + x) < L\), and held stationary.
What is the magnitude of the additional upward force exerted by the liquid on the cylinder as a result of being submerged by the additional depth \(x\)?
A.\(\rho A x g\)
B.\(\rho A (h + x) g\)
C.\(\rho A (L - h - x) g\)
D.\(\frac{1}{2}\rho A x^2 g\)
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Worked solution
By Archimedes' principle, the upthrust equals the weight of the displaced liquid. The initial upthrust balances the weight of the cylinder: \(U_0 = \rho A h g\). When submerged to a depth of \((h + x)\), the total upthrust becomes \(U = \rho A (h + x) g\). The additional upward force is \(\Delta U = U - U_0 = \rho A x g\).
Marking scheme
A [1 mark] for identifying the change in upthrust as \(\rho A x g\).
Question 37 · Multiple Choice
1 marks
An electric vehicle of mass \(500\text{ kg}\) travels at a constant speed of \(12\text{ m s}^{-1}\) up a track inclined at an angle of \(5.0^\circ\) to the horizontal.
A constant resistive force of \(250\text{ N}\) acts against the vehicle's motion.
The electrical power supplied to the vehicle's motor is \(9.0\text{ kW}\).
What is the efficiency of the motor system?
A.\(33\%\)
B.\(57\%\)
C.\(75\%\)
D.\(90\%\)
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Worked solution
The force required to maintain constant speed up the incline is \(F = mg \sin\theta + F_{\text{resistive}} = (500 \times 9.81 \times \sin 5.0^\circ) + 250 = 427.5 + 250 = 677.5\text{ N}\). The useful power output is \(P_{\text{out}} = Fv = 677.5\text{ N} \times 12\text{ m s}^{-1} = 8130\text{ W} = 8.13\text{ kW}\). Efficiency \(= \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{8.13\text{ kW}}{9.0\text{ kW}} \times 100\% = 90.3\% \approx 90\%\).
Marking scheme
D [1 mark] for computing useful power output as \(8.13\text{ kW}\) and efficiency as \(90\%\).
Question 38 · Multiple Choice
1 marks
A stationary observer detects sound of frequency \(680\text{ Hz}\) emitted from a siren on an emergency vehicle that is moving directly away from the observer at a constant speed of \(25\text{ m s}^{-1}\).
The speed of sound in air is \(340\text{ m s}^{-1}\).
What is the frequency of the sound emitted by the siren?
A.\(630\text{ Hz}\)
B.\(680\text{ Hz}\)
C.\(730\text{ Hz}\)
D.\(734\text{ Hz}\)
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Worked solution
Using the Doppler effect formula for a moving source moving away from a stationary observer: \(f_o = f_s \left(\frac{v}{v + v_s}\right)\). Substituting the known values: \(680 = f_s \left(\frac{340}{340 + 25}\right) = f_s \left(\frac{340}{365}\right)\). Thus, \(f_s = 680 \times \frac{365}{340} = 2 \times 365 = 730\text{ Hz}\).
Marking scheme
C [1 mark] for applying \(f_o = f_s \frac{v}{v + v_s}\) to find source frequency \(730\text{ Hz}\).
Question 39 · Multiple Choice
1 marks
Monochromatic light of wavelength \(600\text{ nm}\) is incident normally on a diffraction grating that has \(4.00 \times 10^5\text{ lines m}^{-1}\).
What is the total number of bright maxima (orders) that can be observed on a screen behind the grating?
A.\(4\)
B.\(8\)
C.\(9\)
D.\(10\)
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Worked solution
The grating spacing is \(d = \frac{1}{4.00 \times 10^5\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m}\). Using the diffraction formula \(d \sin\theta = n\lambda\), the maximum possible order occurs when \(\sin\theta \le 1\): \(n_{\text{max}} \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}\text{ m}}{600 \times 10^{-9}\text{ m}} = 4.17\). Since \(n\) must be an integer, \(n_{\text{max}} = 4\). The total number of maxima is \(2n_{\text{max}} + 1 = 2(4) + 1 = 9\) (including the central zeroth-order maximum).
Marking scheme
C [1 mark] for finding \(n_{\text{max}} = 4\) and total maxima \(= 2(4) + 1 = 9\).
Question 40 · Multiple Choice
1 marks
A power supply of electromotive force (e.m.f.) \(12\text{ V}\) and negligible internal resistance is connected in series with a fixed resistor of resistance \(300\ \Omega\) and a light-dependent resistor (LDR).
The intensity of the light incident on the LDR increases.
What happens to the resistance of the LDR and to the potential difference (p.d.) across the \(300\ \Omega\) resistor?
A.Resistance of LDR decreases; p.d. across \(300\ \Omega\) resistor decreases
B.Resistance of LDR decreases; p.d. across \(300\ \Omega\) resistor increases
C.Resistance of LDR increases; p.d. across \(300\ \Omega\) resistor decreases
D.Resistance of LDR increases; p.d. across \(300\ \Omega\) resistor increases
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Worked solution
When the light intensity on an LDR increases, its resistance decreases. Since the total resistance in the series circuit \(R_{\text{total}} = 300\ \Omega + R_{\text{LDR}}\) decreases, the circuit current \(I = \frac{E}{R_{\text{total}}}\) increases. The potential difference across the fixed resistor is \(V = I \times 300\ \Omega\), which consequently increases.
Marking scheme
B [1 mark] for stating that LDR resistance decreases and p.d. across the fixed resistor increases.
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Answer all six questions in the spaces provided. Show all working clearly and state appropriate units for all numerical answers.
6 Question · 60 marks
Question 1 · Data Analysis & Kinematics Graphing
11 marks
(a) Define acceleration.
[1]
(b) A model electric vehicle moves from rest along a horizontal straight track. The motion is divided into three consecutive stages:
- Stage 1 (0 to 4.0 s): The vehicle accelerates with constant acceleration from rest to a velocity of \(12.0\text{ m s}^{-1}\) at \(t = 4.0\text{ s}\). - Stage 2 (4.0 s to 9.0 s): The vehicle travels at a constant velocity of \(12.0\text{ m s}^{-1}\). - Stage 3 (9.0 s to 15.0 s): Brakes are applied, and the vehicle decelerates to rest at \(t = 15.0\text{ s}\).
For Stage 1 (between \(t = 0\) and \(t = 4.0\text{ s}\)): (i) Calculate the acceleration of the vehicle.
(c) During Stage 3, the velocity \(v\) of the vehicle decreases non-uniformly.
(i) State how the acceleration at any instant during Stage 3 can be determined from a velocity-time graph.
[1]
(ii) The data shows that the magnitude of the slope of the velocity-time graph becomes steeper as time increases from \(t = 9.0\text{ s}\) to \(t = 15.0\text{ s}\). State and explain what happens to the magnitude of the deceleration during this stage.
[2]
(d) The measured values of velocity \(v\) during Stage 3 at intervals of \(1.0\text{ s}\) are given in the table below:
(i) Use the data in the table (e.g., using the trapezium rule or average velocity of each 1-second strip) to estimate the distance travelled by the vehicle between \(t = 9.0\text{ s}\) and \(t = 15.0\text{ s}\).
(c)(i) The acceleration is determined by drawing a tangent to the curve at the given time and finding the gradient of this tangent.
(c)(ii) The magnitude of deceleration increases because the gradient of the velocity-time graph becomes progressively steeper (i.e. increases in magnitude), which means the rate at which velocity decreases is increasing with time.
(d)(i) Using the trapezium rule to find the area under the \(v\)-\(t\) graph: $$\text{Area} = h \left[ \frac{v_0 + v_n}{2} + v_1 + v_2 + v_3 + v_4 + v_5 \right]$$ where \(h = 1.0\text{ s}\): $$\text{Area} = 1.0 \times \left[ \frac{12.0 + 0.0}{2} + 11.7 + 10.7 + 9.0 + 6.7 + 3.7 \right]$$ $$\text{Area} = 1.0 \times [6.0 + 41.8] = 47.8\text{ m} \approx 48\text{ m}$$
(a) - B1: rate of change of velocity OR (change in velocity) / (time taken). (Do not accept 'speed' instead of 'velocity').
(b)(i) - A1: \(a = \frac{12.0}{4.0} = 3.0\text{ m s}^{-2}\) (allow 3 or 3.0).
(b)(ii) - C1: \(s = \frac{1}{2}(12.0)(4.0)\) or \(s = \frac{1}{2}(3.0)(4.0)^2\) OR correct method to find area. - A1: \(s = 24\text{ m}\).
(c)(i) - B1: Gradient / slope of the tangent (to the curve).
(c)(ii) - M1: (magnitude of deceleration) increases. - A1: (because) gradient of the line / curve gets steeper OR rate of change of velocity increases.
(d)(i) - C1: Method of summation using trapeziums / strip areas shown, e.g. \(1.0 \times (\frac{12.0+0}{2} + 11.7 + 10.7 + 9.0 + 6.7 + 3.7)\) or sum of strip areas \((11.85 + 11.2 + 9.85 + 7.85 + 5.2 + 1.85)\). - A1: \(47.8\text{ m}\) or \(48\text{ m}\) (allow values in range 47 to 49 m).
(d)(ii) - C1: Calculation of total distance \(= 24 + (12.0 \times 5.0) + (\text{answer to d(i)}) = 131.8\text{ m}\) (allow ecf from (b)(ii) and (d)(i)). - A1: \(\frac{131.8}{15.0} = 8.79\text{ m s}^{-1}\) or \(8.8\text{ m s}^{-1}\).
Question 2 · structured
8 marks
(a) Define the Young modulus.
(b) A uniform horizontal rod \(PQ\) of length \(1.50\text{ m}\) and mass \(6.00\text{ kg}\) is hinged to a vertical wall at end \(P\). The rod is held horizontally in equilibrium by a metal wire attached to end \(Q\) and fixed to the wall at point \(R\) vertically above \(P\). The wire is at an angle of \(30.0^\circ\) to the horizontal rod. A load of mass \(10.0\text{ kg}\) is suspended from the rod at a point \(1.20\text{ m}\) from \(P\).
(i) By taking moments about \(P\), calculate the tension \(T\) in the wire.
(ii) The wire has an unstretched length of \(1.73\text{ m}\) and a cross-sectional area of \(1.80 \times 10^{-6}\text{ m}^2\). The Young modulus of the metal is \(2.10 \times 10^{11}\text{ Pa}\).
Calculate the extension \(\Delta L\) of the wire.
(iii) Calculate the elastic potential energy \(E_{\text{p}}\) stored in the wire.
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Worked solution
(a) The Young modulus is defined as the ratio of tensile stress to tensile strain (or stress / strain).
(b)(i) Weight of the uniform rod: \(W_{\text{rod}} = mg = 6.00 \times 9.81 = 58.86\text{ N}\), acting at the centre of gravity \(0.750\text{ m}\) from \(P\).
Weight of the suspended load: \(W_{\text{load}} = mg = 10.0 \times 9.81 = 98.10\text{ N}\), acting at \(1.20\text{ m}\) from \(P\).
(a) - B1: (tensile) stress / (tensile) strain [Allow symbols if defined, e.g. σ/ε]
(b)(i) - C1: Calculation of either weight (58.9 N or 98.1 N) OR sum of clockwise moments \((6.00 \times 9.81 \times 0.750) + (10.0 \times 9.81 \times 1.20)\) [= 162 N m] - C1: Equating clockwise moments to anticlockwise moment \(T \sin(30^\circ) \times 1.50\) - A1: \(T = 216\text{ N}\) (accept 215 N to 216 N; allow 2 or 3 s.f.)
(b)(ii) - C1: Correct rearrangement or substitution into \(E = \frac{T L}{A \Delta L}\) or \(\Delta L = \frac{T L}{A E}\) - A1: \(\Delta L = 9.88 \times 10^{-4}\text{ m}\) (or \(9.89 \times 10^{-4}\text{ m}\) using 216 N, allow ECF from (b)(i))
(b)(iii) - C1: Formula \(E_{\text{p}} = \frac{1}{2} F x\) or \(\frac{1}{2} T \Delta L\) or \(\frac{1}{2} \frac{E A (\Delta L)^2}{L}\) with values substituted - A1: \(E_{\text{p}} = 0.107\text{ J}\) (accept 0.106 J to 0.107 J, allow ECF from (b)(i) and (b)(ii))
Question 3 · Superposition & Wave Properties
10 marks
(a) State the principle of superposition.
[1]
(b) A sound wave is produced by a loudspeaker placed near the open end of a hollow pipe of length \(0.680\text{ m}\). The opposite end of the pipe is closed. The speed of sound in air is \(340\text{ m s}^{-1}\).
(i) Explain how a stationary sound wave is formed inside the pipe.
[2]
(ii) The frequency of the loudspeaker is gradually increased from \(0\text{ Hz}\) to \(1000\text{ Hz}\). Determine the lowest two frequencies at which resonance occurs in the pipe.
[3]
(c) Light from a laser of wavelength \(632.8\text{ nm}\) is incident normally on a diffraction grating with \(400\text{ lines mm}^{-1}\).
(i) Calculate the angle of diffraction for the second-order maximum.
[2]
(ii) Determine the total number of maxima that can be observed on a screen.
[2]
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Worked solution
(a) The principle of superposition states that when two or more waves meet at a point, the resultant displacement is equal to the algebraic (or vector) sum of the individual displacements.
(b)(i) 1. The progressive sound wave travels along the tube and is reflected at the closed end. 2. The reflected wave travels in the opposite direction and superposes (interferes) with the incident wave to form a stationary wave pattern of nodes and antinodes.
(b)(ii) For a pipe of length \(L\) closed at one end, resonance occurs when an antinode is at the open end and a node is at the closed end. Fundamental mode (lowest frequency): \(L = \frac{\lambda_1}{4} \implies \lambda_1 = 4L = 4 \times 0.680 = 2.72\text{ m}\) \(f_1 = \frac{v}{\lambda_1} = \frac{340}{2.72} = 125\text{ Hz}\)
(c)(ii) For maximum observable order, \(\sin\theta \le 1\): \(n_{\text{max}} \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}}{632.8 \times 10^{-9}} = 3.95\) Thus, highest integer order \(n = 3\). Total number of maxima \(= 2n + 1 = 2(3) + 1 = 7\) (central zeroth-order plus 3 on either side).
Marking scheme
(a) - When two (or more) waves meet at a point, resultant displacement is the (vector/algebraic) sum of individual displacements [B1]
(b)(i) - Wave reflects at the closed end / opposite boundary [B1] - Incident and reflected waves (of same frequency/speed) travel in opposite directions and superpose [B1]
(b)(ii) - Calculation of fundamental wavelength: \(\lambda = 4L = 2.72\text{ m}\) OR formula \(f = v / 4L\) [C1] - Lowest frequency \(f_1 = 125\text{ Hz}\) [A1] - Second lowest frequency \(f_2 = 375\text{ Hz}\) [A1]
(c)(i) - Correct calculation of \(d = 2.50 \times 10^{-6}\text{ m}\) and substitution into \(d \sin\theta = n\lambda\) [C1] - \(\theta = 30.4^{\circ}\) (or \(0.531\text{ rad}\)) [A1]
(c)(ii) - Calculation of highest order \(n = 3\) from \(n \le d/\lambda\) [C1] - Total number of maxima = 7 [A1]
Question 4 · structured
10 marks
(a) State the principle of conservation of energy. [1]
(b) An electric railway carriage of mass \(4.2 \times 10^4\text{ kg}\) is moving up an incline of angle \(2.8^\circ\) to the horizontal.
(i) Calculate the increase in gravitational potential energy of the carriage as it travels a distance of \(80\text{ m}\) along the incline. [2]
(ii) Over this \(80\text{ m}\) distance along the incline, the carriage accelerates uniformly from an initial speed of \(12\text{ m s}^{-1}\) to a final speed of \(18\text{ m s}^{-1}\).
Calculate the increase in kinetic energy of the carriage. [2]
(iii) A constant resistive force of \(6.5\text{ kN}\) opposes the motion of the carriage as it travels the \(80\text{ m}\).
Calculate the total work done by the forward driving force of the carriage over this distance. [2]
(iv) The electrical energy supplied to the motor system of the carriage over this \(80\text{ m}\) distance is \(9.2 \times 10^6\text{ J}\).
Calculate the efficiency of the motor system. [2]
(c) Suggest one reason why the efficiency of the motor system is less than \(100\%\). [1]
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Worked solution
(a) The principle of conservation of energy states that energy cannot be created or destroyed, only converted (transferred) from one form into another, or that the total energy of an isolated (closed) system remains constant.
(c) Energy is dissipated / lost as thermal energy (heat) due to internal electrical resistance in the windings / friction in the mechanical components (e.g. bearings, gears).
Marking scheme
(a) - Energy cannot be created or destroyed, only transferred from one form to another (or total energy of a closed system remains constant) [B1]
(b)(iv) - \(\text{efficiency} = \frac{\text{work done from (b)(iii)}}{9.2 \times 10^6}\) [C1] - \(\text{efficiency} = 0.64\) or \(64\%\) (allow \(0.642\) or \(64.2\%\)) [allow ecf from (b)(iii)] [A1]
(c) - Thermal energy / heat generated in the motor (due to electrical resistance / friction in moving parts) [B1]
Question 5 · structured
12 marks
(a) (i) Define electromotive force (e.m.f.) of an electrical power supply. [2]
(ii) Distinguish between electromotive force (e.m.f.) and potential difference (p.d.). [1]
(b) A uniform cylindrical resistance wire XY has length \(0.80\text{ m}\) and cross-sectional area \(4.5 \times 10^{-7}\text{ m}^2\). The wire is made of a metal of resistivity \(1.8 \times 10^{-6}\ \Omega\text{ m}\). The number density of free electrons in the metal is \(6.2 \times 10^{28}\text{ m}^{-3}\).
(i) Show that the resistance of wire XY is \(3.2\ \Omega\). [2]
(ii) A steady current of \(1.5\text{ A}\) is passed through wire XY. Calculate the average drift speed \(v\) of the conduction electrons in the wire.
(c) Wire XY is connected in a circuit with a battery of e.m.f. \(E = 9.0\text{ V}\) and internal resistance \(r\). A fixed resistor of resistance \(R_1 = 4.8\ \Omega\) is connected in parallel with wire XY. This parallel combination is connected in series with a fixed resistor of resistance \(R_2 = 1.6\ \Omega\) and the battery.
The current flowing through resistor \(R_2\) is \(1.8\text{ A}\).
(i) Calculate the total equivalent resistance of the parallel combination of wire XY and resistor \(R_1\).
(iii) Resistor \(R_1\) is now removed from the circuit.
State and explain the effect on the electrical power dissipated in wire XY. [2]
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Worked solution
(a) (i) Electromotive force (e.m.f.) is the energy converted from chemical (or other) energy to electrical energy per unit charge that passes through the source.
(ii) e.m.f. represents energy transferred into electrical energy per unit charge, whereas potential difference (p.d.) represents energy transferred from electrical energy to other forms of energy per unit charge.
(ii) Total external resistance \(R_{ext} = R_p + R_2 = 1.92 + 1.6 = 3.52\ \Omega\). Using \(E = I(R_{ext} + r)\): \[ 9.0 = 1.8(3.52 + r) \] \[ 3.52 + r = 5.00 \implies r = 1.48\ \Omega \approx 1.5\ \Omega \]
(iii) When \(R_1\) is removed, the total resistance of the circuit increases, so total current in the circuit decreases. However, wire XY now takes the full current of the branch rather than sharing it with \(R_1\). The fraction of circuit resistance across XY increases, causing the p.d. across XY (and the current through XY) to increase (from \(3.46\text{ V}\) to \(4.59\text{ V}\)). Therefore, the power dissipated in wire XY increases.
Marking scheme
(a) (i) B1: Energy converted from other forms (e.g. chemical) to electrical energy B1: Per unit charge (or per coulomb)
(ii) B1: e.m.f. is energy transferred to electrical energy (per unit charge) whereas p.d. is energy transferred from electrical energy (per unit charge)
(b) (i) M1: \(R = \frac{\rho L}{A}\) stated and values substituted: \(R = \frac{1.8 \times 10^{-6} \times 0.80}{4.5 \times 10^{-7}}\) A1: Calculation completed to show \(3.2\ \Omega\) (must show at least 2 significant figures)
(ii) C1: \(v = \frac{I}{n A q}\) or \(1.5 = 4.5 \times 10^{-7} \times 6.2 \times 10^{28} \times 1.60 \times 10^{-19} \times v\) A1: \(v = 3.4 \times 10^{-4}\text{ m s}^{-1}\) (allow \(3.36 \times 10^{-4}\text{ m s}^{-1}\))
(c) (i) A1: \(1.9\ \Omega\) or \(1.92\ \Omega\)
(ii) C1: \(E = I(R_{total} + r)\) or \(V_{terminal} = E - Ir\) with correct substitution of \(R_{ext} = 3.52\ \Omega\) A1: \(r = 1.5\ \Omega\) (or \(1.48\ \Omega\))
(iii) M1: (Electrical power) increases A1: Potential difference across XY (or current through XY) increases / XY now constitutes a larger fraction of the total circuit resistance
Question 6 · structured
9 marks
(a) State what is meant by a hadron and distinguish between a baryon and a meson.
(b) A stationary nitrogen-14 (\(^{14}_{\phantom{0}7}\text{N}\)) nucleus is struck by an \(\alpha\)-particle (\(^4_2\alpha\)) moving at a speed of \(1.80 \times 10^7\text{ m s}^{-1}\). The collision results in a nuclear reaction represented by the equation: \[ ^4_2\alpha + {}^{14}_{\phantom{0}7}\text{N} \rightarrow {}^{17}_{\phantom{0}8}\text{O} + {}^1_1\text{p} \] (i) State the quark composition of a proton and of a neutron.
(ii) In this interaction, the proton (mass \(1.01\text{ u}\)) is ejected along the initial line of motion of the \(\alpha\)-particle (mass \(4.00\text{ u}\)) with a speed of \(2.20 \times 10^7\text{ m s}^{-1}\). The mass of the oxygen-17 nucleus is \(17.0\text{ u}\).
Calculate the velocity of the oxygen-17 nucleus immediately after the interaction.
(c) In a separate process, a stationary free neutron decays into a proton, an electron, and a third particle. State the name of this third particle and write the quark change that occurs during this decay.
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Worked solution
(a) A hadron is a particle composed of quarks (or a particle that experiences the strong nuclear force). A baryon consists of three quarks (or three antiquarks), whereas a meson consists of a quark and an antiquark pair.
(b)(ii) By conservation of linear momentum along the initial direction of motion: \(p_{\text{initial}} = p_{\text{final}}\) \(m_\alpha u_\alpha + m_{\text{N}} u_{\text{N}} = m_{\text{O}} v_{\text{O}} + m_{\text{p}} v_{\text{p}}\)
Since \(E_{k,\text{final}} \neq E_{k,\text{initial}}\) (kinetic energy decreases), kinetic energy is not conserved (the interaction is inelastic).
(c) Particle: antineutrino (or electron antineutrino / \(\bar{\nu}_e\)) Quark change: \(\text{d} \rightarrow \text{u}\) (down quark changes to an up quark)
Marking scheme
(a) - Hadron: particle made of quarks / subject to the strong force [B1] - Baryon: composed of 3 quarks AND Meson: composed of a quark and an antiquark [B1]
(b)(i) - proton: uud AND neutron: udd [B1]
(b)(ii) - (4.00 x 1.80 x 10^7) = (17.0 x v) + (1.01 x 2.20 x 10^7) OR clear use of conservation of momentum [C1] - magnitude: 2.93 x 10^6 m s^-1 (accept 2.9 x 10^6 m s^-1) [A1] - direction: in the same direction as the incident alpha-particle / forward / to the right [B1]
(b)(iii) - Calculation of initial KE (= 1.08 x 10^-12 J or 6.48 x 10^14 u m^2 s^-2) AND final KE (= 5.27 x 10^-13 J or 3.17 x 10^14 u m^2 s^-2) [M1] - Explicit comparison stating kinetic energy is not conserved / collision is inelastic [A1]
(c) - (electron) antineutrino / \(\bar{\nu}\) AND d -> u / down (quark) to up (quark) [B1]
Paper 3 (Advanced Practical Skills)
Complete two experimental investigations within 2 hours. Record raw data systematically with units, plot graph lines of best fit, calculate constants, and evaluate experimental limitations.
(b) Set up the apparatus with the pivot at a distance $d = 15.0\text{ cm}$ from the top suspension point and attach the spring at distance $x = 20.0\text{ cm}$ from the pivot. Displace the bottom end of the bar by a small angle (less than $10^\circ$) and release it so that it oscillates in a vertical plane.
Determine the period $T$ of the oscillations by timing a sufficient number of complete oscillations.
$$\text{Number of oscillations } N = \text{........................}$$ $$\text{Time for } N \text{ oscillations } t_1 = \text{........................}\text{ s}$$ $$\text{Time for } N \text{ oscillations } t_2 = \text{........................}\text{ s}$$ $$T = \text{...................................................}\text{ s}$$
(c) By adjusting the position of the spring attachment, vary the distance $x$ in the range $15.0\text{ cm} \le x \le 50.0\text{ cm}$.
For each value of $x$, measure the time for $N$ oscillations and calculate the period $T$, and the quantity $\dfrac{1}{T^2}$.
Record all your values of $x$, raw timing data, $T$, and $\dfrac{1}{T^2}$ in a single table. Include values of $x^2$ in your table.
(d) (i) On the grid provided, plot a graph of $\dfrac{1}{T^2}$ on the $y$-axis against $x^2$ on the $x$-axis.
(ii) Draw the straight line of best fit.
(e) (i) Determine the gradient of the line of best fit.
#### (d) Graph - Axes: Scale on $x$-axis from $0$ to $2600\text{ cm}^2$ ($1\text{ cm} = 200\text{ cm}^2$), $y$-axis from $0$ to $4.50\text{ s}^{-2}$ ($1\text{ cm} = 0.50\text{ s}^{-2}$). - Points occupy more than $50\%$ of each axis grid area. - All six points accurately plotted to within half a small grid square. - Straight line of best fit drawn smoothly with balanced point distribution.
#### (e) Gradient and Intercept Calculation - Selected coordinates on the line of best fit: $(x_1, y_1) = (200, 0.99)$ and $(x_2, y_2) = (2400, 3.80)$. $$\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3.80 - 0.99}{2400 - 200} = \frac{2.81}{2200} = 1.28 \times 10^{-3}\text{ s}^{-2}\text{ cm}^{-2} \text{ (or } 12.8\text{ s}^{-2}\text{ m}^{-2}\text{)}$$
- Using point $(2400, 3.80)$: $$y\text{-intercept } c = y_2 - m x_2 = 3.80 - (1.277 \times 10^{-3} \times 2400) = 3.80 - 3.065 = 0.735\text{ s}^{-2}$$
#### (f) Determination of Constants Comparing $\dfrac{1}{T^2} = P x^2 + Q$ with $y = m x + c$: - $P = \text{gradient} = 1.28 \times 10^{-3}\text{ s}^{-2}\text{ cm}^{-2}$ - $Q = y\text{-intercept} = 0.735\text{ s}^{-2}$
Marking scheme
(a) [1 Mark] - Value of $L$ recorded to the nearest millimetre ($0.1\text{ cm}$) with correct unit. [1]
(b) [2 Marks] - Evidence of repeated timing for $N \ge 10$ oscillations ($t_1$ and $t_2$). [1] - Period $T$ calculated correctly, $0.5\text{ s} < T < 2.0\text{ s}$, with correct unit (s). [1]
(c) Table of Results [6 Marks] - Range: Values of $x$ include $x_{\text{min}} \le 20.0\text{ cm}$ and $x_{\text{max}} \ge 45.0\text{ cm}$. [1] - Column headings: Each column has a correct quantity name and unit separated by a slash, e.g., $x\text{ / cm}$, $x^2\text{ / cm}^2$, $t_1\text{ / s}$, $t_2\text{ / s}$, $T\text{ / s}$, $1/T^2\text{ / s}^{-2}$. [1] - Consistency: All raw values of $x$ to the nearest $0.1\text{ cm}$ and all raw times to $0.01\text{ s}$ (or consistent $0.1\text{ s}$). [1] - Calculation: Calculated values of $x^2$ and $1/T^2$ are mathematically correct. [1] - Significant figures: Number of significant figures for $1/T^2$ matches or is one more than the s.f. of raw time / $T$. [1] - Number of data sets: At least 6 sets of data recorded systematically. [1]
(d) Graph [4 Marks] - Axes/Scale: Scales are linear, without awkward factors (e.g. 3, 7), points occupy $>50\%$ of the grid in both directions. [1] - Plotting: All points plotted accurately to within $\pm 0.5$ small square. Points must be fine crosses or small encircled dots. [1] - Line of Best Fit: Straight line drawn with balanced point distribution (equal number of points above/below line along its length). [1] - Line quality: Thin, straight line without kinks or excessive thickness ($<0.5\text{ mm}$). [1]
(e) Gradient and Intercept [4 Marks] - Gradient method: Hypotenuse of gradient triangle spans at least half the length of the drawn line, coordinates read correctly from line ($\Delta y / \Delta x$). [1] - Gradient value: Correct calculation matching student's plotted line. [1] - Intercept method: Intercept determined correctly using $c = y - mx$ or directly from axis if $x$-axis starts at $0$. [1] - Intercept value: Correct calculation to 2 or 3 s.f. [1]
(f) Evaluation of Constants [3 Marks] - $P$ equated to gradient and $Q$ equated to intercept. [1] - Correct numerical values of $P$ and $Q$ matching candidate's graph. [1] - Correct units for $P$ (e.g., $\text{s}^{-2}\text{ cm}^{-2}$ or $\text{s}^{-2}\text{ m}^{-2}$) and $Q$ ($\text{s}^{-2}$). [1]
Question 2 · free-response
20 marks
In this experiment, you will investigate the oscillatory motion of a weighted horizontal rod supported by two vertical parallel suspension threads (a bifilar pendulum).
(a) Measure and record the separation $d$ between the two suspension threads attached to the rod.
(ii) Displace the rod through a small horizontal angular displacement about its central vertical axis and release it so that it executes torsional oscillations. Take measurements to determine the period $T$ of these oscillations.
(e) Change the separation between the suspension threads to a new value $d$. Repeat (c)(i), (c)(ii), and (d) to obtain a second set of values for $d$, $h$, $T$, and $k$.
(ii) It is suggested that the relationship $k = \frac{Td}{\sqrt{h}}$ is valid if the percentage difference between the two values of $k$ is within $10\%$.
Explain whether your experimental results support this suggestion. You must include numerical values in your explanation.
(g) Identify four sources of error or limitations of the procedure for this investigation. Suggest four improvements that could be made to the experiment to address these difficulties.
You should state each limitation and the corresponding improvement.
**(e) Second set of measurements and calculation of $k_2$:** * $d_2 = 0.250\text{ m}$ * $h_2 = 0.500\text{ m}$ * Time for $n = 20$ oscillations: $t_1 = 45.3\text{ s}$, $t_2 = 45.5\text{ s} \implies t_{\text{mean}} = 45.4\text{ s}$ * Period $T_2 = \frac{45.4}{20} = 2.27\text{ s}$ * Calculation of $k_2$: $$k_2 = \frac{T_2 d_2}{\sqrt{h_2}} = \frac{2.27 \times 0.250}{\sqrt{0.500}} = \frac{0.5675}{0.7071} = 0.803\text{ m}^{1/2}\text{ s}$$
(f) (i) Significant figures justification: $d$, $h$, and $t$ (and thus $T$) are measured to 3 significant figures (e.g. $d = 0.400\text{ m}$, $h = 0.500\text{ m}$, $T = 1.43\text{ s}$). Therefore, $k$ is quoted to 3 significant figures (or the same number of s.f. as the raw quantity with the least s.f., which is 3 s.f.).
(ii) Evaluation of relationship: $$\text{Percentage difference} = \frac{|k_1 - k_2|}{k_{\text{average}}} \times 100 = \frac{|0.806 - 0.803|}{0.8045} \times 100 = 0.37\%$$ Since the percentage difference ($0.37\%$) is less than the specified criterion ($10\%$) [or within experimental uncertainty], the experimental results support the suggested relationship.
(g) Limitations and Improvements:
1. Limitation: Difficult to judge the exact completion of an oscillation / difficult to count oscillations because the rod moves fast or twists. * Improvement: Place a vertical reference marker / fiducial mark at the equilibrium position (or use a video camera with a timer / light gate connected to a data logger). 2. Limitation: The rod does not solely undergo torsional oscillation (it wobbles / sways from side to side / swings as a linear pendulum). * Improvement: Release the rod carefully from rest using a mechanical release mechanism or guide stops to ensure pure rotational displacement. 3. Limitation: Difficult to measure vertical distance $h$ accurately because it is hard to ensure the ruler is strictly vertical / ruler not perpendicular to bench/clamp. * Improvement: Use a set square against the bench/clamp stand or use a plumb line to ensure vertical alignment of the meter rule. 4. Limitation: Two sets of readings are insufficient to draw a firm conclusion / verify the relationship reliably. * Improvement: Take repeated sets of readings for at least 5 or more different thread separations $d$ and plot a graph of $T$ against $1/d$ (or $Td$ against $\sqrt{h}$).
Marking scheme
(a) [1 mark] * Value of $d$ recorded to the nearest millimeter with appropriate unit (e.g. in $\text{cm}$ or $\text{m}$). [B1]
(b) [1 mark] * Calculation of percentage uncertainty in $d$ using absolute uncertainty in the range $1\text{ mm} \le \Delta d \le 4\text{ mm}$. [B1]
(c)(i) [1 mark] * Value of $h$ recorded with appropriate unit to nearest mm. [B1]
(c)(ii) [2 marks] * Raw times for $n \ge 10$ oscillations recorded, with evidence of repeats (minimum two readings of $n T$). [M1] * Period $T$ calculated correctly from mean time $t/n$, given to 2 or 3 significant figures with unit (s). [A1]
(d) [1 mark] * Value of $k$ calculated correctly with consistent unit (e.g. $\text{m}^{1/2}\text{ s}$ or $\text{cm}^{1/2}\text{ s}$). [B1]
(e) [4 marks] * Second value of $d$ set and recorded correctly. [B1] * Second value of $h$ recorded. [B1] * Second period $T$ determined from repeated timings of $n \ge 10$ oscillations, showing increase in $T$ as $d$ decreases. [B1] * Second value of $k$ calculated correctly. [B1]
(f)(i) [1 mark] * Relates the number of significant figures in $k$ directly to the number of significant figures in the raw measurements ($d$, $h$, and $t$). [B1] (Reject vague statements like 'because my calculator gives this' or 'it is standard').
(f)(ii) [1 mark] * Clear calculation of percentage difference between $k_1$ and $k_2$, followed by a valid comparison against the $10\%$ criterion leading to an explicit concluding statement ('supports' or 'does not support'). [B1]
(g) [8 marks: 4 limitations (L), 4 corresponding improvements (I)] Award 1 mark for each valid limitation (max 4) and 1 mark for each corresponding valid improvement (max 4): * L1: Two sets of readings are not enough to confirm a relationship / draw a firm conclusion. [1] I1: Take at least 5 different values of $d$ (or $h$) and plot a graph of $T$ against $1/d$ (or $Td$ against $\sqrt{h}$). [1] * L2: Difficult to determine the start/end of an oscillation / exact moment of turning point. [1] I2: Use a fiducial marker / reference mark at the equilibrium center position (or light gate and data logger / video recording with frame timer). [1] * L3: Rod wobbles / introduces unwanted pendular swinging / sideways motion during oscillation. [1] I3: Use a release latch / mechanical release mechanism to release smoothly from rest without sideways thrust. [1] * L4: Difficult to measure vertical height $h$ accurately / ruler may not be vertical. [1] I4: Use a plumb line / set square against the bench to ensure rule is vertical. [1] * (Do not credit generic answers such as 'human reaction error', 'air resistance', or 'repeat readings' without specific practical detail). [Total: 8 marks]
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