An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V1) Cambridge International A Level Physics (9702) paper. Not affiliated with or reproduced from Cambridge.
Paper 1 (Multiple Choice)
Answer all 40 questions. Choose the single best answer A, B, C, or D.
40 Question · 40 marks
Question 1 · Multiple Choice
1 marks
What are the SI base units of electrical resistivity?
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Worked solution
Resistivity is defined by \( \rho = \frac{R A}{L} \). Resistance \( R = \frac{V}{I} = \frac{W}{I Q} = \frac{W}{I^2 t} \). In SI base units: Work \( W \) has units \( \text{kg m}^2 \text{s}^{-2} \). Current \( I \) has unit \( \text{A} \) and time \( t \) has unit \( \text{s} \). Therefore, \( R \) has units \( \frac{\text{kg m}^2 \text{s}^{-2}}{\text{A}^2 \text{s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2} \). Multiplying by \( \frac{A}{L} \) (units \( \frac{\text{m}^2}{\text{m}} = \text{m} \)) gives: \( \text{units of } \rho = \text{kg m}^3 \text{s}^{-3} \text{A}^{-2} \).
Marking scheme
A is correct [1]. Base units correctly derived from definition \( \rho = \frac{R A}{L} \) and electrical definitions.
Question 2 · Multiple Choice
1 marks
A small metal sphere is projected horizontally from the top of a vertical cliff of height \( 45.0\text{ m} \) with an initial speed of \( 18.0\text{ m s}^{-1} \). Air resistance is negligible.
What is the speed of the sphere immediately before it strikes the horizontal ground below?
A.\( 29.7\text{ m s}^{-1} \)
B.\( 34.7\text{ m s}^{-1} \)
C.\( 47.7\text{ m s}^{-1} \)
D.\( 1210\text{ m s}^{-1} \)
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Worked solution
The horizontal velocity remains constant: \( v_x = 18.0\text{ m s}^{-1} \). Using the equation of motion for vertical velocity: \( v_y^2 = u_y^2 + 2 g s_y \). \( v_y^2 = 0^2 + 2 \times 9.81 \times 45.0 = 882.9\text{ m}^2\text{ s}^{-2} \). Resultant speed \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{18.0^2 + 882.9} = \sqrt{324 + 882.9} = \sqrt{1206.9} \approx 34.7\text{ m s}^{-1} \).
Marking scheme
B is correct [1]. Correct vector addition of horizontal and vertical velocity components.
Question 3 · Multiple Choice
1 marks
Two trolleys X and Y move towards each other on a frictionless horizontal linear track. Trolley X has mass \( 0.40\text{ kg} \) and moves to the right at \( +3.0\text{ m s}^{-1} \). Trolley Y has mass \( 0.60\text{ kg} \) and moves to the left at \( -2.0\text{ m s}^{-1} \).
The collision between the trolleys is perfectly elastic.
What is the velocity of trolley X after the collision?
A.\( -3.0\text{ m s}^{-1} \)
B.\( -1.0\text{ m s}^{-1} \)
C.\( +1.0\text{ m s}^{-1} \)
D.\( +3.0\text{ m s}^{-1} \)
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Worked solution
Total initial momentum: \( p_i = m_X u_X + m_Y u_Y = (0.40)(+3.0) + (0.60)(-2.0) = 1.20 - 1.20 = 0 \). For an elastic collision, relative speed of separation equals relative speed of approach: \( u_X - u_Y = 3.0 - (-2.0) = 5.0\text{ m s}^{-1} \). Thus, \( v_Y - v_X = 5.0\text{ m s}^{-1} \implies v_Y = v_X + 5.0 \). By conservation of momentum: \( m_X v_X + m_Y v_Y = 0 \implies 0.40 v_X + 0.60(v_X + 5.0) = 0 \). \( 1.00 v_X + 3.0 = 0 \implies v_X = -3.0\text{ m s}^{-1} \).
Marking scheme
A is correct [1]. Application of conservation of momentum and relative speed rule for elastic collisions.
Question 4 · Multiple Choice
1 marks
A solid uniform block of density \( 850\text{ kg m}^{-3} \) is placed in a tank containing a layer of oil of density \( 750\text{ kg m}^{-3} \) floating on water of density \( 1000\text{ kg m}^{-3} \).
The block floats in equilibrium completely submerged across the interface between the two immiscible liquids.
What fraction of the total volume of the block is submerged in the water?
A.0.15
B.0.25
C.0.40
D.0.60
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Worked solution
Let \( V \) be the total volume of the block, \( V_w \) the volume submerged in water, and \( V_o = V - V_w \) the volume in oil. In equilibrium, weight = total upthrust: \( \rho_{\text{block}} V g = \rho_{\text{oil}} (V - V_w) g + \rho_{\text{water}} V_w g \) \( 850 V = 750 (V - V_w) + 1000 V_w \) \( 850 V = 750 V + 250 V_w \) \( 100 V = 250 V_w \implies \frac{V_w}{V} = \frac{100}{250} = 0.40 \).
Marking scheme
C is correct [1]. Correctly setting total upthrust from both liquids equal to total weight.
Question 5 · Multiple Choice
1 marks
A uniform wire of cross-sectional area \( A \) and original length \( L \) is made of a metal with Young modulus \( E \). A tensile force extends the wire elastically by an extension \( \Delta L \) within its limit of proportionality.
Which expression gives the elastic strain energy stored in the wire?
A.\( \frac{E A (\Delta L)^2}{2 L} \)
B.\( \frac{E A \Delta L}{2 L} \)
C.\( \frac{E A^2 \Delta L}{L^2} \)
D.\( \frac{2 E A (\Delta L)^2}{L} \)
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Worked solution
Elastic strain energy stored is \( W = \frac{1}{2} F \Delta L \). From the definition of the Young modulus: \( E = \frac{\sigma}{\varepsilon} = \frac{F / A}{\Delta L / L} \implies F = \frac{E A \Delta L}{L} \). Substituting \( F \) into the strain energy formula: \( W = \frac{1}{2} \left( \frac{E A \Delta L}{L} \right) \Delta L = \frac{E A (\Delta L)^2}{2 L} \).
Marking scheme
A is correct [1]. Correct derivation linking Young modulus, Hooke's law, and area under force-extension graph.
Question 6 · Multiple Choice
1 marks
A train horn emits a sound wave of constant frequency \( 640\text{ Hz} \). The train travels at a constant speed of \( 20\text{ m s}^{-1} \) directly towards a stationary observer standing on a platform. The speed of sound in air is \( 340\text{ m s}^{-1} \).
What is the frequency of the sound detected by the observer?
A.\( 602\text{ Hz} \)
B.\( 640\text{ Hz} \)
C.\( 678\text{ Hz} \)
D.\( 680\text{ Hz} \)
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Worked solution
Using the Doppler effect formula for a moving source approaching a stationary observer: \( f_o = f_s \left( \frac{v}{v - v_s} \right) \) \( f_o = 640 \times \left( \frac{340}{340 - 20} \right) = 640 \times \left( \frac{340}{320} \right) = 2 \times 340 = 680\text{ Hz} \).
Marking scheme
D is correct [1]. Correct substitution into Doppler effect formula for approaching source.
Question 7 · Multiple Choice
1 marks
Monochromatic light of wavelength \( 600\text{ nm} \) is incident normally on a diffraction grating ruled with \( 4.00 \times 10^5\text{ lines m}^{-1} \).
What is the total number of bright interference maxima, including the central zeroth-order maximum, that can be observed?
A.4
B.5
C.8
D.9
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Worked solution
Grating spacing \( d = \frac{1}{N} = \frac{1}{4.00 \times 10^5\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m} \). The diffraction grating equation is \( d \sin \theta = n \lambda \). For the maximum order \( n_{\max} \), \( \sin \theta \le 1 \): \( n \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}}{600 \times 10^{-9}} = 4.17 \). Since \( n \) must be an integer, the highest order is \( n_{\max} = 4 \). The observable maxima correspond to orders \( n = 0, \pm 1, \pm 2, \pm 3, \pm 4 \). Total number of maxima = \( 2 n_{\max} + 1 = 2(4) + 1 = 9 \).
Marking scheme
D is correct [1]. Determination of highest integer order followed by counting both sides plus central maximum.
Question 8 · Multiple Choice
1 marks
A cell of constant electromotive force (e.m.f.) \( E \) and non-zero internal resistance \( r \) is connected in series with a variable resistor of resistance \( R \).
The resistance \( R \) is gradually increased from a non-zero value.
How do the terminal potential difference across the cell and the power dissipated inside the cell change?
A.Terminal p.d. increases; power dissipated inside cell decreases
B.Terminal p.d. increases; power dissipated inside cell increases
C.Terminal p.d. decreases; power dissipated inside cell decreases
D.Terminal p.d. decreases; power dissipated inside cell increases
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Worked solution
The current in the circuit is \( I = \frac{E}{R + r} \). As \( R \) increases, the total resistance \( R + r \) increases, so the circuit current \( I \) decreases. Terminal potential difference is \( V = E - I r \). Since \( I \) decreases, the lost volts \( I r \) decrease, so \( V \) increases. The power dissipated in the internal resistance is \( P_{\text{internal}} = I^2 r \). Since \( I \) decreases, \( P_{\text{internal}} \) decreases.
Marking scheme
A is correct [1]. Correct reasoning for both terminal p.d. increasing and internal power dissipation decreasing.
Question 9 · Multiple Choice
1 marks
Which expression gives the SI base units of electrical resistivity?
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Worked solution
Electrical resistivity \(\rho\) is defined by the formula \(\rho = \frac{R A}{L}\), where \(R\) is resistance, \(A\) is cross-sectional area, and \(L\) is length. Resistance is \(R = \frac{V}{I} = \frac{W}{I Q} = \frac{W}{I^2 t}\). In SI base units, work \(W\) has units of \(\text{kg}\,\text{m}^2\,\text{s}^{-2}\). Therefore, the units of resistance are \(\frac{\text{kg}\,\text{m}^2\,\text{s}^{-2}}{\text{A}^2\,\text{s}} = \text{kg}\,\text{m}^2\,\text{s}^{-3}\,\text{A}^{-2}\). Multiplying by the units of \(\frac{A}{L}\) (which is \(\text{m}\)) gives \(\text{kg}\,\text{m}^3\,\text{s}^{-3}\,\text{A}^{-2}\).
Marking scheme
A [1] (Correct SI base units of electrical resistivity)
Question 10 · Multiple Choice
1 marks
A small ball is projected horizontally with speed \(u\) from the top of a vertical cliff of height \(h\) and lands at a horizontal distance \(d\) from the base of the cliff. An identical ball is projected horizontally from height \(4h\) with speed \(2u\). Air resistance is negligible. What is the horizontal distance travelled by the second ball before landing?
A.\(2d\)
B.\(4d\)
C.\(8d\)
D.\(16d\)
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Worked solution
The time of flight \(t\) for a horizontal projectile is given by \(h = \frac{1}{2}g t^2 \implies t = \sqrt{\frac{2h}{g}}\). The horizontal distance is \(d = u t = u \sqrt{\frac{2h}{g}}\). For the second ball, height is \(4h\), so the time of flight is \(t_2 = \sqrt{\frac{2(4h)}{g}} = 2\sqrt{\frac{2h}{g}} = 2t\). The horizontal distance travelled is \(d_2 = (2u) t_2 = (2u)(2t) = 4(ut) = 4d\).
Marking scheme
B [1] (Horizontal distance is 4d)
Question 11 · Multiple Choice
1 marks
A ball of mass \(m\) falls vertically and hits a horizontal surface with speed \(v_1\). It rebounds vertically upwards with speed \(v_2\). The contact time between the ball and the surface is \(\Delta t\). What is the average normal force exerted on the ball by the surface during the impact?
A.\(\frac{m(v_1 + v_2)}{\Delta t} + mg\)
B.\(\frac{m(v_1 + v_2)}{\Delta t} - mg\)
C.\(\frac{m(v_1 - v_2)}{\Delta t} + mg\)
D.\(\frac{m(v_1 - v_2)}{\Delta t}\)
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Worked solution
Taking the upward direction as positive, the initial momentum is \(-m v_1\) and the final momentum is \(+m v_2\). The rate of change of momentum is \(F_{\text{net}} = \frac{\Delta p}{\Delta t} = \frac{m v_2 - (-m v_1)}{\Delta t} = \frac{m(v_1 + v_2)}{\Delta t}\). The resultant vertical force acting on the ball during contact is \(F_{\text{net}} = N - mg\), where \(N\) is the normal contact force exerted by the surface. Therefore, \(N = F_{\text{net}} + mg = \frac{m(v_1 + v_2)}{\Delta t} + mg\).
Marking scheme
A [1] (Correct expression for the normal contact force accounting for weight)
Question 12 · Multiple Choice
1 marks
A vertical cylinder of uniform cross-sectional area \(A\) floats in a liquid of density \(\rho\). A small object of mass \(m\) is placed on the top of the cylinder, causing the cylinder to sink a further vertical distance \(\Delta d\) into the liquid without becoming completely submerged. What is the additional depth \(\Delta d\)?
A.\(\frac{m}{\rho A}\)
B.\(\frac{\rho m}{A}\)
C.\(\frac{m g}{\rho A}\)
D.\(\frac{m}{\rho A g}\)
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Worked solution
When the object of mass \(m\) is placed on the cylinder, the cylinder sinks further until the increase in upthrust equals the weight of the additional mass. The volume of liquid displaced increases by \(\Delta V = A \Delta d\). The additional upthrust is \(\Delta U = \rho (A \Delta d) g\). Equating this to the additional weight \(mg\): \(\rho A \Delta d g = mg \implies \Delta d = \frac{m}{\rho A}\).
Marking scheme
A [1] (Equating additional upthrust to additional weight gives m / (rho A))
Question 13 · Multiple Choice
1 marks
A metal wire of length \(L\) and circular cross-section of diameter \(D\) is stretched within its limit of proportionality by a tensile force \(F\), storing elastic potential energy \(W\). A second wire made of the same metal has length \(2L\) and diameter \(2D\). It is stretched by the same tensile force \(F\). What is the elastic potential energy stored in the second wire?
A.\(\frac{1}{4}W\)
B.\(\frac{1}{2}W\)
C.\(W\)
D.\(2W\)
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Worked solution
The extension of a wire obeying Hooke's law is \(e = \frac{F L}{A E}\), and the elastic potential energy stored is \(W = \frac{1}{2} F e = \frac{F^2 L}{2 A E}\). The cross-sectional area is \(A = \frac{\pi D^2}{4}\), so \(W \propto \frac{L}{D^2}\). For the second wire, \(L_2 = 2L\) and \(D_2 = 2D\), so the area increases by a factor of \(2^2 = 4\). Therefore, \(W_2 = W \times \frac{2}{4} = \frac{1}{2}W\).
Marking scheme
B [1] (Energy scales as L / D^2, yielding 0.5 W)
Question 14 · Multiple Choice
1 marks
A source of sound emits waves of constant frequency \(f_{\text{s}}\). The source moves directly towards a stationary observer with a constant speed of \(0.20 v\), where \(v\) is the speed of sound in air, and the observer detects a frequency \(f_1\). The source then moves directly away from the stationary observer at the same speed of \(0.20 v\), and the observer detects a frequency \(f_2\). What is the ratio \(\frac{f_1}{f_2}\)?
A.1.20
B.1.25
C.1.44
D.1.50
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Worked solution
By the Doppler effect formula for a moving source: when moving towards the observer, \(f_1 = f_{\text{s}} \left(\frac{v}{v - v_{\text{s}}}\right) = f_{\text{s}} \left(\frac{v}{v - 0.20v}\right) = \frac{f_{\text{s}}}{0.80}\). When moving away from the observer, \(f_2 = f_{\text{s}} \left(\frac{v}{v + v_{\text{s}}}\right) = f_{\text{s}} \left(\frac{v}{v + 0.20v}\right) = \frac{f_{\text{s}}}{1.20}\). The ratio is \(\frac{f_1}{f_2} = \frac{1 / 0.80}{1 / 1.20} = \frac{1.20}{0.80} = 1.50\).
Marking scheme
D [1] (Ratio of Doppler shifted frequencies is 1.50)
Question 15 · Multiple Choice
1 marks
Monochromatic light of wavelength \(520\text{ nm}\) is incident normally on a diffraction grating that has \(4.00 \times 10^5\text{ lines m}^{-1}\). What is the total number of interference maxima that can be observed?
A.4
B.5
C.8
D.9
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Worked solution
The grating spacing is \(d = \frac{1}{4.00 \times 10^5\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m}\). Using the grating equation \(d \sin\theta = n\lambda\), the maximum possible order occurs when \(\sin\theta \le 1\): \(n_{\text{max}} \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}\text{ m}}{520 \times 10^{-9}\text{ m}} \approx 4.81\). Thus, the maximum integer order is \(n = 4\). The visible maxima correspond to orders \(n = 0, \pm 1, \pm 2, \pm 3, \pm 4\), giving a total of \(2(4) + 1 = 9\) maxima.
Marking scheme
D [1] (9 total maxima including central order and +/- 1 to 4)
Question 16 · Multiple Choice
1 marks
A battery of electromotive force (e.m.f.) \(12.0\text{ V}\) and internal resistance \(2.0\,\Omega\) is connected in series with a fixed resistor of resistance \(6.0\,\Omega\) and a variable resistor. The resistance of the variable resistor is changed from \(4.0\,\Omega\) to \(16.0\,\Omega\). What is the change in the potential difference across the variable resistor?
A.\(2.4\text{ V}\)
B.\(4.0\text{ V}\)
C.\(6.0\text{ V}\)
D.\(8.0\text{ V}\)
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Worked solution
Initially, total circuit resistance is \(R_1 = 2.0 + 6.0 + 4.0 = 12.0\,\Omega\). Initial current is \(I_1 = \frac{12.0\text{ V}}{12.0\,\Omega} = 1.0\text{ A}\). The potential difference across the variable resistor is \(V_1 = I_1 R_{\text{var}} = 1.0\text{ A} \times 4.0\,\Omega = 4.0\text{ V}\). Finally, total circuit resistance is \(R_2 = 2.0 + 6.0 + 16.0 = 24.0\,\Omega\). Final current is \(I_2 = \frac{12.0\text{ V}}{24.0\,\Omega} = 0.50\text{ A}\). The potential difference across the variable resistor is \(V_2 = 0.50\text{ A} \times 16.0\,\Omega = 8.0\text{ V}\). The change in potential difference is \(\Delta V = 8.0\text{ V} - 4.0\text{ V} = 4.0\text{ V}\).
Marking scheme
B [1] (Change in p.d. is 4.0 V)
Question 17 · Multiple Choice
1 marks
Which combination of SI base units is equivalent to 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 \(\rho\) is defined by \(\rho = \frac{R A}{L}\). Resistance \(R = \frac{V}{I}\) and potential difference \(V = \frac{W}{Q} = \frac{W}{I t}\). In terms of SI base units: - Work \(W\): \(\text{kg m}^2\text{ s}^{-2}\) - Charge \(Q = I t\): \(\text{A s}\) - Potential difference \(V\): \(\frac{\text{kg m}^2\text{ s}^{-2}}{\text{A s}} = \text{kg m}^2\text{ s}^{-3}\text{ A}^{-1}\) - Resistance \(R\): \(\frac{\text{kg m}^2\text{ s}^{-3}\text{ A}^{-1}}{\text{A}} = \text{kg m}^2\text{ s}^{-3}\text{ A}^{-2}\) - Resistivity \(\rho = \frac{R A}{L}\): \(\frac{(\text{kg m}^2\text{ s}^{-3}\text{ A}^{-2})(\text{m}^2)}{\text{m}} = \text{kg m}^3\text{ s}^{-3}\text{ A}^{-2}\).
Marking scheme
A is correct [1 mark]. B is incorrect as it represents resistance (\(\Omega\)). C is incorrect due to incorrect time/current powers. D is incorrect as the length dimension is not \(\text{m}^3\).
Question 18 · Multiple Choice
1 marks
A stone is projected horizontally at a speed of \(15\text{ m s}^{-1}\) from the top of a vertical cliff of height \(45\text{ m}\) above horizontal ground. Air resistance is negligible.
What is the speed of the stone immediately before it hits the ground?
A.\(29.7\text{ m s}^{-1}\)
B.\(33.3\text{ m s}^{-1}\)
C.\(44.7\text{ m s}^{-1}\)
D.\(54.8\text{ m s}^{-1}\)
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Worked solution
For vertical motion: \(u_y = 0\text{ m s}^{-1}\) \(v_y^2 = u_y^2 + 2 g s = 0 + 2 \times 9.81 \times 45 = 882.9\text{ m}^2\text{ s}^{-2}\)
For horizontal motion: \(v_x = 15\text{ m s}^{-1}\) \(v_x^2 = 225\text{ m}^2\text{ s}^{-2}\)
The final speed \(v\) is given by the magnitude of the resultant velocity: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{225 + 882.9} = \sqrt{1107.9} \approx 33.3\text{ m s}^{-1}\).
Marking scheme
B is correct [1 mark]. A is incorrect as it calculates only the vertical velocity component \(v_y = 29.7\text{ m s}^{-1}\). C and D arise from arithmetic errors or adding speeds algebraically.
Question 19 · Multiple Choice
1 marks
A ball of mass \(0.20\text{ kg}\) travels horizontally with a speed of \(14\text{ m s}^{-1}\). It strikes a vertical wall and rebounds along its original line of motion with a speed of \(11\text{ m s}^{-1}\). The ball is in contact with the wall for \(0.040\text{ s}\).
What is the magnitude of the average force exerted by the wall on the ball?
A.\(15\text{ N}\)
B.\(70\text{ N}\)
C.\(125\text{ N}\)
D.\(250\text{ N}\)
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Worked solution
Taking the initial direction as positive: Initial velocity \(u = +14\text{ m s}^{-1}\) Final velocity \(v = -11\text{ m s}^{-1}\)
Change in momentum \(\Delta p = m(v - u) = 0.20 \times (-11 - 14) = 0.20 \times (-25) = -5.0\text{ N s}\).
Average force \(F = \frac{|\Delta p|}{\Delta t} = \frac{5.0}{0.040} = 125\text{ N}\).
Marking scheme
C is correct [1 mark]. A is incorrect due to neglecting the change in direction (calculating \(0.20 \times (14 - 11) / 0.040 = 15\text{ N}\)). B and D are common arithmetic or conceptual distractors.
Question 20 · Multiple Choice
1 marks
A solid uniform cylinder of cross-sectional area \(A\) and height \(h\) floats upright in a liquid of density \(\rho\). The cylinder is submerged to a depth of \(d\).
What is the upthrust acting on the cylinder?
A.\(\rho g A d\)
B.\(\rho g A h\)
C.\(\rho g A (h - d)\)
D.\(\frac{\rho g A d^2}{h}\)
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Worked solution
By Archimedes' principle, the upthrust equals the weight of the liquid displaced. Volume of liquid displaced \(V = A d\). Mass of liquid displaced \(m = \rho V = \rho A d\). Weight of displaced liquid (upthrust) \(F = m g = \rho g A d\).
Marking scheme
A is correct [1 mark]. B uses total volume \(A h\) instead of displaced volume. C uses un-submerged depth \((h - d)\). D is dimensionally incorrect.
Question 21 · Multiple Choice
1 marks
A wire of length \(L\) and circular cross-section of diameter \(D\) is made of a metal of Young modulus \(E\). When a tensile force \(F\) is applied to the wire within its limit of proportionality, the elastic potential energy stored in the wire is \(E_p\).
A second wire of the same metal has length \(2L\) and diameter \(2D\).
What is the elastic potential energy stored in the second wire when the same tensile force \(F\) is applied?
A.\(0.25 E_p\)
B.\(0.50 E_p\)
C.\(E_p\)
D.\(2.0 E_p\)
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Worked solution
Elastic potential energy \(E_p = \frac{1}{2} F x\). Extension \(x = \frac{F L}{A E} = \frac{4 F L}{\pi D^2 E}\). Thus, \(E_p = \frac{2 F^2 L}{\pi D^2 E}\).
For the second wire with length \(2L\) and diameter \(2D\): \(E_p' = \frac{2 F^2 (2L)}{\pi (2D)^2 E} = \frac{2 F^2 (2L)}{\pi (4 D^2) E} = \frac{1}{2} \left( \frac{2 F^2 L}{\pi D^2 E} \right) = 0.50 E_p\).
Marking scheme
B is correct [1 mark]. A is obtained if diameter scaling is squared without accounting for the doubled length. C and D arise from incorrect ratios of length to area.
Question 22 · Multiple Choice
1 marks
A train travels directly towards a stationary observer at a constant speed of \(28\text{ m s}^{-1}\). The train emits a sound wave of constant frequency \(680\text{ Hz}\). The speed of sound in air is \(340\text{ m s}^{-1}\).
What is the observed frequency of the sound heard by the observer?
A.\(628\text{ Hz}\)
B.\(680\text{ Hz}\)
C.\(736\text{ Hz}\)
D.\(741\text{ Hz}\)
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Worked solution
Using the Doppler formula for a source moving towards a stationary observer: \(f_o = f_s \left( \frac{v}{v - v_s} \right)\)
Substitute the given values: \(f_o = 680 \times \left( \frac{340}{340 - 28} \right) = 680 \times \frac{340}{312} \approx 741\text{ Hz}\).
Marking scheme
D is correct [1 mark]. A is obtained if the source is incorrectly assumed to be moving away: \(680 \times 340 / 368 = 628\text{ Hz}\). B is the emitted frequency unchanged. C is obtained by using the formula for moving observer \(680 \times (340 + 28) / 340 = 736\text{ Hz}\).
Question 23 · Multiple Choice
1 marks
Monochromatic light of wavelength \(600\text{ nm}\) is incident normally on a diffraction grating with \(5.0 \times 10^5\text{ lines m}^{-1}\).
What is the total number of principal maxima that can be observed on a screen placed far from the grating?
A.\(3\)
B.\(6\)
C.\(7\)
D.\(8\)
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Worked solution
Grating spacing \(d = \frac{1}{5.0 \times 10^5\text{ m}^{-1}} = 2.0 \times 10^{-6}\text{ m}\). The condition for diffraction maxima is \(d \sin\theta = n \lambda\). For the maximum possible order \(n\), \(\sin\theta \le 1\): \(n \le \frac{d}{\lambda} = \frac{2.0 \times 10^{-6}\text{ m}}{600 \times 10^{-9}\text{ m}} = 3.33\).
Since \(n\) must be an integer, the highest order is \(n = 3\). The observable orders are \(n = 0, \pm 1, \pm 2, \pm 3\). Total number of maxima \(= 2(3) + 1 = 7\).
Marking scheme
C is correct [1 mark]. A is only the maximum order number on one side \(n = 3\). B omits the central zero-order maximum \(2 \times 3 = 6\). D is an incorrect rounding.
Question 24 · Multiple Choice
1 marks
A cell of electromotive force (e.m.f.) \(E\) and 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 changes to the terminal potential difference across the cell and to the power dissipated in the internal resistance of the cell?
| | Terminal potential difference | Power dissipated in internal resistance | |---|---|---| | A | increases | decreases | | B | increases | increases | | C | decreases | decreases | | D | decreases | increases |
A.Terminal potential difference increases; power dissipated in internal resistance decreases
B.Terminal potential difference increases; power dissipated in internal resistance increases
C.Terminal potential difference decreases; power dissipated in internal resistance decreases
D.Terminal potential difference decreases; power dissipated in internal resistance increases
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Worked solution
Circuit current is \(I = \frac{E}{R + r}\). As \(R\) increases from \(r\) to \(3r\), the total resistance \(R + r\) increases, so the current \(I\) decreases.
Terminal potential difference \(V = E - I r\). Since \(I\) decreases, the lost volts \(I r\) decrease, and thus \(V\) increases (alternatively \(V = E \frac{R}{R + r}\), which increases from \(0.50 E\) to \(0.75 E\)).
Power dissipated in internal resistance is \(P_r = I^2 r\). Since \(I\) decreases, \(P_r\) decreases.
Therefore, terminal p.d. increases and power dissipated in the internal resistance decreases.
Marking scheme
A is correct [1 mark]. B incorrectly identifies the change in internal power dissipation. C incorrectly identifies the change in terminal p.d. D inverts both trends.
Question 25 · Multiple Choice
1 marks
The drag force \(F\) exerted on a small spherical particle of radius \(r\) falling at speed \(v\) through a fluid of dynamic viscosity \(\eta\) is given by \(F = 6 \pi \eta r v\).
What are the SI base units of dynamic viscosity \(\eta\)?
A.\(\text{kg}\text{ m}^{-1}\text{ s}^{-1}\)
B.\(\text{kg}\text{ m}^{-2}\text{ s}^{-1}\)
C.\(\text{kg}\text{ m}\text{ s}^{-1}\)
D.\(\text{kg}\text{ m}^{-1}\text{ s}^{-2}\)
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Worked solution
Rearranging the equation for dynamic viscosity gives: \[\eta = \frac{F}{6\pi r v}\]
Substitute the SI base units for each quantity: - Force \(F\): \(\text{N} = \text{kg}\text{ m}\text{ s}^{-2}\) - Radius \(r\): \(\text{m}\) - Speed \(v\): \(\text{m}\text{ s}^{-1}\) - \(6\pi\) is dimensionless.
Thus, the correct SI base units are \(\text{kg}\text{ m}^{-1}\text{ s}^{-1}\).
Marking scheme
A [1] — 1 mark for correct option selection.
Question 26 · Multiple Choice
1 marks
A projectile is launched from horizontal ground with initial speed \(u\) at an angle \(\theta\) to the horizontal. Air resistance is negligible.
Which expression gives the maximum height \(H\) reached by the projectile above the ground?
A.\(\dfrac{u^2 \sin\theta}{2g}\)
B.\(\dfrac{u^2 \sin^2\theta}{2g}\)
C.\(\dfrac{u^2 \cos^2\theta}{2g}\)
D.\(\dfrac{u^2 \sin^2\theta}{g}\)
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Worked solution
Consider the vertical component of the motion: - Initial vertical velocity \(u_y = u \sin\theta\) - Vertical acceleration \(a_y = -g\) - At maximum height \(H\), the vertical velocity \(v_y = 0\)
Using the equation of uniformly accelerated motion \(v_y^2 = u_y^2 + 2 a_y H\): \[0 = (u \sin\theta)^2 - 2gH\] \[2gH = u^2 \sin^2\theta \implies H = \frac{u^2 \sin^2\theta}{2g}\]
Marking scheme
B [1] — 1 mark for correct option selection.
Question 27 · Multiple Choice
1 marks
Two spheres, P of mass \(2.0\text{ kg}\) and Q of mass \(3.0\text{ kg}\), travel towards each other along a straight horizontal frictionless track.
Sphere P has an initial velocity of \(+4.0\text{ m s}^{-1}\) and sphere Q has an initial velocity of \(-2.0\text{ m s}^{-1}\).
The collision between the spheres is perfectly elastic.
What is the velocity of sphere P after the collision?
A.\(-3.2\text{ m s}^{-1}\)
B.\(-0.8\text{ m s}^{-1}\)
C.\(+0.4\text{ m s}^{-1}\)
D.\(+2.8\text{ m s}^{-1}\)
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Worked solution
For a perfectly elastic collision in one dimension: 1. Relative speed of approach equals relative speed of separation: \[u_1 - u_2 = v_2 - v_1\] \[4.0 - (-2.0) = v_2 - v_1 \implies v_2 - v_1 = 6.0 \implies v_2 = v_1 + 6.0\]
Thus, the velocity of sphere P after the collision is \(-3.2\text{ m s}^{-1}\).
Marking scheme
A [1] — 1 mark for correct option selection.
Question 28 · Multiple Choice
1 marks
A solid uniform cylinder of cross-sectional area \(A\) and height \(h\) is made of a material of density \(\rho_S\). It is placed in a liquid of density \(\rho_L\), where \(\rho_L > \rho_S\).
A downward vertical force \(F\) is applied to hold the cylinder completely submerged in the liquid.
Which expression gives the magnitude of the force \(F\)?
A.\(\rho_L A h g\)
B.\((\rho_L - \rho_S) A h g\)
C.\((\rho_S - \rho_L) A h g\)
D.\((\rho_L + \rho_S) A h g\)
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Worked solution
The forces acting on the submerged cylinder in vertical equilibrium are: - Downward force \(F\) - Downward weight \(W = m g = (\rho_S V) g = \rho_S A h g\) - Upward upthrust \(U = (\rho_L V) g = \rho_L A h g\)
For equilibrium: \[F + W = U\] \[F = U - W = \rho_L A h g - \rho_S A h g = (\rho_L - \rho_S) A h g\]
Marking scheme
B [1] — 1 mark for correct option selection.
Question 29 · Multiple Choice
1 marks
A wire of original length \(L\) and uniform cross-sectional area \(A\) is made of metal with Young modulus \(E\). The wire is stretched within its limit of proportionality by a tensile force such that its extension is \(x\).
What is the elastic potential energy stored in the wire?
A.\(\dfrac{E A x}{2 L}\)
B.\(\dfrac{E A x^2}{L}\)
C.\(\dfrac{E A x^2}{2 L}\)
D.\(\dfrac{2 E A x^2}{L}\)
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Worked solution
The elastic potential energy stored within the limit of proportionality is: \[E_p = \frac{1}{2} F x\]
From the definition of Young modulus \(E\): \[E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\]
Rearranging for force \(F\): \[F = \frac{E A x}{L}\]
Substitute \(F\) into the energy expression: \[E_p = \frac{1}{2} \left(\frac{E A x}{L}\right) x = \frac{E A x^2}{2 L}\]
Marking scheme
C [1] — 1 mark for correct option selection.
Question 30 · Multiple Choice
1 marks
A vehicle with its horn emitting a continuous sound of frequency \(900\text{ Hz}\) travels at a constant speed of \(34.0\text{ m s}^{-1}\) directly towards a stationary observer.
The speed of sound in air is \(340\text{ m s}^{-1}\).
What is the frequency of the sound detected by the observer?
A.\(818\text{ Hz}\)
B.\(900\text{ Hz}\)
C.\(990\text{ Hz}\)
D.\(1000\text{ Hz}\)
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Worked solution
Using the Doppler effect formula for a moving source approaching a stationary observer: \[f_o = f_s \left( \frac{v}{v - v_s} \right)\]
Where: - \(f_s = 900\text{ Hz}\) - \(v = 340\text{ m s}^{-1}\) - \(v_s = 34.0\text{ m s}^{-1}\)
A power supply has an electromotive force (e.m.f.) of \(12.0\text{ V}\) and internal resistance \(2.0\,\Omega\). It is connected in series with a \(10.0\,\Omega\) resistor and a parallel pair of identical resistors, each of resistance \(R\).
The potential difference across the \(10.0\,\Omega\) resistor is \(6.0\text{ V}\).
What is the resistance \(R\) of each of the two identical resistors?
A.\(4.0\,\Omega\)
B.\(8.0\,\Omega\)
C.\(16\,\Omega\)
D.\(32\,\Omega\)
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Worked solution
1. Current \(I\) in the series circuit is the current through the \(10.0\,\Omega\) resistor: \[I = \frac{V_{10}}{10.0} = \frac{6.0\text{ V}}{10.0\,\Omega} = 0.60\text{ A}\]
2. The total resistance of the circuit \(R_{\text{total}}\) is: \[R_{\text{total}} = \frac{E}{I} = \frac{12.0\text{ V}}{0.60\text{ A}} = 20.0\,\Omega\]
3. The total resistance is also given by: \[R_{\text{total}} = r + R_{10} + R_p\] \[20.0 = 2.0 + 10.0 + R_p \implies R_p = 8.0\,\Omega\]
4. Since \(R_p\) is the resistance of two identical resistors in parallel: \[R_p = \frac{R}{2} = 8.0\,\Omega \implies R = 16.0\,\Omega\]
Marking scheme
C [1] — 1 mark for correct option selection.
Question 32 · Multiple Choice
1 marks
A nucleus of carbon-11 (\(^{11}_{\ 6}\text{C}\)) undergoes \(\beta^+\) decay to form a nucleus of boron-11 (\(^{11}_{\ 5}\text{B}\)).
Which statement correctly describes the quark transformation and the additional fundamental particle emitted in this decay?
A.An up quark changes into a down quark, and an electron antineutrino is emitted.
B.An up quark changes into a down quark, and an electron neutrino is emitted.
C.A down quark changes into an up quark, and an electron antineutrino is emitted.
D.A down quark changes into an up quark, and an electron neutrino is emitted.
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Worked solution
In \(\beta^+\) (positron) decay, a proton decays into a neutron, a positron (\(\beta^+\)), and an electron neutrino (\(\nu_e\)): \[p \rightarrow n + e^+ + \nu_e\]
In terms of quark composition: - Proton = \(uud\) - Neutron = \(udd\)
Therefore, an up quark (\(u\)) changes into a down quark (\(d\)), and an electron neutrino is emitted.
Marking scheme
B [1] — 1 mark for correct option selection.
Question 33 · Multiple Choice
1 marks
Which expression gives the SI base units of electrical resistivity?
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Worked solution
Resistivity \(\rho\) is defined by \(\rho = \frac{RA}{L}\). Resistance \(R = \frac{V}{I} = \frac{W}{Q I} = \frac{W}{I^2 t}\). In terms of base units: - Work \(W\) has units \(\text{kg m}^2 \text{s}^{-2}\) - Current squared \(\times\) time \(I^2 t\) has units \(\text{A}^2 \text{s}\)
A is correct [1]. B is the unit for resistance (\(\Omega\)). C has incorrect powers of seconds and amperes. D has an incorrect power of metres.
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 when it reaches a vertical height of \(2.0\text{ m}\) above the ground?
A.\(10.8\text{ m s}^{-1}\)
B.\(21.7\text{ m s}^{-1}\)
C.\(24.2\text{ m s}^{-1}\)
D.\(28.1\text{ m s}^{-1}\)
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Worked solution
Using conservation of energy (or kinematics): \[ v^2 = u^2 - 2gh \] \[ v^2 = 25^2 - 2(9.81)(2.0) = 625 - 39.24 = 585.76 \] \[ v = \sqrt{585.76} \approx 24.2\text{ m s}^{-1} \]
Marking scheme
C is correct [1]. A is the vertical component of velocity only (\(\approx 10.8\text{ m s}^{-1}\)). B is the horizontal component of velocity only (\(25\cos 30^\circ \approx 21.7\text{ m s}^{-1}\)). D is an incorrect calculation.
Question 35 · Multiple Choice
1 marks
Two spheres, P and Q, move towards each other along a straight line on a smooth horizontal surface.
Sphere P has mass \(0.40\text{ kg}\) and moves to the right at \(3.0\text{ m s}^{-1}\). Sphere Q has mass \(0.20\text{ kg}\) and moves to the left at \(6.0\text{ m s}^{-1}\).
The collision between the spheres is perfectly elastic.
What is the velocity of sphere P after the collision?
A.\(1.0\text{ m s}^{-1}\text{ to the left}\)
B.\(3.0\text{ m s}^{-1}\text{ to the left}\)
C.\(3.0\text{ m s}^{-1}\text{ to the right}\)
D.\(6.0\text{ m s}^{-1}\text{ to the left}\)
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Worked solution
Taking motion to the right as positive: Total initial momentum \(p_{\text{total}} = (0.40)(+3.0) + (0.20)(-6.0) = +1.2 - 1.2 = 0\text{ kg m s}^{-1}\).
Since total momentum is zero, after the collision the two spheres must again have equal and opposite momenta: \(0.40 v_{\text{P}} + 0.20 v_{\text{Q}} = 0 \implies v_{\text{Q}} = -2 v_{\text{P}}\).
Since sphere P must bounce back, \(v_{\text{P}} = -3.0\text{ m s}^{-1}\) (i.e. \(3.0\text{ m s}^{-1}\) to the left).
Marking scheme
B is correct [1]. A, C, and D violate conservation of momentum and/or kinetic energy.
Question 36 · Multiple Choice
1 marks
A solid cylinder of uniform cross-sectional area \(4.0 \times 10^{-3}\text{ m}^2\) and height \(0.15\text{ m}\) is completely submerged in a liquid of density \(1200\text{ kg m}^{-3}\). The top circular face is horizontal and located at a depth of \(0.25\text{ m}\) below the surface.
What is the difference in hydrostatic pressure between the bottom and top faces, and what is the upthrust exerted by the liquid on the cylinder?
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Worked solution
The pressure difference between the top and bottom faces depends only on the height \(h\) of the cylinder: \[ \Delta p = \rho g \Delta h = 1200 \times 9.81 \times 0.15 = 1765.8\text{ Pa} \approx 1.8\text{ kPa} \]
The upthrust \(F\) is given by Archimedes' principle: \[ F = \Delta p \times A = 1765.8 \times (4.0 \times 10^{-3}) = 7.06\text{ N} \approx 7.1\text{ N} \]
Marking scheme
A is correct [1]. B uses the total depth to the bottom for the upthrust calculation. C uses total depth \(0.25\text{ m}\) for the pressure difference. D uses total depth for both pressure and force.
Question 37 · Multiple Choice
1 marks
An electric car travels along a straight, horizontal road at a constant speed of \(24\text{ m s}^{-1}\). The electric motor has an overall efficiency of \(80\%\) and consumes an electrical power input of \(36\text{ kW}\).
What is the total resistive force opposing the motion of the car?
A.\(0.96\text{ kN}\)
B.\(1.2\text{ kN}\)
C.\(1.5\text{ kN}\)
D.\(1.9\text{ kN}\)
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Worked solution
The useful mechanical power output delivered to move the car is: \[ P_{\text{out}} = 0.80 \times 36\text{ kW} = 28.8\text{ kW} = 28800\text{ W} \]
Since the vehicle moves at a constant speed, the forward driving force \(F\) equals the resistive force: \[ P = F v \implies F = \frac{P_{\text{out}}}{v} = \frac{28800}{24} = 1200\text{ N} = 1.2\text{ kN} \]
Marking scheme
B is correct [1]. A applies efficiency incorrectly (\(0.80^2 \times 36 / 24\)). C neglects motor efficiency (\(36000 / 24 = 1.5\text{ kN}\)). D divides by efficiency instead of multiplying (\(36000 / (0.80 \times 24) = 1.875\text{ kN}\)).
Question 38 · Multiple Choice
1 marks
A train travels at a constant speed along a straight track directly towards a stationary observer. The horn on the train emits sound of constant frequency \(480\text{ Hz}\). The speed of sound in air is \(340\text{ m s}^{-1}\).
The observer detects sound of frequency \(510\text{ Hz}\).
What is the speed of the train?
A.\(20\text{ m s}^{-1}\)
B.\(21\text{ m s}^{-1}\)
C.\(23\text{ m s}^{-1}\)
D.\(40\text{ m s}^{-1}\)
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Worked solution
Using the Doppler effect formula for a moving source approaching a stationary observer: \[ f_o = f_s \left( \frac{v}{v - v_s} \right) \] \[ 510 = 480 \left( \frac{340}{340 - v_s} \right) \] \[ 340 - v_s = \frac{480 \times 340}{510} = 320\text{ m s}^{-1} \] \[ v_s = 340 - 320 = 20\text{ m s}^{-1} \]
Marking scheme
A is correct [1]. B uses \(\Delta f / f_s \times v = (30/480) \times 340 \approx 21.3\text{ m s}^{-1}\). C and D are calculation errors.
Question 39 · 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 principal intensity maxima that can be observed on a distant semi-circular screen centered on the grating?
A.4
B.8
C.9
D.10
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Worked solution
The grating spacing \(d\) is: \[ d = \frac{1}{4.0 \times 10^5\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m} \]
Using the grating equation \(d \sin \theta = n \lambda\), the maximum possible order \(n\) corresponds to \(\sin \theta \le 1\): \[ n \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}}{590 \times 10^{-9}} = 4.24 \]
Since \(n\) must be an integer, the maximum observable order on each side of the central maximum is \(n = 4\).
The total number of maxima is: \[ N = 2n_{\max} + 1 = 2(4) + 1 = 9 \]
Marking scheme
C is correct [1]. A is the maximum order number \(n = 4\). B omits the zeroth-order central maximum (\(2 \times 4 = 8\)). D rounds the order number up to 5.
Question 40 · Multiple Choice
1 marks
During the radioactive decay of a nucleus by \(\beta^-\text{ (beta-minus)}\) emission, a neutron transforms into a proton.
Which row correctly identifies the change in quark flavour and the leptons emitted in this decay?
$$\begin{array}{|c|c|c|} \hline & \text{Quark transformation} & \text{Leptons emitted} \\ \hline \text{A} & \text{down } (\text{d}) \to \text{up } (\text{u}) & \text{electron and electron antineutrino} \\ \text{B} & \text{down } (\text{d}) \to \text{up } (\text{u}) & \text{positron and electron neutrino} \\ \text{C} & \text{up } (\text{u}) \to \text{down } (\text{d}) & \text{electron and electron antineutrino} \\ \text{D} & \text{up } (\text{u}) \to \text{down } (\text{d}) & \text{positron and electron neutrino} \\ \hline \end{array}$$
A.Row A
B.Row B
C.Row C
D.Row D
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Worked solution
A neutron has quark structure \(\text{udd}\) and a proton has quark structure \(\text{uud}\). Therefore, in \(\beta^-\) decay, a down quark (charge \(-\frac{1}{3}e\)) changes into an up quark (charge \(+\frac{2}{3}e\)): \[ \text{d} \to \text{u} + \text{e}^- + \bar{\nu}_\text{e} \] To conserve lepton number and charge, an electron (\(\beta^-\)) and an electron antineutrino (\(\bar{\nu}_\text{e}\)) are emitted.
Marking scheme
A is correct [1]. B gives the leptons for \(\beta^+\) decay. C has the incorrect quark transformation for \(\beta^-\) decay. D describes \(\beta^+\) decay entirely.
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Answer all questions in the spaces provided. Show all working and use appropriate units.
5 Question · 60 marks
Question 1 · Structured Question
12 marks
(a) (i) Show that the SI base units of power are \(\text{kg}\,\text{m}^2\,\text{s}^{-3}\). [2]
(ii) A spherical particle of radius \(r\) moves at speed \(v\) through a fluid of density \(\rho\). The resistive drag force \(F_D\) acting on the sphere is given by
\[ F_D = \frac{1}{2} C_D \rho \pi r^2 v^2 \]
where \(C_D\) is a dimensionless drag coefficient. Show that this equation is dimensionally homogeneous. [2]
(b) A small ball is launched from the top of a vertical cliff of height \(35.0\text{ m}\) above horizontal ground with an initial velocity of \(18.0\text{ m s}^{-1}\) at an angle of \(30.0^\circ\) above the horizontal. Air resistance is negligible.
(i) Calculate the vertical component \(u_y\) of the initial velocity. [1]
(ii) Calculate the maximum height reached by the ball above the horizontal ground. [2]
(iii) Calculate the time taken for the ball to reach the ground. [3]
(iv) Calculate the speed of the ball immediately before it hits the ground. [2]
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Worked solution
(a) (i) \(\text{Power} = \frac{\text{Work done}}{\text{time}} = \frac{\text{Force} \times \text{distance}}{\text{time}}\) Base units of force \(= \text{kg}\,\text{m}\,\text{s}^{-2}\) Base units of power \(= \frac{\text{kg}\,\text{m}\,\text{s}^{-2} \times \text{m}}{\text{s}} = \text{kg}\,\text{m}^2\,\text{s}^{-3}\)
(ii) LHS: \(F_D\) has base units \(\text{kg}\,\text{m}\,\text{s}^{-2}\) RHS: \(\frac{1}{2}\) and \(\pi\) have no units. \(C_D\) is dimensionless. Base units of \(\rho r^2 v^2 = (\text{kg}\,\text{m}^{-3}) \times (\text{m})^2 \times (\text{m}\,\text{s}^{-1})^2\) \(= \text{kg}\,\text{m}^{-3} \times \text{m}^2 \times \text{m}^2\,\text{s}^{-2} = \text{kg}\,\text{m}\,\text{s}^{-2}\) Since base units of LHS = base units of RHS, the equation is homogeneous.
(b) (i) \(u_y = u \sin(30.0^\circ) = 18.0 \times \sin(30.0^\circ) = 9.00\text{ m s}^{-1}\)
(ii) At maximum height above cliff, \(v_y = 0\): \(v_y^2 = u_y^2 - 2gs_y \implies 0 = 9.00^2 - 2(9.81)s_y\) \(s_y = \frac{81.0}{19.62} = 4.13\text{ m}\) Total height above ground \(H = 35.0 + 4.13 = 39.1\text{ m}\)
(iv) Horizontal component: \(v_x = u \cos(30.0^\circ) = 18.0 \cos(30.0^\circ) = 15.59\text{ m s}^{-1}\) Vertical component before impact: \(v_y = u_y - gt = 9.00 - (9.81 \times 3.742) = -27.71\text{ m s}^{-1}\) \(\text{Speed } v = \sqrt{v_x^2 + v_y^2} = \sqrt{(15.59)^2 + (-27.71)^2} = \sqrt{243.0 + 767.8} = \sqrt{1010.8} = 31.8\text{ m s}^{-1}\)
Marking scheme
(a)(i) - Uses \(P = W/t\) and \(W = F \times d\) with \(F = ma\) or equivalent formula [M1] - Fully shows derivation to \(\text{kg}\,\text{m}^2\,\text{s}^{-3}\) [A1]
(a)(ii) - Shows base units of force as \(\text{kg}\,\text{m}\,\text{s}^{-2}\) [B1] - Correctly evaluates base units of RHS as \(\text{kg}\,\text{m}^{-3} \times \text{m}^2 \times \text{m}^2\,\text{s}^{-2} = \text{kg}\,\text{m}\,\text{s}^{-2}\) and states homogeneity [B1]
(b)(i) - \(u_y = 9.00\text{ m s}^{-1}\) [A1]
(b)(ii) - Correct formula for maximum height above launch: \(s_y = u_y^2 / (2g) = 4.13\text{ m}\) [C1] - Total height \(= 35.0 + 4.13 = 39.1\text{ m}\) [A1]
(b)(iv) - Correct calculation of \(v_x = 15.6\text{ m s}^{-1}\) and \(v_y = -27.7\text{ m s}^{-1}\) or use of conservation of energy \(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 + mgh\) [C1] - \(v = 31.8\text{ m s}^{-1}\) (allow 32 m s^-1) [A1]
Question 2 · Structured Question
12 marks
(a) (i) Define the moment of a force. [1]
(ii) State the principle of moments. [1]
(b) (i) Define upthrust. [1]
(ii) A uniform solid cylinder of mass \(0.500\text{ kg}\), length \(0.800\text{ m}\), and cross-sectional area \(1.00 \times 10^{-3}\text{ m}^2\) is suspended horizontally. Calculate the density of the cylinder. [2]
(c) The uniform cylinder in (b)(ii) is pivoted smoothly at one end P. The cylinder is held horizontally in equilibrium by a vertical string attached to its other end Q. A section of the cylinder extending \(0.200\text{ m}\) from end Q is submerged in water of density \(1.00 \times 10^3\text{ kg m}^{-3}\).
(i) Show that the upthrust acting on the submerged section of the cylinder is \(1.96\text{ N}\). [2]
(ii) State the distance from the pivot P to the line of action of the upthrust. [1]
(iii) By taking moments about P, calculate the tension \(T\) in the vertical string attached at Q. [3]
(iv) Determine the magnitude of the vertical force exerted by the pivot P on the cylinder. [1]
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Worked solution
(a) (i) Moment of a force is defined as the force multiplied by the perpendicular distance from the line of action of the force to the pivot.
(ii) For an object in equilibrium, the sum of clockwise moments about a point is equal to the sum of anticlockwise moments about that same point.
(b) (i) Upthrust is the upward force exerted on an object immersed in a fluid equal to the weight of fluid displaced (or caused by the pressure difference between the bottom and top of the object).
(ii) Volume of cylinder \(V = A \times L = 1.00 \times 10^{-3} \times 0.800 = 8.00 \times 10^{-4}\text{ m}^3\) Density \(\rho = \frac{M}{V} = \frac{0.500}{8.00 \times 10^{-4}} = 625\text{ kg m}^{-3}\)
(ii) The submerged section extends from \(x = 0.600\text{ m}\) to \(x = 0.800\text{ m}\). Its midpoint (line of action of upthrust) is at \(0.600 + 0.100 = 0.700\text{ m}\) from P.
(iii) Weight of the cylinder \(W = Mg = 0.500 \times 9.81 = 4.905\text{ N}\) Since the cylinder is uniform, \(W\) acts at \(0.400\text{ m}\) from P. Taking moments about P: Clockwise moment \(= W \times 0.400 = 4.905 \times 0.400 = 1.962\text{ N m}\) Anticlockwise moment \(= (U \times 0.700) + (T \times 0.800) = (1.962 \times 0.700) + 0.800T = 1.3734 + 0.800T\) Equating clockwise and anticlockwise moments: \(1.962 = 1.3734 + 0.800T\) \(0.800T = 0.5886\) \(T = 0.73575\text{ N} \approx 0.736\text{ N}\)
(iv) Resolving forces vertically for equilibrium: \(F_P + U + T = W\) \(F_P = W - U - T = 4.905 - 1.962 - 0.736 = 2.207\text{ N} \approx 2.21\text{ N}\) (acting upwards)
Marking scheme
(a)(i) - Force \(\times\) perpendicular distance from pivot to line of action of force [B1]
(a)(ii) - (Sum of) clockwise moments equals (sum of) anticlockwise moments about any point for a body in equilibrium [B1]
(b)(i) - Upward force on a body immersed in a fluid (due to pressure difference) [B1]
(iii) Distinguish between elastic deformation and plastic deformation. [2]
(b) A uniform metal wire of original length \(2.40\text{ m}\) and cross-sectional area \(3.80 \times 10^{-7}\text{ m}^2\) is suspended vertically from a rigid support. A mass of \(6.50\text{ kg}\) is attached to the lower end of the wire, causing an extension of \(1.85\text{ mm}\). The wire does not exceed its limit of proportionality.
(i) Calculate the tensile stress in the wire. [2]
(ii) Calculate the tensile strain in the wire. [1]
(iii) Determine the Young modulus \(E\) of the metal. [2]
(iv) Calculate the elastic potential energy stored in the wire. [2]
(c) The mass is increased until the wire deforms plastically. State and explain the change, if any, in the internal energy of the wire when the load is removed compared to before the wire was loaded. [1]
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Worked solution
(a) (i) Young modulus is the ratio of tensile stress to tensile strain.
(ii) Hooke's law states that the extension is directly proportional to the applied load/force, provided the limit of proportionality is not exceeded.
(iii) Elastic deformation: the material returns to its original length/dimensions when the deforming force is removed. Plastic deformation: the material does not return to its original length/dimensions (permanent deformation remains) when the deforming force is removed.
(c) The internal energy increases because work is done to permanently displace planes of atoms (plastic work), which is converted into thermal energy within the wire.
Marking scheme
(a)(i) - Stress / strain [B1]
(a)(ii) - Extension is (directly) proportional to force / load provided limit of proportionality is not exceeded [B1]
(a)(iii) - Elastic: returns to original shape / length when load is removed [B1] - Plastic: permanent deformation / does not return to original shape when load is removed [B1]
(b)(iv) - Formula \(E_p = \frac{1}{2}F\Delta L\) or \(E_p = \frac{1}{2} k x^2\) [C1] - \(E_p = 5.90 \times 10^{-2}\text{ J}\) [A1]
(c) - Internal energy increases AND work done in plastic deformation is converted into thermal energy [B1]
Question 4 · Structured Question
12 marks
(a) (i) State what is meant by the Doppler effect. [1]
(ii) An emergency vehicle travels at a constant speed of \(28.0\text{ m s}^{-1}\) directly towards a stationary observer. The siren on the vehicle emits sound of frequency \(850\text{ Hz}\). The speed of sound in air is \(340\text{ m s}^{-1}\). Calculate the frequency of the sound detected by the observer. [2]
(b) (i) State the principle of superposition. [1]
(ii) State two conditions required for two waves to produce an observable interference pattern. [2]
(c) Light of wavelength \(532\text{ nm}\) is incident normally on a double slit of separation \(d\). An interference pattern is formed on a screen placed parallel to the slits at a distance of \(2.20\text{ m}\) from the slits. The distance across 8 bright fringe separations is measured to be \(19.2\text{ mm}\).
(i) Calculate the fringe separation \(x\). [1]
(ii) Determine the slit separation \(d\). [2]
(iii) The double slit is replaced by a diffraction grating having \(4.00 \times 10^5\text{ lines m}^{-1}\).
1. Calculate the angle of the second-order maximum for this light. [2] 2. Determine the total number of primary maxima observable for this light with the grating. [1]
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Worked solution
(a) (i) The Doppler effect is the observed change in frequency (or wavelength) of a wave caused by the relative motion between the source and the observer.
(ii) Using the Doppler formula for a approaching source: \(f_o = f_s \left(\frac{v}{v - v_s}\right) = 850 \times \left(\frac{340}{340 - 28.0}\right) = 850 \times \frac{340}{312} = 926.3\text{ Hz} \approx 926\text{ Hz}\)
(b) (i) The principle of superposition states that when two or more waves meet at a point, the resultant displacement is equal to the vector sum of the displacements of the individual waves.
(ii) 1. The sources/waves must be coherent (have a constant phase difference). 2. The waves must have approximately equal amplitudes (or be unpolarised / polarised in the same plane).
(c)(iii)2. - Total maxima \(= 9\) (allow \(n_{\text{max}} = 4\)) [B1]
Question 5 · Structured Question
12 marks
(a) (i) Define electric potential difference (p.d.). [1]
(ii) In the equation for electric current \(I = Anvq\), identify the meaning of the symbols \(n\) and \(v\). [2]
(b) A cylindrical metal wire of uniform cross-sectional area \(4.50 \times 10^{-7}\text{ m}^2\) and length \(1.80\text{ m}\) is made of a material with resistivity \(2.80 \times 10^{-8}\ \Omega\text{ m}\). Show that the resistance of the wire is \(0.112\ \Omega\). [2]
(c) A battery of electromotive force (e.m.f.) \(E = 9.00\text{ V}\) and internal resistance \(r = 1.50\ \Omega\) is connected in series with a fixed resistor of resistance \(R_1 = 4.50\ \Omega\) and a negative temperature coefficient (NTC) thermistor of resistance \(R_T\).
A voltmeter of negligible current consumption is connected across the fixed resistor \(R_1\).
(i) Define electromotive force (e.m.f.). [1]
(ii) At a temperature of \(20^\circ\text{C}\), the reading on the voltmeter is \(3.60\text{ V}\).
1. Calculate the current in the circuit. [2] 2. Calculate the resistance \(R_T\) of the thermistor at \(20^\circ\text{C}\). [2] 3. Determine the terminal potential difference across the battery. [1]
(iii) The temperature of the thermistor is increased. State and explain the change, if any, in the voltmeter reading. [1]
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Worked solution
(a) (i) Potential difference is defined as the electrical energy transferred per unit charge (or work done per unit charge in moving charge between two points).
(ii) \(n\): number density of charge carriers (number of free charge carriers per unit volume). \(v\): average / mean drift velocity of charge carriers.
(c) (i) Electromotive force is the energy converted from chemical (or other) forms to electrical energy per unit charge transferred round a complete circuit.
(iii) As the temperature of an NTC thermistor increases, its resistance \(R_T\) decreases. This reduces the total circuit resistance, increasing the circuit current \(I\). Since \(V_1 = I R_1\) and \(R_1\) is constant, the voltmeter reading increases.
Marking scheme
(a)(i) - Energy transferred (from electrical to other forms) per unit charge [B1]
(a)(ii) - \(n\): number of charge carriers per unit volume / number density of free electrons [B1] - \(v\): (mean) drift velocity / speed of charge carriers [B1]
(b) - Substitution \(R = (2.80 \times 10^{-8} \times 1.80) / (4.50 \times 10^{-7})\) [M1] - Evaluation to \(0.112\ \Omega\) showing at least 3 sig figs before rounding (\(0.1120\ \Omega\)) [A1]
(c)(i) - Energy converted from other forms to electrical energy per unit charge [B1]
(c)(ii)1. - \(I = V / R = 3.60 / 4.50\) [C1] - \(I = 0.800\text{ A}\) [A1]
(c)(iii) - Resistance of thermistor decreases, so current increases, causing p.d. across \(R_1\) to increase [B1]
Paper 3 (Advanced Practical Skills)
Perform two experimental investigations. Record raw data, plot graphs, and evaluate uncertainties.
2 Question · 40 marks
Question 1 · practical_investigation
20 marks
In this experiment, you will investigate how the electric current in a circuit depends on the effective length of a resistance wire.
(a) Set up the circuit comprising a power supply, a switch, a fixed resistor, a digital ammeter, and a resistance wire mounted along a metre rule.
Attach the crocodile clip to the wire so that the contact length is \(x = 0.200\text{ m}\). Close the switch, measure and record the current \(I\), and open the switch.
(b) Vary the contact position to obtain values of current \(I\) for at least six different lengths \(x\) in the range \(0.100\text{ m} \leqslant x \leqslant 0.900\text{ m}\).
Record all your values of \(x\), \(I\), and \(\frac{1}{I}\) in a suitable table. Include columns for any repeated readings and calculated values. [9]
(c) (i) On the grid provided, plot a graph of \(\frac{1}{I}\) on the \(y\)-axis against \(x\) on the \(x\)-axis. Draw the straight line of best fit. [4]
(ii) Determine the gradient and the \(y\)-intercept of the line of best fit. [2]
(e) The constant \(P\) is given by \(P = \frac{r}{E}\), where \(r\) is the resistance per unit length of the wire and \(E\) is the electromotive force (e.m.f.) of the power supply. Given that \(E = 3.0\text{ V}\), calculate the value of \(r\). Include an appropriate unit. [1]
- At least 6 sets of readings covering a range \(\Delta x \geqslant 0.600\text{ m}\). - Column headings correctly formatted with quantity name and solidus/brackets with unit. - Raw \(x\) recorded to nearest \(1\text{ mm}\) (\(0.001\text{ m}\)), raw \(I\) to equal precision. - Calculated \(\frac{1}{I}\) values given to the same number of significant figures as (or one more than) raw current \(I\).
(c) Graph Plotting: - Axes and scales: Linear scales selected such that plotted points occupy more than half the grid in both directions; no awkward scales (e.g. no multiples of 3, 7). Marked with variable and unit. - Plotting: All points plotted accurately to within half a small grid square. - Line of best fit: Thin, straight line with balanced distribution of points above and below along the entire line.
(c)(ii) Gradient and Intercept: - Points from line chosen with \(\Delta x\) spanning at least half the length of drawn line, e.g. \((0.100, 4.72)\) and \((0.850, 9.41)\): $$\text{gradient} = \frac{9.41 - 4.72}{0.850 - 0.100} = \frac{4.69}{0.750} = 6.25\text{ A}^{-1}\text{ m}^{-1}$$ - \(y\)-intercept determined either by direct reading if axis starts at \(x = 0\), or using \(c = y_1 - m x_1\): $$y\text{-intercept} = 4.72 - (6.25 \times 0.100) = 4.10\text{ A}^{-1}$$
**(d) Determining \(P\) and \(Q\):** - Equation: \(\frac{1}{I} = P x + Q\) - Comparing with \(y = mx + c\): - \(P = \text{gradient} = 6.25\text{ A}^{-1}\text{ m}^{-1}\) - \(Q = y\text{-intercept} = 4.10\text{ A}^{-1}\)
(a) - [1] Raw value of \(I\) recorded with appropriate unit (\(\text{A}\) or \(\text{mA}\)) and sensible precision for the digital ammeter. - [1] Calculation of \(\frac{1}{I}\) correct and given to 2 or 3 significant figures.
(b) Data Table (9 marks total): - [1] Successful collection of data for at least 6 distinct values of \(x\). - [1] Wide range of \(x\) values explored: difference between largest and smallest \(x\) is at least \(0.600\text{ m}\). - [1] Column headings: every column has quantity and standard unit separated by a solidus or in brackets (e.g., \(x / \text{m}\), \(I / \text{A}\), \(1/I / \text{A}^{-1}\)). - [1] Consistency of raw length measurements: recorded consistently to \(1\text{ mm}\) (or \(0.001\text{ m}\)). - [1] Consistency of raw current measurements: all recorded to the same number of decimal places. - [1] Evidence of repeated measurements of \(I\) for each length \(x\) and correct mean calculated. - [1] Significant figures: values of \(\frac{1}{I}\) given to the same number of s.f. as (or one more than) the least number of s.f. in raw \(I\). - [1] Arithmetic: calculated values of \(\frac{1}{I}\) are mathematically correct. - [1] Quality of data: trend shows \(I\) decreases as \(x\) increases; points lie reasonably close to a straight line.
(c) Graph (6 marks total): - [1] Axes: continuous linear scale with convenient divisions (e.g. 1, 2, 5); plotted data occupies > 50% of the grid area in both \(x\) and \(y\) directions; axes labelled with quantity and unit. - [1] Plotting: all table points plotted accurately to within half a small square, with sharp crosses or circled dots. - [1] Line of best fit: straight line drawn with a ruler, well-balanced points along its full length. - [1] Gradient: uses coordinates from the line (not raw data points unless on the line); the triangle used has hypotenuse > 50% of the drawn line; correct substitution of \(\Delta y / \Delta x\). - [1] \(y\)-intercept: read directly if \(x=0\) is on the axis, or correctly calculated using \(c = y - mx\) with coordinates from the line.
(d) Constants (2 marks total): - [1] Value of \(P\) equated to gradient and value of \(Q\) equated to intercept, both given to 2 or 3 s.f. - [1] Units for both constants correct (\(P\) in \(\text{A}^{-1}\text{ m}^{-1}\) or \(\Omega^{-1}\text{ m}^{-1}\text{ V}^{-1}\); \(Q\) in \(\text{A}^{-1}\)).
(e) Analysis (1 mark total): - [1] Correct calculation of \(r = P \times E\) with appropriate unit (\(\Omega\text{ m}^{-1}\) or \(\text{V A}^{-1}\text{ m}^{-1}\)).
Question 2 · Apparatus Measurement with Error Evaluation & Improvements
20 marks
In this experiment, you will investigate the oscillatory motion of a suspended weighted ruler immersed in water.
(a) Measure and record the width \(w\) and thickness \(t\) of the wooden rule provided.
(c) Displace the top of the ruler horizontally through a small displacement and release it so that it executes small angular oscillations. Determine the period \(T_1\) of the oscillations.
(d) Adjust the clamp so that a greater length \(L_2\) of the ruler is submerged in water. Measure and record \(L_2\) and determine the new period \(T_2\).
(e) It is suggested that the relationship between the period \(T\) and the submerged length \(L\) is given by: \[ T^2 = C \left(1 + \frac{\rho_{\text{w}}}{\rho_{\text{r}}} \frac{L}{L_0} \right) \] where \(C\) is a constant, \(\rho_{\text{w}}\) is the density of water, \(\rho_{\text{r}}\) is the density of the ruler, and \(L_0\) is the total length of the ruler. Assuming \(k = \frac{T^2}{L_0 + \beta L}\) is constant where \(\beta = 1.25\) and \(L_0 = 1.000\text{ m}\):
(i) Calculate the two values of \(k\) for your data.
first value of \(k = \) ........................................................ second value of \(k = \) ........................................................
(ii) State whether the experimental data support the suggested relationship. Justify your conclusion by comparing the percentage difference between the two values of \(k\) with an estimated percentage uncertainty. [2]
(f) (i) Suggest four sources of error or limitations of the procedure for this experiment.
(ii) Suggest four improvements that could be made to the experiment to address the limitations identified in (f)(i). You may suggest the use of other apparatus.
Write your answers in the table provided. [8]
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Worked solution
(a) Representative measurements: \(w = 28.5\text{ mm} = 2.85\text{ cm}\) \(t = 4.2\text{ mm} = 0.42\text{ cm}\) Repeated readings taken and recorded with correct units.
(b) (i) Submerged length: \(L_1 = 12.0\text{ cm} = 0.120\text{ m}\). (ii) Uncertainty in reading the meniscus/submerged level: \(\Delta L \approx \pm 0.2\text{ cm}\). Percentage uncertainty: \(\frac{0.2}{12.0} \times 100\% = 1.7\%\).
For set 2: \(k_2 = \frac{T_2^2}{L_0 + 1.25 L_2} = \frac{1.56^2}{1.000 + 1.25(0.250)} = \frac{2.4336}{1.3125} = 1.85\text{ s}^2\text{ m}^{-1}\)
(ii) Percentage difference: \(\frac{|1.85 - 1.75|}{1.75} \times 100\% = 5.7\%\). Since the percentage difference \((5.7\%)\) is less than the typical experimental uncertainty limit (e.g. \(10\%\)), the experimental data support the suggested relationship.
(f) Limitations and Improvements Table: 1. Limitation: Meniscus and water surface disturbance make reading the submerged length \(L\) inaccurate. Improvement: Add a drop of detergent / use a narrow pointer fixed to the ruler aligned with the flat water surface.
2. Limitation: Rapid damping of oscillations makes counting a large number of cycles difficult. Improvement: Use a light gate connected to a data logger to record time period accurately over fewer oscillations.
3. Limitation: Two sets of readings are not enough to confirm a relationship. Improvement: Take measurements for at least five different submerged lengths \(L\) and plot a graph of \(T^2\) against \(L\).
4. Limitation: Ruler hits the sides or bottom of the beaker during oscillation / water sloshes in the beaker. Improvement: Use a wider and deeper container / tank so wall effects and boundary drag are negligible.
Marking scheme
(a) - B1: Value of \(w\) given to the nearest \(0.1\text{ mm}\) (or \(0.01\text{ cm}\)) and in range \(2.0\text{ cm} - 3.5\text{ cm}\). - B1: Value of \(t\) given to the nearest \(0.1\text{ mm}\) (or \(0.01\text{ cm}\)) with evidence of repeated readings. [2]
(b) - B1: Value of \(L_1\) recorded with appropriate units and precision (nearest \(1\text{ mm}\)). - B1: Percentage uncertainty in \(L_1\) calculated correctly using absolute uncertainty in range \(1\text{ mm} - 3\text{ mm}\). [2]
(c) - M1: Measurement of multiple oscillations (at least \(10\) oscillations timed) shown, with repeats. - A1: \(T_1\) calculated correctly and recorded to \(2\) or \(3\) significant figures with unit \(\text{s}\). [2]
(d) - B1: \(L_2\) recorded with \(L_2 > L_1\). - B1: Raw timing measurements for second oscillation showing repeated trials for multiple swings. - B1: \(T_2\) calculated correctly with \(T_2 > T_1\). - B1: Quality of data: values are physically reasonable. [4]
(e) - C1: Calculation of both values of \(k\) correct to \(2\) or \(3\) s.f. with correct units (\(\text{s}^2\text{ m}^{-1}\)). - A1: Sensible comparison of percentage difference to a stated criterion percentage (e.g. \(10\% - 20\%\)), followed by a valid conclusion consistent with candidate's numbers. [2]
(f) Evaluation and Improvements (maximum 4 pairs, 1 mark per valid limitation [A] and 1 mark per matching improvement [B]): - A1: Difficulty in reading \(L\) due to meniscus / water surface movement. - B1: Use a flat pointer attached to ruler / dye water / use a detergent to eliminate meniscus. - A2: Oscillations damp quickly / difficult to judge completion of a cycle. - B2: Use video recording with a timer / light gate with data logger / position sensor. - A3: Two sets of data are insufficient to draw a valid conclusion. - B3: Take at least 6 different values of \(L\) and plot a graph of \(T^2\) against \(L\). - A4: Oscillations are not strictly planar (ruler twists / hits beaker sides). - B4: Use a wider container and guide/pivot collar to restrict motion to one plane. [Total: 20 marks]
Paper 4 (A Level Structured Questions)
Answer all questions in the spaces provided. Full working and SI units required.
10 Question · 100 marks
Question 1 · theory
10 marks
(a) Define gravitational potential at a point. [2]
(b) A spherical asteroid of radius \(R = 4.2 \times 10^5\text{ m}\) has uniform density and no atmosphere. The gravitational potential at its surface is \(-3.8 \times 10^5\text{ J kg}^{-1}\).
(i) Calculate the mass \(M\) of the asteroid. [2]
(ii) A small probe of mass \(120\text{ kg}\) is launched radially outwards from the surface of the asteroid with speed \(v\). Calculate the minimum launch speed \(v\) required for the probe to reach a distance of \(3R\) from the centre of the asteroid. [3]
(c) Explain why the escape speed from the surface of the asteroid is independent of the mass of the probe. [2]
(d) State the effect, if any, on the gravitational potential energy of the probe as it moves away from the asteroid. [1]
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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) The gravitational potential at the surface of a uniform sphere is given by: $$\phi = -\frac{GM}{R}$$ Rearranging for \(M\): $$M = \frac{|\phi| R}{G} = \frac{3.8 \times 10^5 \times 4.2 \times 10^5}{6.67 \times 10^{-11}} = 2.39 \times 10^{21}\text{ kg} \approx 2.4 \times 10^{21}\text{ kg}$$
(ii) By conservation of energy, the initial kinetic energy per unit mass must equal the change in gravitational potential between \(r = R\) and \(r = 3R\): $$\Delta \phi = \phi(3R) - \phi(R) = -\frac{GM}{3R} - \left(-\frac{GM}{R}\right) = \frac{2}{3}\frac{GM}{R} = \frac{2}{3} \times 3.8 \times 10^5 = 2.533 \times 10^5\text{ J kg}^{-1}$$ $$\frac{1}{2} v^2 = \Delta \phi \implies v = \sqrt{2 \times 2.533 \times 10^5} = 7.12 \times 10^2\text{ m s}^{-1} \approx 7.1 \times 10^2\text{ m s}^{-1}$$
(c) The gravitational potential energy of the probe is proportional to its mass (\(E_p = m\phi\)), and the kinetic energy is also proportional to its mass (\(E_k = \frac{1}{2}mv^2\)). Equating the two quantities causes the mass \(m\) of the probe to cancel out on both sides of the equation.
(d) The gravitational potential energy increases (becomes less negative) as the distance from the asteroid increases.
Marking scheme
(a) - work done per unit mass [B1] - (work done in bringing a test mass) from infinity (to the point) [B1]
(c) - both \(E_k\) and \(\Delta E_p\) are proportional to the mass \(m\) of the probe [B1] - mass \(m\) cancels in the conservation of energy equation (\(\frac{1}{2}mv^2 = m\Delta \phi\)) [B1]
(d) - increases / becomes less negative [B1]
Question 2 · theory
10 marks
(a) State the two conditions required for an oscillation to be simple harmonic. [2]
(b) A light vertical spring has a mass of \(0.35\text{ kg}\) attached to its lower end. The mass is pulled downwards by a distance of \(4.0\text{ cm}\) from its equilibrium position and released from rest at time \(t = 0\). The period of oscillation is \(0.68\text{ s}\). Resistance to the motion is negligible.
(i) Calculate the angular frequency \(\omega\) of the oscillation. [2]
(ii) Determine the magnitude of the maximum acceleration of the mass. [2]
(iii) Calculate the speed of the mass when its displacement is \(2.5\text{ cm}\) from the equilibrium position. [2]
(c) The oscillating mass is now placed into a container of oil.
(i) State what is meant by damped oscillations. [1]
(ii) State the type of damping that causes the mass to return to its equilibrium position in the shortest possible time without oscillating. [1]
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Worked solution
(a) For simple harmonic motion: 1. Acceleration is directly proportional to displacement from a fixed reference point (equilibrium position). 2. Acceleration is always directed towards the equilibrium position (opposite in direction to displacement).
(b)(i) Angular frequency is given by: $$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.68} = 9.24\text{ rad s}^{-1} \approx 9.2\text{ rad s}^{-1}$$
(ii) Maximum acceleration occurs at maximum displacement (amplitude \(x_0 = 4.0\text{ cm} = 0.040\text{ m}\)): $$a_{\max} = \omega^2 x_0 = (9.24)^2 \times 0.040 = 85.37 \times 0.040 = 3.41\text{ m s}^{-2} \approx 3.4\text{ m s}^{-2}$$
(iii) The speed at displacement \(x = 0.025\text{ m}\) is given by: $$v = \omega \sqrt{x_0^2 - x^2} = 9.24 \times \sqrt{0.040^2 - 0.025^2} = 9.24 \times \sqrt{0.0016 - 0.000625} = 9.24 \times \sqrt{0.000975} = 9.24 \times 0.03122 = 0.289\text{ m s}^{-1} \approx 0.29\text{ m s}^{-1}$$
(c)(i) Damped oscillations are oscillations in which resistive forces cause energy to be dissipated from the system, resulting in a gradual decrease in amplitude over time.
(ii) Critical damping.
Marking scheme
(a) - acceleration directly proportional to displacement [B1] - acceleration directed towards equilibrium / fixed point (or in opposite direction to displacement) [B1]
(b)(i) - \(\omega = 2\pi / T\) OR \(2\pi / 0.68\) [C1] - \(\omega = 9.2\text{ rad s}^{-1}\) (allow \(9.24\text{ rad s}^{-1}\)) [A1]
(b)(ii) - \(a = \omega^2 x_0\) OR \((9.24)^2 \times 0.040\) [C1] - \(a = 3.4\text{ m s}^{-2}\) (allow \(3.41\text{ m s}^{-2}\)) [A1]
(b)(iii) - \(v = \omega \sqrt{x_0^2 - x^2}\) OR \(9.24 \times \sqrt{0.040^2 - 0.025^2}\) [C1] - \(v = 0.29\text{ m s}^{-1}\) (allow \(0.289\text{ m s}^{-1}\)) [A1]
(c)(i) - loss of energy / amplitude decreasing due to resistive / frictional forces [B1]
(c)(ii) - critical damping [B1]
Question 3 · theory
10 marks
(a) Define capacitance. [1]
(b) An isolated solid metal sphere in a vacuum has a radius \(r = 18\text{ cm}\) and carries a positive charge \(Q = +3.6\text{ nC}\).
(i) Show that the capacitance of the isolated sphere is approximately \(2.0 \times 10^{-11}\text{ F}\). [2]
(ii) Calculate the electric potential at the surface of the sphere. [2]
(c) A circuit contains a \(470\,\mu\text{F}\) capacitor charged to a potential difference of \(12\text{ V}\). The charged capacitor is disconnected from the power supply and connected across a resistor of resistance \(R = 68\text{ k}\Omega\).
(i) Calculate the time constant \(\tau\) of the discharge circuit. [1]
(ii) Calculate the energy initially stored in the capacitor. [2]
(iii) Calculate the potential difference across the capacitor at time \(t = 45\text{ s}\) after the discharge begins. [2]
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Worked solution
(a) Capacitance is the charge stored per unit potential difference (\(C = Q/V\)).
(b)(i) The capacitance of an isolated conducting sphere of radius \(r\) is: $$C = 4\pi\varepsilon_0 r = 4\pi \times (8.85 \times 10^{-12}) \times 0.18 = 2.002 \times 10^{-11}\text{ F} \approx 2.0 \times 10^{-11}\text{ F}$$
(ii) The electric potential at the surface of the sphere is: $$V = \frac{Q}{4\pi\varepsilon_0 r} = \frac{Q}{C} = \frac{3.6 \times 10^{-9}}{2.002 \times 10^{-11}} = 179.8\text{ V} \approx 180\text{ V}$$
(c)(i) The time constant is: $$\tau = RC = (68 \times 10^3) \times (470 \times 10^{-6}) = 31.96\text{ s} \approx 32\text{ s}$$
(ii) The energy stored initially is: $$E = \frac{1}{2} C V_0^2 = \frac{1}{2} \times (470 \times 10^{-6}) \times (12)^2 = 0.5 \times 470 \times 10^{-6} \times 144 = 0.0338\text{ J} \approx 0.034\text{ J}$$
(iii) The potential difference during discharge is given by: $$V = V_0 e^{-t/RC} = 12 \times e^{-45 / 31.96} = 12 \times e^{-1.408} = 12 \times 0.2446 = 2.94\text{ V} \approx 2.9\text{ V}$$
Marking scheme
(a) - charge / potential (difference) OR ratio of charge to potential (difference) [B1]
(b) A flat circular search coil consisting of 250 turns of wire, each of area \(4.8 \times 10^{-4}\text{ m}^2\), is placed in a uniform magnetic field of flux density \(0.064\text{ T}\). Initially, the plane of the coil is perpendicular to the magnetic field.
(i) Calculate the magnetic flux linkage of the search coil. [2]
(ii) The coil is rotated through an angle of \(90^\circ\) about an axis in its plane in a time interval of \(0.030\text{ s}\). Calculate the magnitude of the average electromotive force (e.m.f.) induced across the coil. [2]
(c) A Hall probe is used to measure magnetic flux density. The semiconductor slice in the probe has a thickness \(t = 0.20\text{ mm}\) parallel to the magnetic field and a charge carrier density \(n = 1.8 \times 10^{23}\text{ m}^{-3}\). A current of \(45\text{ mA}\) passes through the slice.
(i) Calculate the Hall voltage \(V_H\) generated when the probe is placed in a perpendicular magnetic field of flux density \(0.15\text{ T}\). [2]
(ii) State and explain the effect on the Hall voltage if a metal slice of the same dimensions is used instead of the semiconductor slice. [2]
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Worked solution
(a) Magnetic flux density is the force experienced per unit length by a long straight conductor carrying unit current at right angles to the magnetic field.
(b)(i) Magnetic flux linkage is given by: $$\Phi N = B A N = 0.064 \times 4.8 \times 10^{-4} \times 250 = 7.68 \times 10^{-3}\text{ Wb} \approx 7.7 \times 10^{-3}\text{ Wb}$$
(ii) When rotated by \(90^\circ\), the plane becomes parallel to the field, so final flux linkage is zero. By Faraday's law: $$|E| = \frac{\Delta (N\Phi)}{\Delta t} = \frac{7.68 \times 10^{-3} - 0}{0.030} = 0.256\text{ V} \approx 0.26\text{ V}$$
(c)(i) The Hall voltage is given by: $$V_H = \frac{B I}{n t q}$$ Substituting the values: $$V_H = \frac{0.15 \times (45 \times 10^{-3})}{(1.8 \times 10^{23}) \times (0.20 \times 10^{-3}) \times (1.60 \times 10^{-19})} = \frac{6.75 \times 10^{-3}}{5.76} = 1.17 \times 10^{-3}\text{ V} \approx 1.2\text{ mV}$$
(ii) For a metal, the number density \(n\) of free electrons (charge carriers) is much greater (by a factor of \(10^5\) or more). Since \(V_H \propto 1/n\), the Hall voltage generated in a metal is much smaller (often negligibly small).
Marking scheme
(a) - force per unit length [B1] - on a straight conductor carrying unit current perpendicular to the field [B1]
(b)(i) - \(\Phi N = B A N\) OR \(0.064 \times 4.8 \times 10^{-4} \times 250\) [C1] - \(\Phi N = 7.7 \times 10^{-3}\text{ Wb}\) (or \(7.68 \times 10^{-3}\text{ Wb}\)) [A1]
(c)(i) - \(V_H = BI / (ntq)\) with correct substitutions [C1] - \(V_H = 1.2 \times 10^{-3}\text{ V}\) (allow \(1.17\text{ mV}\) or \(1.2\text{ mV}\)) [A1]
(c)(ii) - \(V_H\) decreases / becomes much smaller [B1] - because \(n\) (number density of charge carriers) in a metal is much larger (and \(V_H \propto 1/n\)) [B1]
Question 5 · theory
10 marks
(a) State what is meant by the work function energy of a metal. [2]
(b) In a photoelectric effect experiment, monochromatic light of wavelength \(\lambda = 380\text{ nm}\) is incident on a clean zinc surface. The work function energy of zinc is \(4.31\text{ eV}\).
(i) Calculate the energy of a photon of the incident light in electronvolts (\(\text{eV}\)). [2]
(ii) State and explain whether photoelectrons are emitted from the zinc surface for this incident light. [2]
(iii) Calculate the threshold wavelength of zinc. [2]
(c) In a separate experiment, electrons are accelerated from rest through a uniform potential difference of \(250\text{ V}\) in an evacuated chamber. Calculate the de Broglie wavelength of these accelerated electrons. [2]
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Worked solution
(a) The work function energy is the minimum energy required to remove an electron from the surface of a metal.
(b)(i) The energy of a photon in joules is: $$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{380 \times 10^{-9}} = 5.234 \times 10^{-19}\text{ J}$$ In electronvolts: $$E = \frac{5.234 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.271\text{ eV} \approx 3.27\text{ eV}$$
(ii) No photoelectrons are emitted. For emission to occur, the incident photon energy must be greater than or equal to the work function energy \(\Phi\). Here, \(E = 3.27\text{ eV} < \Phi = 4.31\text{ eV}\), so a single photon cannot provide enough energy to liberate an electron.
(c) Kinetic energy of the electrons is: $$E_k = eV = 1.60 \times 10^{-19} \times 250 = 4.00 \times 10^{-17}\text{ J}$$ Momentum \(p\) of an electron: $$p = \sqrt{2 m_e E_k} = \sqrt{2 \times 9.11 \times 10^{-31} \times 4.00 \times 10^{-17}} = \sqrt{7.288 \times 10^{-47}} = 8.537 \times 10^{-24}\text{ kg m s}^{-1}$$ de Broglie wavelength: $$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{8.537 \times 10^{-24}} = 7.766 \times 10^{-11}\text{ m} \approx 7.77 \times 10^{-11}\text{ m}$$
Marking scheme
(a) - minimum energy required [B1] - to remove / emit an electron from the surface of a metal [B1]
(b)(i) - \(E = hc / \lambda\) OR conversion by dividing by \(1.60 \times 10^{-19}\) [C1] - \(E = 3.27\text{ eV}\) (allow \(3.3\text{ eV}\) or \(5.23 \times 10^{-19}\text{ J}\)) [A1]
(b)(ii) - no / no photoelectrons emitted [B1] - (photon) energy \(3.27\text{ eV}\) is less than the work function \(4.31\text{ eV}\) (or \(f < f_0\) / \(\lambda > \lambda_0\)) [B1]
(b)(iii) - \(\lambda_0 = hc / \Phi\) with \(\Phi\) converted to J (\(4.31 \times 1.60 \times 10^{-19}\)) [C1] - \(\lambda_0 = 2.88 \times 10^{-7}\text{ m}\) (or \(288\text{ nm}\)) [A1]
(a) Define the decay constant \(\lambda\) of a radioactive nuclide. [1]
(b) A sample contains iodine-131 (\(^{131}_{\phantom{0}53}\text{I}\)), which has a half-life of \(8.0\text{ days}\). The initial activity of the sample is \(4.8 \times 10^7\text{ Bq}\).
(i) Show that the decay constant of iodine-131 is approximately \(1.0 \times 10^{-6}\text{ s}^{-1}\). [2]
(ii) Calculate the initial number \(N_0\) of iodine-131 nuclei in the sample. [2]
(iii) Calculate the time taken, in days, for the activity of the sample to decrease to \(3.0 \times 10^6\text{ Bq}\). [2]
(c) In a nuclear reactor, a stationary nucleus of uranium-235 undergoes induced fission by absorbing a slow-moving neutron according to the equation: $$^{235}_{\phantom{0}92}\text{U} + ^{1}_{0}\text{n} \to ^{141}_{\phantom{0}56}\text{Ba} + ^{92}_{36}\text{Kr} + 3\,^{1}_{0}\text{n}$$ The binding energy per nucleon of uranium-235 is \(7.59\text{ MeV}\), of barium-141 is \(8.33\text{ MeV}\), and of krypton-92 is \(8.51\text{ MeV}\).
Calculate the total energy released in this fission reaction in \(\text{MeV}\). [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).
(b)(i) The decay constant is given by: $$\lambda = \frac{\ln 2}{t_{1/2}}$$ Converting the half-life to seconds: $$t_{1/2} = 8.0 \times 24 \times 3600 = 6.912 \times 10^5\text{ s}$$ $$\lambda = \frac{0.69315}{6.912 \times 10^5} = 1.003 \times 10^{-6}\text{ s}^{-1} \approx 1.0 \times 10^{-6}\text{ s}^{-1}$$
(ii) The activity is \(A = \lambda N\). Rearranging for \(N_0\): $$N_0 = \frac{A_0}{\lambda} = \frac{4.8 \times 10^7}{1.003 \times 10^{-6}} = 4.79 \times 10^{13} \approx 4.8 \times 10^{13}$$
(a) State what is meant by the acoustic impedance of a medium. [1]
(b) A parallel beam of ultrasound of initial intensity \(I_0\) travels normally through a layer of muscle of thickness \(3.2\text{ cm}\). The linear attenuation coefficient of muscle for this frequency is \(23\text{ m}^{-1}\).
Calculate the ratio \(I / I_0\) of the transmitted intensity \(I\) to the initial intensity \(I_0\). [2]
(c) The transmitted ultrasound beam now strikes a flat boundary between muscle and bone at normal incidence.
The acoustic impedance of muscle is \(1.70 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}\). The acoustic impedance of bone is \(6.40 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}\).
(i) Calculate the intensity reflection coefficient \(\alpha\) at the muscle-bone boundary. [2]
(ii) Determine the percentage of the incident ultrasound intensity that is transmitted into the bone. [1]
(d) In clinical diagnostic ultrasound imaging, a water-based coupling gel is applied to the skin before placing the transducer.
(i) State and explain the problem that occurs if there is an air gap between the transducer and the patient's skin. [2]
(ii) Explain how the coupling gel resolves this problem. [2]
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Worked solution
(a) Acoustic impedance \(Z\) is the product of the density \(\rho\) of the medium and the speed \(c\) of ultrasound in that medium (\(Z = \rho c\)).
(ii) The fraction of intensity transmitted across the boundary is: $$T = 1 - \alpha = 1 - 0.3367 = 0.6633 \approx 66\%$$
(d)(i) Air has an extremely low acoustic impedance (\(\approx 400\text{ kg m}^{-2}\text{ s}^{-1}\)) compared to skin/soft tissue (\(\approx 1.5 \times 10^6\text{ kg m}^{-2}\text{ s}^{-1}\)). Because of this immense difference, almost \(100\%\) (\(\approx 99.9\%\)) of the ultrasound intensity is reflected at the air-skin boundary, so virtually no ultrasound enters the body.
(ii) The coupling gel has an acoustic impedance that is matched (very close) to that of human skin. By excluding air, it prevents large reflections at the surface and allows maximum transmission of ultrasound into the body.
Marking scheme
(a) - product of density and speed of sound / ultrasound (in the medium) [B1]
(c)(ii) - \(66\%\) (or \(0.66\) or \(66.3\%\)) [B1]
(d)(i) - large difference in acoustic impedance between air and skin / soft tissue [B1] - almost all ultrasound reflected / very little ultrasound penetrates the skin [B1]
(d)(ii) - acoustic impedance of gel is very similar / matched to skin [B1] - air is excluded / minimal reflection so ultrasound is transmitted into the body [B1]
Question 8 · structured
10 marks
(a) Define gravitational potential at a point. [2]
(b) An isolated planet P of mass \(M\) and radius \(R\) may be considered to have its mass concentrated at its centre. A spacecraft of mass \(m = 850\text{ kg}\) is launched vertically upwards from the surface of planet P.
(i) Write down an expression in terms of \(G, M, m, R\) and \(h\) for the change in gravitational potential energy \(\Delta E_p\) of the spacecraft when it reaches a height \(h\) above the surface of the planet. [1]
(ii) Planet P has mass \(M = 4.80 \times 10^{24}\text{ kg}\) and radius \(R = 6.00 \times 10^6\text{ m}\). Calculate the minimum launch speed \(v\) required for the spacecraft to reach a maximum height of \(h = 3.00 \times 10^6\text{ m}\) above the surface. Air resistance is negligible. [4]
(c) Planet P has a small moon orbiting it in a circular orbit of radius \(r = 4.20 \times 10^7\text{ m}\). Calculate the orbital period \(T\) of this moon in hours. [3]
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Worked solution
(a) Gravitational potential is the work done per unit mass in bringing a small test mass from infinity to that point.
(a) Define gravitational potential at a point. [2]
(b) An isolated, spherical planet X of radius \(R_X = 3.40 \times 10^6\text{ m}\) has no atmosphere. A probe of mass \(m = 480\text{ kg}\) is launched vertically upwards from the surface of planet X with an initial speed \(v_0 = 2.95 \times 10^3\text{ m s}^{-1}\). The probe reaches a maximum height \(h = 1.80 \times 10^6\text{ m}\) above the surface before momentarily coming to rest.
(i) Show that the mass of planet X is \(6.41 \times 10^{23}\text{ kg}\). [3]
(ii) Calculate the gravitational field strength \(g\) at the maximum height \(h\) above the surface of planet X. [2]
(c) The probe is now placed into a circular orbit at a distance of \(1.80 \times 10^6\text{ m}\) above the surface of planet X.
Calculate the orbital period \(T\) of the probe in hours. [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) By conservation of mechanical energy: Change in kinetic energy = Change in gravitational potential energy \(\frac{1}{2} m v_0^2 - 0 = -\frac{G M_X m}{r} - \left(-\frac{G M_X m}{R_X}\right)\) where \(r = R_X + h = 3.40 \times 10^6 + 1.80 \times 10^6 = 5.20 \times 10^6\text{ m}\).
(a) State what is meant by the work function energy of a metal surface. [2]
(b) In a photoelectric experiment, monochromatic light of wavelength \(\lambda = 380\text{ nm}\) is incident on the clean surface of a caesium cathode inside a vacuum tube. The work function energy of caesium is \(2.14\text{ eV}\).
(i) Calculate the energy, in \(\text{eV}\), of a single photon of the incident light. [2]
(ii) Determine the maximum kinetic energy \(E_{\text{k,max}}\), in \(\text{J}\), of the emitted photoelectrons. [2]
(iii) Calculate the de Broglie wavelength \(\lambda_{\text{dB}}\) of an electron emitted with this maximum kinetic energy. [2]
(c) The intensity of the incident radiation is doubled while the wavelength is kept constant at \(380\text{ nm}\).
State and explain the effect, if any, on: 1. the maximum kinetic energy of the emitted photoelectrons, [1] 2. the rate of emission of photoelectrons. [1]
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Worked solution
(a) Work function energy is the minimum energy required to release an electron from the surface of a metal.
(c) 1. Maximum kinetic energy remains unchanged because the photon energy (or frequency/wavelength) is unchanged. 2. Rate of emission of photoelectrons doubles because doubling intensity doubles the number of incident photons per unit time.
Marking scheme
(a) - minimum energy [B1] - required to remove / emit an electron from the metal surface [B1]
(c) 1. - no change / unchanged because photon energy / frequency is constant [B1] 2. - doubles / increases by a factor of 2 because the rate of arrival of photons doubles (1 photon interacts with 1 electron) [B1]
Paper 5 (Planning, Analysis and Evaluation)
Answer Question 1 (Planning) and Question 2 (Analysis & Uncertainties).
0 Question · 0 marks
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