An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V4) Cambridge International A Level Physics (9702) paper. Not affiliated with or reproduced from Cambridge.
Paper 1 (Multiple Choice)
Answer all 40 multiple-choice questions. Each question carries one mark.
40 Question · 40 marks
Question 1 · Multiple Choice
1 marks
Which expression gives the SI base units of electrical resistivity?
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 \(\rho = \frac{RA}{L}\). Resistance \(R = \frac{V}{I} = \frac{W}{Q I} = \frac{W}{I^2 t}\). In SI base units: Work \(W\) has base units \(\text{kg m}^2 \text{s}^{-2}\). Therefore, \(R\) has base units \(\frac{\text{kg m}^2 \text{s}^{-2}}{\text{A}^2 \text{s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\). Substituting into \(\rho = \frac{RA}{L}\): \([\rho] = \frac{(\text{kg m}^2 \text{s}^{-3} \text{A}^{-2})(\text{m}^2)}{\text{m}} = \text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\).
Marking scheme
A is correct [1 mark]. B is the base unit of resistance (\(\Omega\)). C and D have incorrect powers of base units.
Question 2 · Multiple Choice
1 marks
A stone is thrown vertically upwards from the top edge of a cliff of height \(45.0\text{ m}\) with an initial velocity of \(15.0\text{ m s}^{-1}\).
Air resistance is negligible.
What is the speed of the stone immediately before it hits the ground at the foot of the cliff?
A.\(25.6\text{ m s}^{-1}\)
B.\(29.7\text{ m s}^{-1}\)
C.\(33.3\text{ m s}^{-1}\)
D.\(36.0\text{ m s}^{-1}\)
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Worked solution
Using the equation of uniformly accelerated motion: \(v^2 = u^2 + 2as\). Taking downwards as positive: \(u = -15.0\text{ m s}^{-1}\) \(a = 9.81\text{ m s}^{-2}\) \(s = +45.0\text{ m}\)
C is correct [1 mark]. A results from incorrect subtraction \(v = \sqrt{2as - u^2}\). B is the speed acquired falling purely from rest through \(45.0\text{ m}\).
Question 3 · Multiple Choice
1 marks
Two trolleys X and Y move towards each other on a horizontal frictionless track.
Trolley X has a mass of \(0.40\text{ kg}\) and travels to the right at a speed of \(2.5\text{ m s}^{-1}\). Trolley Y has a mass of \(0.60\text{ kg}\) and travels to the left at a speed of \(1.5\text{ m s}^{-1}\).
The trolleys collide and coalesce (stick together).
What is the total loss of kinetic energy during the collision?
A.\(0.0050\text{ J}\)
B.\(0.96\text{ J}\)
C.\(1.92\text{ J}\)
D.\(1.93\text{ J}\)
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By conservation of momentum (taking right as positive): \(p_{\text{initial}} = (0.40)(+2.5) + (0.60)(-1.5) = 1.00 - 0.90 = +0.10\text{ kg m s}^{-1}\).
Combined mass \(m_{\text{total}} = 0.40 + 0.60 = 1.00\text{ kg}\). Final velocity \(v = \frac{0.10}{1.00} = 0.10\text{ m s}^{-1}\).
Final kinetic energy: \(E_{k\text{, final}} = \frac{1}{2}(1.00)(0.10)^2 = 0.005\text{ J}\).
Loss in kinetic energy \(\Delta E_k = 1.925 - 0.005 = 1.92\text{ J}\).
Marking scheme
C is correct [1 mark]. A is the remaining kinetic energy after collision. B is calculated by neglecting the vector nature of momentum when finding final velocity.
Question 4 · Multiple Choice
1 marks
A solid cylinder of cross-sectional area \(2.5 \times 10^{-3}\text{ m}^2\) and height \(0.12\text{ m}\) is completely submerged in a liquid of uniform density \(800\text{ kg m}^{-3}\).
The axis of the cylinder is vertical, and its top surface is at a depth of \(0.20\text{ m}\) below the surface of the liquid.
What is the upthrust exerted by the liquid on the cylinder?
A.\(1.57\text{ N}\)
B.\(2.35\text{ N}\)
C.\(3.92\text{ N}\)
D.\(6.28\text{ N}\)
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Worked solution
By Archimedes' principle, upthrust \(F = \rho g V\), where \(V\) is the submerged volume of the cylinder. \(V = A \times h = (2.5 \times 10^{-3}\text{ m}^2)(0.12\text{ m}) = 3.0 \times 10^{-4}\text{ m}^3\). \(F = (800\text{ kg m}^{-3})(9.81\text{ m s}^{-2})(3.0 \times 10^{-4}\text{ m}^3) = 2.3544\text{ N} \approx 2.35\text{ N}\). Note that upthrust is independent of the depth of the top surface as long as the object is completely submerged.
Marking scheme
B is correct [1 mark]. A uses the depth \(0.20\text{ m}\) instead of the height of the cylinder. C uses the total depth to the bottom of the cylinder.
Question 5 · Multiple Choice
1 marks
Two identical light springs, each of spring constant \(160\text{ N m}^{-1}\), are connected in parallel.
A load of \(24\text{ N}\) is hung from the parallel combination, and the springs extend elastically within their limit of proportionality.
What is the total elastic potential energy stored in the combination of springs?
A.\(0.45\text{ J}\)
B.\(0.90\text{ J}\)
C.\(1.80\text{ J}\)
D.\(3.60\text{ J}\)
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Worked solution
For two identical springs in parallel, the effective spring constant is \(k_{\text{eff}} = 2k = 2 \times 160 = 320\text{ N m}^{-1}\). The extension is \(x = \frac{F}{k_{\text{eff}}} = \frac{24\text{ N}}{320\text{ N m}^{-1}} = 0.075\text{ m}\). The total elastic potential energy is \(E_p = \frac{1}{2} F x = \frac{1}{2} (24)(0.075) = 0.90\text{ J}\).
(Alternatively: Each spring supports \(12\text{ N}\) and extends by \(x = \frac{12}{160} = 0.075\text{ m}\). Energy stored in one spring = \(\frac{1}{2}(160)(0.075)^2 = 0.45\text{ J}\). Total energy in both springs = \(2 \times 0.45\text{ J} = 0.90\text{ J}\)).
Marking scheme
B is correct [1 mark]. A is the energy stored in just one of the two springs. C treats the springs as if they were in series.
Question 6 · Multiple Choice
1 marks
An ambulance travels at a constant speed of \(28.0\text{ m s}^{-1}\) along a straight road directly towards a stationary pedestrian. The siren on the ambulance emits sound waves of frequency \(680\text{ Hz}\).
The speed of sound in air is \(340\text{ m s}^{-1}\).
What is the frequency of the sound detected by the pedestrian?
A.\(628\text{ Hz}\)
B.\(680\text{ Hz}\)
C.\(736\text{ Hz}\)
D.\(741\text{ Hz}\)
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Worked solution
For a source moving towards a stationary observer, the Doppler formula is: \(f_o = f_s \left(\frac{v}{v - v_s}\right)\)
D is correct [1 mark]. A is the frequency detected when the ambulance is moving away (\(v + v_s\) in denominator). C uses an approximation \(f_s (1 + v_s/v)\).
Question 7 · Multiple Choice
1 marks
Monochromatic light of wavelength \(532\text{ nm}\) is incident normally on a diffraction grating ruled with \(4.00 \times 10^5\text{ lines m}^{-1}\).
What is the total number of principal intensity maxima (bright fringes) 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 maximum possible order \(n_{\text{max}}\), \(\sin\theta \le 1\): \(n \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}\text{ m}}{532 \times 10^{-9}\text{ m}} = 4.699\).
Since \(n\) must be an integer, the highest order formed is \(n = 4\). The total number of maxima includes the central maximum (\(n = 0\)) and orders on both sides: \(\text{Total} = 2n + 1 = 2(4) + 1 = 9\).
Marking scheme
D is correct [1 mark]. A is the maximum order \(n\). C misses the central zeroth-order maximum.
Question 8 · Multiple Choice
1 marks
A proton decays into a neutron via \(\beta^+\) (positron) emission inside an unstable nucleus.
Which row correctly identifies the change in quark flavour, the emitted charged lepton, and the emitted neutral particle?
$$\begin{array}{|c|c|c|c|}\hline & \text{Quark transformation} & \text{Emitted charged lepton} & \text{Emitted neutral particle} \\hline \text{A} & u \rightarrow d & e^+ & \nu_e \\hline \text{B} & u \rightarrow d & e^+ & \bar{\nu}_e \\hline \text{C} & d \rightarrow u & e^- & \bar{\nu}_e \\hline \text{D} & d \rightarrow u & e^+ & \nu_e \\hline\end{array}$$
A.Row A
B.Row B
C.Row C
D.Row D
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Worked solution
A proton has quark composition \(uud\) and a neutron has quark composition \(udd\). Therefore, in \(\beta^+\) decay, an up quark changes into a down quark (\(u \rightarrow d\)). To conserve charge: \(+\frac{2}{3}e = -\frac{1}{3}e + (+1e) + 0\), so a positron \(e^+\) is emitted. To conserve lepton number (lepton number of \(e^+\) is \(-1\)), an electron neutrino \(\nu_e\) (lepton number \(+1\)) must be emitted.
Marking scheme
A is correct [1 mark]. B has an antineutrino which violates lepton number conservation. C represents \(\beta^-\) decay.
Question 9 · Multiple Choice
1 marks
The viscous drag force \(F\) acting on a small sphere of radius \(r\) moving at speed \(v\) through a fluid is given by the equation
\[F = 6 \pi \eta r v\]
where \(\eta\) is the dynamic viscosity of the fluid.
What are the SI base units of \(\eta\)?
A.\(\text{kg m}^{-1}\text{s}^{-1}\)
B.\(\text{kg m s}^{-2}\)
C.\(\text{kg m}^{-2}\text{s}^{-1}\)
D.\(\text{kg m}^2\text{s}^{-1}\)
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Worked solution
Rearranging the equation for viscosity gives: \[\eta = \frac{F}{6 \pi r v}\]
The constant \(6 \pi\) is dimensionless.
Substituting the SI base units for each quantity: - Force \(F = m a\): \(\text{kg m s}^{-2}\) - Radius \(r\): \(\text{m}\) - Velocity \(v\): \(\text{m s}^{-1}\)
\[[\eta] = \frac{\text{kg m s}^{-2}}{\text{m} \times \text{m s}^{-1}} = \frac{\text{kg m s}^{-2}}{\text{m}^2 \text{s}^{-1}} = \text{kg m}^{-1}\text{s}^{-1}\]
Marking scheme
A is correct [1 mark].
- Method mark (M1): Correctly expressing force in base units as \(\text{kg m s}^{-2}\) and setting up base unit ratio \(\frac{\text{kg m s}^{-2}}{\text{m} \cdot \text{m s}^{-1}}\). - Accuracy mark (A1): Simplifying to \(\text{kg m}^{-1}\text{s}^{-1}\).
Question 10 · Multiple Choice
1 marks
A small stone 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 stone immediately before it hits the horizontal ground below?
A.\(18\text{ m s}^{-1}\)
B.\(30\text{ m s}^{-1}\)
C.\(35\text{ m s}^{-1}\)
D.\(48\text{ m s}^{-1}\)
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Worked solution
1. Horizontal velocity remains constant: \[v_x = 18\text{ m s}^{-1}\]
2. Vertical velocity \(v_y\) just before impact can be determined using: \[v_y^2 = u_y^2 + 2 g s_y\] Since initial vertical velocity \(u_y = 0\): \[v_y^2 = 0 + 2 \times 9.81 \times 45 = 882.9\text{ m}^2\text{ s}^{-2}\]
3. The resultant speed \(v\) is: \[v = \sqrt{v_x^2 + v_y^2} = \sqrt{18^2 + 882.9} = \sqrt{324 + 882.9} = \sqrt{1206.9} \approx 35\text{ m s}^{-1}\]
Marking scheme
C is correct [1 mark].
- Method mark: Determines vertical component \(v_y = \sqrt{2 \times 9.81 \times 45} = 29.7\text{ m s}^{-1}\) and combines with horizontal component via Pythagoras theorem \(v = \sqrt{v_x^2 + v_y^2}\). - Final answer: \(35\text{ m s}^{-1}\).
Question 11 · Multiple Choice
1 marks
Trolley X of mass \(0.40\text{ kg}\) moves along a frictionless track with a velocity of \(+1.5\text{ m s}^{-1}\). It collides head-on and elastically with a stationary trolley Y of mass \(0.60\text{ kg}\).
What are the velocities of trolley X and trolley Y immediately after the collision?
A.velocity of X = \(-0.30\text{ m s}^{-1}\) ; velocity of Y = \(+1.20\text{ m s}^{-1}\)
B.velocity of X = \(+0.30\text{ m s}^{-1}\) ; velocity of Y = \(+0.90\text{ m s}^{-1}\)
C.velocity of X = \(-0.60\text{ m s}^{-1}\) ; velocity of Y = \(+1.50\text{ m s}^{-1}\)
D.velocity of X = \(0\text{ m s}^{-1}\) ; velocity of Y = \(+1.00\text{ m s}^{-1}\)
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Worked solution
For an elastic collision: 1. Relative speed of approach equals relative speed of separation: \[u_X - u_Y = v_Y - v_X\] \[1.5 - 0 = v_Y - v_X \implies v_Y = v_X + 1.5\]
- Uses conservation of momentum: \(0.60 = 0.40 v_X + 0.60 v_Y\). - Uses relative speed condition for elastic collision: \(v_Y - v_X = 1.5\). - Solves simultaneously to find \(v_X = -0.30\text{ m s}^{-1}\) and \(v_Y = +1.20\text{ m s}^{-1}\).
Question 12 · Multiple Choice
1 marks
A solid uniform block of wood has mass \(1.20\text{ kg}\) and density \(600\text{ kg m}^{-3}\). It is held completely submerged under a liquid of density \(850\text{ kg m}^{-3}\) by a light vertical cord attached to the bottom of the container.
What is the tension in the cord?
A.\(2.9\text{ N}\)
B.\(4.9\text{ N}\)
C.\(12\text{ N}\)
D.\(17\text{ N}\)
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Worked solution
1. Find the volume \(V\) of the wooden block: \[V = \frac{m}{\rho_{\text{wood}}} = \frac{1.20\text{ kg}}{600\text{ kg m}^{-3}} = 2.00 \times 10^{-3}\text{ m}^3\]
2. The upthrust \(U\) on the block equals the weight of displaced liquid: \[U = \rho_{\text{liquid}} V g = 850 \times (2.00 \times 10^{-3}) \times 9.81 = 1.70 \times 9.81 = 16.68\text{ N}\]
3. The weight \(W\) of the block is: \[W = m g = 1.20 \times 9.81 = 11.77\text{ N}\]
4. For equilibrium in the vertical direction: \[U = W + T \implies T = U - W = 16.68 - 11.77 = 4.91\text{ N} \approx 4.9\text{ N}\]
Two wires, P and Q, are made of the same metal. Wire P has length \(L\) and circular diameter \(d\). Wire Q has length \(2L\) and circular diameter \(2d\).
Both wires obey Hooke's law and are extended by applying the same tensile force \(F\).
What is the ratio \(\frac{\text{strain energy stored in wire Q}}{\text{strain energy stored in wire P}}\)?
A.\(0.25\)
B.\(0.50\)
C.\(1.0\)
D.\(2.0\)
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Worked solution
1. Elastic strain energy stored is given by \(E = \frac{1}{2} F x\). Since both wires experience the same force \(F\), the ratio of strain energies is equal to the ratio of their extensions: \[\frac{E_Q}{E_P} = \frac{x_Q}{x_P}\]
2. The extension \(x\) of a wire is given by: \[x = \frac{F L}{A E_Y} = \frac{F L}{\left(\frac{\pi d^2}{4}\right) E_Y} \propto \frac{L}{d^2}\]
- Relates strain energy to extension: \(E \propto x\) for constant force. - Uses \(x = \frac{F L}{A E_Y}\) where \(A \propto d^2\) so \(x \propto \frac{L}{d^2}\). - Ratio \(\frac{x_Q}{x_P} = \frac{2}{2^2} = 0.50\).
Question 14 · Multiple Choice
1 marks
A police car emits sound of frequency \(750\text{ Hz}\) from its siren and travels at a constant speed of \(28\text{ m s}^{-1}\) along a straight road towards a stationary observer.
The speed of sound in air is \(340\text{ m s}^{-1}\).
What is the frequency of the sound heard by the observer as the car approaches, and how does the observed frequency change after the car has passed the observer?
A.frequency before passing is \(820\text{ Hz}\) ; after passing it decreases to \(690\text{ Hz}\)
B.frequency before passing is \(820\text{ Hz}\) ; after passing it increases to \(890\text{ Hz}\)
C.frequency before passing is \(690\text{ Hz}\) ; after passing it decreases to \(630\text{ Hz}\)
D.frequency before passing is \(690\text{ Hz}\) ; after passing it increases to \(820\text{ Hz}\)
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Worked solution
1. Using the Doppler formula for a source moving towards a stationary observer: \[f_o = f_s \left(\frac{v}{v - v_s}\right) = 750 \times \left(\frac{340}{340 - 28}\right) = 750 \times \frac{340}{312} \approx 817\text{ Hz} \approx 820\text{ Hz}\]
2. As the car moves away, the source velocity is added in the denominator: \[f_o' = f_s \left(\frac{v}{v + v_s}\right) = 750 \times \left(\frac{340}{340 + 28}\right) = 750 \times \frac{340}{368} \approx 693\text{ Hz} \approx 690\text{ Hz}\]
Thus, the frequency drops to a lower constant value of approximately \(690\text{ Hz}\).
2. Use the diffraction grating equation: \[d \sin \theta = n \lambda\] Since the maximum possible angle \(\theta\) is \(90^\circ\) (where \(\sin \theta = 1\)): \[n_{\text{max}} = \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}\text{ m}}{532 \times 10^{-9}\text{ m}} \approx 4.70\]
3. The maximum integer order \(n\) is \(4\).
4. The total number of maxima is: \[\text{Total} = 2 n_{\text{max}} + 1 = 2(4) + 1 = 9\] (consisting of the central maximum \(n = 0\), and 4 orders on each side: \(n = \pm 1, \pm 2, \pm 3, \pm 4\)).
Marking scheme
D is correct [1 mark].
- Calculates grating spacing \(d = 2.50 \times 10^{-6}\text{ m}\). - Applies \(n \le \frac{d}{\lambda} = 4.70\), giving maximum integer order \(n = 4\). - Total visible maxima \(= 2n + 1 = 2(4) + 1 = 9\).
Question 16 · Multiple Choice
1 marks
A battery of electromotive force (e.m.f.) \(9.0\text{ V}\) and internal resistance \(1.5\ \Omega\) is connected in series with a fixed resistor of resistance \(4.5\ \Omega\) and a variable resistor \(R\).
The resistance of \(R\) is adjusted from \(0\ \Omega\) to \(12.0\ \Omega\).
What is the range of potential difference (p.d.) across the \(4.5\ \Omega\) fixed resistor?
A.\(0\text{ V}\) to \(6.8\text{ V}\)
B.\(2.3\text{ V}\) to \(6.8\text{ V}\)
C.\(2.3\text{ V}\) to \(9.0\text{ V}\)
D.\(3.0\text{ V}\) to \(6.8\text{ V}\)
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Worked solution
1. Total circuit resistance is: \[R_{\text{total}} = r + R_{\text{fixed}} + R = 1.5 + 4.5 + R = 6.0 + R\]
2. The potential difference across the \(4.5\ \Omega\) resistor is: \[V = I R_{\text{fixed}} = \left(\frac{E}{R_{\text{total}}}\right) \times 4.5 = \frac{9.0 \times 4.5}{6.0 + R} = \frac{40.5}{6.0 + R}\]
Thus, the range of p.d. is \(2.3\text{ V}\) to \(6.8\text{ V}\).
Marking scheme
B is correct [1 mark].
- Calculates max p.d. at \(R = 0\ \Omega\): \(I = \frac{9.0}{6.0} = 1.5\text{ A} \implies V = 1.5 \times 4.5 = 6.75\text{ V}\). - Calculates min p.d. at \(R = 12.0\ \Omega\): \(I = \frac{9.0}{18.0} = 0.50\text{ A} \implies V = 0.50 \times 4.5 = 2.25\text{ V}\). - Range is \(2.3\text{ V}\) to \(6.8\text{ V}\).
Question 17 · Multiple Choice
1 marks
Thermal conductivity \(k\) is defined by the rate of heat flow equation:
$$\frac{\Delta Q}{\Delta t} = \frac{k A \Delta T}{L}$$
where \(\Delta Q\) is energy transferred in time \(\Delta t\), \(A\) is cross-sectional area, \(\Delta T\) is temperature difference, and \(L\) is length.
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Worked solution
Rearranging the equation for thermal conductivity gives: $$k = \frac{\left(\frac{\Delta Q}{\Delta t}\right) L}{A \Delta T}$$
Substituting SI base units for each quantity: - Energy \(\Delta Q\): \(\text{J} = \text{kg m}^2 \text{s}^{-2}\) - Rate of heat flow \(\frac{\Delta Q}{\Delta t}\): \(\text{kg m}^2 \text{s}^{-3}\) - Length \(L\): \(\text{m}\) - Area \(A\): \(\text{m}^2\) - Temperature difference \(\Delta T\): \(\text{K}\)
Therefore, the base units of \(k\) are: $$[k] = \frac{(\text{kg m}^2 \text{s}^{-3})(\text{m})}{\text{m}^2 \times \text{K}} = \text{kg m s}^{-3} \text{K}^{-1}$$
Marking scheme
[1 mark] A is the correct answer. B is incorrect as area was not divided out correctly; C and D result from incorrect base units of energy/power.
Question 18 · Multiple Choice
1 marks
A small ball is projected horizontally with an initial speed \(u\) from the edge of a vertical cliff of height \(h\). Air resistance is negligible.
The ball strikes the level ground below at an angle of \(60^\circ\) to the horizontal.
Which expression gives the height \(h\) in terms of \(u\) and the acceleration of free fall \(g\)?
D.\(\frac{3u^2}{g}\)
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Worked solution
The horizontal velocity remains constant throughout the flight: \(v_x = u\).
When the ball strikes the ground at an angle of \(60^\circ\) to the horizontal: $$\tan(60^\circ) = \frac{v_y}{v_x} = \sqrt{3} \implies v_y = u\sqrt{3}$$
Using the vertical equation of motion \(v_y^2 = u_y^2 + 2gh\) with initial vertical velocity \(u_y = 0\): $$(u\sqrt{3})^2 = 0 + 2gh$$ $$3u^2 = 2gh \implies h = \frac{3u^2}{2g}$$
Marking scheme
[1 mark] C is the correct answer. A corresponds to using \(\tan(30^\circ)\); B assumes \(v_y = u\); D omits the factor of 2 in \(v^2 = 2gh\).
Question 19 · Multiple Choice
1 marks
A trolley of mass \(2.0\text{ kg}\) travels with a velocity of \(6.0\text{ m s}^{-1}\) to the right on a horizontal, frictionless track. It makes a head-on, perfectly elastic collision with a stationary trolley of mass \(4.0\text{ kg}\).
What are the velocities of the two trolleys immediately after the collision?
D.\(2.0\text{ kg}\) trolley: \(4.0\text{ m s}^{-1}\) to the left; \(4.0\text{ kg}\) trolley: \(5.0\text{ m s}^{-1}\) to the right
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2. In an elastic collision, relative speed of approach equals relative speed of separation: $$u_1 - u_2 = v_2 - v_1$$ $$6.0 - 0 = v_2 - v_1 \implies v_2 - v_1 = 6.0$$
Adding the two equations: $$(v_1 + 2v_2) + (v_2 - v_1) = 6.0 + 6.0$$ $$3v_2 = 12.0 \implies v_2 = +4.0\text{ m s}^{-1} \text{ (to the right)}$$
Substituting \(v_2\) back: $$v_1 = v_2 - 6.0 = 4.0 - 6.0 = -2.0\text{ m s}^{-1} \text{ (to the left)}$$
Marking scheme
[1 mark] A is the correct answer. B conserves momentum but not kinetic energy; C represents an inelastic collision where mass 1 stops.
Question 20 · Multiple Choice
1 marks
A solid cylinder of uniform density \(\rho_{\text{c}}\), cross-sectional area \(A\) and total height \(H\) floats vertically in a container.
The cylinder is partially submerged in a liquid of density \(\rho_{\text{L}}\) and the remainder is submerged in an immiscible layer of oil of density \(\rho_{\text{oil}}\) resting above the liquid, where \(\rho_{\text{oil}} < \rho_{\text{c}} < \rho_{\text{L}}\).
The cylinder is completely submerged in the two liquids combined, with a height \(h\) in the lower liquid.
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Worked solution
The cylinder is in equilibrium, so its weight is balanced by the sum of the upthrusts from the two liquid layers: $$\text{Weight of cylinder} = \text{Upthrust from liquid} + \text{Upthrust from oil}$$ $$\rho_{\text{c}} A H g = \rho_{\text{L}} A h g + \rho_{\text{oil}} A (H - h) g$$
Cancelling \(A g\) from all terms: $$\rho_{\text{c}} H = \rho_{\text{L}} h + \rho_{\text{oil}} H - \rho_{\text{oil}} h$$ $$(\rho_{\text{c}} - \rho_{\text{oil}}) H = (\rho_{\text{L}} - \rho_{\text{oil}}) h$$
[1 mark] A is the correct answer. B gives the height submerged in the oil \((H - h)\); C and D are algebraically incorrect.
Question 21 · Multiple Choice
1 marks
A uniform metal wire of length \(L\) and cross-sectional area \(A\) is made of a material of Young modulus \(E\). When a tensile force \(F\) is applied within the limit of proportionality, the elastic strain energy stored in the wire is \(U\).
A second wire of the same material has length \(2L\) and cross-sectional area \(2A\).
What is the elastic strain energy stored in the second wire when the same tensile force \(F\) is applied?
D.\(2 U\)
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Worked solution
The extension \(x\) of a wire under tensile force \(F\) is given by: $$x = \frac{F L}{A E}$$
The elastic strain energy stored within Hooke's law is: $$U = \frac{1}{2} F x = \frac{F^2 L}{2 A E}$$
For the second wire with length \(L' = 2L\), cross-sectional area \(A' = 2A\), same Young modulus \(E\), and same force \(F\): $$U' = \frac{F^2 (2L)}{2 (2A) E} = \frac{F^2 L}{2 A E} = U$$
Therefore, the strain energy stored remains \(U\).
Marking scheme
[1 mark] C is the correct answer. Doubling both length and area keeps the ratio \(L/A\) and hence the extension and strain energy unchanged.
Question 22 · Multiple Choice
1 marks
A car horn emits sound of constant frequency \(680\text{ Hz}\). The car travels at a constant velocity of \(25\text{ m s}^{-1}\) directly towards a stationary observer.
The speed of sound in air is \(340\text{ m s}^{-1}\).
What is the frequency of the sound observed by the stationary observer?
D.\(734\text{ Hz}\)
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Worked solution
Using the Doppler effect formula for a source moving towards a stationary observer: $$f_o = f_s \left(\frac{v}{v - v_s}\right)$$
Substituting the given values: - \(f_s = 680\text{ Hz}\) - \(v = 340\text{ m s}^{-1}\) - \(v_s = 25\text{ m s}^{-1}\)
[1 mark] D is the correct answer. A corresponds to a source moving away \(f_o = 680(340/365) = 633\text{ Hz}\); C corresponds to using \(f_o = f_s (v + v_s)/v\).
Question 23 · Multiple Choice
1 marks
A cell of electromotive force (e.m.f.) \(E\) and constant internal resistance \(r\) is connected in series with an external variable resistor of resistance \(R\).
The resistance \(R\) is increased from \(R = r\) to \(R = 3r\).
Which row correctly describes the change in the terminal potential difference across the cell and the change in the total power dissipated in the complete circuit?
D.Terminal p.d.: increases | Total power: increases
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Worked solution
1. Terminal potential difference \(V\): $$V = E \left(\frac{R}{R + r}\right)$$ - When \(R = r\): \(V_1 = E \left(\frac{r}{2r}\right) = 0.50 E\) - When \(R = 3r\): \(V_2 = E \left(\frac{3r}{4r}\right) = 0.75 E\) Thus, the terminal p.d. increases.
2. Total power dissipated in the complete circuit \(P_{\text{total}}\): $$P_{\text{total}} = E I = \frac{E^2}{R + r}$$ - When \(R = r\): \(P_1 = \frac{E^2}{2r}\) - When \(R = 3r\): \(P_2 = \frac{E^2}{4r}\) Thus, the total power dissipated decreases.
Marking scheme
[1 mark] C is the correct answer. As \(R\) increases, current decreases so lost volts \(Ir\) decreases and terminal p.d. \(E - Ir\) increases; total power \(EI\) decreases because current decreases.
Question 24 · Multiple Choice
1 marks
In a \(\beta^+\) (positron) radioactive decay, a proton in an unstable nucleus transforms into a neutron.
Which statement correctly describes the quark change and names the other lepton emitted in this decay?
D.a down quark changes to an up quark, and an electron neutrino is emitted
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Worked solution
A proton has quark composition \(uud\) and a neutron has quark composition \(udd\).
During \(\beta^+\) decay, a proton changes into a neutron: $$u \rightarrow d + \text{e}^+ + \nu_{\text{e}}$$
Hence, an up (\(u\)) quark changes into a down (\(d\)) quark, and a positron (\(\text{e}^+\)) and an electron neutrino (\(\nu_{\text{e}}\)) are emitted.
Marking scheme
[1 mark] B is the correct answer. In \(\beta^+\) decay, \(u \rightarrow d + \text{e}^+ + \nu_{\text{e}}\); antineutrinos are emitted in \(\beta^-\) decay where \(d \rightarrow u\).
Question 25 · Multiple Choice
1 marks
Which unit, expressed in SI base units, is equivalent to the unit of power?
A.\(\text{kg}\,\text{m}\,\text{s}^{-2}\)
B.\(\text{kg}\,\text{m}^{2}\,\text{s}^{-2}\)
C.\(\text{kg}\,\text{m}^{2}\,\text{s}^{-3}\)
D.\(\text{kg}\,\text{m}^{3}\,\text{s}^{-3}\)
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Worked solution
Power is defined as work done per unit time: \(P = \frac{W}{t}\). Work done has the unit of force multiplied by displacement: \(W = F \times s\). Force in base units is \(\text{kg}\times \text{m s}^{-2} = \text{kg}\,\text{m}\,\text{s}^{-2}\). Therefore, work done is \(\text{kg}\,\text{m}\,\text{s}^{-2} \times \text{m} = \text{kg}\,\text{m}^{2}\,\text{s}^{-2}\). Dividing by time (\(\text{s}\)) gives the SI base units of power as \(\text{kg}\,\text{m}^{2}\,\text{s}^{-3}\).
Marking scheme
C is the correct answer [1 mark]. A represents force (\(\text{N}\)). B represents energy/work done (\(\text{J}\)). D is dimensionally incorrect.
Question 26 · Multiple Choice
1 marks
A small ball is projected horizontally at a speed of \(15\text{ m s}^{-1}\) from the top of a vertical cliff of height \(45\text{ m}\). Air resistance is negligible.
What is the speed of the ball immediately before it strikes the horizontal ground below?
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
The horizontal component of velocity remains constant: \(v_x = 15\text{ m s}^{-1}\). The vertical component of velocity \(v_y\) is given by: \(v_y^2 = u_y^2 + 2 g h = 0 + 2(9.81)(45) = 882.9\text{ m}^2\text{ s}^{-2}\). The resultant speed \(v\) is: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 882.9} = \sqrt{225 + 882.9} = \sqrt{1107.9} \approx 33.3\text{ m s}^{-1}\).
Marking scheme
B is the correct answer [1 mark]. A gives only the vertical velocity component (\(29.7\text{ m s}^{-1}\)). C is the scalar sum of horizontal and vertical speeds (\(15 + 29.7 = 44.7\text{ m s}^{-1}\)). D uses an incorrect kinematic relationship.
Question 27 · Multiple Choice
1 marks
A trolley of mass \(0.80\text{ kg}\) travels at a speed of \(3.0\text{ m s}^{-1}\) along a frictionless horizontal track. It collides with and sticks to a stationary trolley of mass \(1.20\text{ kg}\).
What is the loss in total kinetic energy of the trolleys as a result of the collision?
A.\(1.4\text{ J}\)
B.\(2.2\text{ J}\)
C.\(2.4\text{ J}\)
D.\(3.6\text{ J}\)
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Worked solution
Initial kinetic energy: \(E_{k1} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2}(0.80)(3.0)^2 = 3.60\text{ J}\). By conservation of linear momentum: \(m_1 u_1 = (m_1 + m_2) v \implies (0.80)(3.0) = (0.80 + 1.20) v \implies 2.40 = 2.00 v \implies v = 1.20\text{ m s}^{-1}\). Final kinetic energy: \(E_{k2} = \frac{1}{2}(m_1 + m_2) v^2 = \frac{1}{2}(2.00)(1.20)^2 = 1.44\text{ J}\). Loss in kinetic energy: \(\Delta E_k = E_{k1} - E_{k2} = 3.60 - 1.44 = 2.16\text{ J} \approx 2.2\text{ J}\).
Marking scheme
B is the correct answer [1 mark]. A is the final kinetic energy (\(1.44\text{ J}\)). C is the total initial momentum (\(2.4\text{ N s}\)). D is the initial kinetic energy (\(3.6\text{ J}\)).
Question 28 · Multiple Choice
1 marks
A uniform solid cylinder of cross-sectional area \(4.0 \times 10^{-3}\text{ m}^2\) and height \(0.25\text{ m}\) is completely submerged in a liquid of density \(1.2 \times 10^3\text{ kg m}^{-3}\). The circular top face of the cylinder is horizontal and located at a depth of \(0.50\text{ m}\) below the surface of the liquid.
What is the upthrust exerted by the liquid on the cylinder?
A.\(5.9\text{ N}\)
B.\(12\text{ N}\)
C.\(35\text{ N}\)
D.\(47\text{ N}\)
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Worked solution
Upthrust is equal to the weight of liquid displaced (Archimedes' principle): \(F = \rho g V\). The volume of the cylinder is \(V = A \times h = (4.0 \times 10^{-3}\text{ m}^2)(0.25\text{ m}) = 1.0 \times 10^{-3}\text{ m}^3\). Therefore: \(F = (1.2 \times 10^3\text{ kg m}^{-3})(9.81\text{ m s}^{-2})(1.0 \times 10^{-3}\text{ m}^3) = 11.772\text{ N} \approx 12\text{ N}\).
Marking scheme
B is the correct answer [1 mark]. A corresponds to calculating \(\rho g A h / 2\). C corresponds to the downward force on the top face (\(23.5\text{ N}\)) plus upthrust. D corresponds to the upward hydrostatic force acting on the bottom face (\(35.3\text{ N} + 11.8\text{ N}\)).
Question 29 · Multiple Choice
1 marks
A wire of original length \(L\) and uniform cross-sectional area \(A\) is made of metal of Young modulus \(E\). When a tensile force \(F\) is applied within the limit of proportionality, the wire stores elastic potential energy \(W\).
A second wire made of the same metal has original length \(2L\) and cross-sectional area \(2A\). It is subjected to a tensile force of \(2F\) within its limit of proportionality.
What is the elastic potential energy stored in the second wire?
A.\(W\)
B.\(2W\)
C.\(4W\)
D.\(8W\)
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Worked solution
The elastic potential energy stored in a wire is \(W = \frac{1}{2} F e = \frac{F^2 L}{2 A E}\). For the second wire, force is \(2F\), length is \(2L\), and area is \(2A\): \(W_2 = \frac{(2F)^2 (2L)}{2 (2A) E} = \frac{4 F^2 \cdot 2 L}{4 A E} = \frac{2 F^2 L}{A E} = 4 \left(\frac{F^2 L}{2 A E}\right) = 4W\).
Marking scheme
C is the correct answer [1 mark]. A assumes no change in stored energy. B assumes energy scales linearly with force. D assumes extension scales as \(2^3\).
Question 30 · Multiple Choice
1 marks
A train whistle emits a sound wave of constant frequency \(680\text{ Hz}\). The speed of sound in air is \(340\text{ m s}^{-1}\).
The train travels directly towards a stationary observer at a constant speed of \(20\text{ m s}^{-1}\).
What is the frequency of the sound detected by the observer?
A.\(642\text{ Hz}\)
B.\(680\text{ Hz}\)
C.\(720\text{ Hz}\)
D.\(723\text{ Hz}\)
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Worked solution
Using the Doppler effect formula for a moving source approaching a stationary observer: \(f_o = f_s \left( \frac{v}{v - v_s} \right)\) Substituting the values: \(f_o = 680 \times \left( \frac{340}{340 - 20} \right) = 680 \times \left( \frac{340}{320} \right) = 680 \times 1.0625 = 722.5\text{ Hz} \approx 723\text{ Hz}\).
Marking scheme
D is the correct answer [1 mark]. A is the frequency if the source were moving away (\(680 \times 340 / 360 = 642\text{ Hz}\)). B assumes no frequency shift. C is rounded incorrectly or obtained from \(f_s (1 + v_s / v) = 680 \times 1.0588 = 720\text{ Hz}\).
Question 31 · Multiple Choice
1 marks
Monochromatic light of wavelength \(590\text{ nm}\) is incident normally on a diffraction grating with \(4.0 \times 10^5\text{ lines m}^{-1}\).
What is the total number of diffraction maxima (bright spots) that can be observed on a screen placed far beyond the grating?
A.4
B.5
C.8
D.9
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Worked solution
The grating spacing is \(d = \frac{1}{N} = \frac{1}{4.0 \times 10^5\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m}\). Using the grating equation \(d \sin\theta = n \lambda\), the maximum order \(n_{\text{max}}\) corresponds to \(\sin\theta \le 1\): \(n_{\text{max}} \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}\text{ m}}{590 \times 10^{-9}\text{ m}} \approx 4.24\). Since \(n\) must be an integer, \(n_{\text{max}} = 4\). The total number of observed maxima includes the central maximum (\(n=0\)) and orders \(n = 1, 2, 3, 4\) on either side: \(\text{Total} = 2 n_{\text{max}} + 1 = 2(4) + 1 = 9\).
Marking scheme
D is the correct answer [1 mark]. A gives only \(n_{\text{max}}\). B gives the number of maxima on one side plus the centre (\(n_{\text{max}} + 1\)). C excludes the central zero-order maximum (\(2 n_{\text{max}}\)).
Question 32 · Multiple Choice
1 marks
In a \(\beta^-\)-decay process, a neutron decays into a proton, an electron, and an electron antineutrino.
Which row correctly identifies the change in quark flavour and the net change in the total lepton number of the system?
A.Quark change: \(\text{u} \rightarrow \text{d}\) ; Total lepton number change: increases by 1
B.Quark change: \(\text{u} \rightarrow \text{d}\) ; Total lepton number change: no change
C.Quark change: \(\text{d} \rightarrow \text{u}\) ; Total lepton number change: increases by 1
D.Quark change: \(\text{d} \rightarrow \text{u}\) ; Total lepton number change: no change
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Worked solution
A neutron has quark composition \(\text{udd}\) and a proton has quark composition \(\text{uud}\). Therefore, one down quark changes to an up quark: \(\text{d} \rightarrow \text{u}\). Prior to decay, the neutron is a baryon with lepton number \(L = 0\). After decay, the proton has \(L = 0\), the electron has \(L = +1\), and the electron antineutrino has \(L = -1\). The total lepton number after decay is \(0 + 1 + (-1) = 0\). Thus, the total lepton number is conserved (net change is 0 / no change).
Marking scheme
D is the correct answer [1 mark]. A and B incorrectly identify the quark transformation as \(\text{u} \rightarrow \text{d}\) (which occurs in \(\beta^+\) decay). C incorrectly states that lepton number increases by 1 (neglecting the antineutrino with \(L = -1\)).
Question 33 · Multiple Choice
1 marks
The drag force \(F\) acting on a small sphere of radius \(r\) moving at speed \(v\) through a fluid of dynamic viscosity \(\eta\) is given by the equation
\[F = 6 \pi \eta r v\]
What are the SI base units of \(\eta\)?
A.\(\text{kg m}^{-1} \text{s}^{-1}\)
B.\(\text{kg m s}^{-1}\)
C.\(\text{kg m}^{-1} \text{s}^{-2}\)
D.\(\text{kg m}^2 \text{s}^{-1}\)
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Worked solution
Rearranging the equation for dynamic viscosity gives: \[\eta = \frac{F}{6 \pi r v}\] The constant \(6 \pi\) is dimensionless. Substituting the SI base units of each quantity: - Force \(F\): \(\text{kg m s}^{-2}\) - Radius \(r\): \(\text{m}\) - Speed \(v\): \(\text{m s}^{-1}\)
\[[\eta] = \frac{\text{kg m s}^{-2}}{\text{m} \times \text{m s}^{-1}} = \frac{\text{kg m s}^{-2}}{\text{m}^2 \text{s}^{-1}} = \text{kg m}^{-1} \text{s}^{-1}\]
Therefore, option A is correct.
Marking scheme
A [1 mark]: Correct identification of base units \(\text{kg m}^{-1} \text{s}^{-1}\).
Question 34 · Multiple Choice
1 marks
A projectile is launched from horizontal ground with an initial speed of \(25\text{ m s}^{-1}\) at an angle of \(30^\circ\) above the horizontal. Air resistance is negligible.
What is the speed of the projectile at time \(t = 1.8\text{ s}\) after launch?
A.\(5.2\text{ m s}^{-1}\)
B.\(17\text{ m s}^{-1}\)
C.\(22\text{ m s}^{-1}\)
D.\(27\text{ m s}^{-1}\)
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Worked solution
Resolve the initial velocity into horizontal and vertical components: - Horizontal component: \(u_x = 25 \cos 30^\circ = 21.65\text{ m s}^{-1}\) - Vertical component: \(u_y = 25 \sin 30^\circ = 12.50\text{ m s}^{-1}\)
At time \(t = 1.8\text{ s}\): - The horizontal velocity remains constant: \(v_x = 21.65\text{ m s}^{-1}\) - The vertical velocity is given by \(v_y = u_y - gt = 12.50 - (9.81 \times 1.8) = 12.50 - 17.66 = -5.16\text{ m s}^{-1}\)
The resultant speed \(v\) is: \[v = \sqrt{v_x^2 + v_y^2} = \sqrt{(21.65)^2 + (-5.16)^2} = \sqrt{468.7 + 26.6} = \sqrt{495.3} \approx 22\text{ m s}^{-1}\]
Therefore, option C is correct.
Marking scheme
C [1 mark]: Correct resultant speed of \(22\text{ m s}^{-1}\).
Question 35 · Multiple Choice
1 marks
Glider X of mass \(0.40\text{ kg}\) moves along a frictionless linear air track with a velocity of \(+1.5\text{ m s}^{-1}\). It collides head-on and elastically with a stationary glider Y of mass \(0.60\text{ kg}\).
What is the velocity of glider X immediately after the collision?
A.\(0.30\text{ m s}^{-1}\) in the direction opposite to its initial motion
B.\(0.30\text{ m s}^{-1}\) in the same direction as its initial motion
C.\(0.60\text{ m s}^{-1}\) in the direction opposite to its initial motion
D.\(1.2\text{ m s}^{-1}\) in the same direction as its initial motion
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Worked solution
For an elastic collision, the relative speed of approach equals the relative speed of separation: \[u_X - u_Y = v_Y - v_X\] \[1.5 - 0 = v_Y - v_X \implies v_Y = v_X + 1.5\]
The negative sign indicates that glider X rebounds at \(0.30\text{ m s}^{-1}\) in the direction opposite to its initial motion. Therefore, option A is correct.
Marking scheme
A [1 mark]: Correct velocity of \(0.30\text{ m s}^{-1}\) opposite to initial direction.
Question 36 · Multiple Choice
1 marks
A solid uniform cylinder of cross-sectional area \(4.0 \times 10^{-3}\text{ m}^2\) and height \(0.15\text{ m}\) is suspended vertically from a light string. The cylinder is completely submerged in a liquid of density \(850\text{ kg m}^{-3}\).
The tension in the string is \(3.2\text{ N}\).
What is the density of the material of the cylinder?
A.\(540\text{ kg m}^{-3}\)
B.\(850\text{ kg m}^{-3}\)
C.\(1400\text{ kg m}^{-3}\)
D.\(2200\text{ kg m}^{-3}\)
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Worked solution
The volume of the cylinder is: \[V = A h = (4.0 \times 10^{-3}) \times 0.15 = 6.0 \times 10^{-4}\text{ m}^3\]
The upthrust \(U\) on the cylinder by Archimedes' principle is: \[U = \rho_{\text{liquid}} g V = 850 \times 9.81 \times 6.0 \times 10^{-4} = 5.003\text{ N}\]
For vertical equilibrium: \[W = T + U = 3.2 + 5.003 = 8.203\text{ N}\]
The mass of the cylinder is: \[m = \frac{W}{g} = \frac{8.203}{9.81} = 0.8362\text{ kg}\]
The density of the cylinder is: \[\rho_{\text{cylinder}} = \frac{m}{V} = \frac{0.8362}{6.0 \times 10^{-4}} \approx 1400\text{ kg m}^{-3}\]
Therefore, option C is correct.
Marking scheme
C [1 mark]: Correct density calculation of \(1400\text{ kg m}^{-3}\).
Question 37 · Multiple Choice
1 marks
Two wires, P and Q, are made of the same metal and obey Hooke's law. Wire P has length \(L\) and diameter \(d\). Wire Q has length \(2L\) and diameter \(2d\).
Both wires are stretched by the same tensile force \(F\).
What is the ratio \(\frac{\text{strain energy in wire P}}{\text{strain energy in wire Q}}\mathrm{?}\)
A.\(\frac{1}{2}\)
B.\(1\)
C.\(2\)
D.\(4\)
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Worked solution
The strain energy \(E_s\) stored in a wire within its elastic limit is given by: \[E_s = \frac{1}{2} F x = \frac{1}{2} F \left(\frac{F L}{A E}\right) = \frac{F^2 L}{2 A E}\] Since \(A = \frac{\pi d^2}{4}\), we have: \[E_s \propto \frac{L}{d^2}\]
For wire P: \[E_{s,\text{P}} \propto \frac{L}{d^2}\]
Thus, the ratio is: \[\frac{E_{s,\text{P}}}{E_{s,\text{Q}}} = \frac{1}{\frac{1}{2}} = 2\]
Therefore, option C is correct.
Marking scheme
C [1 mark]: Correct strain energy ratio of 2.
Question 38 · Multiple Choice
1 marks
An emergency vehicle travels at a constant speed of \(28.0\text{ m s}^{-1}\) along a straight road while emitting a siren tone of frequency \(720\text{ Hz}\). A stationary observer stands by the side of the road.
The speed of sound in air is \(336\text{ m s}^{-1}\).
What is the change in the observed frequency as the vehicle passes the observer and recedes into the distance?
A.\(55.4\text{ Hz}\)
B.\(65.5\text{ Hz}\)
C.\(121\text{ Hz}\)
D.\(144\text{ Hz}\)
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Worked solution
Using the Doppler equation for a moving source: 1. Approaching observer: \[f_o = f_s \left(\frac{v}{v - v_s}\right) = 720 \times \left(\frac{336}{336 - 28.0}\right) = 720 \times \frac{336}{308} = 785.5\text{ Hz}\]
3. Total change in frequency: \[\Delta f = 785.5 - 664.6 = 120.9\text{ Hz} \approx 121\text{ Hz}\]
Therefore, option C is correct.
Marking scheme
C [1 mark]: Correct frequency shift \(\Delta f = 121\text{ Hz}\).
Question 39 · Multiple Choice
1 marks
Monochromatic light of wavelength \(532\text{ nm}\) is incident normally on a plane transmission diffraction grating with \(4.00 \times 10^5\text{ lines m}^{-1}\).
What is the total number of bright intensity maxima produced by the grating?
A.4
B.5
C.8
D.9
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Worked solution
The grating spacing \(d\) is: \[d = \frac{1}{N} = \frac{1}{4.00 \times 10^5} = 2.50 \times 10^{-6}\text{ m}\]
The diffraction grating equation is \(d \sin \theta = n \lambda\). For the maximum order \(n_{\text{max}}\), \(\sin \theta \le 1\): \[n_{\text{max}} = \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}}{532 \times 10^{-9}} \approx 4.70\]
Since \(n\) must be an integer, \(n_{\text{max}} = 4\).
The total number of bright maxima observable includes: - 1 central zeroth-order maximum (\(n = 0\)) - 4 maxima on one side (\(n = 1, 2, 3, 4\)) - 4 maxima on the opposite side (\(n = 1, 2, 3, 4\))
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Worked solution
In \(\beta^+\) decay, a proton decays into a neutron, a positron, and an electron neutrino: \[\text{p} \rightarrow \text{n} + \text{e}^+ + \nu_e\] In terms of quarks: - A proton has quark composition \(\text{uud}\). - A neutron has quark composition \(\text{udd}\). Thus, an up quark changes into a down quark (\(\text{u} \rightarrow \text{d}\)).
To conserve lepton number (since the positron \(\text{e}^+\) has lepton number \(L = -1\)), an electron neutrino (\(\nu_e\), with \(L = +1\)) must be emitted.
Therefore, row A is correct.
Marking scheme
A [1 mark]: Correct quark transformation \(\text{u} \rightarrow \text{d}\) and lepton \(\nu_e\).
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Answer all questions. Show all working and appropriate units.
7 Question · 59.99 marks
Question 1 · Structured Calculation & Explanation
8.57 marks
(a) Define what is meant by a derived quantity and give one example of an SI base unit.
(b) The drag force \(F_D\) acting on a small sphere of radius \(r\) moving at speed \(v\) through a fluid of dynamic viscosity \(\eta\) is given by Stokes' law: \[ F_D = 6\pi \eta r v \]
(i) Show that the SI base units of dynamic viscosity \(\eta\) are \(\text{kg}\text{ m}^{-1}\text{ s}^{-1}\).
(ii) In an experiment, the radius \(r\) of a sphere is measured as \((3.50 \pm 0.07)\text{ mm}\) and its terminal speed \(v\) is measured as \((0.80 \pm 0.04)\text{ m}\text{ s}^{-1}\). The viscous drag force \(F_D\) at this terminal speed is \((1.20 \pm 0.06)\times 10^{-3}\text{ N}\). Calculate the value of \(\eta\) and determine its absolute uncertainty. State your final answer to an appropriate number of significant figures.
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Worked solution
(a) A derived quantity is a physical quantity that is defined as a combination (product and/or quotient) of base quantities. An example of an SI base unit is the kilogram (kg), metre (m), second (s), ampere (A), or kelvin (K).
(b)(i) Rearranging the equation for dynamic viscosity: \[ \eta = \frac{F_D}{6\pi r v} \] Base units of \(F_D = \text{N} = \text{kg m s}^{-2}\) Base units of \(r = \text{m}\) Base units of \(v = \text{m s}^{-1}\) Substituting base units: \[ [\eta] = \frac{\text{kg m s}^{-2}}{\text{m} \times \text{m s}^{-1}} = \frac{\text{kg m s}^{-2}}{\text{m}^2 \text{ s}^{-1}} = \text{kg m}^{-1}\text{ s}^{-1} \]
(a) B1: A physical quantity that consists of / is formed from combinations of base quantities. B1: Any one valid SI base unit: kg, m, s, A, K, mol, cd.
(b)(i) M1: Identifies base units of force as \(\text{kg m s}^{-2}\). A1: Shows full substitution into \(\frac{[F]}{[r][v]}\) leading correctly to \(\text{kg m}^{-1}\text{ s}^{-1}\).
(b)(ii) C1: Correct calculation of central value \(\eta = 0.0227\text{ kg m}^{-1}\text{ s}^{-1}\) (or \(2.27 \times 10^{-2}\)). C1: Calculation of individual fractional/percentage uncertainties (5%, 2%, 5%) and sums them to get 12% (or 0.12). A1: Correct determination of absolute uncertainty \(\Delta \eta = 0.0027\text{ kg m}^{-1}\text{ s}^{-1}\) (or 0.003). A1: Final value stated with matching precision / appropriate significant figures (e.g. \(0.023 \pm 0.003\) or \((2.3 \pm 0.3) \times 10^{-2}\text{ kg m}^{-1}\text{ s}^{-1}\)).
Question 2 · Structured Calculation & Explanation
8.57 marks
(a) State the principle of conservation of linear momentum.
(b) A trolley \(A\) of mass \(0.75\text{ kg}\) travels along a horizontal frictionless track with a constant velocity of \(2.4\text{ m s}^{-1}\). It collides head-on with a stationary trolley \(B\) of mass \(1.25\text{ kg}\). After the collision, trolley \(A\) rebounds in the opposite direction with a speed of \(0.60\text{ m s}^{-1}\).
(i) Calculate the velocity \(v_B\) of trolley \(B\) immediately after the collision.
(ii) Determine, by calculation of kinetic energies or relative speeds, whether the collision is elastic or inelastic.
(c) During the collision, the contact time between the two trolleys is \(45\text{ ms}\). Calculate the average magnitude of the force exerted by trolley \(A\) on trolley \(B\).
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Worked solution
(a) Total linear momentum of an isolated system remains constant provided no external resultant force acts.
(b)(i) Let the initial direction of motion of trolley \(A\) be positive. Initial momentum: \[ p_i = m_A u_A + m_B u_B = (0.75)(2.4) + 0 = 1.80 \text{ kg m s}^{-1} \] Final momentum: \[ p_f = m_A v_A + m_B v_B = (0.75)(-0.60) + (1.25)(v_B) = -0.45 + 1.25 v_B \] Applying conservation of momentum: \[ 1.80 = -0.45 + 1.25 v_B \implies 1.25 v_B = 2.25 \implies v_B = 1.80 \text{ m s}^{-1} \]
Method 2 (Relative Speed): Relative speed of approach = \(2.4 - 0 = 2.4\text{ m s}^{-1}\) Relative speed of separation = \(1.80 - (-0.60) = 2.40\text{ m s}^{-1}\) Since relative speed of approach equals relative speed of separation, the collision is elastic.
(c) Force is rate of change of momentum: \[ F = \frac{\Delta p}{\Delta t} = \frac{m_B v_B - 0}{\Delta t} = \frac{1.25 \times 1.80}{45 \times 10^{-3}} = \frac{2.25}{0.045} = 50 \text{ N} \]
Marking scheme
(a) B1: Total linear momentum of a closed/isolated system remains constant (or sum of momentum before equals sum of momentum after in absence of external force).
(b)(i) C1: \(m_A u_A = m_A v_A + m_B v_B\) with correct signs taken into account (e.g. \(0.75 \times 2.4 = 0.75 \times (-0.60) + 1.25 v_B\)). A1: \(v_B = 1.8\text{ m s}^{-1}\) (or \(1.80\text{ m s}^{-1}\)).
(b)(ii) M1: Calculates total initial \(E_k\) (\(2.16\text{ J}\)) and total final \(E_k\) (\(2.16\text{ J}\)) OR calculates relative speed of approach (\(2.4\text{ m s}^{-1}\)) and relative speed of separation (\(2.4\text{ m s}^{-1}\)). A1: States that kinetic energy is conserved (or relative speeds are equal) AND concludes that the collision is elastic.
(c) C1: Uses \(F = \frac{\Delta p}{\Delta t}\) or \(F = m a\) with \(a = \frac{v-u}{t}\). A1: \(F = 50\text{ N}\).
Question 3 · Structured Calculation & Explanation
8.57 marks
(a) State the two conditions required for a rigid body to be in static equilibrium.
(b) A uniform horizontal beam \(AB\) of length \(2.40\text{ m}\) and weight \(180\text{ N}\) is hinged smoothly to a vertical wall at end \(A\). The beam is supported by a light steel cable attached to end \(B\) at an angle of \(35.0^\circ\) to the horizontal, with the other end of the cable fixed to the wall above \(A\). A load of weight \(320\text{ N}\) is suspended from the beam at a distance of \(1.80\text{ m}\) from \(A\).
(i) By taking moments about hinge \(A\), calculate the tension \(T\) in the cable.
(ii) Calculate the magnitude and direction of the force \(R\) exerted by the hinge on the beam at \(A\).
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Worked solution
(a) The conditions for equilibrium are: 1. The resultant force acting on the body in any direction is zero (\(\Sigma F = 0\)). 2. The resultant torque (moment of forces) about any pivot point is zero (\(\Sigma \tau = 0\)).
(b)(i) Since the beam is uniform, its weight acts at its midpoint, \(d_w = 1.20\text{ m}\) from \(A\). Taking moments about hinge \(A\): Clockwise moments = Counter-clockwise moments \[ (180 \times 1.20) + (320 \times 1.80) = (T \sin 35.0^\circ) \times 2.40 \] \[ 216 + 576 = 2.40 \times T \sin 35.0^\circ \] \[ 792 = 1.3765 T \implies T = \frac{792}{1.3765} = 575.39 \approx 575\text{ N} \text{ (or } 579\text{ N if using full precision: } 792 / (2.4 \sin 35^\circ) = 575.39\text{ N}) \] Let's recompute: \(2.4 \times \sin(35^\circ) = 2.4 \times 0.573576 = 1.37658\) \(T = 792 / 1.37658 = 575.34\text{ N} \approx 575\text{ N}\).
(b)(ii) Resolve forces horizontally and vertically on the beam: Let horizontal and vertical components of hinge force be \(R_x\) and \(R_y\). Horizontal equilibrium: \(\Sigma F_x = 0 \implies R_x - T \cos 35.0^\circ = 0\) \(R_x = 575.34 \times \cos 35.0^\circ = 575.34 \times 0.81915 = 471.3\text{ N}\) (acting away from the wall, to the right).
Angle \(\theta\) with the horizontal: \[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{170.0}{471.3}\right) = 19.8^\circ \text{ above the horizontal} \]
Marking scheme
(a) B1: Resultant force is zero / no net force. B1: Resultant moment (or torque) is zero / principle of moments holds.
(b)(i) C1: Moment of weight of beam \(= 180 \times 1.20 = 216\text{ N m}\) AND moment of load \(= 320 \times 1.80 = 576\text{ N m}\). C1: Sets total clockwise moment equal to \(T \sin(35^\circ) \times 2.40\). A1: \(T = 575\text{ N}\) (or \(580\text{ N}\)).
(b)(ii) C1: Resolves horizontally: \(R_x = T \cos(35^\circ) = 471\text{ N}\). C1: Resolves vertically: \(R_y = 500 - T \sin(35^\circ) = 170\text{ N}\). A1: Magnitude \(R = 501\text{ N}\) (accept \(500\text{ N}\) to \(505\text{ N}\)). A1: Direction \(\theta = 19.8^\circ\) (or \(20^\circ\)) above the horizontal (or to the horizontal).
Question 4 · Structured Calculation & Explanation
8.57 marks
(a) Distinguish between elastic deformation and plastic deformation.
(b) A uniform metal wire of original length \(2.20\text{ m}\) and uniform cross-sectional area \(4.50 \times 10^{-7}\text{ m}^2\) is suspended vertically from a rigid support. A mass of \(15.0\text{ kg}\) is attached to the lower end of the wire, causing it to extend elastically by \(1.60\text{ mm}\).
(i) Calculate the tensile stress in the wire.
(ii) Calculate the Young modulus \(E\) of the metal.
(iii) Calculate the elastic potential energy (strain energy) stored in the wire.
(c) A second wire made of the same metal has half the length and twice the diameter of the original wire. Calculate the extension produced in this second wire when supporting the same \(15.0\text{ kg}\) mass.
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Worked solution
(a) Elastic deformation: the object returns to its original length / shape when the deforming load is removed. Plastic deformation: the object does not return to its original length / shape and suffers permanent deformation when the load is removed.
(c) Extension is given by \(\Delta L = \frac{F L}{A E}\). For the second wire: \(L_2 = \frac{1}{2} L_1\) \(d_2 = 2 d_1 \implies A_2 = 4 A_1\) Force \(F\) and material \(E\) are the same. \[ \Delta L_2 = \frac{F (L_1 / 2)}{(4 A_1) E} = \frac{1}{8} \left(\frac{F L_1}{A_1 E}\right) = \frac{1}{8} \Delta L_1 = \frac{1.60 \text{ mm}}{8} = 0.200 \text{ mm} \]
Marking scheme
(a) B1: Elastic deformation: returns to original shape/length when load is removed. B1: Plastic deformation: permanent extension / does not return to original shape/length when load is removed.
(b)(iii) A1: \(E_p = \frac{1}{2} F x = \frac{1}{2} \times 147.15 \times 1.60 \times 10^{-3} = 0.118\text{ J}\).
(c) C1: Identifies that area increases by a factor of 4 (\(A \propto d^2\)). A1: \(\Delta L_2 = \frac{1.60}{8} = 0.200\text{ mm}\) (or \(2.00 \times 10^{-4}\text{ m}\)).
Question 5 · Structured Calculation & Explanation
8.57 marks
(a) State what is meant by the Doppler effect.
(b) An ambulance emits sound of frequency \(680\text{ Hz}\) from its siren. The ambulance travels at a constant speed of \(28.0\text{ m s}^{-1}\) along a straight road directly towards a stationary observer. The speed of sound in air is \(340\text{ m s}^{-1}\).
(i) Calculate the frequency of the sound heard by the stationary observer.
(ii) State and explain the change, if any, in the observed frequency after the ambulance passes the observer and recedes at the same speed.
(c) Monochromatic light of wavelength \(532\text{ nm}\) is incident normally on a diffraction grating. The second-order diffraction maximum is observed at an angle of \(38.6^\circ\) to the central maximum.
(i) Calculate the line spacing \(d\) of the diffraction grating.
(ii) Determine the total number of bright maxima that can be observed on a distant screen.
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Worked solution
(a) The Doppler effect is the observed change in frequency (or wavelength) of a wave when there is relative motion between the source and the observer.
(b)(ii) After passing, the source moves away from the observer. The observed frequency decreases to a value below \(680\text{ Hz}\) (specifically \(680 \times \frac{340}{368} = 628\text{ Hz}\)). This is because the sound waves are stretched out / the distance between successive wavefronts arriving at the observer increases, leading to a longer observed wavelength and hence a lower observed frequency.
(c)(ii) Maximum angle \(\theta = 90^\circ\), so \(\sin\theta \le 1\): \[ n_{\text{max}} = \frac{d}{\lambda} = \frac{1.7054 \times 10^{-6}}{532 \times 10^{-9}} = 3.205 \] Since \(n\) must be an integer, the highest order visible is \(n = 3\). Total number of maxima \(= 2 n_{\text{max}} + 1 = 2(3) + 1 = 7\) (orders \(n = 0, \pm 1, \pm 2, \pm 3\)).
Marking scheme
(a) B1: The observed change in frequency (or wavelength) due to relative motion between source and observer.
(b)(i) M1: Substitution into Doppler formula \(f_o = 680 \times \frac{340}{340 - 28}\). A1: \(f_o = 741\text{ Hz}\).
(b)(ii) B1: States that frequency decreases (to less than \(680\text{ Hz}\) / to \(628\text{ Hz}\)). B1: Wavefronts are spaced further apart / observed wavelength increases.
(c)(ii) C1: Calculates \(n_{\text{max}} = \frac{d}{\lambda} = 3.2\) so max order \(n = 3\). A1: Total number of maxima \(= 7\).
Question 6 · Structured Calculation & Explanation
8.57 marks
(a) State Kirchhoff's second law and the physical principle on which it is based.
(b) A battery of electromotive force (e.m.f.) \(E = 9.0\text{ V}\) and internal resistance \(r = 1.5\ \Omega\) is connected in series with a fixed resistor of resistance \(R_1 = 6.5\ \Omega\) and a uniform resistance wire \(XY\) of length \(1.00\text{ m}\) and resistance \(R_{XY} = 12.0\ \Omega\).
(i) Show that the potential difference across the wire \(XY\) is \(5.4\text{ V}\).
(ii) A cell of e.m.f. \(E_x = 2.16\text{ V}\) with internal resistance \(0.80\ \Omega\) is connected in series with a sensitive galvanometer and connected across a length \(L\) of the wire \(XY\) from end \(X\) using a sliding contact. Calculate the balance length \(L\) for which the galvanometer reads zero current.
(iii) Explain why the internal resistance of the cell \(E_x\) does not affect the balance length \(L\).
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Worked solution
(a) Kirchhoff's second law states that the sum of the electromotive forces (e.m.f.s) around any closed loop in a circuit is equal to the sum of the potential differences (p.d.s) across the components in that loop. It is based on the principle of conservation of energy.
(b)(i) Total resistance of the primary circuit: \[ R_{\text{total}} = r + R_1 + R_{XY} = 1.5 + 6.5 + 12.0 = 20.0\ \Omega \] Current in the circuit: \[ I = \frac{E}{R_{\text{total}}} = \frac{9.0}{20.0} = 0.45\text{ A} \] Potential difference across \(XY\): \[ V_{XY} = I R_{XY} = 0.45 \times 12.0 = 5.40\text{ V} \]
(b)(ii) Potential gradient along the wire: \[ k = \frac{V_{XY}}{L_{XY}} = \frac{5.40\text{ V}}{1.00\text{ m}} = 5.40\text{ V m}^{-1} \] At balance, p.d. across length \(L\) equals \(E_x\): \[ V_L = k L = E_x \] \[ 5.40 \times L = 2.16 \implies L = \frac{2.16}{5.40} = 0.400\text{ m} \text{ (or } 40.0\text{ cm)} \]
(b)(iii) At the balance point, the galvanometer registers zero current, meaning there is zero current flowing through the branch containing cell \(E_x\). Consequently, the potential drop across its internal resistance \(I r_x = 0\), and the terminal p.d. of the cell is equal to its e.m.f. \(E_x\).
Marking scheme
(a) B1: Sum of e.m.f.s = sum of p.d.s around any closed circuit/loop. B1: Conservation of energy.
(b)(i) C1: Calculation of total resistance \(= 20.0\ \Omega\) and current \(I = 0.45\text{ A}\) (or potential divider ratio \(\frac{12.0}{20.0} \times 9.0\)). A1: Correct multiplication leading to \(V_{XY} = 5.4\text{ V}\).
(b)(ii) C1: Correct formula for balance \(\frac{L}{1.00} \times 5.40 = 2.16\) or finds potential gradient \(5.40\text{ V m}^{-1}\). A1: \(L = 0.40\text{ m}\) (or \(40\text{ cm}\)).
(b)(iii) B1: Zero current flows through the galvanometer / cell at balance point. B1: No potential difference / lost volts across the internal resistance (so terminal p.d. equals e.m.f.).
Question 7 · Structured Calculation & Explanation
8.57 marks
(a) In an \(\alpha\)-particle scattering experiment, a beam of \(\alpha\)-particles is directed at a thin sheet of gold foil in an evacuated chamber. State two observations from this experiment and the conclusion deduced from each observation about the structure of the atom.
(b) A stationary nucleus of carbon-14 (\(^{14}_{\ 6}\text{C}\)) undergoes \(\beta^-\)-decay to form nitrogen-14 (\(^{14}_{\ 7}\text{N}\)), an electron (\(^0_{-1}\text{e}\)), and an antineutrino (\(\bar{\nu}_e\)).
(i) Write the complete nuclear equation for this decay.
(ii) State the change in quark flavour that occurs during this \(\beta^-\)-decay process.
(iii) Explain why the emitted \(\beta^-\)-particles have a continuous range of kinetic energies up to a maximum value, whereas \(\alpha\)-particles emitted in radioactive decay have discrete energies.
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Worked solution
(a) Observation 1: Most \(\alpha\)-particles pass straight through the gold foil with little or no deflection. Conclusion 1: Most of the atom is empty space. Observation 2: A small fraction of \(\alpha\)-particles are deflected through very large angles (greater than \(90^\circ\)). Conclusion 2: The atom contains a tiny, dense, positively charged region of mass concentrated at the center (the nucleus).
(b)(ii) A down quark converts into an up quark (\(d \rightarrow u\)).
(b)(iii) In \(\beta^-\)-decay, the total energy released (the decay energy) is shared between the emitted \(\beta^-\)-particle and the antineutrino (\(\bar{\nu}_e\)). Because the energy can be partitioned between the electron and the antineutrino in a continuous distribution of proportions, the electron has a continuous range of kinetic energies. In \(\alpha\)-decay, it is a two-body decay (nucleus and \(\alpha\)-particle), so conservation of momentum and energy uniquely determines the kinetic energy of the \(\alpha\)-particle, resulting in discrete energy values.
Marking scheme
(a) B1: Most pass through undeflected \(\rightarrow\) atom mostly empty space. B1: Small fraction deflected through large angles \(\rightarrow\) mass and positive charge concentrated in a tiny dense core (nucleus).
(b)(ii) B1: Down quark (\(d\)) changes to up quark (\(u\)).
(b)(iii) B1: Energy released is shared between the \(\beta\)-particle and the antineutrino. B1: The sharing occurs in variable/continuous proportions (so kinetic energy of \(\beta\) varies). B1: In \(\alpha\)-decay, there is only a two-body emission so energy is fixed by conservation laws.
Paper 3 (Advanced Practical Skills)
Complete two 1-hour laboratory practical experiments involving data acquisition, table formatting, and graphical analysis.
2 Question · 40 marks
Question 1 · Practical Experiment & Data Analysis
20 marks
In this experiment, you will investigate the oscillations of a loaded metre rule suspended horizontally by two vertical threads.
(a) (i) Set up the apparatus as shown in Fig. 1.1. Two vertical strings of equal length $L = 0.500\text{ m}$ are attached to the rule at distances $x$ from the centre of the rule ($50.0\text{ cm}$ mark). A mass $M = 200\text{ g}$ is taped securely to the centre of the rule.
(ii) Adjust the positions of the two suspension strings so that the separation $2x$ is approximately $30\text{ cm}$. Measure and record the distance $x$.
(b) Displace the rule through a small angle about a vertical axis passing through its centre and release it so that it executes torsional oscillations in a horizontal plane.
Determine the period $T$ of the oscillations by timing a suitable number of complete oscillations.
(c) Vary $x$ in the range $0.100\text{ m} \leqslant x \leqslant 0.450\text{ m}$ and repeat (b) to obtain values of $T$ for at least six different values of $x$.
Record all your values of $x$, raw timing data, $T$, and $\dfrac{1}{x^2}$ in a table. Include values of $\dfrac{1}{T^2}$ in your table.
(d) (i) Plot a graph of $\dfrac{1}{T^2}$ on the $y$-axis against $\dfrac{1}{x^2}$ on the $x$-axis.
(ii) Draw the straight line of best fit.
(iii) Determine the gradient and $y$-intercept of this line.
(c) Table of results with 6 sets of readings across the specified range: - Columns: $x / \text{m}$, $t_1 / \text{s}$, $t_2 / \text{s}$, $t_{\text{mean}} / \text{s}$, $T / \text{s}$, $1/x^2 / \text{m}^{-2}$, $1/T^2 / \text{s}^{-2}$. - Values covering $x = 0.100\text{ m}$ to $0.450\text{ m}$. - Consistent raw decimal places for $x$ and $t$, correct s.f. for calculated quantities ($1/x^2$ to 3 s.f., $1/T^2$ to 3 s.f.).
(d) (i) Graph axes scaled so points occupy $>50\%$ of grid in both directions, correctly labeled with quantity and unit. (ii) Line of best fit drawn with points balanced evenly above and below line. (iii) Gradient calculated using points on line with $\Delta x > 50\%$ of drawn length: $\text{gradient} = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{4.50 - 0.55}{95.0 - 9.0} = 0.0459\text{ m}^2\text{ s}^{-2}$. $y$-intercept read directly or calculated from $y = mx + c$: $c = 0.55 - (0.0459)(9.0) = 0.137\text{ s}^{-2}$.
Part (a) & (b) [3 marks] - Value of $x$ recorded with unit and to nearest millimetre. [1] - Timing of multiple oscillations recorded (at least 10 oscillations timed), repeated timing shown, $T$ calculated correctly. [1] - Quality of data: $T$ lies in the expected range $1.0\text{ s} \leqslant T \leqslant 4.0\text{ s}$. [1]
Part (c) Table of results [6 marks] - At least 6 sets of readings recorded covering a range of $x \geqslant 0.25\text{ m}$. [1] - Column headings: each column has a correct label and unit separated by a solidus or brackets (e.g. $1/x^2\ /\ \text{m}^{-2}$ or $1/T^2\ (\text{s}^{-2})$). [1] - Raw measurements of length and time recorded to consistent and appropriate precision (length to $1\text{ mm}$, time to $0.01\text{ s}$ or $0.1\text{ s}$). [1] - Significant figures: values of $1/x^2$ and $1/T^2$ given to the same number of (or one more than) the s.f. of the raw data. [1] - Calculated values of $1/x^2$ and $1/T^2$ are mathematically correct. [1] - Quality of data: points show a consistent trend with little scatter. [1]
Part (d) Graph [6 marks] - Axes: sensible scales chosen (no awkward ratios such as 3, 6, 7; data occupies $>50\%$ of the grid). [1] - Plotting: points plotted accurately to within half a small square. [1] - Line of best fit: straight line showing balanced distribution of points. [1] - Gradient: hypotenuse of the triangle used $\geqslant 50\%$ of the plotted line, read-offs within half a small square, calculation correct. [1] - $y$-intercept: correctly determined by direct reading from $x=0$ axis or substitution of point on line into $y = mx + c$. [1]
Part (e) Analysis [5 marks] - Value of $A$ equated to gradient and value of $B$ equated to $y$-intercept. [1] - Values of $A$ and $B$ given to 2 or 3 s.f. with correct signs. [1] - Correct unit for $A$: $\text{s}^{-2}\text{ m}^2$ or $\text{m}^2\text{ s}^{-2}$. [1] - Correct unit for $B$: $\text{s}^{-2}$. [1] - Working clearly shown and methodologically valid. [1]
Question 2 · Practical Experiment & Data Analysis
20 marks
In this experiment, you will investigate the equilibrium of a weighted wooden rod floating upright in a cylinder of liquid.
(a) (i) You are provided with a uniform circular rod of diameter $d$ and total length $L$. Measure and record $d$ using the micrometer screw gauge.
(ii) Attach a mass $m_1 = 20\text{ g}$ to the bottom end of the rod. Place the rod vertically into the cylinder containing water and allow it to float freely.
Measure and record the depth of immersion $h_1$ of the rod beneath the water surface.
(c) Remove the rod, dry it, and replace the mass with $m_2 = 50\text{ g}$. Place the rod back into the cylinder and measure the new depth of immersion $h_2$.
(iii) Explain whether your results support the suggested relationship. Justify your answer by comparing the percentage difference between your $k$ values with an appropriate experimental uncertainty limit.
(e) (i) Describe four sources of error or limitations of the procedure for this experiment.
(ii) Describe four improvements that could be made to this experiment to address the limitations identified in (e)(i). You may suggest the use of other apparatus.
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Worked solution
(a) (i) Measurement of $d$ with micrometer: $d = 12.42\text{ mm}$ (repeated at multiple orientations, mean taken). (ii) Absolute uncertainty in $d = \pm 0.02\text{ mm}$ (or half range of repeats). $\text{Percentage uncertainty} = \dfrac{0.02}{12.42} \times 100\% \approx 0.16\%$.
(b) (i) $L = 30.0\text{ cm}$ to nearest mm ($0.1\text{ cm}$). (ii) Submerged depth for $m_1 = 20\text{ g}$: $h_1 = 12.1\text{ cm}$.
(d) (i) For $m_1 = 20\text{ g}$ and $m_0 = 35\text{ g}$: $k_1 = \dfrac{h_1}{m_1 + m_0} = \dfrac{12.1}{20 + 35} = \dfrac{12.1}{55} = 0.220\text{ cm}\text{ g}^{-1}$. (ii) For $m_2 = 50\text{ g}$ and $m_0 = 35\text{ g}$: $k_2 = \dfrac{h_2}{m_2 + m_0} = \dfrac{18.7}{50 + 35} = \dfrac{18.7}{85} = 0.220\text{ cm}\text{ g}^{-1}$. (iii) $\text{Percentage difference} = \dfrac{|k_1 - k_2|}{\text{mean } k} \times 100\% = \dfrac{|0.220 - 0.220|}{0.220} \times 100\% = 0.0\%$. Since the percentage difference is less than the percentage uncertainty in measuring the submerged depth (estimated at $\pm 2\text{ mm}$ in $12\text{ cm} \approx 5\% - 10\%$), the suggested relationship is supported.
(e) Limitations and Improvements: 1. Limitation: Meniscus of water makes it difficult to locate the exact water level on the rod. Improvement: Add a drop of surfactant/detergent to reduce surface tension and flatten meniscus / use viewing aid. 2. Limitation: Rod tilts / does not float completely vertically, causing error in measured vertical depth. Improvement: Use a loose guide ring / collar around the rod to keep it vertical without adding friction. 3. Limitation: Water absorbs into the wooden rod, changing its mass $m_0$ over time. Improvement: Coat rod in varnish / waterproof sealant or use non-porous material (e.g. acrylic/plastic rod). 4. Limitation: Two sets of readings are insufficient to confirm the linear relationship definitively. Improvement: Take measurements with at least 6 different masses and plot a graph of $h$ against $(m + m_0)$.
Marking scheme
Part (a) [2 marks] - Raw value of $d$ recorded to $0.01\text{ mm}$ with unit, with evidence of repeated readings. [1] - Percentage uncertainty in $d$ correctly calculated using absolute uncertainty of $0.01\text{ mm}$ to $0.05\text{ mm}$ (or half range). [1]
Part (b) & (c) [3 marks] - Value of $L$ recorded to nearest millimetre with unit. [1] - First value of $h_1$ recorded to nearest millimetre with unit. [1] - Second value of $h_2$ recorded to nearest millimetre, with $h_2 > h_1$. [1]
Part (d) [5 marks] - Correct calculation of $k_1$ with appropriate significant figures (2 or 3 s.f.). [1] - Correct calculation of $k_2$ with appropriate significant figures. [1] - Calculation of percentage difference between $k_1$ and $k_2$. [1] - Sensible estimate of experimental uncertainty percentage stated (e.g. 5% to 15%). [1] - Clear conclusion comparing the percentage difference to the specified percentage criterion, stating whether the relationship is supported. [1]
Part (e) Limitations and Improvements [10 marks: 4 for limitations, 4 for improvements, 2 for detail/matching] Award 1 mark for each valid limitation (max 4) and 1 mark for each corresponding improvement (max 4): - L1: Difficulty in observing the exact water surface on the rod due to meniscus / parallax error. I1: Add a surfactant/detergent to flatten the meniscus / use a set square aligned with water level. - L2: Rod does not stay strictly vertical / touches the sides of the measuring cylinder creating friction. I2: Use a wider cylinder or a guide loop / frictionless collar to maintain vertical orientation. - L3: Wood absorbs water during the experiment, altering $m_0$. I3: Varnish / waterproof the rod or use a non-porous solid (e.g. perspex/plastic cylinder). - L4: Two values of mass are not enough to confirm a valid relationship. I4: Take at least 6 different values of $m$ and plot a graph of $h$ against $(m + m_0)$.
Paper 4 (A Level Structured)
Answer all questions spanning advanced mechanics, thermal, fields, electromagnetism, and modern physics.
10 Question · 100 marks
Question 1 · structured
10 marks
(a) Define gravitational potential at a point in a gravitational field. [2]
(b) A spherical planet of mass $M$ and radius $R = 3.40 \times 10^6\text{ m}$ has no atmosphere. A space probe of mass $m = 480\text{ kg}$ is launched vertically upwards from the surface of the planet with an initial kinetic energy of $2.09 \times 10^9\text{ J}$. The probe reaches a maximum height $h = 1.80 \times 10^6\text{ m}$ above the surface before momentarily coming to rest.
(i) By considering the change in gravitational potential energy, show that the mass $M$ of the planet is approximately $6.4 \times 10^{23}\text{ kg}$. [4]
(ii) Calculate the escape speed of the probe from the surface of the planet. [2]
(iii) A second probe of mass $2m$ is launched vertically upwards from the surface with the same initial kinetic energy of $2.09 \times 10^9\text{ J}$. State and explain how the maximum height reached compares with $h$. [2]
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Worked solution
(a) Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point.
(b)(i) At maximum height, all initial kinetic energy has been converted into gravitational potential energy: $$\Delta E_p = E_k = 2.09 \times 10^9\text{ J}$$ The initial distance from the centre of the planet is $r_1 = R = 3.40 \times 10^6\text{ m}$. The final distance from the centre of the planet is $r_2 = R + h = 3.40 \times 10^6 + 1.80 \times 10^6 = 5.20 \times 10^6\text{ m}$.
(ii) Escape speed $v_{\text{esc}}$ satisfies $\frac{1}{2} m v_{\text{esc}}^2 = \frac{G M m}{R}$: $$v_{\text{esc}} = \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (6.41 \times 10^{23})}{3.40 \times 10^6}}$$ $$v_{\text{esc}} = \sqrt{2.515 \times 10^7} = 5.01 \times 10^3\text{ m s}^{-1}$$
(iii) The gain in gravitational potential energy is equal to the initial kinetic energy ($2.09 \times 10^9\text{ J}$). Since $\Delta E_p = m \Delta \phi$, doubling the mass means the change in gravitational potential $\Delta \phi$ is halved. Because gravitational potential increases with distance from the planet, a smaller $\Delta \phi$ corresponds to a smaller maximum height reached.
Marking scheme
(a) - Work done per unit mass [B1] - in bringing a small test mass from infinity (to the point) [B1]
(b)(i) - Recognition that $\Delta E_p = G M m (1/r_1 - 1/r_2)$ or $\Delta \phi = G M (1/r_1 - 1/r_2)$ [C1] - Correct substitution of radii $r_1 = 3.40 \times 10^6\text{ m}$ and $r_2 = 5.20 \times 10^6\text{ m}$ [C1] - Complete substitution of values: $2.09 \times 10^9 = 6.67 \times 10^{-11} \times M \times 480 \times (1/(3.40 \times 10^6) - 1/(5.20 \times 10^6))$ [M1] - Evaluation leading to $M = 6.41 \times 10^{23}\text{ kg}$ (at least 2 s.f. shown before rounding) [A1]
(b)(ii) - Formula $v_{\text{esc}} = \sqrt{2GM/R}$ or equating kinetic energy to $\frac{GMm}{R}$ [C1] - $v_{\text{esc}} = 5.01 \times 10^3\text{ m s}^{-1}$ (accept $5.0 \times 10^3\text{ m s}^{-1}$) [A1]
(b)(iii) - Maximum height is less / smaller [B1] - Explanation: $\Delta E_p = m \Delta \phi$ so $\Delta \phi$ is halved (or gain in PE per unit mass is less) [B1]
Question 2 · structured
10 marks
(a) State two basic assumptions of the kinetic theory of ideal gases. [2]
(b) An ideal monatomic gas consists of $0.450\text{ mol}$ of particles enclosed in a cylinder fitted with a movable piston. Initially, the gas is at pressure $p_1 = 1.20 \times 10^5\text{ Pa}$ and temperature $T_1 = 300\text{ K}$.
(i) Calculate the initial volume $V_1$ occupied by the gas. [2]
(ii) The gas is now heated at constant pressure until its temperature increases to $T_2 = 450\text{ K}$.
1. Calculate the work done $W$ by the gas during this expansion. [2]
2. The mean translational kinetic energy $E_k$ of an atom in an ideal gas is given by $E_k = \frac{3}{2} k T$. Calculate the change in internal energy $\Delta U$ of the gas during this heating process. [2]
3. Determine the quantity of thermal energy $q$ transferred to the gas during this process. [2]
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Worked solution
(a) Any two standard assumptions: 1. The volume of the gas particles/molecules is negligible compared to the total volume occupied by the gas. 2. Collisions between molecules and between molecules and the container walls are perfectly elastic. (Alternatively: Intermolecular forces between particles are negligible except during collisions; time of collision is negligible compared to time between collisions; large number of molecules in continuous random motion).
(b)(i) Using the ideal gas equation $p_1 V_1 = n R T_1$: $$V_1 = \frac{n R T_1}{p_1} = \frac{0.450 \times 8.31 \times 300}{1.20 \times 10^5} = 9.349 \times 10^{-3}\text{ m}^3 = 9.35 \times 10^{-3}\text{ m}^3$$
(b)(ii) 1. At constant pressure, $W = p \Delta V = n R \Delta T$: $$\Delta T = 450 - 300 = 150\text{ K}$$ $$W = n R \Delta T = 0.450 \times 8.31 \times 150 = 560.9\text{ J} \approx 561\text{ J}$$
2. For a monatomic ideal gas, internal energy is entirely translational kinetic energy: $$U = N \times \left(\frac{3}{2} k T\right) = \frac{3}{2} n R T$$ $$\Delta U = \frac{3}{2} n R \Delta T = 1.5 \times (0.450 \times 8.31 \times 150) = 841.4\text{ J} \approx 841\text{ J}$$
3. Applying the first law of thermodynamics, $\Delta U = q + W_{\text{on}}$ where $W_{\text{on}} = -W_{\text{by}}$: $$q = \Delta U + W_{\text{by}} = 841.4 + 560.9 = 1402.3\text{ J} \approx 1.40 \times 10^3\text{ J}$$ (or $1400\text{ J}$).
Marking scheme
(a) - Any two correct assumptions from: negligible volume of molecules compared to gas volume / perfectly elastic collisions / no intermolecular forces (except during collision) / time of collision negligible / particles in continuous random motion [B2] (1 mark each)
(b)(i) - $p V = n R T$ used with correct substitutions [C1] - $V_1 = 9.35 \times 10^{-3}\text{ m}^3$ (or $9.34 \times 10^{-3}\text{ m}^3$) [A1]
(b)(ii) 1. - $W = p \Delta V$ or $W = n R \Delta T$ [C1] - $W = 561\text{ J}$ (accept $560\text{ J}$) [A1]
(b)(ii) 2. - $\Delta U = \frac{3}{2} n R \Delta T$ or $\Delta U = \frac{3}{2} N k \Delta T$ [C1] - $\Delta U = 841\text{ J}$ (accept $840\text{ J}$) [A1]
(b)(ii) 3. - First law of thermodynamics $q = \Delta U + W$ [C1] - $q = 1400\text{ J}$ or $1.40 \times 10^3\text{ J}$ (allow ECF from (b)(ii) 1 and 2) [A1]
Question 3 · structured
10 marks
(a) State the two defining conditions for a body to undergo simple harmonic motion. [2]
(b) A block of mass $m = 0.350\text{ kg}$ is attached to a light horizontal spring of spring constant $k$ and rests on a frictionless horizontal surface. The other end of the spring is fixed. The block is pulled aside and released so that it executes simple harmonic motion with a frequency of $2.50\text{ Hz}$ and an amplitude of $4.20\text{ cm}$.
(i) Calculate the spring constant $k$ of the spring. [3]
(ii) Calculate the speed of the block when its displacement from equilibrium is $2.50\text{ cm}$. [2]
(c) The block is now placed in a viscous medium which provides light damping.
(i) State what is meant by damping. [1]
(ii) An external periodic driving force of variable frequency $f_{\text{D}}$ is applied to the block. State the effect of increased damping on: 1. the resonant frequency, 2. the amplitude of oscillation at resonance. [2]
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Worked solution
(a) The two conditions are: 1. Acceleration is directly proportional to displacement from the equilibrium position. 2. Acceleration is always directed towards the equilibrium position (opposite in direction to displacement).
(b)(i) Angular frequency $\omega = 2 \pi f = 2 \pi \times 2.50 = 15.708\text{ rad s}^{-1}$. For a mass-spring system, $\omega^2 = \frac{k}{m}$, hence: $$k = m \omega^2 = 0.350 \times (15.708)^2 = 0.350 \times 246.74 = 86.36\text{ N m}^{-1} \approx 86.4\text{ N m}^{-1}$$
(b)(ii) The velocity at displacement $x$ is: $$v = \omega \sqrt{x_0^2 - x^2}$$ $$v = 15.708 \times \sqrt{(0.0420)^2 - (0.0250)^2} = 15.708 \times \sqrt{0.001764 - 0.000625}$$ $$v = 15.708 \times \sqrt{0.001139} = 15.708 \times 0.03375 = 0.530\text{ m s}^{-1}$$
(c)(i) Damping is the continuous removal/loss of mechanical energy from an oscillating system as a result of resistive/frictional forces.
(c)(ii) 1. The resonant frequency decreases (shifts to a lower frequency). 2. The amplitude of oscillation at resonance decreases (the resonance peak becomes broader and lower).
Marking scheme
(a) - Acceleration is directly proportional to displacement [B1] - Acceleration is directed towards a fixed point / opposite to displacement [B1]
(b)(i) - $\omega = 2 \pi f = 15.7\text{ rad s}^{-1}$ [C1] - $k = m \omega^2$ or $T = 2\pi \sqrt{m/k}$ [C1] - $k = 86.4\text{ N m}^{-1}$ [A1]
(b)(ii) - $v = \omega \sqrt{x_0^2 - x^2}$ with correct substitutions [C1] - $v = 0.530\text{ m s}^{-1}$ (or $0.53\text{ m s}^{-1}$) [A1]
(c)(i) - Loss of mechanical/total energy from the system due to resistive/opposing forces [B1]
(c)(ii) - 1. Resonant frequency decreases / shifts lower [B1] - 2. Peak amplitude at resonance decreases / lowers [B1]
Question 4 · structured
10 marks
(a) Define the capacitance of a capacitor. [1]
(b) Two capacitors of capacitances $C_1 = 47\ \mu\text{F}$ and $C_2 = 22\ \mu\text{F}$ are connected in parallel across a $12.0\text{ V}$ d.c. power supply.
(i) Calculate the total charge stored in the capacitor combination. [2]
(ii) Calculate the total electrical energy stored in the combination. [2]
(c) The fully charged parallel combination is disconnected from the supply at time $t = 0$ and allowed to discharge through a resistor of resistance $R = 150\text{ k}\Omega$.
(i) Calculate the time constant $\tau$ of the discharge circuit. [2]
(ii) Calculate the time $t$ taken for the potential difference across the parallel combination to fall from $12.0\text{ V}$ to $3.00\text{ V}$. [3]
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Worked solution
(a) Capacitance is the charge stored on one plate of a capacitor per unit potential difference across the plates ($C = Q/V$).
(b)(i) For capacitors in parallel, the total capacitance is: $$C_{\text{total}} = C_1 + C_2 = 47\ \mu\text{F} + 22\ \mu\text{F} = 69\ \mu\text{F} = 6.90 \times 10^{-5}\text{ F}$$ Total charge stored: $$Q_{\text{total}} = C_{\text{total}} V = 6.90 \times 10^{-5} \times 12.0 = 8.28 \times 10^{-4}\text{ C}$$
(ii) Total energy stored: $$E_{\text{total}} = \frac{1}{2} C_{\text{total}} V^2 = \frac{1}{2} \times 6.90 \times 10^{-5} \times (12.0)^2 = 4.968 \times 10^{-3}\text{ J} \approx 4.97 \times 10^{-3}\text{ J}$$
(b) In a velocity selector, an electric field of strength $E = 4.80 \times 10^4\text{ V m}^{-1}$ and a uniform magnetic field of flux density $B_1 = 0.160\text{ T}$ act at right angles to each other and to the path of a beam of singly-charged positive ions ($q = +e$).
(i) Explain why only ions of a specific speed $v_0$ emerge undeflected from the velocity selector. [2]
(ii) Calculate $v_0$. [2]
(c) The undeflected ions enter a region of uniform magnetic field $B_2 = 0.250\text{ T}$ directed perpendicular to their velocity. The ions travel in a semicircular path of diameter $d = 0.180\text{ m}$ before striking a detector.
(i) Show that the mass $m$ of one ion is $1.20 \times 10^{-26}\text{ kg}$. [3]
(ii) A different isotope with ions of the same mass but charge $+2e$ enters the region of magnetic field $B_2$ at the same speed $v_0$. State the new diameter of the semicircular path. [1]
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Worked solution
(a) Magnetic flux density is defined as the force per unit length on a long straight conductor carrying unit current placed at right angles to the magnetic field ($B = F / (I L)$).
(b)(i) The electric force on an ion is $F_E = q E$ and the magnetic force is $F_B = q v B_1$. The fields are oriented so that these forces act in opposite directions. For the path to remain straight (no deflection), the forces must be equal in magnitude: $q E = q v B_1 \implies v = \frac{E}{B_1}$. Any ion with speed different from $v_0$ will experience a resultant force and be deflected.
(ii) $$v_0 = \frac{E}{B_1} = \frac{4.80 \times 10^4}{0.160} = 3.00 \times 10^5\text{ m s}^{-1}$$
(c)(i) In the second magnetic field $B_2$, the magnetic force provides the centripetal force: $$q v_0 B_2 = \frac{m v_0^2}{r}$$ $$m = \frac{q B_2 r}{v_0}$$ The radius of the semicircle is $r = \frac{d}{2} = \frac{0.180}{2} = 0.0900\text{ m}$. Substituting the values: $$m = \frac{(1.60 \times 10^{-19}) \times 0.250 \times 0.0900}{3.00 \times 10^5} = \frac{3.60 \times 10^{-21}}{3.00 \times 10^5} = 1.20 \times 10^{-26}\text{ kg}$$
(ii) Since $d = 2r = \frac{2 m v_0}{q B_2}$, doubling the charge $q$ while keeping $m, v_0, B_2$ unchanged halves the diameter: $$d_{\text{new}} = \frac{0.180}{2} = 0.0900\text{ m} = 9.00\text{ cm}$$
Marking scheme
(a) - Force per unit length [B1] - per unit current for a wire perpendicular to the field (or $F = B I L \sin\theta$ with terms defined) [B1]
(b)(i) - Electric force $F_E = qE$ and magnetic force $F_B = qvB$ act in opposite directions [B1] - Forces balance ($qE = qvB$) only when $v = E/B$ [B1]
(a) State Faraday's law of electromagnetic induction. [2]
(b) A flat circular search coil of $N = 350$ turns and radius $r = 4.00\text{ cm}$ is placed inside a long solenoid. The plane of the coil is perpendicular to the longitudinal axis of the solenoid. The alternating current in the solenoid creates a uniform magnetic flux density $B$ along its axis given by: $$B = B_0 \sin(\omega t)$$ where $B_0 = 4.50 \times 10^{-2}\text{ T}$ and $\omega = 314\text{ rad s}^{-1}$.
(i) Calculate the maximum magnetic flux linkage through the search coil. [3]
(ii) Determine the maximum electromotive force (e.m.f.) induced in the search coil. [3]
(c) Explain how Lenz's law is an application of the principle of conservation of energy. [2]
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Worked solution
(a) Faraday's law states that the magnitude of the induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage.
(b)(i) The cross-sectional area of the search coil is: $$A = \pi r^2 = \pi \times (0.0400)^2 = 5.0265 \times 10^{-3}\text{ m}^2$$ The maximum magnetic flux through one turn is: $$\Phi_0 = B_0 A = (4.50 \times 10^{-2}) \times (5.0265 \times 10^{-3}) = 2.262 \times 10^{-4}\text{ Wb}$$ Maximum flux linkage: $$\Phi N = N B_0 A = 350 \times 2.262 \times 10^{-4} = 7.917 \times 10^{-2}\text{ Wb} \approx 7.92 \times 10^{-2}\text{ Wb}$$
(ii) The magnetic flux linkage at time $t$ is $\lambda(t) = N B_0 A \sin(\omega t)$. By Faraday's law, the magnitude of induced e.m.f. is: $$E = \left| \frac{\mathrm{d}(\lambda)}{\mathrm{d}t} \right| = \omega N B_0 A \cos(\omega t)$$ The maximum induced e.m.f. is therefore: $$E_{\text{max}} = \omega (N B_0 A) = 314 \times 7.917 \times 10^{-2} = 24.86\text{ V} \approx 24.9\text{ V}$$
(c) Lenz's law states that the direction of the induced current is such that it opposes the change producing it. Consequently, an opposing force is produced against the motion causing the change in flux. Mechanical work must be done against this opposing force, and this work done is transformed into electrical energy in the circuit. If it did not oppose the change, energy would be created from nothing, violating the conservation of energy.
Marking scheme
(a) - Induced e.m.f. is directly proportional to [B1] - the rate of change of magnetic flux linkage [B1]
(b)(ii) - Recognition that $E_{\text{max}} = \omega \times (\text{flux linkage}_{\text{max}})$ or $E = \frac{\mathrm{d}}{\mathrm{d}t}(N B A)$ [C1] - Substitution: $E_{\text{max}} = 314 \times 7.917 \times 10^{-2}$ [M1] - $E_{\text{max}} = 24.9\text{ V}$ (allow ECF from (b)(i)) [A1]
(c) - Induced current produces an opposing force/field against the motion/change [B1] - Work must be done against this force to produce electrical energy (preventing creation of energy without work) [B1]
Question 7 · structured
10 marks
(a) Define the radioactive decay constant $\lambda$. [1]
(b) A sample of phosphorus-32 ($^{32}_{15}\text{P}$) is a pure $\beta^-$ emitter with a half-life of $14.3\text{ days}$. The initial activity of the sample is $3.60 \times 10^6\text{ Bq}$.
(i) Calculate the decay constant $\lambda$ of phosphorus-32 in $\text{s}^{-1}$. [2]
(ii) Calculate the initial number $N_0$ of phosphorus-32 nuclei present in the sample. [2]
(iii) Determine the activity of the sample after a time of $40.0\text{ days}$. [2]
(c) In a fusion process, two deuterium nuclei ($^2_1\text{H}$) fuse to form a helium-3 nucleus ($^3_2\text{He}$) and a neutron ($^1_0\text{n}$): $$^2_1\text{H} + ^2_1\text{H} \rightarrow ^3_2\text{He} + ^1_0\text{n}$$ Data for the masses of the particles are given below: - mass of $^2_1\text{H} = 2.01410\text{ u}$ - mass of $^3_2\text{He} = 3.01603\text{ u}$ - mass of $^1_0\text{n} = 1.00866\text{ u}$
Calculate the energy released, in $\text{MeV}$, in this single fusion reaction. [3]
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Worked solution
(a) The decay constant is the probability per unit time of the decay of a nucleus (or the fraction of undecayed nuclei decaying per unit time).
(c) - $\Delta m = 2(2.01410) - (3.01603 + 1.00866) = 0.00351\text{ u}$ [C1] - Conversion to energy via $\Delta E = \Delta m c^2$ or $\times 931.5\text{ MeV u}^{-1}$ [M1] - Energy released $= 3.27\text{ MeV}$ (accept $3.28\text{ MeV}$ or $5.24 \times 10^{-13}\text{ J}$) [A1]
Question 8 · structured
10 marks
(a) (i) Describe the principles of the generation of ultrasound using a piezoelectric crystal. [2]
(ii) In medical ultrasound scanning, a coupling medium (gel) is applied between the transducer and the patient's skin. Explain the necessity for this coupling medium. [2]
(b) A parallel beam of ultrasound is incident normally on a boundary between muscle and bone. - Specific acoustic impedance of muscle $Z_1 = 1.70 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}$ - Specific acoustic impedance of bone $Z_2 = 6.40 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}$
Calculate the ratio of the reflected intensity $I_R$ to the incident intensity $I_0$ at the boundary. [2]
(c) In positron emission tomography (PET), a positron emitted by a radiotracer annihilates with an electron in surrounding tissue, producing two identical gamma-ray photons travelling in opposite directions.
(i) Explain why two photons travelling in opposite directions must be emitted rather than a single photon. [2]
(ii) Calculate the frequency of each gamma-ray photon, assuming the initial kinetic energy of the electron and positron prior to annihilation is negligible. [2]
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Worked solution
(a)(i) An alternating potential difference (voltage) is applied across opposite faces of a piezoelectric crystal (e.g. quartz). The crystal contracts and expands alternately at the frequency of the applied p.d. When this frequency matches the natural frequency of the crystal, resonance occurs, producing high-frequency ultrasound waves.
(ii) Air trapped between the transducer and the skin has a specific acoustic impedance $Z$ that is very different from that of skin/soft tissue. This large difference causes almost total reflection of ultrasound at the air-skin boundary. The acoustic impedance of the gel matches that of skin, eliminating air pockets and allowing maximum transmission of ultrasound into the body.
(c)(i) Linear momentum must be conserved. Prior to annihilation, the electron-positron system is approximately at rest, so the total initial momentum is approximately zero. A single photon carries momentum ($p = E/c$), which would violate momentum conservation. Two photons of equal energy emitted in opposite directions have equal and opposite momenta, so their vector sum is zero.
(ii) Each photon receives the rest mass energy of one particle ($m_e = 9.11 \times 10^{-31}\text{ kg}$): $$E = m_e c^2 = (9.11 \times 10^{-31}) \times (3.00 \times 10^8)^2 = 8.199 \times 10^{-14}\text{ J}$$ The frequency $f$ of each photon is: $$f = \frac{E}{h} = \frac{8.199 \times 10^{-14}}{6.63 \times 10^{-34}} = 1.2366 \times 10^{20}\text{ Hz} \approx 1.24 \times 10^{20}\text{ Hz}$$
Marking scheme
(a)(i) - Alternating potential difference applied across the crystal faces [B1] - Causes crystal to vibrate at its resonant/driving frequency, emitting ultrasound [B1]
(a)(ii) - Air and skin have very different acoustic impedances leading to large reflection [B1] - Gel has acoustic impedance matching skin, so transmission is maximised / reflection minimised [B1]
(c)(i) - Initial momentum of the system is zero [B1] - Two photons emitted in opposite directions have equal and opposite momenta so total momentum remains zero (momentum conservation) [B1]
(a) Define gravitational potential at a point in a gravitational field. [2]
(b) A spherical moon of mass \(M\) and radius \(R\) has no atmosphere. A space probe of mass \(m\) is launched vertically upwards from the surface of the moon. (i) Show that the escape speed \(v_{\text{esc}}\) from the surface of the moon is given by the expression \[ v_{\text{esc}} = \sqrt{\frac{2GM}{R}} \] where \(G\) is the gravitational constant. [2] (ii) The mass of the moon is \(7.35 \times 10^{22} \text{ kg}\) and its radius is \(1.74 \times 10^{6} \text{ m}\). Calculate \(v_{\text{esc}}\). [2]
(c) The probe is launched vertically from the moon's surface with an initial speed of \(1.20 \times 10^3 \text{ m s}^{-1}\). (i) State, with a reason, whether the probe escapes the moon's gravitational field. [1] (ii) Calculate the maximum height \(h\) above the surface of the moon reached by the probe. [3]
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Worked solution
(a) Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to the point.
(b)(i) For an object of mass \(m\) to escape from the surface of a spherical mass \(M\) of radius \(R\), its total energy at infinity must be at least zero. By conservation of mechanical energy: \[ \frac{1}{2}m v_{\text{esc}}^2 - \frac{GMm}{R} = 0 \] \[ \frac{1}{2} v_{\text{esc}}^2 = \frac{GM}{R} \] \[ v_{\text{esc}} = \sqrt{\frac{2GM}{R}} \]
(c)(i) The launch speed \(v_0 = 1.20 \times 10^3 \text{ m s}^{-1}\) is less than the escape speed \(v_{\text{esc}} = 2.37 \times 10^3 \text{ m s}^{-1}\), so the probe has insufficient kinetic energy to reach infinity and does not escape.
(a) - Work done per unit mass [M1] - in bringing a (small/test) mass from infinity (to the point) [A1]
(b)(i) - Equating initial kinetic energy to the change in gravitational potential energy / stating total mechanical energy \(= 0\) at infinity: \(\frac{1}{2}mv^2 = \frac{GMm}{R}\) [M1] - Correct rearrangement leading clearly to \(v_{\text{esc}} = \sqrt{\frac{2GM}{R}}\) [A1]
(c)(i) - Probe does not escape AND valid reason comparing \(v_0\) with \(v_{\text{esc}}\) (e.g. \(1.20 \times 10^3 < 2.37 \times 10^3\)) [B1]
(c)(ii) - Statement or formulation of conservation of energy: \(\frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{r_{\text{max}}}\) (or \(\Delta E_k = \Delta E_p\)) [C1] - Correct calculation of \(r_{\text{max}} = 2.34 \times 10^6 \text{ m}\) [C1] - \(h = 5.97 \times 10^5 \text{ m}\) (accept range \(5.95 \times 10^5\) to \(6.00 \times 10^5 \text{ m}\)) [A1]
Question 10 · structured
10 marks
(a) State what is meant by the work function energy of a metal. [2]
(b) In a photoelectric experiment, monochromatic light of wavelength \(\lambda = 3.80 \times 10^{-7} \text{ m}\) is incident on a clean potassium surface inside an evacuated tube. The work function energy \(\Phi\) of potassium is \(2.26 \text{ eV}\).
(i) Calculate the energy of an incident photon in electronvolts (\(\text{eV}\)). [3]
(ii) Determine the maximum kinetic energy \(E_{\text{k,max}}\), in \(\text{J}\), of the photoelectrons emitted from the potassium surface. [2]
(iii) Calculate the de Broglie wavelength \(\lambda_{\text{dB}}\) of an electron emitted with this maximum kinetic energy. [3]
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Worked solution
(a) The work function energy is the minimum amount of energy required to remove an electron from the surface of a metal.
(b)(i) Photon energy is given by: \[ E = \frac{hc}{\lambda} \] \[ E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.80 \times 10^{-7}} = 5.234 \times 10^{-19} \text{ J} \] Converting to electronvolts: \[ E = \frac{5.234 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.271 \text{ eV} \approx 3.27 \text{ eV} \]
(b)(iii) Linear momentum \(p\) of the electron is related to kinetic energy by: \[ p = \sqrt{2 m_e E_{\text{k,max}}} \] \[ p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.618 \times 10^{-19}} = \sqrt{2.948 \times 10^{-49}} = 5.430 \times 10^{-25} \text{ kg m s}^{-1} \] The de Broglie wavelength is: \[ \lambda_{\text{dB}} = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.430 \times 10^{-25}} = 1.22 \times 10^{-9} \text{ m} \]
Marking scheme
(a) - Minimum energy [B1] - required to remove/emit an electron from the surface of a metal [B1]
(b)(i) - Formula used: \(E = \frac{hc}{\lambda}\) [C1] - Energy in Joules: \(E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.80 \times 10^{-7}} = 5.234 \times 10^{-19} \text{ J}\) [C1] - Division by \(1.60 \times 10^{-19}\) to obtain \(E = 3.27 \text{ eV}\) (allow 3.27 or 3.271 eV) [A1]
(b)(iii) - Using \(p = \sqrt{2 m_e E_k}\) or \(v = \sqrt{\frac{2E_k}{m_e}}\) with \(m_e = 9.11 \times 10^{-31} \text{ kg}\) [C1] - Calculating momentum \(p = 5.43 \times 10^{-25} \text{ kg m s}^{-1}\) (or \(v = 5.96 \times 10^5 \text{ m s}^{-1}\)) [C1] - Using \(\lambda = \frac{h}{p}\) to obtain \(\lambda_{\text{dB}} = 1.22 \times 10^{-9} \text{ m}\) (accept \(1.21 \times 10^{-9}\) to \(1.23 \times 10^{-9} \text{ m}\)) [A1]
Paper 5 (Planning, Analysis and Evaluation)
Answer Question 1 (Planning experiment) and Question 2 (Data analysis, uncertainties, and worst-fit line graph).
0 Question · 0 marks
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