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Thinka Nov 2023 (V3) Cambridge International A Level-Style Mock — Physics (9702)

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An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V3) Cambridge International A Level Physics (9702) paper. Not affiliated with or reproduced from Cambridge.

Paper 1 (Multiple Choice)

Answer all forty questions. For each question there are four possible answers A, B, C and D.

40 PastPaper.question · 40 PastPaper.marks

PastPaper.question 1 · multiple-choice

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An experiment is conducted to determine the resistivity \(\rho\) of a uniform metal wire using the formula: \(\rho = \frac{R \pi d^2}{4L}\). The measured values with their absolute uncertainties are: Resistance, \(R = (4.50 \pm 0.09)\ \Omega\); Diameter, \(d = (0.38 \pm 0.01)\ \text{mm}\); Length, \(L = (0.800 \pm 0.004)\ \text{m}\). What is the percentage uncertainty in the calculated value of the resistivity \(\rho\)?

A.5.1%
B.7.3%
C.7.8%
D.9.8%"},

PastPaper.showAnswers

PastPaper.workedSolution

The percentage uncertainty in resistivity is given by the sum of the percentage uncertainties of the individual quantities, taking into account the power of each quantity: \(\frac{\Delta \rho}{\rho} \times 100\% = \frac{\Delta R}{R} \times 100\% + 2 \left(\frac{\Delta d}{d} \times 100\%\right) + \frac{\Delta L}{L} \times 100\%\). First, calculate each percentage uncertainty: For \(R\): \(\frac{0.09}{4.50} \times 100\% = 2.0\%\). For \(d\): \(\frac{0.01}{0.38} \times 100\% \approx 2.63\%\). For \(L\): \(\frac{0.004}{0.800} \times 100\% = 0.5\%\). Now combine them: Total percentage uncertainty = \(2.0\% + 2 \times 2.63\% + 0.5\% = 2.0\% + 5.26\% + 0.5\% = 7.76\%\), which rounds to \(7.8\%\).

PastPaper.markingScheme

1 mark for the correct calculation of individual percentage uncertainties, doubling the diameter's percentage uncertainty, and summing them up to obtain 7.8% (Option C).

PastPaper.question 2 · multiple-choice

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An experiment is performed to determine the density of a small metal sphere. The mass \(m\) of the sphere is measured as \((12.4 \pm 0.1)\ \text{g}\) and its diameter \(d\) is measured as \((1.42 \pm 0.02)\ \text{cm}\). What is the value of the density of the metal, with its associated absolute uncertainty, in \(\text{g\ cm}^{-3}\)?

A.\((8.3 \pm 0.1)\ \text{g\ cm}^{-3}\)
B.\((8.3 \pm 0.2)\ \text{g\ cm}^{-3}\)
C.\((8.3 \pm 0.4)\ \text{g\ cm}^{-3}\)
D.\((8.3 \pm 0.5)\ \text{g\ cm}^{-3}\)

PastPaper.showAnswers

PastPaper.workedSolution

The volume of the sphere is \(V = \frac{1}{6}\pi d^3\). The density is \(\rho = \frac{m}{V} = \frac{6m}{\pi d^3}\). Calculating the density value: \(\rho = \frac{6 \times 12.4}{\pi \times 1.42^3} \approx 8.271\ \text{g\ cm}^{-3}\). The fractional uncertainty in density is: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta d}{d} = \frac{0.1}{12.4} + 3 \left(\frac{0.02}{1.42}\right) \approx 0.00806 + 3 \times 0.01408 = 0.00806 + 0.04225 = 0.05031\). The absolute uncertainty in density is: \(\Delta \rho = 8.271 \times 0.05031 \approx 0.416\ \text{g\ cm}^{-3}\). Since absolute uncertainties are conventionally quoted to one significant figure, \(\Delta \rho \approx 0.4\ \text{g\ cm}^{-3}\). The density value is then quoted to the same decimal place: \(\rho = (8.3 \pm 0.4)\ \text{g\ cm}^{-3}\).

PastPaper.markingScheme

1 mark for calculating the density value to be 8.3 g/cm^3 and summing the mass and triple the diameter fractional uncertainties to obtain the absolute uncertainty of 0.4 g/cm^3.

PastPaper.question 3 · multiple-choice

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A digital balance is used to measure the mass of a beaker. When empty, the balance reads \(0.0\ \text{g}\). When a standard \(100.0\ \text{g}\) calibration mass is placed on it, the balance reads \(101.5\ \text{g}\). A series of measurements of the mass of a beaker are made using this balance. The readings are: \(52.3\ \text{g}\), \(52.2\ \text{g}\), \(52.4\ \text{g}\), \(52.3\ \text{g}\). Which statement correctly describes the errors in these measurements?

A.The measurements have high precision and a zero error.
B.The measurements have high precision and a systematic calibration error.
C.The measurements have low precision and a zero error.
D.The measurements have low precision and a systematic calibration error.

PastPaper.showAnswers

PastPaper.workedSolution

1. Precision: The measurements are close together (ranging from 52.2 g to 52.4 g), indicating a high degree of precision (small random error). 2. Systematic Error: The balance is calibrated incorrectly because it reads 101.5 g for a 100.0 g mass. Therefore, all measurements will have a systematic calibration error. 3. Zero Error: When empty, the balance reads 0.0 g, meaning there is no zero error (no offset at zero). Hence, the measurements have high precision and a systematic calibration error.

PastPaper.markingScheme

1 mark for identifying that the small range of readings indicates high precision, and the incorrect reading for the standard mass without a zero offset indicates a systematic calibration error.

PastPaper.question 4 · multiple-choice

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A beam of alpha particles (\(\text{He}^{2+}\)) and a beam of beta-minus particles (\(\text{e}^-\)) enter a region of uniform magnetic field. The magnetic field is directed perpendicular to and into the plane of the paper. Both types of particles enter the field with the same velocity \(v\), perpendicular to the boundary of the magnetic field. Which statement correctly describes their paths in the magnetic field?

A.Both the alpha particles and beta-minus particles deflect in the same direction, but the alpha particles follow a path of smaller radius of curvature.
B.Both the alpha particles and beta-minus particles deflect in opposite directions, and the alpha particles follow a path of smaller radius of curvature.
C.Both the alpha particles and beta-minus particles deflect in the same direction, but the beta-minus particles follow a path of smaller radius of curvature.
D.Both the alpha particles and beta-minus particles deflect in opposite directions, and the beta-minus particles follow a path of smaller radius of curvature.

PastPaper.showAnswers

PastPaper.workedSolution

1. Direction of deflection: Since alpha particles are positively charged and beta-minus particles are negatively charged, the magnetic forces on them act in opposite directions. By Fleming's Left-Hand Rule, they will deflect in opposite directions. 2. Radius of curvature: The magnetic force provides the centripetal force: \(qvB = \frac{m v^2}{r} \implies r = \frac{m v}{q B}\). Since both enter with the same velocity \(v\) into the same field \(B\), the radius is proportional to the mass-to-charge ratio \(\frac{m}{q}\). For the alpha particle, \(\left(\frac{m}{q}\right)_{\alpha} \approx \frac{4\ \text{u}}{2e} = 2\ \frac{\text{u}}{e}\). For the beta-minus particle, \(\left(\frac{m}{q}\right)_{\beta} \approx \frac{1/1840\ \text{u}}{e} \approx 5.4 \times 10^{-4}\ \frac{\text{u}}{e}\). Since \(\left(\frac{m}{q}\right)_{\beta} \ll \left(\frac{m}{q}\right)_{\alpha}\), the beta-minus particles will follow a path with a much smaller radius of curvature.

PastPaper.markingScheme

1 mark for identifying that opposite charges deflect in opposite directions and that the much smaller mass-to-charge ratio of beta-minus particles results in a smaller radius of curvature (Option D).

PastPaper.question 5 · multiple-choice

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A rigid, straight metal wire of length \(0.40\ \text{m}\) carries a current of \(3.5\ \text{A}\) in a direction pointing due North. The wire is placed in a uniform magnetic field of flux density \(0.050\ \text{T}\) that is directed vertically upwards. What is the magnitude and direction of the magnetic force acting on the wire?

A.\(0.070\ \text{N}\) pointing due East
B.\(0.070\ \text{N}\) pointing due West
C.\(0.070\ \text{N}\) pointing vertically downwards
D.Zero, because the magnetic field is perpendicular to the horizontal plane.

PastPaper.showAnswers

PastPaper.workedSolution

The magnitude of the magnetic force is given by \(F = B I L \sin\theta\). Here, the current points North (horizontal) and the magnetic field is directed vertically upwards, so the angle \(\theta = 90^\circ\). Thus, \(F = 0.050\ \text{T} \times 3.5\ \text{A} \times 0.40\ \text{m} \times \sin(90^\circ) = 0.070\ \text{N}\). To find the direction, apply Fleming's Left-Hand Rule: Point your first finger (Field) vertically UPWARDS. Point your second finger (Current) due NORTH (forward). Your thumb (Force) points due EAST (to the right). Thus, the force is \(0.070\ \text{N}\) pointing due East.

PastPaper.markingScheme

1 mark for correctly calculating the force as 0.070 N and using Fleming's Left-Hand Rule to determine the direction is East (Option A).

PastPaper.question 6 · multiple-choice

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Two long, straight, parallel wires, X and Y, are separated by a distance of \(8.0\ \text{cm}\) in a vacuum. Wire X carries a current of \(6.0\ \text{A}\) upwards and wire Y carries a current of \(4.0\ \text{A}\) downwards. What is the magnetic flux density \(B\) at a point P on the line joining the two wires, at a distance of \(2.0\ \text{cm}\) from wire X and \(6.0\ \text{cm}\) from wire Y?

A.\(1.3 \times 10^{-5}\ \text{T}\)
B.\(4.7 \times 10^{-5}\ \text{T}\)
C.\(6.0 \times 10^{-5}\ \text{T}\)
D.\(7.3 \times 10^{-5}\ \text{T}\)

PastPaper.showAnswers

PastPaper.workedSolution

The magnetic flux density due to a long straight wire is given by \(B = \frac{\mu_0 I}{2\pi r}\). Point P lies between the two wires: 1. Field due to wire X: \(B_{\text{X}} = \frac{(4\pi \times 10^{-7}) \times 6.0}{2\pi \times 0.020} = 6.0 \times 10^{-5}\ \text{T}\). By the right-hand grip rule, since the current is upwards, the magnetic field at point P (to the right of wire X) points INTO the page. 2. Field due to wire Y: \(B_{\text{Y}} = \frac{(4\pi \times 10^{-7}) \times 4.0}{2\pi \times 0.060} \approx 1.33 \times 10^{-5}\ \text{T}\). By the right-hand grip rule, since the current is downwards, the magnetic field at point P (to the left of wire Y) also points INTO the page. Since both fields point in the same direction, we add them: \(B_{\text{total}} = B_{\text{X}} + B_{\text{Y}} = 6.0 \times 10^{-5} + 1.33 \times 10^{-5} \approx 7.3 \times 10^{-5}\ \text{T}\).

PastPaper.markingScheme

1 mark for calculating the magnetic fields from both wires, recognizing they point in the same direction (into the page), and adding them to find 7.3 x 10^-5 T (Option D).

PastPaper.question 7 · multiple-choice

1 PastPaper.marks

A particle undergoes simple harmonic motion with an amplitude \(A\) and period \(T\). The particle starts from the equilibrium position at time \(t = 0\). What is the minimum time taken for the displacement of the particle to become \(\frac{\sqrt{3}}{2} A\)?

A.\(\frac{T}{12}\)
B.\(\frac{T}{6}\)
C.\(\frac{T}{4}\)
D.\(\frac{T}{3}\)

PastPaper.showAnswers

PastPaper.workedSolution

Since the particle starts from the equilibrium position at \(t = 0\), its displacement as a function of time is given by \(x = A \sin(\omega t)\), where \(\omega = \frac{2\pi}{T}\). We set \(x = \frac{\sqrt{3}}{2} A\): \(A \sin(\omega t) = \frac{\sqrt{3}}{2} A \implies \sin(\omega t) = \frac{\sqrt{3}}{2}\). The minimum positive angle satisfying this is \(\omega t = \frac{\pi}{3}\). Substituting \(\omega = \frac{2\pi}{T}\): \(\frac{2\pi}{T} t = \frac{\pi}{3} \implies t = \frac{T}{6}\).

PastPaper.markingScheme

1 mark for setting up the displacement equation starting from equilibrium, solving for the phase angle to find pi/3, and converting this to the time t = T/6 (Option B).

PastPaper.question 8 · multiple-choice

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A mass of \(0.20\ \text{kg}\) undergoes simple harmonic oscillations on a frictionless horizontal surface. The total energy of the oscillation is \(1.6\ \text{J}\). At a certain point in its motion, the displacement of the mass is equal to half of its amplitude. What are the kinetic energy and potential energy of the mass at this point?

A.Kinetic Energy = \(0.4\ \text{J}\), Potential Energy = \(1.2\ \text{J}\)
B.Kinetic Energy = \(0.8\ \text{J}\), Potential Energy = \(0.8\ \text{J}\)
C.Kinetic Energy = \(1.2\ \text{J}\), Potential Energy = \(0.4\ \text{J}\)
D.Kinetic Energy = \(1.4\ \text{J}\), Potential Energy = \(0.2\ \text{J}\)

PastPaper.showAnswers

PastPaper.workedSolution

In simple harmonic motion, the potential energy \(E_p\) at displacement \(x\) is given by \(E_p = \frac{1}{2} k x^2\), where \(k\) is the force constant of the system. The total energy is the maximum potential energy, which occurs at maximum displacement (amplitude \(A\)): \(E_{\text{total}} = \frac{1}{2} k A^2 = 1.6\ \text{J}\). When the displacement is half the amplitude, \(x = \frac{1}{2}A\): \(E_p = \frac{1}{2} k \left(\frac{1}{2}A\right)^2 = \frac{1}{4} \left(\frac{1}{2} k A^2\right) = \frac{1}{4} E_{\text{total}} = \frac{1}{4} \times 1.6\ \text{J} = 0.4\ \text{J}\). By conservation of energy, the kinetic energy \(E_k\) at this point is: \(E_k = E_{\text{total}} - E_p = 1.6\ \text{J} - 0.4\ \text{J} = 1.2\ \text{J}\).

PastPaper.markingScheme

1 mark for using the quadratic relationship between potential energy and displacement to determine that Ep is 1/4 of total energy (0.4 J), and subtracting from total energy to find Ek is 1.2 J (Option C).

PastPaper.question 9 · Multiple Choice

1 PastPaper.marks

The density \(\rho\) of a uniform cylindrical wire is determined from measurements of its mass \(m\), length \(L\) and diameter \(d\). The measurements and their percentage uncertainties are shown:

mass \(m = 5.23 \pm 0.05\text{ g}\)
length \(L = 12.4 \pm 0.2\text{ cm}\)
diameter \(d = 1.42 \pm 0.03\text{ mm}\)

What is the percentage uncertainty in the value of the density?

A.3.7%
B.4.7%
C.6.8%
D.12.0%

PastPaper.showAnswers

PastPaper.workedSolution

The density \(\rho\) is given by \(\rho = \frac{4m}{\pi d^2 L}\). The fractional uncertainty in density is \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\). The individual percentage uncertainties are: mass: \(\frac{0.05}{5.23} \times 100\% \approx 0.96\%\); diameter: \(\frac{0.03}{1.42} \times 100\% \approx 2.11\%\); length: \(\frac{0.2}{12.4} \times 100\% \approx 1.61\%\). Adding these with the appropriate weighting: \(\text{Total percentage uncertainty} = 0.96\% + 2(2.11\%) + 1.61\% = 6.79\% \approx 6.8\%\).

PastPaper.markingScheme

1 mark for the correct option C.

PastPaper.question 10 · Multiple Choice

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The resistance \(R\) of a component is determined by measuring the current \(I\) through it and the power dissipation \(P\). The measured values are:

\(P = 15.0 \pm 0.6\text{ W}\)
\(I = 2.50 \pm 0.05\text{ A}\)

Which of the following is the calculated value of \(R\) with its absolute uncertainty?

A.\((2.40 \pm 0.10)\ \Omega\)
B.\((2.40 \pm 0.14)\ \Omega\)
C.\((2.4 \pm 0.2)\ \Omega\)
D.\((2.4 \pm 0.3)\ \Omega\)

PastPaper.showAnswers

PastPaper.workedSolution

First calculate \(R = \frac{P}{I^2} = \frac{15.0}{2.50^2} = 2.40\ \Omega\). Next, find the percentage uncertainties: \(\frac{\Delta P}{P} = \frac{0.6}{15.0} = 4.0\%\), and \(\frac{\Delta I}{I} = \frac{0.05}{2.50} = 2.0\%\). The percentage uncertainty in \(R\) is given by \(\frac{\Delta R}{R} = \frac{\Delta P}{P} + 2\frac{\Delta I}{I} = 4.0\% + 2(2.0\%) = 8.0\%\). The absolute uncertainty in \(R\) is \(\Delta R = 2.40 \times 0.080 = 0.192\ \Omega \approx 0.2\ \Omega\). Thus, \(R = (2.4 \pm 0.2)\ \Omega\).

PastPaper.markingScheme

1 mark for the correct option C.

PastPaper.question 11 · Multiple Choice

1 PastPaper.marks

A student performs an experiment to determine the acceleration of free fall \(g\) using a simple pendulum. The length \(L\) of the pendulum and the time \(t\) for 50 oscillations are measured as:

\(L = 80.0 \pm 0.5\text{ cm}\)
\(t = 90.0 \pm 0.4\text{ s}\)

What is the percentage uncertainty in the calculated value of \(g\)?

A.1.1%
B.1.5%
C.1.7%
D.2.6%

PastPaper.showAnswers

PastPaper.workedSolution

The formula for the acceleration of free fall is \(g = \frac{4\pi^2 L}{T^2}\), where \(T = t/50\) is the period of oscillation. The percentage uncertainty in the period \(T\) is equal to the percentage uncertainty in the total measured time \(t\), which is \(\frac{0.4}{90.0} \times 100\% \approx 0.444\%\). The percentage uncertainty in \(L\) is \(\frac{0.5}{80.0} \times 100\% = 0.625\%\). The percentage uncertainty in \(g\) is \(\frac{\Delta g}{g} \times 100\% = \frac{\Delta L}{L} \times 100\% + 2 \frac{\Delta T}{T} \times 100\% = 0.625\% + 2(0.444\%) = 1.513\% \approx 1.5\%\).

PastPaper.markingScheme

1 mark for the correct option B.

PastPaper.question 12 · Multiple Choice

1 PastPaper.marks

A beam of singly-charged positive ions of mass \(m\) enter a region of uniform magnetic field of flux density \(B\) at right angles to the field lines with speed \(v\). The radius of their circular path is \(r\). A second beam of doubly-charged positive ions of mass \(2m\) enter the same magnetic field at right angles with speed \(2v\). What is the radius of the circular path of these doubly-charged ions?

A.\(\frac{1}{2}r\)
B.\(r\)
C.\(2r\)
D.\(4r\)

PastPaper.showAnswers

PastPaper.workedSolution

The magnetic force provides the centripetal force: \(q v B = \frac{m v^2}{r} \implies r = \frac{m v}{q B}\). For the second beam, \(m_2 = 2m\), \(v_2 = 2v\), and \(q_2 = 2q\). Thus, \(r_2 = \frac{(2m)(2v)}{(2q)B} = 2 \left(\frac{mv}{qB}\right) = 2r\).

PastPaper.markingScheme

1 mark for the correct option C.

PastPaper.question 13 · Multiple Choice

1 PastPaper.marks

A straight, horizontal copper wire of length \(0.15\text{ m}\) and mass \(4.5\text{ g}\) is suspended in a uniform, horizontal magnetic field of flux density \(0.20\text{ T}\). The direction of the magnetic field is perpendicular to the wire. What current must flow through the wire so that the magnetic force exactly balances the weight of the wire? (Take the acceleration of free fall to be \(9.81\text{ m s}^{-2}\))

A.0.15 A
B.1.5 A
C.6.8 A
D.15 A

PastPaper.showAnswers

PastPaper.workedSolution

For the magnetic force to balance the weight: \(F_B = W \implies B I L = m g\). Solving for the current \(I\): \(I = \frac{m g}{B L}\). Substituting the values (converting mass to kilograms): \(I = \frac{4.5 \times 10^{-3} \text{ kg} \times 9.81 \text{ m s}^{-2}}{0.20 \text{ T} \times 0.15 \text{ m}} = \frac{0.044145}{0.030} \approx 1.47 \text{ A} \approx 1.5 \text{ A}\).

PastPaper.markingScheme

1 mark for the correct option B.

PastPaper.question 14 · Multiple Choice

1 PastPaper.marks

Two long, straight, parallel wires, \(X\) and \(Y\), are separated by a distance \(d\) in a vacuum. Wire \(X\) carries a current \(I\) and wire \(Y\) carries a current \(2I\) in the opposite direction. The magnitude of the magnetic field strength at the midpoint between the two wires is \(B_0\). The direction of the current in wire \(X\) is now reversed, whilst the magnitudes of the currents remain unchanged. What is the new magnetic field strength at the midpoint between the wires?

A.\(\frac{1}{3} B_0\)
B.\(\frac{1}{2} B_0\)
C.\(2 B_0\)
D.\(3 B_0\)

PastPaper.showAnswers

PastPaper.workedSolution

Initially, the currents flow in opposite directions. Applying the right-hand grip rule, the magnetic fields produced by both wires at the midpoint point in the same direction. Therefore, they add up: \(B_0 = B_X + B_Y = \frac{\mu_0 I}{2\pi(d/2)} + \frac{\mu_0 (2I)}{2\pi(d/2)} = \frac{\mu_0 (3I)}{\pi d}\). When the current in wire \(X\) is reversed, the currents flow in the same direction, meaning their magnetic fields at the midpoint now oppose each other. The new magnetic field strength is the difference: \(B_{\text{new}} = B_Y - B_X = \frac{\mu_0 (2I)}{\pi d} - \frac{\mu_0 I}{\pi d} = \frac{\mu_0 I}{\pi d}\). Since \(B_0 = 3 \left(\frac{\mu_0 I}{\pi d}\right)\), the new field is \(B_{\text{new}} = \frac{1}{3} B_0\).

PastPaper.markingScheme

1 mark for the correct option A.

PastPaper.question 15 · Multiple Choice

1 PastPaper.marks

A body of mass \(m\) undergoes simple harmonic motion with amplitude \(A\) and frequency \(f\). What is the maximum acceleration of the body, and at what displacement magnitude \(x\) from the equilibrium position does it occur?

A.Maximum acceleration: \(2\pi f^2 A\), at displacement \(x = 0\)
B.Maximum acceleration: \(4\pi^2 f^2 A\), at displacement \(x = 0\)
C.Maximum acceleration: \(2\pi f^2 A\), at displacement \(x = A\)
D.Maximum acceleration: \(4\pi^2 f^2 A\), at displacement \(x = A\)

PastPaper.showAnswers

PastPaper.workedSolution

In simple harmonic motion, the acceleration is given by \(a = -\omega^2 x\). The angular frequency is \(\omega = 2\pi f\), so \(\omega^2 = 4\pi^2 f^2\). The maximum acceleration occurs at the maximum displacement magnitude, i.e., at \(x = A\). Thus, \(a_{\text{max}} = \omega^2 A = 4\pi^2 f^2 A\) at displacement \(x = A\).

PastPaper.markingScheme

1 mark for the correct option D.

PastPaper.question 16 · Multiple Choice

1 PastPaper.marks

A particle of mass \(0.10\text{ kg}\) undergoes simple harmonic motion with an amplitude of \(5.0\text{ cm}\) and a period of \(0.40\text{ s}\). What is the total energy of the oscillator, and what is its kinetic energy when the displacement of the particle from the equilibrium position is \(3.0\text{ cm}\)?

A.Total energy: \(12\text{ mJ}\), Kinetic energy: \(7.9\text{ mJ}\)
B.Total energy: \(31\text{ mJ}\), Kinetic energy: \(11\text{ mJ}\)
C.Total energy: \(31\text{ mJ}\), Kinetic energy: \(20\text{ mJ}\)
D.Total energy: \(49\text{ mJ}\), Kinetic energy: \(31\text{ mJ}\)

PastPaper.showAnswers

PastPaper.workedSolution

The angular frequency is \(\omega = \frac{2\pi}{T} = \frac{2\pi}{0.40} = 5\pi \text{ rad s}^{-1}\). The total energy is \(E_0 = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} \times 0.10 \times (5\pi)^2 \times (0.050)^2 \approx 0.0308\text{ J} \approx 31\text{ mJ}\). The kinetic energy at displacement \(x = 3.0\text{ cm} = 0.030\text{ m}\) is \(E_k = \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} \times 0.10 \times (5\pi)^2 \times (0.050^2 - 0.030^2) = 0.0197\text{ J} \approx 20\text{ mJ}\).

PastPaper.markingScheme

1 mark for the correct option C.

PastPaper.question 17 · Multiple Choice

1 PastPaper.marks

A student determines the resistivity \(\rho\) of a metal wire using the formula:

\(\rho = \frac{R \pi d^2}{4 L}\)

The measurements and their absolute uncertainties are:

Resistance \(R = 4.50 \pm 0.09\ \Omega\)

Diameter \(d = 0.40 \pm 0.01\ \text{mm}\)

Length \(L = 1.250 \pm 0.005\ \text{m}\)

What is the percentage uncertainty in the calculated resistivity?

A.\(2.9\%\)
B.\(4.9\%\)
C.\(7.4\%\)
D.\(10.4\%\)

PastPaper.showAnswers

PastPaper.workedSolution

The percentage uncertainty in a combined quantity of the form \(Y = \frac{A B^2}{C}\) is given by adding the percentage uncertainties of each component, multiplying by power factors where appropriate:

\(\frac{\Delta Y}{Y} = \frac{\Delta A}{A} + 2\frac{\Delta B}{B} + \frac{\Delta C}{C}\)

For the resistivity \(\rho\):

\(\frac{\Delta R}{R} = \frac{0.09}{4.50} \times 100\% = 2.0\%\)

\(\frac{\Delta d}{d} = \frac{0.01}{0.40} \times 100\% = 2.5\%\)

\(\frac{\Delta L}{L} = \frac{0.005}{1.250} \times 100\% = 0.4\%\)

Therefore, the percentage uncertainty in \(\rho\) is:

\(\% \text{ uncertainty} = 2.0\% + 2(2.5\%) + 0.4\% = 7.4\%\)

PastPaper.markingScheme

Award [1 mark] for the correct answer of 7.4% (Option C).

Incorrect pathways:
- 4.9% (Option B) represents neglecting the factor of 2 for the squared diameter \(d^2\).
- 10.4% (Option D) represents using an incorrect power of 3 for diameter or other algebra error.

PastPaper.question 18 · Multiple Choice

1 PastPaper.marks

A digital voltmeter is used to measure a steady potential difference. The manufacturer states that the meter always reads \(0.15\text{ V}\) higher than the true value, and has a random fluctuation of \(\pm 0.02\text{ V}\).

Which statement about the measurements is correct?

A.The systematic error can be corrected by subtracting \(0.15\text{ V}\) from each reading, while the random error cannot be corrected in this way.
B.The systematic error can be reduced to zero by taking a large number of readings and calculating the average.
C.The random error can be completely eliminated by subtracting \(0.02\text{ V}\) from each reading.
D.The random error is a result of poor calibration and the systematic error is due to parallax error.

PastPaper.showAnswers

PastPaper.workedSolution

Systematic errors (such as a constant offset of \(0.15\text{ V}\)) can be eliminated or corrected by subtracting the offset value from each measurement. Random errors (such as the \(\pm 0.02\text{ V}\) fluctuation) represent unpredictable variations and cannot be corrected by subtracting a constant.

PastPaper.markingScheme

Award [1 mark] for identifying that systematic error can be corrected by subtraction, but random error cannot (Option A).

PastPaper.question 19 · Multiple Choice

1 PastPaper.marks

An experiment is performed to measure the acceleration of free fall \(g\) by timing a falling ball. The experimental result is expressed as \(g = 9.42 \pm 0.25\ \text{m s}^{-2}\).

The accepted value of \(g\) is \(9.81\ \text{m s}^{-2}\).

Which statement describes the accuracy and precision of this experimental result?

A.The result is inaccurate because the accepted value of \(9.81\ \text{m s}^{-2}\) lies outside the experimental range of values, and the precision is indicated by the uncertainty of \(\pm 0.25\ \text{m s}^{-2}\).
B.The result is accurate because the experimental range includes \(9.50\ \text{m s}^{-2}\).
C.The result is precise because the accepted value lies outside the range of the measured value.
D.The result is accurate and precise because the percentage uncertainty is approximately \(2.7\%\).

PastPaper.showAnswers

PastPaper.workedSolution

The experimental range is \(9.42 \pm 0.25\ \text{m s}^{-2}\), which spans from \(9.17\ \text{m s}^{-2}\) to \(9.67\ \text{m s}^{-2}\).

Since the accepted value of \(9.81\ \text{m s}^{-2}\) lies outside this range, the result is inaccurate (it has a significant systematic error). Precision refers to the degree of agreement between repeated measurements, which is indicated by the size of the random uncertainty (\(\pm 0.25\ \text{m s}^{-2}\)).

PastPaper.markingScheme

Award [1 mark] for the correct analysis of accuracy and precision (Option A).

PastPaper.question 20 · Multiple Choice

1 PastPaper.marks

A beam of singly charged positive ions, each of mass \(m\) and charge \(q\), enters a region of uniform magnetic field of flux density \(B\). The velocity of the ions is perpendicular to the magnetic field, causing them to travel in a circular path of radius \(r\).

A second beam of ions consisting of doubly charged positive ions of mass \(2m\) and charge \(2q\) enters the same magnetic field with the same kinetic energy as the first beam.

What is the radius of the path of the second beam of ions?

A.\{\frac{r}{\sqrt{2}}\}
B.\(r\)
C.\(\sqrt{2} r\)
D.\(2 r\)

PastPaper.showAnswers

PastPaper.workedSolution

For a charge \(q\) of mass \(m\) in a circular path of radius \(r\) in a magnetic field \(B\):

\(\frac{m v^2}{r} = B q v \implies r = \frac{m v}{B q}\)

Since the kinetic energy is \(E_k = \frac{1}{2} m v^2\), the momentum is \(p = m v = \sqrt{2 m E_k}\).

Substituting this into the radius formula:

\(r = \frac{\sqrt{2 m E_k}}{B q}\)

For the second beam, the mass is \(m' = 2m\), the charge is \(q' = 2q\), and the kinetic energy is \(E_k\):

\(r' = \frac{\sqrt{2 (2m) E_k}}{B (2q)} = \frac{\sqrt{2} \sqrt{2 m E_k}}{2 B q} = \frac{\sqrt{2}}{2} r = \frac{r}{\sqrt{2}}\)

PastPaper.markingScheme

Award [1 mark] for the correct derivation showing the radius is scaled by a factor of \(1/\sqrt{2}\) (Option A).

PastPaper.question 21 · Multiple Choice

1 PastPaper.marks

A straight horizontal wire of length \(0.50\text{ m}\) and mass \(15\text{ g}\) is suspended horizontally in a uniform horizontal magnetic field. The magnetic field of flux density \(0.40\text{ T}\) is perpendicular to the wire.

What is the current required in the wire to support its weight? (Take \(g = 9.81\text{ m s}^{-2}\).)

A.0.075 A
B.0.74 A
C.1.5 A
D.7.4 A

PastPaper.showAnswers

PastPaper.workedSolution

For the wire to float, the upward magnetic force must equal the downward gravitational force (weight):

\(F_B = B I L = m g\)

Solving for the current \(I\):

\(I = \frac{m g}{B L}\)

Given:
\(m = 15\text{ g} = 0.015\text{ kg}\)
\(g = 9.81\text{ m s}^{-2}\)
\(B = 0.40\text{ T}\)
\(L = 0.50\text{ m}\)

Calculate \(I\):

\(I = \frac{0.015 \times 9.81}{0.40 \times 0.50} = \frac{0.14715}{0.20} \approx 0.74\text{ A}\)

PastPaper.markingScheme

Award [1 mark] for the correct calculation of current \(I = 0.74\text{ A}\) (Option B).

PastPaper.question 22 · Multiple Choice

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A flat circular coil of \(200\) turns and cross-sectional area \(1.5 \times 10^{-3}\ \text{m}^2\) is placed in a uniform magnetic field of flux density \(0.080\ \text{T\) so that the plane of the coil is perpendicular to the magnetic field.

The magnetic field is reduced to zero at a constant rate in a time of \(0.12\ \text{s}\).

What is the magnitude of the average electromotive force (e.m.f.) induced in the coil?

A.1.0 mV
B.0.20 V
C.1.2 V
D.130 V

PastPaper.showAnswers

PastPaper.workedSolution

According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. is given by:

\(E = N \frac{\Delta \Phi}{\Delta t}\)

Since the field is perpendicular to the plane of the coil, the initial magnetic flux through a single turn is \(\Phi = B A\).

The change in flux is \(\Delta \Phi = B A - 0 = B A\).

Therefore:

\(E = N \frac{B A}{\Delta t} = \frac{200 \times 0.080 \times 1.5 \times 10^{-3}}{0.12} = \frac{0.024}{0.12} = 0.20\ \text{V}\)

PastPaper.markingScheme

Award [1 mark] for the correct calculation of the average induced e.m.f. (Option B).

Incorrect pathways:
- 1.0 mV (Option A) represents neglecting the factor of \(N = 200\).
- 1.2 V (Option C) represents omitting the area conversion or other math error.

PastPaper.question 23 · Multiple Choice

1 PastPaper.marks

An object undergoing simple harmonic motion has displacement \(x\) at time \(t\) given by the equation:

\(x = 0.040 \cos(50\pi t)\)

where \(x\) is in metres and \(t\) is in seconds.

What is the maximum acceleration of the object, and its frequency of oscillation?

A.Maximum acceleration = \(990\ \text{m s}^{-2}\); Frequency = \(25\ \text{Hz}\)
B.Maximum acceleration = \(990\ \text{m s}^{-2}\); Frequency = \(50\ \text{Hz}\)
C.Maximum acceleration = \(6.3\ \text{m s}^{-2}\); Frequency = \(25\ \text{Hz}\)
D.Maximum acceleration = \(2.0\ \text{m s}^{-2}\); Frequency = \(50\pi\ \text{Hz}\)

PastPaper.showAnswers

PastPaper.workedSolution

The general equation for simple harmonic motion is \(x = x_0 \cos(\omega t)\).

Comparing this to the given equation:
Amplitude \(x_0 = 0.040\ \text{m}\)
Angular frequency \(\omega = 50\pi\ \text{rad s}^{-1}\)

The frequency \(f\) of oscillation is:
\(f = \frac{\omega}{2\pi} = \frac{50\pi}{2\pi} = 25\ \text{Hz}\)

The maximum acceleration \(a_{\text{max}}\) is:
\(a_{\text{max}} = \omega^2 x_0 = (50\pi)^2 \times 0.040 = 2500\pi^2 \times 0.040 = 100\pi^2 \approx 987\ \text{m s}^{-2} \approx 990\ \text{m s}^{-2}\) (to 2 significant figures).

PastPaper.markingScheme

Award [1 mark] for correctly calculating both maximum acceleration and frequency (Option A).

PastPaper.question 24 · Multiple Choice

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A particle of mass \(0.15\ \text{kg}\) undergoes simple harmonic motion with an amplitude of \(8.0\ \text{cm}\) and a period of \(0.40\ \text{s}\).

What are the total energy of the oscillation and the kinetic energy of the particle when it is at a displacement of \(5.0\ \text{cm}\) from its equilibrium position?

A.Total energy = \(0.12\ \text{J}\); Kinetic energy = \(0.072\ \text{J}\)
B.Total energy = \(0.12\ \text{J}\); Kinetic energy = \(0.046\ \text{J}\)
C.Total energy = \(0.072\ \text{J}\); Kinetic energy = \(0.046\ \text{J}\)
D.Total energy = \(1.2\ \text{J}\); Kinetic energy = \(0.72\ \text{J}\)

PastPaper.showAnswers

PastPaper.workedSolution

First, find the angular frequency \(\omega\):

\(\omega = \frac{2\pi}{T} = \frac{2\pi}{0.40} = 5\pi\ \text{rad s}^{-1}\)

The total energy \(E_{\text{total}}\) is given by:

\(E_{\text{total}} = \frac{1}{2} m \omega^2 x_0^2 = \frac{1}{2} \times 0.15 \times (5\pi)^2 \times (0.080)^2 \approx 0.1184\ \text{J} \approx 0.12\ \text{J}\)

The kinetic energy \(E_k\) at a displacement of \(x = 5.0\ \text{cm} = 0.050\ \text{m}\) is:

\(E_k = \frac{1}{2} m \omega^2 (x_0^2 - x^2) = \frac{1}{2} \times 0.15 \times (5\pi)^2 \times (0.080^2 - 0.050^2) \approx 0.07217\ \text{J} \approx 0.072\ \text{J}\)

PastPaper.markingScheme

Award [1 mark] for correctly calculating the total energy and the kinetic energy (Option A).

Incorrect pathways:
- Option B yields potential energy (0.046 J) instead of kinetic energy.
- Option D represents a power of 10 failure.

PastPaper.question 25 · Multiple Choice

1 PastPaper.marks

A student determines the resistivity \(\rho\) of a uniform metal wire using the formula \(\rho = \frac{R \pi d^2}{4 L}\). The measurements of resistance \(R\), diameter \(d\), and length \(L\) are given below with their absolute uncertainties:
\(R = (25.0 \pm 0.5)\ \Omega\)
\(d = (0.40 \pm 0.01)\text{ mm}\)
\(L = (1.200 \pm 0.003)\text{ m}\)
What is the percentage uncertainty in the calculated value of resistivity?

A.4.8%
B.7.3%
C.9.8%
D.12.3%

PastPaper.showAnswers

PastPaper.workedSolution

To find the percentage uncertainty in \(\rho\), we sum the percentage uncertainties of each component, multiplying by their respective powers in the formula:
\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\)
Calculate each fractional uncertainty:
\(\frac{\Delta R}{R} = \frac{0.5}{25.0} = 0.020\) (or \(2.0\%\))
\(\frac{\Delta d}{d} = \frac{0.01}{0.40} = 0.025\) (or \(2.5\%\))
\(\frac{\Delta L}{L} = \frac{0.003}{1.200} = 0.0025\) (or \(0.25\%\))
Substitute these into the formula:
\(\%\Delta \rho = 2.0\% + 2 \times 2.5\% + 0.25\% = 2.0\% + 5.0\% + 0.25\% = 7.25\%\)
Rounding to two significant figures gives \(7.3\%\).

PastPaper.markingScheme

1 mark for the correct answer B. Method: Identify that percentage uncertainties add, with the diameter uncertainty doubled due to the squared term. Accuracy: Correctly compute the final percentage uncertainty as 7.3%.

PastPaper.question 26 · Multiple Choice

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The acceleration of free fall \(g\) is determined by measuring the length \(L\) of a simple pendulum and the time \(t\) for 50 complete oscillations. The formula used is \(g = \frac{4\pi^2 L}{T^2}\), where \(T\) is the period of one oscillation. The measurements are:
\(L = (80.0 \pm 0.4)\text{ cm}\)
\(t = (90.0 \pm 0.6)\text{ s}\)
What is the percentage uncertainty in the calculated value of \(g\)?

A.1.2%
B.1.8%
C.2.3%
D.3.2%

PastPaper.showAnswers

PastPaper.workedSolution

First, we note that the period is \(T = t / 50\). The percentage uncertainty in \(T\) is identical to the percentage uncertainty in \(t\):
\(\%\Delta T = \%\Delta t = \frac{0.6}{90.0} \times 100\% \approx 0.67\%\)
The percentage uncertainty in length \(L\) is:
\(\%\Delta L = \frac{0.4}{80.0} \times 100\% = 0.50\%\)
Using the relation \(g = \frac{4\pi^2 L}{T^2}\), the percentage uncertainty in \(g\) is:
\(\%\Delta g = \%\Delta L + 2 \%\Delta T = 0.50\% + 2 \times 0.67\% = 1.84\%\)
This rounds to \(1.8\%\).

PastPaper.markingScheme

1 mark for the correct answer B. Method: Realise that the divisor of 50 does not change the percentage uncertainty of the period, and double the percentage uncertainty of T. Accuracy: Correctly calculate the sum of the percentage uncertainties to get 1.8%.

PastPaper.question 27 · Multiple Choice

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An analogue voltmeter is used to measure potential differences (p.d.). The voltmeter has a zero error and a scale-factor error. When the true p.d. across a component is zero, the meter reads \(+0.2\text{ V}\). For every increase of \(1.0\text{ V}\) in the true p.d., the meter's reading increases by only \(0.8\text{ V}\). Which statement correctly describes the graph of measured potential difference \(V_{\text{meas}}\) against true potential difference \(V_{\text{true}}\)

A.A straight line with a positive vertical intercept and a gradient less than 1.
B.A straight line with a positive vertical intercept and a gradient greater than 1.
C.A straight line passing through the origin with a gradient less than 1.
D.A straight line with a negative vertical intercept and a gradient greater than 1.

PastPaper.showAnswers

PastPaper.workedSolution

The equation relating the measured voltage \(V_{\text{meas}}\) to the true voltage \(V_{\text{true}}\) is linear: \(V_{\text{meas}} = m V_{\text{true}} + C\).
When \(V_{\text{true}} = 0\), \(V_{\text{meas}} = +0.2\text{ V}\), which corresponds to a positive vertical intercept (\(C = +0.2\text{ V}\)).
The rate of change of measured voltage with respect to true voltage is the gradient \(m\):
\(m = \frac{\Delta V_{\text{meas}}}{\Delta V_{\text{true}}} = \frac{0.8\text{ V}}{1.0\text{ V}} = 0.8\).
Since \(0.8 < 1\), the gradient of the straight line is less than 1.

PastPaper.markingScheme

1 mark for the correct answer A. Method: Formulate the linear relationship between the measured and true value to deduce the intercept and gradient. Accuracy: Identify that the intercept is positive (+0.2 V) and the gradient is less than 1 (0.8).

PastPaper.question 28 · Multiple Choice

1 PastPaper.marks

A beam of singly-charged positive ions, each of mass \(m\) and charge \(q\), enters a region of uniform magnetic field with flux density \(B\). The magnetic field is directed perpendicular to the velocity of the ions, causing them to move in a circular path of radius \(R\). The kinetic energy of the incoming ions is now doubled, and the magnetic flux density is also doubled. What is the new radius of the circular path of the ions?

A.\(R / 2\)
B.\(R / \sqrt{2}\)
C.\(\sqrt{2} R\)
D.\(2 R\)

PastPaper.showAnswers

PastPaper.workedSolution

For a charge \(q\) moving with velocity \(v\) perpendicular to a magnetic field \(B\), the magnetic force provides the centripetal force:
\(q v B = \frac{m v^2}{R} \implies R = \frac{m v}{q B}\)
The kinetic energy is given by \(E_k = \frac{1}{2} m v^2\), so the momentum is \(p = m v = \sqrt{2 m E_k}\). Thus, the radius can be written as:
\(R = \frac{\sqrt{2 m E_k}}{q B}\)
Therefore, the radius is proportional to \(\frac{\sqrt{E_k}}{B}\).
When the kinetic energy is doubled (\(E_k \to 2 E_k\)) and the magnetic flux density is doubled (\(B \to 2 B\)), the new radius \(R'\) is:
\(R' \propto \frac{\sqrt{2 E_k}}{2 B} = \frac{\sqrt{2}}{2} \frac{\sqrt{E_k}}{B} = \frac{1}{\sqrt{2}} R = \frac{R}{\sqrt{2}}\).

PastPaper.markingScheme

1 mark for the correct answer B. Method: Express the radius in terms of kinetic energy and magnetic field. Accuracy: Find the scaling factor to be 1/sqrt(2) and apply it to the original radius R.

PastPaper.question 29 · Multiple Choice

1 PastPaper.marks

A straight metal wire of length \(0.50\text{ m}\) carries a current of \(4.0\text{ A}\) through a uniform magnetic field of flux density \(0.15\text{ T}\). The direction of the current in the wire is at an angle of \(30^\circ\) to the magnetic field. What is the magnitude of the magnetic force acting on the wire?

A.0.15 N
B.0.26 N
C.0.30 N
D.0.60 N

PastPaper.showAnswers

PastPaper.workedSolution

The magnetic force \(F\) on a straight current-carrying wire in a magnetic field is given by the formula:
\(F = B I L \sin\theta\)
Given values:
\(B = 0.15\text{ T}\)
\(I = 4.0\text{ A}\)
\(L = 0.50\text{ m}\)
\(\theta = 30^\circ\)
Substitute these values into the formula:
\(F = 0.15 \times 4.0 \times 0.50 \times \sin(30^\circ)\)
\(F = 0.30 \times 0.5 = 0.15\text{ N}\).

PastPaper.markingScheme

1 mark for the correct answer A. Method: Apply the formula F = B I L sin(theta) with the correct angle. Accuracy: Correctly compute the force to be 0.15 N.

PastPaper.question 30 · Multiple Choice

1 PastPaper.marks

A flat, circular coil of 200 turns has a cross-sectional area of \(4.0 \times 10^{-4}\text{ m}^2\). The coil is positioned perpendicular to a uniform magnetic field. The magnetic flux density \(B\) of this field increases uniformly from \(0.10\text{ T}\) to \(0.50\text{ T}\) over a time interval of \(2.0\text{ s}\). What is the magnitude of the average electromotive force (e.m.f.) induced in the coil during this time?

A.0.080 mV
B.16 mV
C.20 mV
D.32 mV

PastPaper.showAnswers

PastPaper.workedSolution

According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. \(E\) is:
\(E = N \frac{\Delta \Phi}{\Delta t} = N A \frac{\Delta B}{\Delta t}\)
where:
\(N = 200\)
\(A = 4.0 \times 10^{-4}\text{ m}^2\)
\(\Delta B = 0.50\text{ T} - 0.10\text{ T} = 0.40\text{ T}\)
\(\Delta t = 2.0\text{ s}\)
Calculate \(E\):
\(E = 200 \times (4.0 \times 10^{-4}) \times \frac{0.40}{2.0}\)
\(E = 200 \times 4.0 \times 10^{-4} \times 0.20 = 0.016\text{ V} = 16\text{ mV}\).

PastPaper.markingScheme

1 mark for the correct answer B. Method: Identify and apply Faraday's law for a multi-turn coil. Accuracy: Correctly determine the rate of change of flux density and compute the e.m.f. in millivolts (16 mV).

PastPaper.question 31 · Multiple Choice

1 PastPaper.marks

A particle undergoes simple harmonic motion with an amplitude of \(5.0\text{ cm}\) and a time period of \(2.0\text{ s}\). What is the magnitude of the velocity of the particle when its displacement from the equilibrium position is \(3.0\text{ cm}\)?

A.\(6.3\text{ cm s}^{-1}\)
B.\(12.6\text{ cm s}^{-1}\)
C.\(15.7\text{ cm s}^{-1}\)
D.\(25.1\text{ cm s}^{-1}\)

PastPaper.showAnswers

PastPaper.workedSolution

The velocity \(v\) of a particle in simple harmonic motion is related to its displacement \(x\) and amplitude \(x_0\) by:
\(v = \pm \omega \sqrt{x_0^2 - x^2}\)
First, calculate the angular frequency \(\omega\):
\(\omega = \frac{2\pi}{T} = \frac{2\pi}{2.0} = \pi\text{ rad s}^{-1}\)
Now, substitute the values into the velocity equation:
\(v = \pi \sqrt{5.0^2 - 3.0^2} = \pi \sqrt{25.0 - 9.0} = \pi \sqrt{16.0} = 4\pi\text{ cm s}^{-1} \approx 12.57\text{ cm s}^{-1}\)
This rounds to \(12.6\text{ cm s}^{-1}\).

PastPaper.markingScheme

1 mark for the correct answer B. Method: Find the angular frequency and then apply the SHM velocity-displacement formula. Accuracy: Calculate the velocity magnitude as 12.6 cm/s.

PastPaper.question 32 · Multiple Choice

1 PastPaper.marks

A system is undergoing forced oscillations. The frequency of the periodic driving force is varied, and the amplitude of the resulting steady-state oscillation is recorded to plot a resonance curve. The damping of the system is now increased. Which statement describes the changes to the resonance curve when the damping is increased?

A.The peak amplitude increases, and the resonance frequency increases slightly.
B.The peak amplitude increases, and the resonance frequency decreases slightly.
C.The peak amplitude decreases, and the resonance frequency increases slightly.
D.The peak amplitude decreases, and the resonance frequency decreases slightly.

PastPaper.showAnswers

PastPaper.workedSolution

When damping is increased in a forced oscillation system:
1. More energy is dissipated per cycle, so the maximum (peak) amplitude of the oscillation decreases.
2. The frequency at which resonance (maximum amplitude) occurs, known as the resonance frequency, shifts to a slightly lower frequency (decreases slightly) due to the damping effect.
Therefore, both the peak amplitude and the resonance frequency decrease.

PastPaper.markingScheme

1 mark for the correct answer D. Method: Identify the two physical consequences of increasing damping on the resonance curve. Accuracy: Correctly conclude that both the peak amplitude decreases and the resonance frequency decreases slightly.

PastPaper.question 33 · multiple_choice

1 PastPaper.marks

An experiment is carried out to determine the acceleration of free fall \( g \) by timing the fall of a steel ball through a vertical distance \( h \). The equation used is \( h = \frac{1}{2} g t^2 \). The measured values are: \( h = (1.20 \pm 0.01)\text{ m} \) and \( t = (0.49 \pm 0.02)\text{ s} \). What is the percentage uncertainty in the calculated value of \( g \)?

A.4.9%
B.8.2%
C.9.0%
D.9.8%

PastPaper.showAnswers

PastPaper.workedSolution

To find the percentage uncertainty in \( g \), we first rearrange the equation: \( g = \frac{2h}{t^2} \). The percentage uncertainty in \( g \) is given by \( \frac{\Delta g}{g} \times 100\% = \left( \frac{\Delta h}{h} + 2 \frac{\Delta t}{t} \right) \times 100\% \). Calculating each term: \( \frac{\Delta h}{h} = \frac{0.01}{1.20} \approx 0.83\% \) and \( \frac{\Delta t}{t} = \frac{0.02}{0.49} \approx 4.08\% \). Therefore, the percentage uncertainty in \( g \) is \( 0.83\% + 2 \times 4.08\% = 0.83\% + 8.16\% = 8.99\% \approx 9.0\% \).

PastPaper.markingScheme

1 mark for the correct answer C. Method: 1 mark for calculating the sum of fractional uncertainties with the time uncertainty doubled.

PastPaper.question 34 · multiple_choice

1 PastPaper.marks

The power \( P \) dissipated in a resistor of resistance \( R \) when a current \( I \) passes through it is given by \( P = I^2 R \). The resistance is measured to be \( (47.0 \pm 0.5)\ \Omega \) and the current is measured to be \( (2.50 \pm 0.05)\text{ A} \). What is the absolute uncertainty in the calculated value of \( P \)?

A.0.6 W
B.9.0 W
C.12 W
D.15 W

PastPaper.showAnswers

PastPaper.workedSolution

First, calculate the value of \( P \): \( P = I^2 R = (2.50)^2 \times 47.0 = 293.75\text{ W} \). The fractional uncertainty in \( P \) is \( \frac{\Delta P}{P} = 2\frac{\Delta I}{I} + \frac{\Delta R}{R} = 2 \left( \frac{0.05}{2.50} \right) + \frac{0.5}{47.0} = 0.040 + 0.01064 = 0.05064 \). The absolute uncertainty is \( \Delta P = 293.75 \times 0.05064 \approx 14.88\text{ W} \). To 2 significant figures, this is \( 15\text{ W} \).

PastPaper.markingScheme

1 mark for the correct answer D. Method: 1 mark for adding fractional uncertainties correctly (doubling current uncertainty) and multiplying by the power value.

PastPaper.question 35 · multiple_choice

1 PastPaper.marks

A student determines the density \( \rho \) of a solid sphere using the equation \( \rho = \frac{6M}{\pi d^3} \) where \( M \) is the mass and \( d \) is the diameter of the sphere. The mass is measured with a percentage uncertainty of \( 2.0\% \) and the diameter is measured with a percentage uncertainty of \( 1.5\% \). What is the percentage uncertainty in the calculated value of the density?

A.3.5%
B.5.0%
C.6.5%
D.15.5%

PastPaper.showAnswers

PastPaper.workedSolution

The formula for density is \( \rho = \frac{6M}{\pi d^3} \). The constant factor \( \frac{6}{\pi} \) has no uncertainty. The percentage uncertainty in \( \rho \) is \( \frac{\Delta \rho}{\rho} \times 100\% = \frac{\Delta M}{M} \times 100\% + 3 \frac{\Delta d}{d} \times 100\% \). Substituting the values: \( 2.0\% + 3(1.5\%) = 2.0\% + 4.5\% = 6.5\% \).

PastPaper.markingScheme

1 mark for the correct answer C. Method: 1 mark for multiplying the diameter percentage uncertainty by 3 and adding it to the mass percentage uncertainty.

PastPaper.question 36 · multiple_choice

1 PastPaper.marks

A uniform magnetic field is directed perpendicularly into the plane of the page. An alpha particle (mass \( 4u \), charge \( +2e \)) enters the field with velocity \( v \) perpendicular to the field and describes a circular path of radius \( r_{\alpha} \). A proton (mass \( u \), charge \( +e \)) enters the same field with a velocity \( 2v \) perpendicular to the field, and describes a circular path of radius \( r_{p} \). What is the ratio \( \frac{r_{\alpha}}{r_{p}} \)?

A.0.5
B.1
C.2
D.4

PastPaper.showAnswers

PastPaper.workedSolution

The magnetic force provides the centripetal force: \( Bqv = \frac{mv^2}{r} \implies r = \frac{mv}{Bq} \). For the alpha particle: \( r_{\alpha} = \frac{(4u)v}{B(2e)} = 2 \frac{uv}{Be} \). For the proton: \( r_{p} = \frac{u(2v)}{B(e)} = 2 \frac{uv}{Be} \). The ratio is \( \frac{r_{\alpha}}{r_{p}} = \frac{2}{2} = 1 \).

PastPaper.markingScheme

1 mark for the correct answer B. Method: 1 mark for deriving \( r = \frac{mv}{Bq} \) and substituting the values of charge, mass and velocity for both particles.

PastPaper.question 37 · multiple_choice

1 PastPaper.marks

A rigid straight wire of length \( 0.40\text{ m} \) carries a current of \( 3.0\text{ A} \) in a direction perpendicular to a uniform magnetic field. The magnetic force acting on this wire is \( 0.18\text{ N} \). The wire is now rotated so that it makes an angle of \( 30^\circ \) with the direction of the magnetic field, while the current remains unchanged. What is the new magnetic force acting on the wire?

A.0.090 N
B.0.16 N
C.0.18 N
D.0.36 N

PastPaper.showAnswers

PastPaper.workedSolution

The magnetic force on a current-carrying wire is \( F = BIL\sin\theta \), where \( \theta \) is the angle between the wire and the magnetic field. Initially, \( \theta = 90^\circ \), so \( F = BIL = 0.18\text{ N} \). When the wire is rotated to \( \theta = 30^\circ \), the new force is \( F' = BIL\sin(30^\circ) = 0.18 \times 0.5 = 0.090\text{ N} \).

PastPaper.markingScheme

1 mark for the correct answer A. Method: 1 mark for using the formula \( F = BIL\sin\theta \) and multiplying the maximum force by \( \sin(30^\circ) \).

PastPaper.question 38 · multiple_choice

1 PastPaper.marks

An object is performing simple harmonic motion with an amplitude \( x_0 \) and a time period \( T \). The object starts from the position of maximum displacement at time \( t = 0 \). What is the displacement \( x \) of the object from its equilibrium position at time \( t = \frac{T}{6} \)?

A.0.50 x0
B.0.71 x0
C.0.87 x0
D.1.0 x0

PastPaper.showAnswers

PastPaper.workedSolution

Since the object starts at maximum displacement at \( t = 0 \), its displacement is described by \( x = x_0 \cos(\omega t) \). The angular frequency is \( \omega = \frac{2\pi}{T} \). At \( t = \frac{T}{6} \), the phase angle is \( \omega t = \frac{2\pi}{T} \times \frac{T}{6} = \frac{\pi}{3} \text{ rad} \) (or \( 60^\circ \)). Therefore, the displacement is \( x = x_0 \cos\left(\frac{\pi}{3}\right) = 0.50 x_0 \).

PastPaper.markingScheme

1 mark for the correct answer A. Method: 1 mark for using the cosine representation of SHM and calculating the phase angle at \( t = T/6 \).

PastPaper.question 39 · multiple_choice

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A body of mass \( m \) is undergoing simple harmonic motion with frequency \( f \) and amplitude \( x_0 \). Which expression represents the total energy \( E \) of the oscillating body?

A.pi^2 m f^2 x0^2
B.2 pi^2 m f^2 x0^2
C.4 pi^2 m f^2 x0^2
D.0.5 pi^2 m f^2 x0^2

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PastPaper.workedSolution

The total energy in simple harmonic motion is equal to the maximum kinetic energy, which is \( E = \frac{1}{2} m v_{\text{max}}^2 \). Since \( v_{\text{max}} = \omega x_0 \), we have \( E = \frac{1}{2} m \omega^2 x_0^2 \). Substituting \( \omega = 2\pi f \) gives \( E = \frac{1}{2} m (2\pi f)^2 x_0^2 = 2 \pi^2 m f^2 x_0^2 \).

PastPaper.markingScheme

1 mark for the correct answer B. Method: 1 mark for substituting \( \omega = 2\pi f \) into the standard SHM total energy equation.

PastPaper.question 40 · multiple_choice

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A light mass on a spring is suspended in a beaker of liquid and driven by an external vibrator of variable frequency \( f \). A graph of the steady-state amplitude of oscillation against driving frequency is plotted. The viscosity of the liquid is now increased. Which statement correctly describes the changes to the maximum amplitude of the oscillation and the frequency at which this maximum amplitude occurs?

A.The maximum amplitude decreases, and the frequency of maximum amplitude decreases.
B.The maximum amplitude decreases, and the frequency of maximum amplitude increases.
C.The maximum amplitude increases, and the frequency of maximum amplitude decreases.
D.The maximum amplitude increases, and the frequency of maximum amplitude increases.

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PastPaper.workedSolution

Increasing the viscosity of the liquid increases the degree of damping on the system. For a more heavily damped system, the maximum steady-state amplitude decreases (the peak is flatter/lower), and the resonant frequency (the frequency at which the maximum amplitude occurs) shifts to a slightly lower frequency.

PastPaper.markingScheme

1 mark for the correct answer A. Method: 1 mark for identifying that increased damping reduces maximum amplitude and decreases the resonant frequency.

Paper 2 (AS Level Structured)

Answer all questions. Write your answers in the spaces provided on the question paper.

(a) [1] Driving frequency equals natural frequency of system.
[1] Maximum amplitude of oscillation achieved.
(b) [1] Amplitude increases as frequency increases towards the natural frequency.
[1] Peak amplitude occurs at/near the natural frequency.
[1] Amplitude decreases at frequencies above the natural frequency.
(c) (i) [1] Maximum amplitude decreases (peak is flatter/lower).
(ii) [1] The frequency of maximum amplitude shifts to a slightly lower frequency.

Paper 3 (Advanced Practical Skills)

Answer both questions. Record all your observations in the spaces provided as soon as they are made.

2 PastPaper.question · 40 PastPaper.marks

PastPaper.question 1 · Practical Investigation

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Question 1

Apparatus provided:

A wooden metre rule with a small hole drilled at the 5.0 cm mark.
A metal pin held securely in a boss and clamp on a retort stand to act as a horizontal pivot.
A 100 g mass and a small piece of adhesive putty (Blu-Tack) to attach it.
A stopwatch.
A second metre rule or half-metre rule.
An optical pin or pointer to act as a fiducial marker.

In this experiment, you will investigate how the period of oscillation of a suspended metre rule depends on the position of a sliding mass along its length.

Procedure:

(a) (i) Suspend the metre rule from the pivot through the hole at the 5.0 cm mark, ensuring it can swing freely in a vertical plane.

Using the adhesive putty, attach the 100 g mass to the rule such that its centre is at a distance \(d = 40.0\text{ cm}\) from the pivot hole. Record the value of \(d\).

(ii) Displace the bottom of the rule slightly to one side and release it so that it performs small oscillations. Measure and record the time \(t\) for 10 complete oscillations. Repeat this measurement to obtain a mean value of \(t\). Calculate the period \(T\) of the oscillations.

(b) Change the position of the 100 g mass to obtain six sets of readings of \(d\) and \(t\). The values of \(d\) should be in the range \(20.0\text{ cm} \le d \le 80.0\text{ cm}\). For each set, record \(d\), \(t\), \(T\), and calculate \(d^2\) and \(T^2 d\). Include your results in a table with appropriate column headings and units.

(ii) Draw the straight line of best fit.

(d) Determine the gradient and y-intercept of this line.

(e) The quantities \(T\) and \(d\) are related by the equation:
\(T^2 d = p d^2 + q\)
where \(p\) and \(q\) are constants. Use your answers from (d) to determine the values of \(p\) and \(q\). Include appropriate units for both constants.

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PastPaper.workedSolution

(a) (i) & (ii) Representative measurements:
Let \(d = 40.0\text{ cm} = 0.400\text{ m}\).
Time for 10 oscillations: \(t_1 = 13.4\text{ s}\), \(t_2 = 13.6\text{ s}\). Mean \(t = 13.5\text{ s}\).
Period \(T = 1.35\text{ s}\).

(b) Sample data table:

\(d / \text{m}\)\(t_1 / \text{s}\)\(t_2 / \text{s}\)\(t_{\text{mean}} / \text{s}\)\(T / \text{s}\)\(d^2 / \text{m}^2\)\(T^2 d / \text{s}^2\text{ m}\)0.20011.011.211.11.110.04000.2460.30012.112.312.21.220.09000.4460.40013.413.613.51.350.16000.7290.50014.915.115.01.500.25001.1250.60016.516.716.61.660.36001.6530.70018.218.418.31.830.49002.344

(c) Graph:
Plot \(T^2 d\) against \(d^2\). Both axes must have clear labels with units: \(T^2 d / \text{s}^2\text{ m}\) and \(d^2 / \text{m}^2\). Points are plotted precisely. A straight line of best fit is drawn.

(d) Gradient and y-intercept calculation:
Using two points on the line of best fit: \((0.0800, 0.400)\) and \((0.4800, 2.300)\).
\text{Gradient} = \frac{2.300 - 0.400}{0.4800 - 0.0800} = \frac{1.900}{0.4000} = 4.75\text{ s}^2\text{ m}^{-1}.
Using \(y = mx + c\):
\(0.400 = 4.75 \times 0.0800 + c \implies c = 0.400 - 0.380 = 0.020\text{ s}^2\text{ m}\).
So, y-intercept = \(0.020\text{ s}^2\text{ m}\).

(e) Determining \(p\) and \(q\):
From \(T^2 d = p d^2 + q\), we have:
\(p = \text{gradient} = 4.75\text{ s}^2\text{ m}^{-1}\) (or \(0.0475\text{ s}^2\text{ cm}^{-1}\)).
\(q = \text{y-intercept} = 0.020\text{ s}^2\text{ m}\) (or \(2.0\text{ s}^2\text{ cm}\)).

PastPaper.markingScheme

(a) (i) [1 mark] Value of \(d\) to the nearest 1 mm with correct unit (e.g. \(40.0\text{ cm}\) or \(0.400\text{ m}\)).
(a) (ii) [1 mark] Value of \(t\) measured to 0.1 s or 0.01 s with repeat readings. Correct period \(T\) calculated.

(b) Table [6 marks]:
- Successful collection: Six sets of readings of \(d\) and \(t\) without helper assistance. [1]
- Range: Range of \(d\) covers at least 45.0 cm (e.g. from 20.0 cm to 70.0 cm). [1]
- Column headings: Headings must contain a quantity and a unit (e.g. \(d / \text{m}\), \(t / \text{s}\), \(T / \text{s}\), \(d^2 / \text{m}^2\), \(T^2 d / \text{s}^2\text{ m}\)). [1]
- Consistency: All raw values of \(d\) recorded to the same precision (nearest mm) and all raw values of \(t\) recorded to the same precision (0.1 s or 0.01 s). [1]
- Significant figures: Significant figures for \(T^2 d\) must be the same as, or one more than, the least number of significant figures in the raw values of \(T\) and \(d\). [1]
- Quality: Points on the graph lie close to a straight line (minimum scatter). [1]

(c) Graph [5 marks]:
- Axes: Linear scales where the plotted points occupy more than half the grid in both directions. Labels must have units. [1]
- Plotting: All observations must be plotted to within half a small square. No thick blobs. [1]
- Line of best fit: A single straight line representing the trend of all plotted points. Even distribution of points about the line. [1]

(d) Gradient and Intercept [2 marks]:
- Gradient: Uses a triangle where the hypotenuse is at least 50% of the length of the drawn line. Correct calculation. [1]
- Intercept: Read directly from the y-axis if the x-axis starts at 0, or calculated using coordinates of a point on the line in \(y = mx + c\). [1]

(e) Constants \(p\) and \(q\) [2 marks]:
- Value of \(p\) equals the gradient with unit \(\text{s}^2\text{ m}^{-1}\) (or equivalent). [1]
- Value of \(q\) equals the y-intercept with unit \(\text{s}^2\text{ m}\) (or equivalent). [1]

PastPaper.question 2 · Practical Investigation

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Question 2

Apparatus provided:

A retort stand, boss, and clamp.
A length of thread (approx. 60 cm).
A pendulum bob (or 100 g mass hanger).
Two rectangular pieces of cardboard:
- Card A of width \(w_1 \approx 4.0\text{ cm}\) and height \(h \approx 10.0\text{ cm}\)
- Card B of width \(w_2 \approx 8.0\text{ cm}\) and height \(h \approx 10.0\text{ cm}\)
Adhesive tape.
A metre rule.

In this experiment, you will investigate how the damping of a simple pendulum depends on the width of a cardboard card attached to it.

Procedure:

(a) (i) Set up a simple pendulum with a length of approximately 50.0 cm from the clamp to the centre of the bob. Tape Card A of width \(w_1\) vertically to the thread just above the bob, ensuring that the plane of the card is perpendicular to the direction of swing. Measure and record the width \(w_1\) of Card A.

(ii) Position the metre rule horizontally on the bench directly behind the pendulum bob to act as a scale. Displace the bob horizontally from its equilibrium position by an initial amplitude \(x_0 = 15.0\text{ cm}\). Release the bob. Measure and record the amplitude \(x\) of the swing (maximum displacement from equilibrium) after exactly 5 complete oscillations.

(iii) Estimate the percentage uncertainty in your value of \(x\), showing your working.

(b) Remove Card A. Attach Card B to the thread just above the bob in the same manner.

(i) Measure and record the width \(w_2\) of Card B.

(ii) Displace the bob horizontally by the same initial amplitude \(x_0 = 15.0\text{ cm}\). Release the bob. Measure and record the amplitude \(x\) after exactly 5 complete oscillations.

(c) (i) It is suggested that the relationship between \(x\) and \(w\) is given by:
\frac{x_0 - x}{x} = k w
where \(k\) is a constant. Use your data to calculate two values of \(k\).

(ii) Explain whether your results support the suggested relationship. State a criterion for your judgment and show your reasoning clearly.

(d) Describe four sources of uncertainty or systematic/random errors in this experiment. For each, describe an improvement that could be made to improve the accuracy of the experiment. You may suggest the use of other apparatus or different procedures.

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PastPaper.workedSolution

(a) (i) & (ii) Representative measurements for Card A:
Width \(w_1 = 4.0\text{ cm} = 0.040\text{ m}\).
Initial amplitude \(x_0 = 15.0\text{ cm}\).
Amplitude after 5 complete oscillations, \(x_1 = 11.2\text{ cm}\).
(a) (iii) Percentage uncertainty calculation:
Absolute uncertainty in measuring the amplitude \(x\) is typically \(\Delta x = 0.2\text{ cm}\) (due to reading a moving bob on a ruler).
\text{Percentage uncertainty} = \frac{0.2}{11.2} \times 100\% \approx 1.8\%.

(b) Representative measurements for Card B:
Width \(w_2 = 8.0\text{ cm} = 0.080\text{ m}\).
Amplitude after 5 complete oscillations, \(x_2 = 8.4\text{ cm}\).

(c) (i) Calculation of \(k\):
For Card A:
\(k_1 = \frac{x_0 - x_1}{x_1 w_1} = \frac{15.0 - 11.2}{11.2 \times 4.0} = \frac{3.8}{44.8} = 0.0848\text{ cm}^{-1}\) (or \(8.48\text{ m}^{-1}\)).
For Card B:
\(k_2 = \frac{x_0 - x_2}{x_2 w_2} = \frac{15.0 - 8.4}{8.4 \times 8.0} = \frac{6.6}{67.2} = 0.0982\text{ cm}^{-1}\) (or \(9.82\text{ m}^{-1}\)).

(c) (ii) Comparison and evaluation:
Percentage difference between \(k_1\) and \(k_2\):
\% \text{ difference} = \frac{|0.0982 - 0.0848|}{0.0848} \times 100\% = \frac{0.0134}{0.0848} \times 100\% \approx 15.8\%.
Since the percentage difference of 15.8% is less than the typical experimental uncertainty limit of 20%, the results support the suggested relationship.

(d) Evaluation of limitations and improvements:
1. Limitation: Difficult to release the pendulum without lateral movement or a slight shove, which alters the initial amplitude \(x_0\).
Improvement: Use a mechanical release mechanism, such as holding the bob with a thread and cutting the thread with scissors.
2. Limitation: Judging the exact maximum turning point (amplitude) after 5 oscillations is difficult as the bob is in motion for only a brief moment.
Improvement: Record a video of the oscillations with a high-speed camera placed parallel to the scale, and analyze frame-by-frame to find the turning point.
3. Limitation: Two sets of readings are insufficient to conclusively prove a mathematical relationship.
Improvement: Repeat the experiment for five or more cards of different widths and plot a graph of \(\frac{x_0 - x}{x}\) against \(w\).
4. Limitation: The cardboard card is flexible and bends or flutters during swing, altering the drag coefficient and the effective area.
Improvement: Use a rigid plastic sheet of the same dimensions that does not deform under air resistance.

PastPaper.markingScheme

(a) (i) [1 mark] Value of \(w_1\) recorded to the nearest 1 mm with unit.
(a) (ii) [1 mark] Value of \(x_1\) recorded to the nearest 1 mm with unit, where \(x_1 < x_0\).
(a) (iii) [1 mark] Correct calculation of percentage uncertainty in \(x_1\), showing an absolute uncertainty of \(0.1\text{ cm}\) to \(0.5\text{ cm}\) (usually \(0.2\text{ cm}\) is standard for a moving scale).
(b) (i) [1 mark] Value of \(w_2\) recorded to the nearest 1 mm, with \(w_2 > w_1\).
(b) (ii) [1 mark] Value of \(x_2\) recorded to the nearest 1 mm, where \(x_2 < x_1\).

(c) (i) [1 mark] Correct calculations of \(k_1\) and \(k_2\) with correct units (e.g., \(\text{cm}^{-1}\) or \(\text{m}^{-1}\)).
(c) (ii) [1 mark] Valid conclusion based on the percentage difference between \(k_1\) and \(k_2\), compared to a stated threshold (e.g. 20%). If % difference \(\le\) threshold, relationship is supported; if not, it is not supported.

(d) Limitations and Improvements [8 marks total - 1 mark for each valid limitation, 1 mark for each corresponding improvement up to a maximum of 4 pairs]:
1. Release consistency:
- Limitation: Human error in releasing the bob without giving it a side push / hand movement affecting \(x_0\). [1]
- Improvement: Release method using electromagnet or cutting a holding string. [1]
2. Measuring amplitude:
- Limitation: Difficult to read amplitude at the extreme of the swing because the bob is moving / parallax error. [1]
- Improvement: Use video camera with frame-by-frame playback / smart sensor / scale positioned very close behind the bob. [1]
3. Range of data:
- Limitation: Two readings of \(w\) are not enough to confirm the relationship. [1]
- Improvement: Collect more data for different widths and plot a graph of \(\frac{x_0 - x}{x}\) against \(w\). [1]
4. Card stability:
- Limitation: Card bends / flutters during oscillation, changing air resistance. [1]
- Improvement: Use a rigid sheet of plastic / metal of same dimensions that does not bend. [1]
5. Background damping:
- Limitation: Pendulum thread and bob experience air resistance even without the card. [1]
- Improvement: Measure damping without any card attached and subtract this background damping value. [1]

Paper 4 (A Level Structured)

Answer all questions. You should show all your working and use appropriate units.

(a)
- Deduces \(a = -(k/m)x\) from \(F = ma\) and \(F = -kx\), explaining why this represents SHM [1]
- Calculates \(\omega = \sqrt{45/0.35} = 11.3\text{ rad s}^{-1}\) [1]

(b) (i)
- Use of \(E_T = \frac{1}{2} k x_0^2\) (or \(E_T = \frac{1}{2} m \omega^2 x_0^2\)) [1]
- Calculates \(0.14\text{ J}\) (or \(0.144\text{ J}\)) [1]

(ii)
- Relates \(E_p = \frac{1}{2} E_T\) or uses \(x = x_0 / \sqrt{2}\) [1]
- Calculates \(5.7\text{ cm}\) (or \(0.057\text{ m}\), accept \(5.66\text{ cm}\)) [1]

(c) (i)
- Use of \(v = \omega \sqrt{x_0^2 - x^2}\) [1]
- Calculates \(0.79\text{ m s}^{-1}\) (or \(0.78\text{ m s}^{-1}\)) [1]

(ii)
- Use of \(x = x_0 \cos(\omega t)\) to obtain \(\omega t = \pi/3\) [1]
- Calculates \(0.092\text{ s}\) (or \(0.093\text{ s}\)) [1]

Paper 5 (Planning, Analysis and Evaluation)

Answer both questions. Write your answers in the spaces provided.

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PastPaper.question 1 · planning

15 PastPaper.marks

A student is investigating the electromagnetic induction between a long solenoid and a small coaxial search coil placed at the centre of the solenoid. An alternating current of peak value \( I_0 \) and frequency \( f \) is passed through the solenoid. This induces an alternating electromotive force (e.m.f.) of peak value \( V \) in the search coil. It is suggested that \( V \) is related to \( f \) by the equation: \( V = \frac{\mu_0 \pi^2 d^2 N_1 N_2 I_0 f}{L} \) where: \( d \) is the diameter of the search coil, \( N_1 \) is the number of turns of the solenoid, \( L \) is the length of the solenoid, \( N_2 \) is the number of turns of the search coil, and \( \mu_0 \) is the permeability of free space. Design a laboratory experiment to investigate how \( V \) depends on \( f \). You should draw a diagram showing the arrangement of your apparatus. In your description, pay particular attention to: the independent and dependent variables, the control of variables, the apparatus to be used, how the peak current \( I_0 \) is kept constant, how the frequency \( f \) and peak e.m.f. \( V \) are measured, how \( \mu_0 \) is determined from the results, including any graph to be plotted, and safety precautions.

PastPaper.showAnswers

PastPaper.workedSolution

To investigate the relationship \( V = \frac{\mu_0 \pi^2 d^2 N_1 N_2 I_0 f}{L} \): 1. Arrange a long solenoid in series with a signal generator, an AC ammeter, and a rheostat. 2. Place a small circular search coil coaxially at the centre of the solenoid. 3. Connect the terminals of the search coil to the input of a cathode-ray oscilloscope (CRO). 4. Measure the physical parameters of the coils: - Measure the diameter \( d \) of the search coil using a micrometer screw gauge. Measure at different angles and average the readings. - Measure the length \( L \) of the solenoid winding using a metre rule. - Record the number of turns \( N_1 \) of the solenoid and \( N_2 \) of the search coil. 5. Set the frequency \( f \) of the signal generator to a low value. Adjust the rheostat until the peak current reaches a designated value \( I_0 \) (as measured on the AC ammeter or calculated from the peak-to-peak voltage across a known series resistor on the CRO). 6. Observe the induced voltage waveform on the CRO. Measure the peak-to-peak voltage \( V_{\text{p-p}} \) and calculate the peak e.m.f. \( V = V_{\text{p-p}} / 2 \). 7. Measure the period \( T \) of the alternating current waveform on the CRO and calculate the frequency \( f = 1 / T \). 8. Vary the frequency \( f \) of the signal generator to obtain several different values. At each frequency, adjust the rheostat to ensure the peak current \( I_0 \) remains constant, and measure the corresponding peak e.m.f. \( V \). 9. Plot a graph of \( V \) on the y-axis against \( f \) on the x-axis. A straight line passing through the origin confirms the suggested relationship. 10. The gradient \( m \) of the line of best fit is: \( m = \frac{\mu_0 \pi^2 d^2 N_1 N_2 I_0}{L} \). The permeability of free space is determined as: \( \mu_0 = \frac{m L}{\pi^2 d^2 N_1 N_2 I_0} \). 11. Safety: Solenoids carry high currents and can become very hot. Turn off the power supply between taking measurements to prevent overheating and potential burns.

PastPaper.markingScheme

Defining the problem (2 marks):
- M1: Identify \( f \) as the independent variable and \( V \) as the dependent variable.
- M2: State that peak current \( I_0 \) is kept constant (accept also that \( d \), \( N_1 \), \( N_2 \), and \( L \) are kept constant).

Methods of data collection (4 marks):
- M3: Draw a clear, labelled circuit diagram showing a signal generator connected in series with a solenoid, an AC ammeter, and a variable resistor; and the coaxial search coil connected directly to a cathode-ray oscilloscope (CRO).
- M4: Describe how the peak e.m.f. \( V \) is measured using a CRO (e.g., measuring peak-to-peak height and dividing by 2).
- M5: Describe how frequency \( f \) is measured using the CRO time-base (e.g., measuring period \( T \) and calculating \( f = 1/T \)).
- M6: Describe measuring the diameter \( d \) of the search coil using a micrometer screw gauge or Vernier caliper, taking measurements along multiple diameters and taking the average.

Method of analysis (3 marks):
- M7: Plot a graph of \( V \) against \( f \).
- M8: State that the relationship is valid if the graph is a straight line through the origin.
- M9: Express \( \mu_0 \) in terms of the gradient \( m \) as \( \mu_0 = \frac{m L}{\pi^2 d^2 N_1 N_2 I_0} \).

Safety considerations (1 mark):
- M10: Precaution to avoid skin burns by switching off the current/signal generator between readings to prevent the solenoid from overheating.

Additional details (5 marks max):
- A1: Use a variable resistor (rheostat) to adjust and keep the current constant as frequency is varied (since inductive reactance varies with frequency).
- A2: Align the central axis of the search coil coaxially with the axis of the solenoid to ensure maximum magnetic flux linkage.
- A3: Use shielded / coaxial cables for connections to the CRO to reduce high-frequency external noise / interference.
- A4: Use a CRO with high input impedance to prevent loading of the search coil.
- A5: Measure the length \( L \) of the solenoid winding using a metre rule.

PastPaper.question 2 · practical-analysis

15 PastPaper.marks

A student investigates how the amplitude of swing of a magnetic pendulum is damped due to electromagnetic induction in a nearby conductor. A small aluminum plate is suspended from a rigid support and swings as a pendulum between the poles of an electromagnet. The electromagnet is connected to a variable direct current (d.c.) power supply. The amplitude \( A \) of the oscillations of the pendulum after a fixed time \( t = 10.0\text{ s} \) is measured for different currents \( I \) in the electromagnet. The relationship between \( A \) and \( I \) is suggested to be: \( A = A_0 e^{-k I^2} \) where \( A_0 \) is the initial amplitude at \( I = 0 \), and \( k \) is a constant.\
\
(a) State the equation of the straight line graph that should be plotted to test this relationship. Explain how the values of \( A_0 \) and \( k \) can be determined from the y-intercept and gradient of this graph. (2 marks)\
\
(b) Values of \( I \) and \( A \) are given in the table below. The absolute uncertainty in \( A \) is \( \pm 0.2\text{ cm} \). Calculate and record values of \( I^2 / \text{A}^2 \) and \( \ln(A / \text{cm}) \) in the table. Include the absolute uncertainties in \( \ln(A / \text{cm}) \). (3 marks)\
\
| \( I / \text{A} \) | \( I^2 / \text{A}^2 \) | \( A / \text{cm} \) | \( \ln(A / \text{cm}) \) |\
| :---: | :---: | :---: | :---: |\
| 0.50 | | 11.6 | |\
| 1.00 | | 10.3 | |\
| 1.50 | | 8.6 | |\
| 2.00 | | 6.6 | |\
| 2.50 | | 4.7 | |\
| 3.00 | | 3.1 | |\
\
(c) (i) Plot a graph of \( \ln(A / \text{cm}) \) against \( I^2 / \text{A}^2 \). Include error bars for \( \ln(A / \text{cm}) \). (2 marks)\
(ii) Draw the straight line of best fit and a worst acceptable straight line on your graph. Both lines should be clearly labelled. (1 mark)\
(iii) Determine the gradient of the line of best fit and the uncertainty in this gradient. (2 marks)\
(iv) Determine the y-intercept of the line of best fit and the uncertainty in this y-intercept. (2 marks)\
\
(d) Using your answers from (c)(iii) and (c)(iv), determine the values of \( k \) and \( A_0 \). Include appropriate units and absolute uncertainties for both constants. (3 marks)

PastPaper.showAnswers

PastPaper.workedSolution

(a) Taking natural logarithms: \( \ln(A) = -k I^2 + \ln(A_0) \). By plotting \( \ln(A / \text{cm}) \) on the y-axis against \( I^2 / \text{A}^2 \) on the x-axis, we obtain a straight line where: - Gradient \( m = -k \) (or \( k = -m \)) - y-intercept \( c = \ln(A_0 / \text{cm}) \) (or \( A_0 = e^c \)).\
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(b) The completed table values: - For \( I = 0.50 \): \( I^2 = 0.25 \), \( A = 11.6 \), \( \ln(A / \text{cm}) = 2.45 \pm 0.02 \) (uncertainty: \( \ln(11.8) - \ln(11.6) = 0.017 \approx 0.02 \)) - For \( I = 1.00 \): \( I^2 = 1.00 \), \( A = 10.3 \), \( \ln(A / \text{cm}) = 2.33 \pm 0.02 \) (uncertainty: \( \ln(10.5) - \ln(10.3) = 0.019 \approx 0.02 \)) - For \( I = 1.50 \): \( I^2 = 2.25 \), \( A = 8.6 \), \( \ln(A / \text{cm}) = 2.15 \pm 0.02 \) (uncertainty: \( \ln(8.8) - \ln(8.6) = 0.023 \approx 0.02 \)) - For \( I = 2.00 \): \( I^2 = 4.00 \), \( A = 6.6 \), \( \ln(A / \text{cm}) = 1.89 \pm 0.03 \) (uncertainty: \( \ln(6.8) - \ln(6.6) = 0.030 \approx 0.03 \)) - For \( I = 2.50 \): \( I^2 = 6.25 \), \( A = 4.7 \), \( \ln(A / \text{cm}) = 1.55 \pm 0.04 \) (uncertainty: \( \ln(4.9) - \ln(4.7) = 0.042 \approx 0.04 \)) - For \( I = 3.00 \): \( I^2 = 9.00 \), \( A = 3.1 \), \( \ln(A / \text{cm}) = 1.13 \pm 0.06 \) (uncertainty: \( \ln(3.3) - \ln(3.1) = 0.063 \approx 0.06 \)).\
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(c) (i) Plotting: Points plotted accurately at (0.25, 2.45), (1.00, 2.33), (2.25, 2.15), (4.00, 1.89), (6.25, 1.55), (9.00, 1.13). Vertical error bars drawn of length \( \pm 0.02 \) to \( \pm 0.06 \).\
(ii) Lines: The best-fit line is drawn smoothly. The worst acceptable line is drawn passing through the bottom of the first error bar and the top of the last error bar (or vice versa).\
(iii) Gradient: - Gradient of best-fit line: \( m = \frac{1.13 - 2.45}{9.00 - 0.25} = \frac{-1.32}{8.75} = -0.151\text{ A}^{-2} \). - Worst acceptable gradient: \( m_{\text{worst}} = \frac{1.19 - 2.43}{9.00 - 0.25} = \frac{-1.24}{8.75} = -0.142\text{ A}^{-2} \). - Uncertainty in gradient: \( \Delta m = |m_{\text{best}} - m_{\text{worst}}| = 0.151 - 0.142 = 0.009\text{ A}^{-2} \) (or \( \pm 0.01\text{ A}^{-2} \).\
(iv) y-intercept: - y-intercept of best-fit line: \( c = 2.45 - (-0.151 \times 0.25) = 2.49 \). - y-intercept of worst acceptable line: \( c_{\text{worst}} = 2.43 - (-0.142 \times 0.25) = 2.47 \). - Uncertainty in y-intercept: \( \Delta c = |c_{\text{best}} - c_{\text{worst}}| = 2.49 - 2.47 = 0.02 \).\
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(d) Determination of constants: - \( k = -m = 0.151 \pm 0.009\text{ A}^{-2} \) (or \( 0.15 \pm 0.01\text{ A}^{-2} \)). - \( A_0 = e^c = e^{2.49} = 12.1\text{ cm} \). - Uncertainty in \( A_0 \): \( \Delta A_0 = e^{c + \Delta c} - e^c = e^{2.51} - e^{2.49} = 12.3 - 12.1 = 0.2\text{ cm} \) (or \( \pm 0.3\text{ cm} \) using worst intercept difference). Thus, \( A_0 = 12.1 \pm 0.3\text{ cm} \).

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Part (a) (2 marks):
- M1: State the equation \( \ln(A) = -k I^2 + \ln(A_0) \).
- M2: State that \( k = -\text{gradient} \) and \( A_0 = e^{\text{y-intercept}} \).

Part (b) (3 marks):
- M3: All values of \( I^2 \) calculated correctly: 0.25, 1.00, 2.25, 4.00, 6.25, 9.00 (allow 2 or 3 decimal places).
- M4: All values of \( \ln(A / \text{cm}) \) calculated correctly: 2.45, 2.33, 2.15, 1.89, 1.55, 1.13.
- M5: Absolute uncertainties in \( \ln(A / \text{cm}) \) calculated correctly: \( \pm 0.02, \pm 0.02, \pm 0.02, \pm 0.03, \pm 0.04, \pm 0.06 \).

Part (c) (7 marks):
- M6: Plotting: All 6 points plotted correctly within half a small square. Error bars plotted correctly on the y-axis with correct lengths.
- M7: Line of Best Fit: Drawn as a single straight line passing through all error bars.
- M8: Worst Acceptable Line: Drawn as a straight line passing through the top of the first error bar and the bottom of the last error bar, or vice versa.
- M9: Gradient of best-fit line calculated correctly using points on the line separated by at least half the length of the line. Expected value: \( -0.151 \pm 0.005 \).
- M10: Uncertainty in gradient calculated correctly as \( |\text{best gradient} - \text{worst gradient}| \). Expected value: \( \pm 0.009 \) (or \( \pm 0.01 \)).
- M11: y-intercept of best-fit line determined correctly. Expected value: \( 2.49 \pm 0.03 \).
- M12: Uncertainty in y-intercept determined correctly as \( |\text{best intercept} - \text{worst intercept}| \). Expected value: \( \pm 0.02 \) to \( \pm 0.03 \).

Part (d) (3 marks):
- M13: Value of \( k \) given with unit: \( 0.151 \pm 0.009\text{ A}^{-2} \) (or \( 0.15 \pm 0.01\text{ A}^{-2} \)). Unit must be \( \text{A}^{-2} \).
- M14: Value of \( A_0 \) calculated correctly: \( 12.1\text{ cm} \) (allow range \( 11.9 \) to \( 12.2 \)).
- M15: Uncertainty in \( A_0 \) calculated correctly using \( e^{c_{\text{best}}} - e^{c_{\text{worst}}} \) or equivalent. Expected value: \( \pm 0.3\text{ cm} \) (allow range \( \pm 0.2 \) to \( \pm 0.4\text{ cm} \)).

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