An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V1) Cambridge International AS Level Physics (9702) paper. Not affiliated with or reproduced from Cambridge.
Paper 1 (Multiple Choice)
Answer all forty questions on the Multiple Choice answer sheet. Choose the single correct option A, B, C, or D.
40 Question · 40 marks
Question 1 · Multiple Choice
1 marks
Which combination of SI base units is equivalent to the ohm (\(\Omega\))?
A.\(\text{kg m}^{2} \text{s}^{-3} \text{A}^{-2}\)
B.\(\text{kg m}^{2} \text{s}^{-2} \text{A}^{-1}\)
C.\(\text{kg m}^{2} \text{s}^{-1} \text{A}^{-2}\)
D.\(\text{kg m}^{3} \text{s}^{-3} \text{A}^{-2}\)
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Worked solution
Resistance \(R = \frac{V}{I}\). Potential difference \(V = \frac{W}{Q} = \frac{W}{I t}\). Therefore, \(R = \frac{W}{I^2 t}\). In base units: work \(W\) is \(\text{N m} = \text{kg m}^2 \text{s}^{-2}\), current \(I\) is \(\text{A}\), and time \(t\) is \(\text{s}\). Thus, the unit of \(R\) is \(\frac{\text{kg m}^2 \text{s}^{-2}}{\text{A}^2 \text{s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\).
Marking scheme
A is correct [1]. \(R = V/I = W/(I^2 t) = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\).
Question 2 · Multiple Choice
1 marks
A stone is projected horizontally at a speed of \(15\text{ m s}^{-1}\) from the top of a vertical cliff of height \(45\text{ m}\) above horizontal ground. Air resistance is negligible.
What is the speed of the stone immediately before it strikes the ground?
A.\(29.7\text{ m s}^{-1}\)
B.\(33.3\text{ m s}^{-1}\)
C.\(44.7\text{ m s}^{-1}\)
D.\(47.4\text{ m s}^{-1}\)
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Worked solution
Horizontal component of velocity remains constant: \(v_x = 15\text{ m s}^{-1}\). Vertical component of velocity \(v_y\) is given by \(v_y^2 = u_y^2 + 2gh = 0 + 2(9.81)(45) = 882.9\text{ m}^2\text{ s}^{-2}\). Final speed \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 882.9} = \sqrt{225 + 882.9} = \sqrt{1107.9} \approx 33.3\text{ m s}^{-1}\).
Marking scheme
B is correct [1]. Speed \(v = \sqrt{15^2 + 2 \times 9.81 \times 45} = 33.3\text{ m s}^{-1}\).
Question 3 · Multiple Choice
1 marks
A trolley of mass \(0.40\text{ kg}\) travels along a straight, horizontal, frictionless track at a velocity of \(2.5\text{ m s}^{-1}\). It collides head-on with a stationary trolley of mass \(0.60\text{ kg}\). The two trolleys stick together after the collision.
What is the loss in kinetic energy during the collision?
A.\(0.50\text{ J}\)
B.\(0.75\text{ J}\)
C.\(1.00\text{ J}\)
D.\(1.25\text{ J}\)
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Worked solution
Initial momentum \(p = m_1 u_1 = 0.40 \times 2.5 = 1.0\text{ kg m s}^{-1}\). By conservation of momentum, combined velocity \(V = \frac{1.0}{0.40 + 0.60} = 1.0\text{ m s}^{-1}\). Initial kinetic energy \(E_{\text{k, initial}} = \frac{1}{2}(0.40)(2.5)^2 = 1.25\text{ J}\). Final kinetic energy \(E_{\text{k, final}} = \frac{1}{2}(0.40 + 0.60)(1.0)^2 = 0.50\text{ J}\). Loss in kinetic energy \(= 1.25 - 0.50 = 0.75\text{ J}\).
Marking scheme
B is correct [1]. Initial \(E_k = 1.25\text{ J}\), final \(E_k = 0.50\text{ J}\), difference \(= 0.75\text{ J}\).
Question 4 · Multiple Choice
1 marks
A solid cylinder of uniform cross-sectional area \(4.0 \times 10^{-3}\text{ m}^{2}\) and height \(0.15\text{ m}\) is completely submerged in a liquid of density \(1.2 \times 10^{3}\text{ kg m}^{-3}\). The top flat face of the cylinder is horizontal and at a depth of \(0.25\text{ m}\) below the liquid surface.
What is the upthrust exerted by the liquid on the cylinder?
A.\(1.8\text{ N}\)
B.\(7.1\text{ N}\)
C.\(12\text{ N}\)
D.\(19\text{ N}\)
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Worked solution
Upthrust is equal to the weight of liquid displaced: \(F = \rho g V\). Volume of cylinder \(V = A h = 4.0 \times 10^{-3} \times 0.15 = 6.0 \times 10^{-4}\text{ m}^3\). Upthrust \(F = 1.2 \times 10^3 \times 9.81 \times 6.0 \times 10^{-4} = 7.06\text{ N} \approx 7.1\text{ N}\). The depth of the top surface does not affect the upthrust on a fully submerged object.
Marking scheme
B is correct [1]. \(F = \rho g V = 1200 \times 9.81 \times (4.0 \times 10^{-3} \times 0.15) = 7.1\text{ N}\).
Question 5 · Multiple Choice
1 marks
A metal wire of original length \(2.0\text{ m}\) and circular cross-section of diameter \(0.80\text{ mm}\) is stretched elastically within its limit of proportionality by a tensile force of \(150\text{ N}\). The Young modulus of the metal is \(1.2 \times 10^{11}\text{ Pa}\).
What is the strain energy stored in the wire?
A.\(0.19\text{ J}\)
B.\(0.37\text{ J}\)
C.\(0.75\text{ J}\)
D.\(1.5\text{ J}\)
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B is correct [1]. Calculation of extension \(x \approx 4.97\text{ mm}\) and \(E_p = \frac{1}{2}Fx = 0.37\text{ J}\).
Question 6 · Multiple Choice
1 marks
A siren on a vehicle emits sound of frequency \(800\text{ Hz}\). The vehicle travels at a constant speed along a straight road directly towards a stationary observer. The speed of sound in air is \(340\text{ m s}^{-1}\). The observer detects a frequency of \(860\text{ Hz}\).
What is the speed of the vehicle?
A.\(21.2\text{ m s}^{-1}\)
B.\(23.7\text{ m s}^{-1}\)
C.\(25.5\text{ m s}^{-1}\)
D.\(27.2\text{ m s}^{-1}\)
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Worked solution
Using the Doppler effect formula for a moving source approaching a stationary observer: \(f_{\text{o}} = f_{\text{s}} \left(\frac{v}{v - v_{\text{s}}}\right)\). \(860 = 800 \left(\frac{340}{340 - v_{\text{s}}}\right)\) \(\frac{860}{800} = 1.075 = \frac{340}{340 - v_{\text{s}}}\) \(340 - v_{\text{s}} = \frac{340}{1.075} \approx 316.28\text{ m s}^{-1}\) \(v_{\text{s}} = 340 - 316.28 = 23.7\text{ m s}^{-1}\).
Marking scheme
B is correct [1]. Rearranging \(f_o = f_s v / (v - v_s)\) yields \(v_s = 23.7\text{ m s}^{-1}\).
Question 7 · Multiple Choice
1 marks
Monochromatic light of wavelength \(630\text{ nm}\) is incident normally on a diffraction grating that has \(400\text{ lines mm}^{-1}\).
What is the total number of principal maxima that can be observed?
A.3
B.4
C.6
D.7
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Worked solution
Grating spacing \(d = \frac{10^{-3}\text{ m}}{400} = 2.50 \times 10^{-6}\text{ m}\). For principal maxima, \(d \sin\theta = n \lambda\). Since \(\sin\theta \le 1\), the maximum order \(n_{\text{max}} = \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}}{630 \times 10^{-9}} = 3.97\). Therefore, the highest integer order is \(n = 3\). The total number of observed maxima is \(2n + 1 = 2(3) + 1 = 7\) (orders \(n = 0, \pm 1, \pm 2, \pm 3\)).
Marking scheme
D is correct [1]. Maximum integer order \(n = 3\); total count \(= 2(3) + 1 = 7\).
Question 8 · multiple_choice
1 marks
A student determines the density \(\rho\) of a solid sphere by measuring its mass \(m\) and diameter \(d\).
The recorded values are: \[m = (56.0 \pm 0.5)\text{ g}\] \[d = (2.40 \pm 0.02)\text{ cm}\]
What is the percentage uncertainty in the calculated density?
A.1.7%
B.2.5%
C.3.4%
D.4.2%
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Worked solution
The density \(\rho\) of a sphere is given by: \[\rho = \frac{m}{V} = \frac{m}{\frac{4}{3}\pi (d/2)^3} = \frac{6m}{\pi d^3}\]
The fractional uncertainty in \(\rho\) is: \[\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\left(\frac{\Delta d}{d}\right)\]
C [1 mark]: Correctly identifies that the fractional uncertainty in diameter is multiplied by 3 and summed with the fractional uncertainty in mass to give 3.4%.
Question 9 · Multiple Choice
1 marks
Which combination of SI base units is equivalent to the ohm metre (\(\Omega\,\text{m}\))?
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Worked solution
Resistance \(R = \frac{V}{I} = \frac{P}{I^2} = \frac{\text{work done}}{I^2 \times t}\). In SI base units, \(R = \frac{\text{kg}\,\text{m}^2\,\text{s}^{-2}}{\text{A}^2 \times \text{s}} = \text{kg}\,\text{m}^2\,\text{s}^{-3}\,\text{A}^{-2}\). Resistivity unit is \(\Omega\,\text{m} = (\text{kg}\,\text{m}^2\,\text{s}^{-3}\,\text{A}^{-2}) \times \text{m} = \text{kg}\,\text{m}^3\,\text{s}^{-3}\,\text{A}^{-2}\).
Marking scheme
B [1 mark]: Correct base-unit derivation leading to \(\text{kg}\,\text{m}^3\,\text{s}^{-3}\,\text{A}^{-2}\).
Question 10 · Multiple Choice
1 marks
A small sphere is projected horizontally from the top of a vertical cliff of height \(45\text{ m}\) with an initial speed of \(15\text{ m s}^{-1}\). Air resistance is negligible.
What is the speed of the sphere immediately before it hits the horizontal ground below?
A.\(15\text{ m s}^{-1}\)
B.\(30\text{ m s}^{-1}\)
C.\(33\text{ m s}^{-1}\)
D.\(45\text{ m s}^{-1}\)
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C [1 mark]: Correct combination of perpendicular velocity components \(v = \sqrt{v_x^2 + v_y^2} = 33\,\text{m s}^{-1}\).
Question 11 · Multiple Choice
1 marks
Trolley X of mass \(0.60\text{ kg}\) moves to the right at a speed of \(2.0\text{ m s}^{-1}\) on a horizontal frictionless track. Trolley Y of mass \(0.40\text{ kg}\) moves to the left towards trolley X at a speed of \(3.0\text{ m s}^{-1}\).
The two trolleys collide and stick together.
What is the total loss of kinetic energy in the collision?
A.\(0\text{ J}\)
B.\(1.2\text{ J}\)
C.\(1.8\text{ J}\)
D.\(3.0\text{ J}\)
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Worked solution
Total initial momentum taking right as positive: \(p_i = (0.60)(2.0) + (0.40)(-3.0) = 1.2 - 1.2 = 0\,\text{N s}\). Because total initial momentum is zero, the combined mass is stationary after the collision (\(v = 0\)). Initial kinetic energy \(E_k = \frac{1}{2}(0.60)(2.0)^2 + \frac{1}{2}(0.40)(3.0)^2 = 1.2\,\text{J} + 1.8\,\text{J} = 3.0\,\text{J}\). Final kinetic energy is \(0\,\text{J}\), so energy lost is \(3.0\,\text{J}\).
Marking scheme
D [1 mark]: Calculation of initial kinetic energy \(3.0\,\text{J}\) and recognition that final momentum is zero, hence all \(3.0\,\text{J}\) is lost.
Question 12 · Multiple Choice
1 marks
A solid uniform cube of edge length \(0.050\text{ m}\) and density \(2400\text{ kg m}^{-3}\) is completely submerged in a liquid of density \(800\text{ kg m}^{-3}\).
The cube is suspended by a light vertical cord attached to its top face.
What is the tension in the cord?
A.\(0.98\text{ N}\)
B.\(1.5\text{ N}\)
C.\(2.0\text{ N}\)
D.\(2.9\text{ N}\)
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Worked solution
Volume \(V = (0.050)^3 = 1.25 \times 10^{-4}\,\text{m}^3\). Weight \(W = \rho_s V g\), Upthrust \(U = \rho_l V g\). Tension \(T = W - U = (\rho_s - \rho_l) V g = (2400 - 800) \times (1.25 \times 10^{-4}) \times 9.81 = 1600 \times 1.25 \times 10^{-4} \times 9.81 = 0.200 \times 9.81 \approx 2.0\,\text{N}\).
Marking scheme
C [1 mark]: Correct calculation using \(T = (\rho_s - \rho_l)Vg = 2.0\,\text{N}\).
Question 13 · Multiple Choice
1 marks
A uniform metal wire of original length \(L\) and cross-sectional area \(A\) is extended by a tensile force \(F\) within its limit of proportionality. The elastic strain energy stored in the wire is \(E_p\).
A second wire of the same metal has original length \(2L\) and cross-sectional area \(2A\). It is stretched by a tensile force of \(2F\).
What is the elastic strain energy stored in the second wire?
A.\(E_p\)
B.\(2E_p\)
C.\(4E_p\)
D.\(8E_p\)
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Worked solution
Young modulus \(E = \frac{F/A}{x/L} \implies x = \frac{FL}{AE}\). Elastic strain energy \(E_p = \frac{1}{2}Fx = \frac{F^2 L}{2AE}\). For the second wire: \(E_{p2} = \frac{(2F)^2(2L)}{2(2A)E} = \frac{4F^2 \times 2L}{4AE} = 2 \times \frac{F^2 L}{2AE} = 4 E_p\) (alternatively, \(x_2 = \frac{(2F)(2L)}{(2A)E} = 2x\), so energy \(= \frac{1}{2}(2F)(2x) = 2Fx = 4 E_p\)).
Marking scheme
C [1 mark]: Correct scaling of strain energy yielding \(4 E_p\).
Question 14 · Multiple Choice
1 marks
A train travels at a constant speed of \(25\text{ m s}^{-1}\) directly towards a stationary observer while sounding a horn of frequency \(660\text{ Hz}\). The speed of sound in air is \(330\text{ m s}^{-1}\).
What is the frequency of the sound heard by the observer?
A.\(614\text{ Hz}\)
B.\(660\text{ Hz}\)
C.\(714\text{ Hz}\)
D.\(720\text{ Hz}\)
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Worked solution
Using the Doppler effect formula for a moving source approaching a stationary observer: \(f_o = f_s \frac{v}{v - v_s} = 660 \times \frac{330}{330 - 25} = 660 \times \frac{330}{305} \approx 714\,\text{Hz}\).
Marking scheme
C [1 mark]: Correct substitution into the Doppler formula giving \(714\,\text{Hz}\).
Question 15 · Multiple Choice
1 marks
A battery of constant electromotive force (e.m.f.) and negligible internal resistance is connected in series with a fixed resistor and a light-dependent resistor (LDR). A high-resistance voltmeter is connected across the LDR.
The intensity of the light incident on the LDR is increased.
Which row correctly describes the change in resistance of the LDR and the change in the voltmeter reading?
A.resistance of LDR decreases; voltmeter reading decreases
B.resistance of LDR decreases; voltmeter reading increases
C.resistance of LDR increases; voltmeter reading decreases
D.resistance of LDR increases; voltmeter reading increases
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Worked solution
As light intensity increases, the resistance of the LDR decreases. By the potential divider principle, the potential difference across the LDR is given by \(V_{\text{LDR}} = E \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\). As \(R_{\text{LDR}}\) decreases, the proportion of total e.m.f. dropped across it decreases, so the voltmeter reading decreases.
Marking scheme
A [1 mark]: Identification of decreasing LDR resistance and correspondingly decreasing voltmeter reading.
Question 16 · Multiple Choice
1 marks
In a \(\beta^-\)-decay process, a stationary nucleus decays by changing a neutron into a proton.
Which pair of particles is emitted from the nucleus during this decay?
A.a positron and an electron neutrino
B.a positron and an electron antineutrino
C.an electron and an electron neutrino
D.an electron and an electron antineutrino
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Worked solution
In \(\beta^-\)-decay, a down quark transforms into an up quark: \(\text{d} \rightarrow \text{u} + \text{e}^- + \bar{\nu}_e\). Thus, the emitted particles are an electron and an electron antineutrino.
Marking scheme
D [1 mark]: Correct identification of an electron and an electron antineutrino.
Question 17 · Multiple Choice
1 marks
Which combination of SI base units is equivalent to the electrical unit ohm (\(\Omega\))?
A.\(\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\)
B.\(\text{kg m}^2 \text{s}^{-2} \text{A}^{-1}\)
C.\(\text{kg m s}^{-3} \text{A}^{-2}\)
D.\(\text{kg m}^2 \text{s}^{-1} \text{A}^{-2}\)
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Worked solution
Resistance is defined by \(R = \frac{V}{I}\).
Potential difference is work done per unit charge: \(V = \frac{W}{Q} = \frac{W}{I t}\).
Therefore, \(R = \frac{W}{I^2 t}\).
In SI base units:
- Work (energy) \(W\) has base units \(\text{kg m}^2 \text{s}^{-2}\)
- Current \(I\) has base unit \(\text{A}\)
- Time \(t\) has base unit \(\text{s}\)
Substituting these base units gives:
\([R] = \frac{\text{kg m}^2 \text{s}^{-2}}{\text{A}^2 \text{s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\).
Marking scheme
A [1]
Derivation from \(R = \frac{V}{I} = \frac{W}{I^2 t}\) or \(P = I^2 R\).
Question 18 · Multiple Choice
1 marks
A small stone is projected horizontally at a speed of \(15\text{ m s}^{-1}\) from the top of a vertical cliff of height \(45\text{ m}\) above horizontal ground. Air resistance is negligible.
What is the speed of the stone immediately before it strikes the ground?
A.\(15\text{ m s}^{-1}\)
B.\(30\text{ m s}^{-1}\)
C.\(33\text{ m s}^{-1}\)
D.\(48\text{ m s}^{-1}\)
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Worked solution
Consider the vertical motion under uniform acceleration:
\(v_y^2 = u_y^2 + 2 g h = 0^2 + 2 \times 9.81 \times 45 = 882.9\text{ m}^2\text{ s}^{-2}\)
The horizontal velocity remains constant:
\(v_x = 15\text{ m s}^{-1}\)
The resultant speed \(v\) on impact is given by:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 882.9} = \sqrt{225 + 882.9} = \sqrt{1107.9} \approx 33\text{ m s}^{-1}\).
Marking scheme
C [1]
Calculation of vertical velocity component and combination with horizontal component via Pythagoras.
Question 19 · Multiple Choice
1 marks
A trolley of mass \(0.60\text{ kg}\) travels with a speed of \(4.0\text{ m s}^{-1}\) along a horizontal frictionless track. It collides head-on with a stationary trolley of mass \(0.40\text{ kg}\).
After the collision, the \(0.60\text{ kg}\) trolley continues in the same direction at a reduced speed of \(1.0\text{ m s}^{-1}\).
What is the total kinetic energy lost in this collision?
A.\(0.45\text{ J}\)
B.\(0.75\text{ J}\)
C.\(4.35\text{ J}\)
D.\(4.80\text{ J}\)
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A [1]
Correct application of conservation of momentum followed by finding difference in kinetic energies.
Question 20 · Multiple Choice
1 marks
A solid cylinder of uniform cross-sectional area \(A\) and height \(h\) is fully submerged in a liquid of density \(\rho\). The upper face of the cylinder is located at depth \(d\) below the surface of the liquid. The atmospheric pressure is \(p_0\) and the acceleration of free fall is \(g\).
Which expression gives the magnitude of the upthrust acting on the cylinder?
A.\(\rho g h A\)
B.\((\rho g d + p_0) A\)
C.\(\rho g (d + h) A\)
D.\((p_0 + \rho g h) A\)
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Worked solution
Upthrust is the resultant upward force exerted by the fluid due to the difference in hydrostatic pressure on the lower and upper faces of the cylinder.
Pressure on upper face: \(p_{\text{top}} = p_0 + \rho g d\)
Pressure on lower face: \(p_{\text{bottom}} = p_0 + \rho g (d + h)\)
Upthrust \(F = (p_{\text{bottom}} - p_{\text{top}}) A = [\rho g (d + h) - \rho g d] A = \rho g h A\).
This is equal to the weight of the displaced fluid and is independent of the depth \(d\) and atmospheric pressure \(p_0\).
Marking scheme
A [1]
Recognition that upthrust is equal to \(\rho g V = \rho g h A\).
Question 21 · Multiple Choice
1 marks
A tensile force \(F\) applied to a uniform metal wire of length \(L\) and circular cross-section of diameter \(d\) produces an extension \(x\) within the limit of proportionality.
A second wire of the same metal has a length \(2L\) and a diameter \(2d\).
What tensile force is required to produce the same extension \(x\) in the second wire?
A.\(0.5F\)
B.\(F\)
C.\(2F\)
D.\(4F\)
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Worked solution
From the definition of the Young modulus \(E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\).
Rearranging for force: \(F = \frac{E A x}{L} = \frac{E (\frac{\pi d^2}{4}) x}{L} \propto \frac{d^2}{L}\).
For the second wire:
- Diameter becomes \(2d\), so cross-sectional area becomes \((2)^2 A = 4A\).
- Length becomes \(2L\).
- Extension \(x\) and Young modulus \(E\) are the same.
Therefore, the new force \(F_2\) is:
\(F_2 = \frac{E (4A) x}{2L} = 2 \left(\frac{E A x}{L}\right) = 2F\).
Marking scheme
C [1]
Correct scaling of area by factor of 4 and length by factor of 2 leading to ratio \(4/2 = 2\).
Question 22 · Multiple Choice
1 marks
A train sounds a horn emitting a sound of constant frequency \(650\text{ Hz}\). The train moves at a constant speed of \(28\text{ m s}^{-1}\) directly towards a stationary observer.
The speed of sound in air is \(340\text{ m s}^{-1}\).
What is the frequency of the sound detected by the observer?
A.\(600\text{ Hz}\)
B.\(650\text{ Hz}\)
C.\(701\text{ Hz}\)
D.\(708\text{ Hz}\)
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Worked solution
For a source moving towards a stationary observer, the observed Doppler frequency is given by:
\(f_{\text{o}} = f_{\text{s}} \left(\frac{v}{v - v_{\text{s}}}\right)\)
D [1]
Correct substitution into the Doppler effect equation for a source approaching a stationary observer.
Question 23 · Multiple Choice
1 marks
Monochromatic light of wavelength \(590\text{ nm}\) is incident normally on a diffraction grating. The first-order maximum is observed at an angle of \(17.2^\circ\) to the normal.
What is the total number of intensity maxima that can be observed for this wavelength?
A.\(3\)
B.\(6\)
C.\(7\)
D.\(8\)
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Worked solution
Using the grating equation \(d \sin \theta = n \lambda\) for \(n = 1\):
\(d = \frac{1 \times 590 \times 10^{-9}}{\sin(17.2^\circ)} = \frac{590 \times 10^{-9}}{0.2957} \approx 1.995 \times 10^{-6}\text{ m}\)
The maximum possible order \(n_{\text{max}}\) occurs when \(\sin \theta \le 1\):
\(n_{\text{max}} = \frac{d}{\lambda} = \frac{1}{\sin(17.2^\circ)} = \frac{1}{0.2957} \approx 3.38\)
Since \(n\) must be an integer, the highest observable order is \(n = 3\).
The total number of maxima includes the central zeroth-order maximum and orders up to \(n = 3\) on both sides:
\(\text{Total maxima} = 2 n_{\text{max}} + 1 = 2(3) + 1 = 7\).
Marking scheme
C [1]
Finding \(n_{\text{max}} = 3\) and calculating total observable orders as \(2n + 1 = 7\).
Question 24 · Multiple Choice
1 marks
A battery of electromotive force (e.m.f.) \(12\text{ V}\) and internal resistance \(2.0\ \Omega\) is connected to a variable load resistor of resistance \(R\).
What is the value of \(R\) when the power dissipated in the load resistor is a maximum, and what is the value of this maximum power?
A.\(R = 1.0\ \Omega\text{ and Power} = 18\text{ W}\)
B.\(R = 2.0\ \Omega\text{ and Power} = 18\text{ W}\)
C.\(R = 2.0\ \Omega\text{ and Power} = 36\text{ W}\)
D.\(R = 4.0\ \Omega\text{ and Power} = 18\text{ W}\)
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Worked solution
The power transferred to an external load resistor \(R\) is maximum when the load resistance equals the internal resistance of the power source (maximum power theorem):
\(R = r = 2.0\ \Omega\)
The current in the circuit under this condition is:
\(I = \frac{E}{R + r} = \frac{12}{2.0 + 2.0} = 3.0\text{ A}\)
The power dissipated in the load resistor is:
\(P = I^2 R = (3.0)^2 \times 2.0 = 18\text{ W}\).
Marking scheme
B [1]
Matching load resistance to internal resistance \(R = 2.0\ \Omega\) and evaluating maximum power as \(18\text{ W}\).
Question 25 · multiple_choice
1 marks
What is the unit of electrical resistivity expressed in SI base units?
A.\(\text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\)
B.\(\text{kg m}^2 \text{s}^{-3} \text{A}^{-1}\)
C.\(\text{kg m}^3 \text{s}^{-2} \text{A}^{-2}\)
D.\(\text{kg m} \text{s}^{-3} \text{A}^{-2}\)
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Worked solution
Resistivity is defined by \(\rho = \frac{R A}{L}\). Resistance is \(R = \frac{V}{I} = \frac{P}{I^2}\), where power \(P = \frac{W}{t}\) has base units \(\text{kg m}^2 \text{s}^{-3}\). Therefore, resistance has base units \(\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\). Multiplying by area (\(\text{m}^2\)) and dividing by length (\(\text{m}\)) gives \(\text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\).
Marking scheme
A is correct [1]. Base units for resistivity derived from R = V/I = W/(I^2 t) and rho = R A / L.
Question 26 · multiple_choice
1 marks
A small steel ball is projected horizontally from the top of a vertical cliff with a speed of \(18\text{ m s}^{-1}\). The ball strikes the horizontal ground below at an angle of \(45^\circ\) to the horizontal. Air resistance is negligible. What is the height of the cliff?
A.\(8.3\text{ m}\)
B.\(17\text{ m}\)
C.\(33\text{ m}\)
D.\(66\text{ m}\)
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Worked solution
The horizontal velocity remains constant at \(v_x = 18\text{ m s}^{-1}\). At ground impact, \(\tan 45^\circ = \frac{v_y}{v_x} = 1\), so \(v_y = 18\text{ m s}^{-1}\). Using \(v_y^2 = 2gh\), \(h = \frac{18^2}{2 \times 9.81} = \frac{324}{19.62} \approx 17\text{ m}\).
Marking scheme
B is correct [1]. v_y = v_x = 18 m s^-1 at 45 degrees, h = v_y^2 / (2g) = 16.5 m rounded to 17 m.
Question 27 · multiple_choice
1 marks
A glider of mass \(m\) moves along a frictionless linear air track with kinetic energy \(E_0\). It makes an elastic head-on collision with a stationary glider of mass \(3m\). What is the kinetic energy of the glider of mass \(m\) after the collision in terms of \(E_0\)?
A.\(\frac{1}{16}E_0\)
B.\(\frac{1}{8}E_0\)
C.\(\frac{1}{4}E_0\)
D.\(\frac{1}{2}E_0\)
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Worked solution
In a 1D elastic collision with a stationary target, the velocity of mass \(m_1\) after collision is \(v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) u\). Here \(v_1 = \left(\frac{m - 3m}{m + 3m}\right) u = -\frac{1}{2}u\). The kinetic energy of the glider of mass \(m\) becomes \(E = \frac{1}{2}m v_1^2 = \frac{1}{2}m \left(-\frac{1}{2}u\right)^2 = \frac{1}{4}\left(\frac{1}{2}mu^2\right) = \frac{1}{4}E_0\).
Marking scheme
C is correct [1]. Fraction of initial speed retained is |(m - 3m)/(m + 3m)| = 1/2, so fraction of kinetic energy retained is (1/2)^2 = 1/4.
Question 28 · multiple_choice
1 marks
A uniform cylinder floats upright in a container with its volume distributed across two immiscible liquids, X and Y. Liquid X has density \(750\text{ kg m}^{-3}\) and floats on top of liquid Y, which has density \(1200\text{ kg m}^{-3}\). The density of the cylinder is \(930\text{ kg m}^{-3}\). What percentage of the cylinder's volume is submerged in liquid Y?
A.\(24\%\)
B.\(40\%\)
C.\(60\%\)
D.\(78\%\)
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Worked solution
For floating equilibrium, total upthrust equals weight: \(V_X \rho_X g + V_Y \rho_Y g = V_{\text{total}} \rho_{\text{cylinder}} g\). Letting \(f = \frac{V_Y}{V_{\text{total}}}\), \((1 - f)\rho_X + f \rho_Y = \rho_{\text{cylinder}}\). Substituting values: \(750(1 - f) + 1200f = 930 \implies 450f = 180 \implies f = 0.40 = 40\%\).
Marking scheme
B is correct [1]. Equate weight to total buoyancy: 750(1 - f) + 1200f = 930 giving f = 0.40 (40%).
Question 29 · multiple_choice
1 marks
A metal wire of cross-sectional area \(A\) and length \(L\) extends by an amount \(x\) when subjected to a tensile force \(F\) within its elastic limit. A second wire made of the same metal has cross-sectional area \(3A\) and length \(2L\). What tensile force produces an extension of \(3x\) in the second wire?
A.\(1.5F\)
B.\(2.0F\)
C.\(4.5F\)
D.\(9.0F\)
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Worked solution
Using Young modulus \(E = \frac{F L}{A x}\), the required force is \(F = \frac{E A x}{L}\). For the second wire, \(F_2 = \frac{E (3A) (3x)}{2L} = \frac{9}{2} \left(\frac{E A x}{L}\right) = 4.5F\).
Marking scheme
C is correct [1]. F is proportional to (A * x) / L. Ratio is (3 * 3) / 2 = 4.5.
Question 30 · multiple_choice
1 marks
An emergency vehicle emits sound from its siren at a constant frequency of \(720\text{ Hz}\). The vehicle travels at a constant speed of \(30\text{ m s}^{-1}\) directly towards a stationary observer. The speed of sound in air is \(330\text{ m s}^{-1}\). What frequency of sound is detected by the observer?
A.\(655\text{ Hz}\)
B.\(660\text{ Hz}\)
C.\(785\text{ Hz}\)
D.\(792\text{ Hz}\)
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Worked solution
Using the Doppler effect formula for a moving source approaching a stationary observer: \(f_o = f_s \left(\frac{v}{v - v_s}\right) = 720 \times \left(\frac{330}{330 - 30}\right) = 720 \times \frac{330}{300} = 792\text{ Hz}\).
Marking scheme
D is correct [1]. f_o = f_s * v / (v - v_s) = 720 * 330 / 300 = 792 Hz.
Question 31 · multiple_choice
1 marks
A cell of electromotive force (e.m.f.) \(E\) and internal resistance \(r\) is connected to a variable external resistor. The resistance of the variable resistor is gradually increased from \(R = r\) to \(R = 3r\). Which statement correctly describes the changes to the power dissipated in the internal resistance and the potential difference across the variable resistor?
A.Power in internal resistance decreases; potential difference across resistor decreases
B.Power in internal resistance decreases; potential difference across resistor increases
C.Power in internal resistance increases; potential difference across resistor decreases
D.Power in internal resistance increases; potential difference across resistor increases
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Worked solution
Current in the circuit is \(I = \frac{E}{R + r}\). As \(R\) increases from \(r\) to \(3r\), total resistance increases from \(2r\) to \(4r\), so current \(I\) decreases. Power in internal resistance is \(P_{\text{int}} = I^2 r\), which decreases. The terminal p.d. across the variable resistor is \(V = E - Ir\), which increases as \(I\) decreases.
Marking scheme
B is correct [1]. Total resistance increases so current falls; I^2 * r decreases and terminal p.d. V = E - Ir increases.
Question 32 · multiple_choice
1 marks
A stationary nucleus of magnesium-23 (\(^{23}_{12}\text{Mg}\)) undergoes \(\beta^+\) decay to produce a sodium (Na) nucleus, a positron, and another particle X. What are the nucleon number and proton number of the sodium nucleus, and what is the identity of particle X?
A.Nucleon number = 22, Proton number = 11, Particle X = electron antineutrino
B.Nucleon number = 23, Proton number = 11, Particle X = electron neutrino
C.Nucleon number = 23, Proton number = 13, Particle X = electron neutrino
D.Nucleon number = 23, Proton number = 11, Particle X = electron antineutrino
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Worked solution
In \(\beta^+\) decay, a proton decays into a neutron, a positron, and an electron neutrino: \(p \rightarrow n + \beta^+ + \nu_e\). Nucleon number is conserved (23). Proton number decreases by 1 (from 12 to 11). Thus, the daughter nucleus is \(^{23}_{11}\text{Na}\) and particle X is an electron neutrino.
Marking scheme
B is correct [1]. Conserving A gives 23, conserving Z gives 12 - 1 = 11, beta-plus emission is accompanied by an electron neutrino.
Question 33 · Multiple Choice
1 marks
Which combination of SI base units is equivalent to the ohm (\(\Omega\))?
A.\(\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\)
B.\(\text{kg m}^2 \text{s}^{-2} \text{A}^{-1}\)
C.\(\text{kg m s}^{-3} \text{A}^{-2}\)
D.\(\text{kg m}^2 \text{s}^{-3} \text{A}^{-1}\)
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Worked solution
Resistance \(R = \frac{V}{I}\). The potential difference \(V = \frac{W}{Q} = \frac{\text{work}}{\text{charge}}\), and charge \(Q = I t\). Therefore, \(R = \frac{\text{work}}{I^2 t}\). In SI base units, work has units \(\text{kg m}^2 \text{s}^{-2}\), current has base unit \(\text{A}\), and time has base unit \(\text{s}\). Thus, the base units of resistance are \(\frac{\text{kg m}^2 \text{s}^{-2}}{\text{A}^2 \text{s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\).
A stone is projected horizontally at a speed of \(15\text{ m s}^{-1}\) from the top of a vertical cliff of height \(45\text{ m}\) above horizontal ground. Air resistance is negligible. What is the speed of the stone immediately before it hits the ground?
A.\(15\text{ m s}^{-1}\)
B.\(30\text{ m s}^{-1}\)
C.\(33\text{ m s}^{-1}\)
D.\(45\text{ m s}^{-1}\)
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Worked solution
The horizontal component of velocity remains constant: \(v_x = 15\text{ m s}^{-1}\). The vertical component of velocity \(v_y\) is given by \(v_y^2 = u_y^2 + 2 g h = 0 + 2 \times 9.81 \times 45 = 882.9\text{ m}^2\text{ s}^{-2}\). The resultant speed \(v\) is \(\sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 882.9} = \sqrt{225 + 882.9} = \sqrt{1107.9} \approx 33.3\text{ m s}^{-1}\), which rounds to \(33\text{ m s}^{-1}\).
Marking scheme
[1] C: \(33\text{ m s}^{-1}\)
Question 35 · Multiple Choice
1 marks
A ball of mass \(0.20\text{ kg}\) travels horizontally at a speed of \(18\text{ m s}^{-1}\) and strikes a vertical wall. It rebounds horizontally in the opposite direction at a speed of \(12\text{ m s}^{-1}\). The ball is in contact with the wall for a time of \(0.040\text{ s}\). What is the magnitude of the average force exerted on the ball by the wall during the collision?
A.\(30\text{ N}\)
B.\(60\text{ N}\)
C.\(150\text{ N}\)
D.\(300\text{ N}\)
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Worked solution
Taking the initial direction as positive: change in momentum \(\Delta p = m v - m u = 0.20 \times (-12) - 0.20 \times 18 = -2.4 - 3.6 = -6.0\text{ N s}\). The magnitude of the average force is \(F = \frac{|\Delta p|}{\Delta t} = \frac{6.0}{0.040} = 150\text{ N}\).
Marking scheme
[1] C: \(150\text{ N}\)
Question 36 · Multiple Choice
1 marks
A solid cylinder of uniform cross-sectional area \(2.5 \times 10^{-3}\text{ m}^2\) and length \(0.40\text{ m}\) is completely submerged in a liquid of density \(1.2 \times 10^3\text{ kg m}^{-3}\). What is the upthrust acting on the cylinder?
A.\(1.2\text{ N}\)
B.\(4.7\text{ N}\)
C.\(12\text{ N}\)
D.\(29\text{ N}\)
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Worked solution
The volume of the cylinder \(V = A \times L = 2.5 \times 10^{-3} \times 0.40 = 1.0 \times 10^{-3}\text{ m}^3\). By Archimedes' principle, the upthrust \(F = \rho g V = (1.2 \times 10^3) \times 9.81 \times (1.0 \times 10^{-3}) = 11.772\text{ N} \approx 12\text{ N}\).
Marking scheme
[1] C: \(12\text{ N}\)
Question 37 · Multiple Choice
1 marks
A metal wire of original length \(2.0\text{ m}\) has a uniform cross-sectional area of \(4.0 \times 10^{-7}\text{ m}^2\). When a tensile force of \(80\text{ N}\) is applied, the wire extends elastically by \(1.6\text{ mm}\). What is the Young modulus of the metal?
A.\(1.6 \times 10^8\text{ Pa}\)
B.\(6.4 \times 10^9\text{ Pa}\)
C.\(1.3 \times 10^{11}\text{ Pa}\)
D.\(2.5 \times 10^{11}\text{ Pa}\)
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Worked solution
The Young modulus is \(E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{\Delta L / L} = \frac{F L}{A \Delta L}\). Substituting the values: \(E = \frac{80 \times 2.0}{(4.0 \times 10^{-7}) \times (1.6 \times 10^{-3})} = \frac{160}{6.4 \times 10^{-10}} = 2.5 \times 10^{11}\text{ Pa}\).
Marking scheme
[1] D: \(2.5 \times 10^{11}\text{ Pa}\)
Question 38 · Multiple Choice
1 marks
A train travels at a constant speed of \(24.0\text{ m s}^{-1}\) directly towards a stationary observer. The train sounds a whistle that emits sound waves of frequency \(660\text{ Hz}\). The speed of sound in air is \(330\text{ m s}^{-1}\). What is the observed frequency of the sound heard by the observer?
A.\(615\text{ Hz}\)
B.\(660\text{ Hz}\)
C.\(708\text{ Hz}\)
D.\(712\text{ Hz}\)
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Worked solution
Using the Doppler effect formula for a moving source approaching a stationary observer: \(f_o = f_s \left(\frac{v}{v - v_s}\right) = 660 \left(\frac{330}{330 - 24.0}\right) = 660 \left(\frac{330}{306}\right) = 711.8\text{ Hz} \approx 712\text{ Hz}\).
Marking scheme
[1] D: \(712\text{ Hz}\)
Question 39 · Multiple Choice
1 marks
Light of wavelength \(600\text{ nm}\) is incident normally on a diffraction grating. The second-order diffraction maximum is observed at an angle of \(36.0^\circ\) to the central normal. What is the number of lines per millimetre of the grating?
A.\(245\text{ mm}^{-1}\)
B.\(490\text{ mm}^{-1}\)
C.\(980\text{ mm}^{-1}\)
D.\(2040\text{ mm}^{-1}\)
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Worked solution
Using the grating equation \(d \sin\theta = n \lambda\), the grating spacing is \(d = \frac{n \lambda}{\sin\theta} = \frac{2 \times 600 \times 10^{-9}}{\sin(36.0^\circ)} = \frac{1.20 \times 10^{-6}}{0.5878} = 2.042 \times 10^{-6}\text{ m}\). The number of lines per metre is \(N = \frac{1}{d} = 4.898 \times 10^5\text{ m}^{-1}\). Thus, the number of lines per millimetre is \(\frac{4.898 \times 10^5}{10^3} \approx 490\text{ mm}^{-1}\).
Marking scheme
[1] B: \(490\text{ mm}^{-1}\)
Question 40 · Multiple Choice
1 marks
A battery of electromotive force (e.m.f.) \(12.0\text{ V}\) and internal resistance \(1.5\ \Omega\) is connected in series with an external resistor of resistance \(4.5\ \Omega\). What is the potential difference across the terminals of the battery?
A.\(3.0\text{ V}\)
B.\(6.0\text{ V}\)
C.\(9.0\text{ V}\)
D.\(12.0\text{ V}\)
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Worked solution
The total resistance of the circuit is \(R_{\text{total}} = R + r = 4.5 + 1.5 = 6.0\ \Omega\). The current in the circuit is \(I = \frac{E}{R_{\text{total}}} = \frac{12.0}{6.0} = 2.0\text{ A}\). The terminal potential difference is \(V = E - I r = 12.0 - (2.0 \times 1.5) = 12.0 - 3.0 = 9.0\text{ V}\) (or \(V = I R = 2.0 \times 4.5 = 9.0\text{ V}\)).
Marking scheme
[1] C: \(9.0\text{ V}\)
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(b) A trolley \( A \) of mass \( 0.80\text{ kg} \) travels at a velocity of \( 3.5\text{ m s}^{-1} \) along a horizontal, frictionless track. Trolley \( A \) collides head-on with a stationary trolley \( B \) of mass \( 1.20\text{ kg} \).
The two trolleys couple together during the collision and move forward with a common speed \( V \).
(i) Calculate the speed \( V \) of the coupled trolleys immediately after the collision.
\( V = \) ................................................... \(\text{m s}^{-1}\) [2]
(ii) Show that the kinetic energy lost during the collision is \( 2.9\text{ J} \).
[3]
(c) After the collision, the coupled trolleys move along the track and pass across a horizontal rough section of length \( 1.2\text{ m} \). A constant resistive force of \( 0.80\text{ N} \) acts on the trolleys while they cross this rough section.
(i) Determine the speed of the trolleys as they leave the rough section.
(ii) After leaving the rough section, the trolleys travel along a frictionless part of the track and collide with a horizontal spring of spring constant \( 50\text{ N m}^{-1} \), compressing the spring until they are brought momentarily to rest.
Calculate the maximum compression \( x \) of the spring.
\( x = \) ................................................... \(\text{m}\) [2]
(d) During the collision between trolley \( A \) and trolley \( B \) in (b), the average force exerted on trolley \( B \) by trolley \( A \) has a magnitude of \( 28\text{ N} \).
(i) State the magnitude and direction of the average force exerted on trolley \( A \) by trolley \( B \), and name the law of motion that applies.
(ii) Calculate the time duration \( \Delta t \) of the collision.
\( \Delta t = \) ................................................... \(\text{s}\) [1]
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Worked solution
(a) (i) Linear momentum is the product of mass and velocity (\( p = mv \)).
(ii) The total momentum of a system remains constant / is conserved, provided no external resultant force acts on the system (or in an isolated / closed system).
(iii) The collision must be (perfectly) elastic.
(b) (i) By conservation of linear momentum: \( m_A u_A + m_B u_B = (m_A + m_B) V \) \( (0.80)(3.5) + (1.20)(0) = (0.80 + 1.20) V \) \( 2.80 = 2.00 V \) \( V = 1.4\text{ m s}^{-1} \)
(ii) Initial kinetic energy before collision: \( E_{k1} = \frac{1}{2} m_A u_A^2 = \frac{1}{2} \times 0.80 \times (3.5)^2 = 4.90\text{ J} \)
Final kinetic energy after collision: \( E_{k2} = \frac{1}{2} (m_A + m_B) V^2 = \frac{1}{2} \times 2.00 \times (1.4)^2 = 1.96\text{ J} \)
Loss in kinetic energy: \( \Delta E_k = 4.90 - 1.96 = 2.94\text{ J} \approx 2.9\text{ J} \)
(c) (i) Method 1: Work-energy theorem Work done against resistive force \( W = F \times d = 0.80 \times 1.2 = 0.96\text{ J} \) Kinetic energy leaving rough section \( E_{k3} = E_{k2} - W = 1.96 - 0.96 = 1.00\text{ J} \) \( \frac{1}{2} M v^2 = 1.00 \implies \frac{1}{2} \times 2.00 \times v^2 = 1.00 \implies v = 1.0\text{ m s}^{-1} \)
Method 2: Kinematics \( a = -\frac{F}{M} = -\frac{0.80}{2.00} = -0.40\text{ m s}^{-2} \) \( v^2 = u^2 + 2as = 1.4^2 + 2(-0.40)(1.2) = 1.96 - 0.96 = 1.00 \implies v = 1.0\text{ m s}^{-1} \)
(ii) By conservation of energy, the kinetic energy is converted to elastic potential energy in the spring: \( E_k = \frac{1}{2} k x^2 \) \( 1.00 = \frac{1}{2} \times 50 \times x^2 \) \( x^2 = \frac{2.00}{50} = 0.0400 \) \( x = 0.20\text{ m} \)
(d) (i) Magnitude is \( 28\text{ N} \), directed in the opposite direction (to the left / opposite to the initial motion of trolley A), by Newton's third law of motion.
(ii) Change in momentum of trolley B: \( \Delta p = m_B V - 0 = 1.20 \times 1.4 = 1.68\text{ N s} \) \( F = \frac{\Delta p}{\Delta t} \implies \Delta t = \frac{1.68}{28} = 0.060\text{ s} \)
Marking scheme
(a)(i) product of mass and velocity [1] (a)(ii) total momentum of a system remains constant / is conserved [1] provided no resultant external force acts / in a closed or isolated system [1] (a)(iii) the collision is (perfectly) elastic [1]
(b)(i) substitution into momentum conservation: \( 0.80 \times 3.5 = (0.80 + 1.20) V \) [1] \( V = 1.4\text{ m s}^{-1} \) [1] (b)(ii) initial kinetic energy \( = 4.90\text{ J} \) (or \( 4.9\text{ J} \)) [1] final kinetic energy \( = 1.96\text{ J} \) (or \( 2.0\text{ J} \)) [1] loss in KE \( = 4.90 - 1.96 = 2.94\text{ J} \) leading to \( 2.9\text{ J} \) shown [1]
(c)(i) calculation of work done against friction \( = 0.80 \times 1.2 = 0.96\text{ J} \) OR acceleration \( a = -0.40\text{ m s}^{-2} \) [1] subtraction of work from KE: \( E_k = 1.96 - 0.96 = 1.00\text{ J} \) OR use of \( v^2 = 1.4^2 + 2(-0.40)(1.2) \) [1] \( v = 1.0\text{ m s}^{-1} \) (allow ecf from (b)(i)) [1] (c)(ii) equating kinetic energy to \( \frac{1}{2} k x^2 \): \( 1.00 = \frac{1}{2} (50) x^2 \) [1] \( x = 0.20\text{ m} \) (allow ecf from (c)(i)) [1]
(d)(i) \( 28\text{ N} \) in opposite direction (or to the left) AND Newton's third law [1] (d)(ii) \( \Delta t = \frac{\Delta p}{F} = \frac{1.68}{28} = 0.060\text{ s} \) (allow ecf from (b)(i)) [1]
(ii) State Hooke's law. ................................................................................................................................................... [1]
(b) A uniform metal wire of original length \(2.40\text{ m}\) and cross-sectional area \(4.50 \times 10^{-7}\text{ m}^2\) is suspended vertically from a rigid support. A load of mass \(18.0\text{ kg}\) is attached to the free lower end of the wire so that it hangs in equilibrium.
The Young modulus of the metal is \(1.20 \times 10^{11}\text{ Pa}\). Assume that the wire obeys Hooke's law and that the mass of the wire is negligible.
(i) Show that the extension of the wire is \(7.85 \times 10^{-3}\text{ m}\).
(ii) Calculate the elastic potential energy (strain energy) stored in the wire.
elastic potential energy = ..................................................... \(\text{J}\) [2]
(c) The load is increased until the wire is stretched beyond its elastic limit, and it is then completely unloaded.
(i) State what is meant by plastic deformation. ................................................................................................................................................... [1]
(ii) On the axes below, sketch a graph to show the variation of the applied tensile force \(F\) with the extension \(x\) of the wire as it is loaded beyond its elastic limit and then unloaded.
(iii) Explain how the graph in (c)(ii) can be used to determine the thermal energy dissipated in permanently deforming the wire. ................................................................................................................................................... ................................................................................................................................................... [1]
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Worked solution
(a) (i) 1. Tensile stress is defined as the force per unit cross-sectional area: \[ \sigma = \frac{F}{A} \]
2. Tensile strain is defined as the extension (change in length) per unit original length: \[ \varepsilon = \frac{\Delta L}{L_0} \]
(ii) Hooke's law states that the extension of a body is directly proportional to the applied force provided the limit of proportionality is not exceeded.
(b) (i) The tension force \(F\) in the wire equals the weight of the load: \[ F = mg = 18.0 \times 9.81 = 176.58\text{ N} \]
Young modulus \(E\) is given by: \[ E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{x / L_0} = \frac{F L_0}{A x} \]
(ii) Elastic potential energy \(E_p\) within Hooke's law is given by: \[ E_p = \frac{1}{2} F x = \frac{1}{2} \times 176.58\text{ N} \times 7.848 \times 10^{-3}\text{ m} = 0.6929\text{ J} \approx 0.693\text{ J} \]
(c) (i) Plastic deformation means that the material does not return to its original shape/length when the applied force is removed (it undergoes permanent deformation / permanent set).
(ii) The sketch should show: - Loading curve: starts at the origin, a straight line of constant gradient representing the elastic region, curving with decreasing gradient beyond the limit of proportionality. - Unloading curve: starts from the maximum extension, decreases linearly parallel to the initial linear loading section, and intersects the extension axis at a positive (non-zero) extension value (permanent set).
(iii) The thermal energy dissipated / work done in permanently deforming the wire is given by the area enclosed by the loading and unloading loops (the hysteresis loop area).
Marking scheme
(a) (i) 1. force / (cross-sectional) area [B1] 2. extension / original length OR change in length / original length [B1]
(a) (ii) force is directly proportional to extension provided the limit of proportionality / elastic limit is not exceeded [B1] (Allow: \(F \propto x\) with terms and condition defined)
(b) (ii) \(E_p = \frac{1}{2} F x\) OR \(E_p = \frac{1}{2} k x^2\) [C1] \(E_p = 0.5 \times 176.58 \times 7.848 \times 10^{-3} = 0.693\text{ J}\) (or \(0.69\text{ J}\)) [A1] (Allow ecf from (b)(i))
(c) (i) wire does not return to its original length / shape when the load / force is removed (or undergoes permanent deformation) [B1]
(c) (ii) Loading line showing initial straight line through origin, then curving to a lower gradient [B1] Unloading line straight and parallel to the initial linear section, reaching zero force at a non-zero positive extension [B1]
(c) (iii) area between the loading and unloading lines / curves (area enclosed by loop) [B1]
(b) A parallel beam of monochromatic light from a laser is incident normally on a pair of narrow parallel slits \( S_1 \) and \( S_2 \) separated by a distance of \( 0.35\text{ mm} \). An interference pattern of bright and dark fringes is formed on a screen placed parallel to the slits at a distance of \( 2.40\text{ m} \).
(i) The separation between the centres of adjacent bright fringes on the screen is \( 4.2\text{ mm} \).
Calculate the wavelength \( \lambda \) of the light.
(ii) The laser is replaced with another source so that the intensity of light passing through slit \( S_1 \) remains constant, but the intensity of light passing through slit \( S_2 \) is decreased.
State and explain the effect of this change on the appearance of the fringes on the screen.
(c) The double slit is now replaced by a diffraction grating with \( 450\text{ lines mm}^{-1} \). A beam of light of wavelength \( 580\text{ nm} \) is incident normally on the grating.
(i) Determine the total number of primary maxima (bright spots) that can be observed.
number of maxima = ..................................................... [3]
(ii) Calculate the angle between the two second-order maxima.
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Worked solution
(a)(i) According to the principle of superposition, when two or more waves meet at a point, the resultant displacement is equal to the algebraic (or vector) sum of the individual displacements of each wave.
(a)(ii) Coherent waves are waves that maintain a constant phase difference over time.
(b)(i) Using the double-slit interference formula: \[ \lambda = \frac{a x}{D} \] where: \( a = 0.35\text{ mm} = 0.35 \times 10^{-3}\text{ m} \) \( x = 4.2\text{ mm} = 4.2 \times 10^{-3}\text{ m} \) \( D = 2.40\text{ m} \)
(b)(ii) - The fringe spacing \( x = \frac{\lambda D}{a} \) depends only on \( \lambda, D, \) and \( a \), so the separation between fringes remains unchanged. - Because the amplitudes from the two slits are no longer equal, destructive interference is no longer complete at the minima (intensity at minima is greater than zero / minima are no longer completely dark) and constructive interference has lower peak intensity (bright fringes are less bright). Thus, the contrast between bright and dark fringes is reduced.
(c)(i) Grating spacing \( d \): \[ d = \frac{1.00 \times 10^{-3}\text{ m}}{450} = 2.222 \times 10^{-6}\text{ m} \]
Using \( d \sin\theta = n\lambda \): For maximum order \( n_{\text{max}} \), \( \sin\theta \le 1 \): \[ n \le \frac{d}{\lambda} = \frac{2.222 \times 10^{-6}}{580 \times 10^{-9}} = 3.83 \] Since \( n \) must be an integer, the highest order observed is \( n = 3 \).
Total number of maxima = \( 2n + 1 = 2(3) + 1 = 7 \) (the central maximum plus 3 orders on each side).
(c)(ii) For \( n = 2 \): \[ \sin\theta_2 = \frac{n\lambda}{d} = \frac{2 \times 580 \times 10^{-9}}{2.222 \times 10^{-6}} = 0.5220 \] \[ \theta_2 = \arcsin(0.5220) = 31.47^\circ \] The angle between the two second-order maxima is \( 2\theta_2 \): \[ 2\theta_2 = 2 \times 31.47^\circ = 62.9^\circ \]
(c)(iii) The maxima produced by a diffraction grating are much sharper (narrower), brighter, and spaced further apart, reducing the percentage uncertainty in measuring the angular position.
Marking scheme
(a)(i) - [B1] (When two or more waves meet at a point,) resultant displacement is the (vector/algebraic) sum of individual displacements. (Do not credit 'amplitude' in place of 'displacement')
(a)(ii) - [B1] (Waves have a) constant phase difference.
(b)(i) - [M1] Formula quoted or substituted: \( \lambda = \frac{ax}{D} \) with values \( a = 0.35 \times 10^{-3} \), \( x = 4.2 \times 10^{-3} \), \( D = 2.40 \) - [A1] \( \lambda = 6.1 \times 10^{-7}\text{ m} \) or \( 6.13 \times 10^{-7}\text{ m} \) (allow 2 or 3 sig figs)
(b)(ii) - [B1] Fringe spacing / fringe separation remains unchanged / the same. - [B1] Contrast is reduced / minima are not completely dark / bright fringes are less bright / difference between maximum and minimum intensity decreases.
(c)(i) - [C1] \( d = \frac{1}{450\times 10^3} = 2.22 \times 10^{-6}\text{ m} \) - [C1] \( n = \frac{d}{\lambda} = \frac{2.22 \times 10^{-6}}{580 \times 10^{-9}} = 3.83 \) leading to maximum order \( n = 3 \) - [A1] Total number = 7 (must be an odd integer, allow ecf from integer value of n)
(c)(iii) - [B1] Maxima are much sharper / narrower / brighter / more widely separated, leading to smaller uncertainty in position/angle measurement.
Question 4 · structured
12 marks
(a) Define electric potential difference (p.d.). [1]
(b) A uniform resistance wire of length \( 1.60\text{ m} \) and uniform cross-sectional area \( 2.40 \times 10^{-7}\text{ m}^2 \) is made of a metal alloy of resistivity \( 1.20 \times 10^{-6}\ \Omega\text{ m} \).
(i) Show that the resistance of the wire is \( 8.00\ \Omega \). [2]
(ii) The number density of free electrons in the wire is \( 6.50 \times 10^{28}\text{ m}^{-3} \). A potential difference of \( 4.80\text{ V} \) is applied across the wire.
Calculate the average drift speed \( v \) of the free electrons in the wire.
\( v = \) ..................................................... \(\text{m s}^{-1}\) [3]
(c) The wire in (b) is used as the slide wire \( XY \) of length \( 1.60\text{ m} \) and resistance \( 8.00\ \Omega \) in a potentiometer circuit. The wire \( XY \) is connected in series with a fixed resistor of resistance \( 4.00\ \Omega \) and a power supply of electromotive force (e.m.f.) \( 6.00\text{ V} \) with negligible internal resistance.
A test cell of e.m.f. \( E_2 \) and internal resistance \( 1.50\ \Omega \) is connected in series with a sensitive galvanometer across a length \( XP \) of the wire, where \( P \) is a moveable sliding contact.
(i) Calculate the potential difference across the wire \( XY \).
(iii) State and explain the effect, if any, of the internal resistance of the test cell on the balance length \( XP \). [2]
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Worked solution
(a) Potential difference is defined as the energy transferred per unit charge from electrical energy to other forms of energy (or work done per unit charge).
(b)(i) Using the formula for resistance: \[ R = \frac{\rho L}{A} \] Substitute the given values: \[ R = \frac{1.20 \times 10^{-6} \times 1.60}{2.40 \times 10^{-7}} = \frac{1.92 \times 10^{-6}}{2.40 \times 10^{-7}} = 8.00\ \Omega \]
(b)(ii) First, determine the current \( I \) in the wire: \[ I = \frac{V}{R} = \frac{4.80\text{ V}}{8.00\ \Omega} = 0.600\text{ A} \] Using the transport equation \( I = A n v q \), where \( q = e = 1.60 \times 10^{-19}\text{ C} \): \[ v = \frac{I}{n A e} = \frac{0.600}{6.50 \times 10^{28} \times 2.40 \times 10^{-7} \times 1.60 \times 10^{-19}} \] \[ v = \frac{0.600}{2496} = 2.4038 \times 10^{-4}\text{ m s}^{-1} \approx 2.40 \times 10^{-4}\text{ m s}^{-1} \]
(c)(i) Total resistance in the primary circuit: \[ R_{\text{total}} = 4.00\ \Omega + 8.00\ \Omega = 12.00\ \Omega \] Using the potential divider equation: \[ V_{XY} = \left( \frac{R_{XY}}{R_{\text{total}}} \right) \times E_1 = \left( \frac{8.00}{12.00} \right) \times 6.00\text{ V} = 4.00\text{ V} \]
(c)(ii) Since the wire is uniform, the potential gradient along \( XY \) is: \[ \text{potential gradient} = \frac{V_{XY}}{L} = \frac{4.00\text{ V}}{1.60\text{ m}} = 2.50\text{ V m}^{-1} \] At the balance point \( XP = 0.960\text{ m} \), the potential difference across \( XP \) is: \[ V_{XP} = 2.50\text{ V m}^{-1} \times 0.960\text{ m} = 2.40\text{ V} \] Because the galvanometer indicates a null reading, no current flows through the test cell circuit, meaning \( E_2 = V_{XP} = 2.40\text{ V} \).
(c)(iii) There is no effect / the balance length remains unchanged. At the balance point, the current through the test cell and galvanometer is zero, so there is no lost volts / potential difference across the internal resistance \( (Ir = 0) \).
Marking scheme
(a) - Energy transferred from electrical to other forms per unit charge OR work done per unit charge. `[1]` (Do not credit just 'p.d. = V/I' or 'work done on charge' without 'per unit charge')
(b)(i) - Correct formula stated and substituted: \( R = \frac{1.20 \times 10^{-6} \times 1.60}{2.40 \times 10^{-7}} \) `[1]` - Final value evaluated clearly to at least 3 s.f. with unit: \( 8.00\ \Omega \) `[1]`
(b)(ii) - Calculation of current: \( I = \frac{4.80}{8.00} = 0.600\text{ A} \) `[1]` - Substitution of values into \( I = n A v q \) with \( e = 1.60 \times 10^{-19}\text{ C} \) `[1]` - Correct final answer: \( 2.40 \times 10^{-4}\text{ m s}^{-1} \) (allow \( 2.4 \times 10^{-4}\text{ m s}^{-1} \)) `[1]`
(c)(i) - Correct method shown: \( I = \frac{6.00}{12.00} = 0.500\text{ A} \) and \( V = 0.500 \times 8.00 \) OR \( V = 6.00 \times \frac{8.00}{8.00 + 4.00} \) `[1]` - Final answer: \( 4.00\text{ V} \) `[1]`
(c)(ii) - Correct proportion or potential gradient calculation: \( \frac{0.960}{1.60} \times 4.00 \) OR \( 2.50 \times 0.960 \) `[1]` - Final answer: \( 2.40\text{ V} \) `[1]`
(c)(iii) - State: (No effect / length remains the) same / unchanged `[1]` - Explain: Current in the galvanometer / test cell branch is zero (at balance), so there is no potential difference across the internal resistance / no 'lost volts' `[1]`
Question 5 · structured
8 marks
(a) State one difference between a hadron and a lepton.
(b) A stationary nucleus of sodium-22 (\(^{22}_{11}\text{Na}\)) decays by emitting a \(\beta^+\) particle and an electron neutrino (\(\nu_{\text{e}}\)) to form a stable nucleus of neon (\(\text{Ne}\)).
(i) State the change in quark flavour that occurs during \(\beta^+\) decay.
(c) (i) Explain why the emitted \(\beta^+\) particles have a continuous distribution of kinetic energies, whereas \(\alpha\) particles emitted in radioactive decay have discrete energies.
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Worked solution
(a) A hadron is composed of quarks (it is not a fundamental particle), whereas a lepton is a fundamental particle (not composed of quarks). Alternatively, hadrons experience the strong nuclear force, whereas leptons do not.
(b)(i) In \(\beta^+\) decay, a proton changes into a neutron: \(p \rightarrow n + \text{e}^+ + \nu_{\text{e}}\). In terms of quarks: \(uud \rightarrow udd + \text{e}^+ + \nu_{\text{e}}\). Therefore, an up quark (\(u\)) changes into a down quark (\(d\)).
(c)(i) In \(\beta^+\) decay, a fixed total amount of energy (decay energy) is released. Because this energy is shared between three products (the daughter nucleus, the positron, and the neutrino), the neutrino carries away a variable portion of the energy, resulting in a continuous spectrum of kinetic energy for the positron. In \(\alpha\) decay, there are only two products (the alpha particle and the daughter nucleus), so momentum and energy conservation constrain the alpha particle to have a single, discrete kinetic energy.
(c)(ii) Physical quantities conserved during nuclear decays include: - Electric charge (proton number) - Nucleon number (or baryon number) - Linear momentum - Lepton number
Marking scheme
(a) [1] Hadron is composed of quarks / is not fundamental AND lepton is not composed of quarks / is fundamental OR Hadrons experience the strong force / strong interaction AND leptons do not.
(b)(i) [1] Up (quark) changes to down (quark) / \(u \rightarrow d\).
(b)(ii) [1] \(^{22}_{10}\text{Ne}\) (both nucleon number 22 and proton number 10 correct with symbol Ne) [1] \(^0_{+1}\text{e}\) or \(^0_{+1}\beta\) or \(\beta^+\) with correct nucleon and proton numbers.
(c)(i) [1] (Total / decay) energy is shared / distributed between the positron and neutrino (and daughter nucleus). [1] The neutrino carries away a variable amount / fraction of energy (leading to a range of positron energies).
(c)(ii) [1] Any one from: (electric) charge / proton number, nucleon number / baryon number, linear momentum, lepton number. [1] A second valid conserved quantity from the list.
Paper 3 (Advanced Practical Skills)
Complete two 1-hour experimental investigations using provided laboratory apparatus. Record tables, graphs, calculations, uncertainties, limitations, and improvements.
In this experiment, you will investigate the equilibrium of a pivoted wooden rule supported by a spring.
(a) (i) Set up the apparatus as shown in Fig. 1.1, with a metre rule supported horizontally on a pivot at distance \(d_0 = 10.0\text{ cm}\) from end A. (ii) Attach a helical spring vertically between a clamp and the rule at a distance \(x = 80.0\text{ cm}\) from end A. (iii) Suspend a mass hanger with a total mass \(m = 100\text{ g}\) at distance \(y\) from end A so that the rule remains horizontal. (iv) Measure and record the position \(y\) on the metre rule where the mass hanger is suspended.
(b) (i) Adjust the position of the spring to a new distance \(x = 70.0\text{ cm}\). (ii) Adjust the position \(y\) of the suspended mass until the rule is once again horizontal. (iii) Measure and record \(y\).
(c) Vary the position \(x\) of the spring in the range \(40.0\text{ cm} \le x \le 90.0\text{ cm}\). For each value of \(x\), adjust \(y\) so that the rule is balanced horizontally. Record \(x\) and \(y\) in a table. Include values of \((x - d_0)\) and \((d_0 - y)\) in your table of results. [6]
(d) (i) On the grid provided, plot a graph of \((d_0 - y)\) on the \(y\)-axis against \((x - d_0)\) on the \(x\)-axis. (ii) Draw the straight line of best fit. [4]
(e) Determine the gradient and \(y\)-intercept of the line of best fit.
(iii) State the correct unit for \(P\) and for \(Q\).
\(\text{unit of } P = \text{...................................................}\) \(\text{unit of } Q = \text{...................................................}\) [3]
(g) (i) State one significant source of random error when adjusting \(y\) to achieve horizontal equilibrium. (ii) Suggest one improvement to reduce this error. [2]
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Worked solution
Sample experimental data and calculations:
(a) Value of \(y\) recorded to the nearest millimetre, e.g., \(y = 4.2\text{ cm}\) (or \(0.042\text{ m}\)).
(b) Value of \(y\) recorded, e.g., \(y = 4.9\text{ cm}\). \(x - d_0 = 70.0 - 10.0 = 60.0\text{ cm}\) \(d_0 - y = 10.0 - 4.9 = 5.1\text{ cm}\)
(f) (i) \(P = \text{gradient} = 0.070\) (ii) \(Q = y\text{-intercept} = 0.90\text{ cm}\) (or \(9.0 \times 10^{-3}\text{ m}\)) (iii) Unit of \(P\): none / dimensionless (or \(\text{cm cm}^{-1}\) / \(\text{m m}^{-1}\)); Unit of \(Q\): \(\text{cm}\) (or \(\text{m}\)).
(g) (i) Difficult to judge exactly when the rule is horizontal by eye. (ii) Use a spirit level placed on the rule / measure height above the bench at both ends of the rule using a set square and metre rule.
Marking scheme
Mark Breakdown (20 marks total):
(a) Value of \(y\) recorded to the nearest 1 mm with appropriate unit. [1]
(b) Value of \(y\) recorded with \((x - d_0)\) and \((d_0 - y)\) calculated correctly. [2] - 1 mark: Raw \(y\) recorded to nearest 1 mm. - 1 mark: Correct subtraction for \((x - d_0)\) and \((d_0 - y)\).
(c) Table of results [6]: - Range and number of values: At least 6 sets of readings recorded without assistance covering \(x\) range \(\ge 40.0\text{ cm}\). [1] - Column headings: Each column must have quantity and unit separated by a solidus or in brackets (e.g. \(x\text{ / cm}\), \((x - d_0)\text{ / cm}\), \(y\text{ / cm}\), \((d_0 - y)\text{ / cm}\)). [1] - Consistency of raw data: All raw values of \(x\) and \(y\) recorded to same degree of precision (nearest 0.1 cm / 1 mm). [1] - Significant figures: Calculated values of \((x - d_0)\) and \((d_0 - y)\) given to appropriate s.f. consistent with raw data. [1] - Calculation arithmetic: Calculated values correct for given raw data. [1] - Quality of data: Points lie close to a straight line with little scatter. [1]
(d) Graph [4]: - Axes: Scales must be chosen such that plotted points occupy at least half the graph grid in both \(x\) and \(y\) directions; axes labelled with quantity and unit. No awkward scales (e.g. multiples of 3, 7). [1] - Plotting of points: All points plotted accurately within half a small square. [1] - Line of best fit: Straight line drawn with balanced distribution of points on either side; line must be thin and continuous (no kinks or multiple strokes). [1] - Trend: Correct sign of gradient (positive slope). [1]
(e) Gradient and intercept [2]: - Gradient: Hypotenuse of read-off triangle must be greater than half the length of the drawn line; coordinates correctly read to within half a small square and substituted into \(\frac{\Delta y}{\Delta x}\). [1] - Intercept: Read directly from graph if \(x = 0\) is on grid, or calculated correctly using a point on the line \(c = y_1 - m x_1\). [1]
(f) Constants \(P\) and \(Q\) and units [3]: - Value of \(P\) equated to gradient with appropriate s.f. [1] - Value of \(Q\) equated to intercept with appropriate s.f. [1] - Correct units for \(P\) (no unit / dimensionless) and \(Q\) (cm or m, consistent with values). [1]
(g) Uncertainty & evaluation [2]: - Relevant source of random error (e.g., difficulty in identifying true horizontal position by eye / friction at pivot). [1] - Valid practical improvement (e.g., use of spirit level / fiducial mark / measure height of both ends above bench with vertical rule). [1]
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