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The horizontal velocity remains constant at \(v_x = u\). At the ground, the angle of impact is \(45^\circ\), which means the vertical component of velocity \(v_y\) is equal to the horizontal component \(v_x\), so \(v_y = u\). Using the equation of motion for the vertical direction, \(v_y^2 = u_y^2 + 2g h\) where the initial vertical velocity \(u_y = 0\), we get \(u^2 = 2g h\). Solving for \(h\) gives \(h = \frac{u^2}{2g}\).

1 mark for the correct answer A. Reject other options.

Since the wires are connected in series, the electrical current \(I\) is the same in both wires. The current is given by \(I = n A v_d q\), where \(n\) is the charge carrier density, \(A\) is the cross-sectional area, \(v_d\) is the drift velocity, and \(q\) is the elementary charge. Since both wires are made of copper, \(n\) and \(q\) are constant. The cross-sectional area of wire X is \(A_X = \pi \left(\frac{d}{2}\right)^2\) and for wire Y is \(A_Y = \pi \left(\frac{2d}{2}\right)^2 = 4 A_X\). Since \(I\) is constant, \(A_X v = A_Y v_Y\), which gives \(v_Y = v \left(\frac{A_X}{A_Y}\right) = \frac{v}{4}\).

1 mark for the correct answer A. Reject other options.

Rearranging the formula for viscosity gives \(\eta = \frac{F}{6\pi r v}\). In terms of SI base units, the unit of force \(F\) is \(\text{kg m s}^{-2}\), the unit of radius \(r\) is \(\text{m}\), and the unit of velocity \(v\) is \(\text{m s}^{-1}\). Substituting these into the formula yields the units of \(\eta\) as \(\frac{\text{kg m s}^{-2}}{\text{m} \times \text{m s}^{-1}} = \text{kg m}^{-1}\text{s}^{-1}\).

1 mark for the correct answer A. Reject other options.

The elastic strain energy stored in a wire is given by \(W = \frac{1}{2} F x\), where \(F\) is the tension force and \(x\) is the extension. From the definition of Young modulus, \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{x/L} = \frac{F L}{A x}\). Rearranging this for force gives \(F = \frac{E A x}{L}\). Substituting this expression for \(F\) into the energy equation gives \(W = \frac{1}{2} \left(\frac{E A x}{L}\right) x = \frac{E A x^2}{2L}\).

1 mark for the correct answer A. Reject other options.

The forces acting on the submerged sphere in equilibrium are the upward upthrust \(U\), the downward weight of the sphere \(W\), and the downward tension \(T\) from the string. Therefore, the equilibrium equation is \(U = W + T\), which rearranges to \(T = U - W\). By Archimedes' principle, the upthrust is the weight of the displaced liquid: \(U = \rho_l V g\). The weight of the sphere is \(W = \rho_s V g\). Substituting these yields \(T = \rho_l V g - \rho_s V g = (\rho_l - \rho_s)Vg\).

1 mark for the correct answer A. Reject other options.

Power is defined as the rate of energy transfer, so the total work done on the car by the engine in time \(t\) is \(W = P t\). Since there are no resistive forces, all of this work is converted into the kinetic energy of the car. Therefore, \(P t = \frac{1}{2} m v^2\). Rearranging for \(v\) gives \(v^2 = \frac{2 P t}{m}\), which simplifies to \(v = \sqrt{\frac{2 P t}{m}}\).

1 mark for the correct answer A. Reject other options.

The relationship between terminal potential difference \(V\), e.m.f. \(\mathcal{E}\), current \(I\), and internal resistance \(r\) is given by the equation \(V = \mathcal{E} - I r\). This can be rearranged into the standard straight-line form \(y = mx + c\) as \(V = (-r)I + \mathcal{E}\). Comparing terms, \(V\) is on the vertical axis, \(I\) is on the horizontal axis, the gradient \(m\) is \(-r\) (a negative gradient of magnitude \(r\)), and the vertical intercept \(c\) is \(\mathcal{E}\).

1 mark for the correct answer A. Reject other options.

The volume \(V\) of a cylinder of length \(L\) and cross-sectional area \(A\) is \(V = A L\). Since the volume remains constant when the wire is stretched to a length of \(2L\), the new cross-sectional area \(A'\) must satisfy \(V = A' (2L) = A L\), which gives \(A' = \frac{A}{2}\). The initial resistance is \(R = \frac{\rho L}{A}\). The new resistance \(R'\) is given by \(R' = \frac{\rho (2L)}{A'} = \frac{\rho (2L)}{A/2} = 4\left(\frac{\rho L}{A}\right) = 4R\).

1 mark for the correct answer A. Reject other options.

For part (a), resolve the initial velocity into its vertical component: \(u_y = 24.0 \times \sin(35.0^\circ) = 13.77\text{ m s}^{-1}\). At the maximum height, the vertical velocity \(v_y = 0\). Using \(v_y^2 = u_y^2 + 2as\), we find the vertical displacement \(s = \frac{0 - (13.77)^2}{2 \times (-9.81)} = 9.66\text{ m}\). The maximum height above the ground is \(5.00\text{ m} + 9.66\text{ m} = 14.66\text{ m}\), which is approximately \(14.7\text{ m}\). For part (b), use conservation of energy: Total Initial Energy = Total Final Energy. \(\frac{1}{2} m u^2 + m g h = \frac{1}{2} m v^2\). Dividing by mass \(m\) gives \(v = \sqrt{u^2 + 2 g h} = \sqrt{24.0^2 + 2 \times 9.81 \times 5.00} = \sqrt{576 + 98.1} = 25.96\text{ m s}^{-1}\), which rounds to \(26.0\text{ m s}^{-1}\).

Mark 1: Vertical component of initial velocity calculated correctly (\(13.8\text{ m s}^{-1}\)). Mark 2: Correct kinematic equation selected and substituted to find vertical displacement (\(9.66\text{ m}\)). Mark 3: Adds the initial 5.00 m height to find total height (\(14.7\text{ m}\)). Mark 4: Correct energy conservation equation or vertical/horizontal kinematic breakdown setup. Mark 5: Identifies correct horizontal velocity component remains constant (\(19.7\text{ m s}^{-1}\)) and solves for final vertical component (\(17.0\text{ m s}^{-1}\)) OR sets up the speed formula using scalar energy conservation. Mark 6: Correct calculation of final speed as \(26.0\text{ m s}^{-1}\) (allow 2 or 3 s.f.). Mark 0.85: Quality of written communication, clarity of steps, and correct units throughout.

We use the relationship \(E = I(R + r)\), where \(E\) is the EMF, \(I\) is the current, \(R\) is the external resistance, and \(r\) is the internal resistance. For the first case: \(E = 0.320(4.50 + r)\). For the second case: \(E = 0.160(9.50 + r)\). Since the EMF is constant, we can equate the two expressions: \(0.320(4.50 + r) = 0.160(9.50 + r)\). Dividing both sides by 0.160 gives: \(2(4.50 + r) = 9.50 + r \Rightarrow 9.00 + 2r = 9.50 + r \Rightarrow r = 0.50\ \Omega\). Substitute \(r\) back into either equation to find \(E\): \(E = 0.320(4.50 + 0.50) = 0.320 \times 5.00 = 1.60\text{ V}\).

Mark 1: State or use the formula \(E = I(R + r)\) or \(E = V + Ir\). Mark 2: Set up two simultaneous equations with the given values. Mark 3: Equate the equations to eliminate \(E\). Mark 4: Correctly solve for internal resistance \(r = 0.50\ \Omega\). Mark 5: Substitute \(r\) back to solve for EMF. Mark 6: Correct value of EMF \(E = 1.60\text{ V}\). Mark 0.85: Correct units (ohms and volts) stated clearly with consistent 2 or 3 significant figures.

First, calculate the cross-sectional area \(A = \pi d^2 / 4 = \pi (0.800 \times 10^{-3})^2 / 4 = 5.027 \times 10^{-7}\text{ m}^2\). The tensile force \(F = m g = 5.50 \times 9.81 = 53.96\text{ N}\). Using Young Modulus \(E = \frac{F L}{A \Delta L}\), we rearrange to find extension \(\Delta L = \frac{F L}{A E} = \frac{53.96 \times 2.40}{5.027 \times 10^{-7} \times 1.20 \times 10^{11}} = 2.146 \times 10^{-3}\text{ m} = 2.15\text{ mm}\). The elastic strain energy stored is \(E_{\text{str}} = \frac{1}{2} F \Delta L = 0.5 \times 53.96 \times 2.146 \times 10^{-3} = 0.0579\text{ J}\), which rounds to \(0.058\text{ J}\).

Mark 1: Calculate correct cross-sectional area \(5.03 \times 10^{-7}\text{ m}^2\). Mark 2: Calculate tension force \(F = 53.96\text{ N}\). Mark 3: Recall and rearrange Young modulus formula for extension. Mark 4: Obtain correct extension of \(2.15\text{ mm}\) (or \(2.1\text{ mm}\)). Mark 5: Recall formula for elastic strain energy \(\frac{1}{2} F \Delta L\). Mark 6: Obtain correct strain energy value of \(0.058\text{ J}\). Mark 0.85: Show consistent use of SI units and final answers to appropriate significant figures.

At terminal velocity, the forces are balanced: Upward forces (Upthrust + Drag) = Downward force (Weight). Thus, \(U + F_d = W \Rightarrow F_d = W - U\). Drag force is given by Stokes' Law: \(F_d = 6 \pi \eta r v\). Weight \(W = \rho_s V g\) and Upthrust \(U = \rho_f V g\). Therefore, \(6 \pi \eta r v = \frac{4}{3} \pi r^3 (\rho_s - \rho_f) g\). Rearranging for velocity: \(v = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}\). Substituting the values: \(v = \frac{2 \times (1.50 \times 10^{-3})^2 \times (7800 - 950) \times 9.81}{9 \times 0.250} = \frac{2 \times (2.25 \times 10^{-6}) \times 6850 \times 9.81}{2.25} = 2 \times 10^{-6} \times 6850 \times 9.81 = 0.1344\text{ m s}^{-1}\).

Mark 1: Identify that at terminal velocity, weight equals the sum of upthrust and drag. Mark 2: Use formulas for sphere volume \(\frac{4}{3}\pi r^3\) and mass to express weight and upthrust. Mark 3: Correctly recall or use Stokes' Law formula \(F_d = 6 \pi \eta r v\). Mark 4: Substitute densities and radius correctly into the balanced force equation. Mark 5: Rearrange equation to make terminal velocity the subject. Mark 6: Calculate the terminal velocity as \(0.134\text{ m s}^{-1}\) (or \(0.13\text{ m s}^{-1}\)). Mark 0.85: Explicitly state the assumption that flow must be laminar for Stokes' Law to hold.

First, calculate the cross-sectional area \(A = \pi d^2 / 4 = \pi (0.46 \times 10^{-3})^2 / 4 = 1.662 \times 10^{-7}\text{ m}^2\). Resistivity is given by \(\rho = \frac{R A}{L} = \frac{4.2 \times 1.662 \times 10^{-7}}{1.50} = 4.65 \times 10^{-7}\ \Omega\text{ m}\). To find the percentage uncertainty, find the individual percentage uncertainties: \(\%\Delta L = \frac{0.01}{1.50} \times 100 = 0.67\%\), \(\%\Delta R = \frac{0.1}{4.2} \times 100 = 2.38\%\), \(\%\Delta d = \frac{0.02}{0.46} \times 100 = 4.35\%\). Since \(A \propto d^2\), the percentage uncertainty in area \(A\) is \(2 \times \%\Delta d = 8.70\%\). The total percentage uncertainty in resistivity is \(\%\Delta \rho = \%\Delta R + \%\Delta A + \%\Delta L = 2.38\% + 8.70\% + 0.67\% = 11.75\% \approx 12\%\). Thus, resistivity \(\rho = 4.7 \times 10^{-7}\ \Omega\text{ m}\) with an uncertainty of \(12\%\).

Mark 1: Calculate the cross-sectional area correctly. Mark 2: Use the resistivity formula \(\rho = R A / L\) to find the absolute value (\(4.7 \times 10^{-7}\ \Omega\text{ m}\)). Mark 3: Find percentage uncertainty in length (\(0.67\%\)) and resistance (\(2.38\%\)). Mark 4: Find percentage uncertainty in diameter (\(4.35\%\)). Mark 5: Double the percentage uncertainty of diameter to find the percentage uncertainty in area (\(8.70\%\)). Mark 6: Sum all percentage uncertainties to obtain approximately \(12\%\). Mark 0.85: Express the final resistivity and uncertainty to a consistent and appropriate number of significant figures.

For part (a), when unpolarized light of intensity \(I_0\) passes through an ideal polarizer, its intensity is halved because only one plane of polarization is transmitted: \(I_1 = \frac{1}{2} I_0 = \frac{1}{2} \times 320 = 160\text{ W m}^{-2}\). For part (b), apply Malus's Law for the second filter: \(I_2 = I_1 \cos^2(\theta)\). Rearranging gives: \(\cos^2(\theta) = \frac{I_2}{I_1} = \frac{40.0}{160} = 0.25\). Taking the square root: \(\cos(\theta) = \sqrt{0.25} = 0.50\). Therefore, the angle \(\theta = \arccos(0.50) = 60.0^\circ\) (or \(\pi/3\) radians).

Mark 1: State that unpolarized light intensity is halved upon passing through the first polarizer. Mark 2: Calculate \(I_1 = 160\text{ W m}^{-2}\). Mark 3: State or write Malus's Law: \(I_2 = I_1 \cos^2(\theta)\). Mark 4: Substitute the values correctly into Malus's Law. Mark 5: Solve for \(\cos^2(\theta) = 0.25\) or \(\cos(\theta) = 0.50\). Mark 6: Calculate angle \(\theta = 60.0^\circ\). Mark 0.85: Show clear steps and include correct units for intensity and angle.

For part (a), the speed of a transverse wave on a stretched string is given by \(v = \sqrt{\frac{T}{\mu}}\), where \(T\) is the tension and \(\mu\) is the mass per unit length. \(v = \sqrt{\frac{110}{4.50 \times 10^{-3}}} = \sqrt{24444.4} = 156.3\text{ m s}^{-1}\). For part (b), the fundamental mode of oscillation has nodes at both fixed ends and a single antinode in the middle. The wavelength of the fundamental mode is \(\lambda = 2L = 2 \times 0.650\text{ m} = 1.30\text{ m}\). The fundamental frequency is \(f = \frac{v}{\lambda} = \frac{156.3}{1.30} = 120.2\text{ Hz}\), which rounds to \(120\text{ Hz}\).

Mark 1: State or select formula \(v = \sqrt{T/\mu}\). Mark 2: Calculate wave speed correctly as \(156\text{ m s}^{-1}\). Mark 3: State the relationship between wavelength and string length for the fundamental mode (\(\lambda = 2L\)). Mark 4: Calculate the fundamental wavelength as \(1.30\text{ m}\). Mark 5: Recall and use wave equation \(v = f \lambda\). Mark 6: Calculate fundamental frequency as \(120\text{ Hz}\). Mark 0.85: Write clear, logical steps with correct units for speed (\(\text{m s}^{-1}\)) and frequency (\(\text{Hz}\)).

**(a)** Original length: Metre rule or tape measure, because the length is about \( 2 \text{ m} \) (beyond the capacity of micrometers or calipers) and a resolution of \( 1 \text{ mm} \) gives a very low percentage uncertainty (about \( 0.05\% \)).
Diameter: Micrometer screw gauge, because the diameter is very small (\( 0.46 \text{ mm} \)) and requires a high resolution of \( 0.01 \text{ mm} \) or \( 0.001 \text{ mm} \) to keep the percentage uncertainty low.

**(b)** Cross-sectional area:
\( A = \frac{\pi d^2}{4} = \frac{\pi (0.46 \times 10^{-3} \text{ m})^2}{4} = 1.662 \times 10^{-7} \text{ m}^2 \) (or \( 1.66 \times 10^{-7} \text{ m}^2 \)).
Percentage uncertainty in diameter \( d \):
\( \frac{0.01}{0.46} \times 100\% = 2.17\% \).
Since \( A \propto d^2 \), the percentage uncertainty in area \( A \) is:
\( 2 \times 2.17\% = 4.35\% \) (accept \( 4.3\% \) or \( 4.4\% \)).

**(c)** Young modulus equation:
\( E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta x / L} = \left( \frac{F}{\Delta x} \right) \frac{L}{A} \).
Since \( \frac{F}{\Delta x} \) is the gradient of the force-extension graph:
\( E = \text{gradient} \times \frac{L}{A} = (1.38 \times 10^4 \text{ N m}^{-1}) \times \frac{2.030 \text{ m}}{1.662 \times 10^{-7} \text{ m}^2} = 1.686 \times 10^{11} \text{ Pa} \).
Rounding to 3 significant figures: \( E = 1.69 \times 10^{11} \text{ Pa} \) (or \( \text{N m}^{-2} \)).

**(d)** If the wire slips or the support bends, the measured extension \( \Delta x \) will be larger than the actual extension of the wire for any given load.
This will cause the calculated gradient \( \frac{F}{\Delta x} \) to be lower than it should be.
Consequently, the calculated Young modulus \( E \) will be underestimated (lower than the true value).
To minimize wire slip, a double-clamp arrangement or Searle's apparatus can be used, or a reference marker (fiducial marker) can be placed on the wire near the clamp to measure extension relative to that marker.

**(a)**
- Award 1 mark for identifying a metre rule/tape measure for length AND a micrometer screw gauge for diameter.
- Award 1 mark for correct justification based on resolution and range (e.g., metre rule has appropriate range for 2 m, micrometer has 0.01 mm resolution suitable for thin wire).

**(b)**
- Award 1 mark for correct calculation of cross-sectional area: \( 1.66 \times 10^{-7} \text{ m}^2 \).
- Award 1 mark for calculating the percentage uncertainty in diameter: \( 2.17\% \) (or \( 2.2\% \)).
- Award 1 mark for doubling the percentage uncertainty in diameter to find the percentage uncertainty in area: \( 4.3\% \) or \( 4.4\% \) (accept \( 4\% \) if 1 sig fig is used).

**(c)**
- Award 1 mark for recalling \( E = \frac{FL}{A\Delta x} \) or recognizing that \( E = \text{gradient} \times \frac{L}{A} \).
- Award 1 mark for substituting their values of gradient, \( L \text{ (2.030 m)} \), and \( A \) correctly into the equation.
- Award 1 mark for calculating a final value in the range \( 1.68 \times 10^{11} \) to \( 1.70 \times 10^{11} \).
- Award 1 mark for the correct unit: \( \text{Pa} \) or \( \text{N m}^{-2} \).

**(d)**
- Award 1 mark for stating that slipping or bending increases the measured extension \( \Delta x \).
- Award 1 mark for explaining that this decreases the gradient / underestimates the Young modulus.
- Award 1 mark for describing a valid method to minimize (e.g., using Searle's apparatus, measuring relative to a control wire, or using a fiducial marker).

**(a)** The circuit diagram must show:
- The cell, an ammeter, a variable resistor (rheostat), and a switch connected in a single series loop.
- A voltmeter connected in parallel across either the cell (and its internal resistance) or the variable resistor.

**(b)** Derivation:
Start with: \( \varepsilon = I(R + r) = IR + Ir \).
Since the terminal potential difference \( V \) across the external load resistance \( R \) is given by \( V = IR \), we can substitute this:
\( \varepsilon = V + Ir \).
Rearranging to make \( V \) the subject:
\( V = -rI + \varepsilon \).
Comparing this with the equation of a straight line, \( y = mx + c \):
- \( V \) is plotted on the vertical y-axis and \( I \) on the horizontal x-axis.
- The gradient of the graph is equal to \( -r \), so the internal resistance \( r = -\text{gradient} \).
- The y-intercept of the graph is equal to the e.m.f. \( \varepsilon \).

**(c)** From the line of best fit \( V = -1.45I + 1.52 \):
Experimental internal resistance \( r = 1.45\ \Omega \).
Manufacturer's internal resistance \( r_{\text{nominal}} = 1.20\ \Omega \).
Percentage difference:
\( \frac{|1.45 - 1.20|}{1.20} \times 100\% = 20.8\% \) (or \( 21\% \)).
Practical reason for higher value:
- The cell has been discharged/used over time, which naturally increases its internal resistance.
- The temperature of the cell increased during the experiment, raising the resistance of the electrolytes.
- The resistance of the connecting wires and contacts is included in the total resistance measured, adding to the apparent internal resistance.

**(d)** If a low-resistance voltmeter is used, a significant current will flow through the voltmeter even when the external switch is open (or the main loop current is very small).
This causes a continuous voltage drop ('lost volts') across the internal resistance of the cell.
As a result, the measured terminal potential difference \( V \) is always lower than the true e.m.f. of the cell, leading to an underestimate of the e.m.f. (the y-intercept of the graph shifts downwards).

**(a)**
- Award 1 mark for drawing a correct series loop containing the cell, variable resistor, ammeter, and switch.
- Award 1 mark for drawing the voltmeter in parallel across the cell or across the variable resistor (using correct standard circuit symbols).

**(b)**
- Award 1 mark for substituting \( V = IR \) to obtain \( \varepsilon = V + Ir \).
- Award 1 mark for rearranging to \( V = -rI + \varepsilon \).
- Award 1 mark for identifying that \( r = -\text{gradient} \) (or magnitude of the gradient).
- Award 1 mark for identifying that \( \varepsilon = \text{y-intercept} \).

**(c)**
- Award 1 mark for calculating the percentage difference correctly: \( 20.8\% \) (or \( 21\% \)).
- Award 1 mark for a valid practical reason (e.g., cell has been used/deteriorated, temperature increase, or wire/contact resistance included in measured value).

**(d)**
- Award 1 mark for stating that current flows through the voltmeter / voltmeter is not drawing negligible current.
- Award 1 mark for explaining that this results in 'lost volts' / voltage drop across the internal resistance.
- Award 1 mark for concluding that the measured terminal potential difference is lower, leading to an underestimate of the e.m.f.

Unpolarised light has its intensity halved upon passing through the first polarizing filter, so the intensity becomes \(I_1 = \frac{1}{2} I_0\). When this polarised light passes through the second filter, Malus's Law applies: \(I_2 = I_1 \cos^2(\theta)\). Here, \(\theta = 30^\circ\), so \(I_2 = \frac{1}{2} I_0 \cos^2(30^\circ) = \frac{1}{2} I_0 \times 0.75 = 0.375 I_0\) which is approximately \(0.38 I_0\).

Correct answer is B (1 mark). Reject other options due to incorrect application of Malus's law or omitting the initial polarisation effect.

The Young's modulus is given by \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta x / L}\), which rearranges to extension \(\Delta x = \frac{FL}{AE}\). Since the cross-sectional area \(A = \pi \frac{d^2}{4}\), the extension is proportional to \(\frac{L}{d^2}\) under the same force and material. For wire P, \(\Delta x_P \propto \frac{L}{d^2}\). For wire Q, \(\Delta x_Q \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = 0.5 \frac{L}{d^2}\). Therefore, the ratio \(\frac{\Delta x_P}{\Delta x_Q} = \frac{1}{0.5} = 2.0\).

Correct answer is C (1 mark). Reject A, B, and D because they result from incorrect proportional reasoning of area or length scaling.

The relationship for the critical angle \(\theta_c\) at the boundary between two media is \(\sin\theta_c = \frac{n_{\text{less}}}{n_{\text{more}}}\). Here, glass is the denser medium (\(n_{\text{more}} = 1.60\)) and the liquid is the less dense medium (\(n_{\text{less}} = n\)). Substituting the values: \(\sin(55^\circ) = \frac{n}{1.60}\), which gives \(n = 1.60 \times \sin(55^\circ) = 1.60 \times 0.8192 = 1.31\).

Correct answer is A (1 mark). Reject other options due to incorrect inversion of the formula or misuse of trigonometric functions.

At the moment of release, the velocity of the sphere is zero, meaning there is no viscous drag. The initial acceleration is at its maximum (determined by gravity and upthrust). As the sphere speeds up, the viscous drag force increases, which decreases the net downward force and thus the acceleration. This decrease occurs at a decreasing rate because the acceleration itself is decreasing, leading to an asymptotic approach to zero acceleration at terminal velocity.

Correct answer is B (1 mark). Reject options describing constant or increasing acceleration, as well as linear decreases which do not account for the velocity-dependent nature of drag.

For a string of length \(L\) fixed at both ends, the third harmonic consists of three complete loops, which correspond to three half-wavelengths: \(3 \times \left(\frac{\lambda}{2}\right) = L\). This gives a wavelength of \(\lambda = \frac{2L}{3}\). The distance between adjacent nodes is equal to half a wavelength, which is \(\frac{\lambda}{2} = \frac{L}{3}\).

Correct answer is A (1 mark). Reject B (wavelength), C (quarter wavelength), and D (incorrect harmonic relation).

Using Einstein's photoelectric equation, \(E_{\text{max}} = \frac{hc}{\lambda} - \phi\). This can be rewritten as \(\frac{hc}{\lambda} = E_{\text{max}} + \phi\). If the wavelength is halved, the energy of each incoming photon becomes \(\frac{hc}{\lambda/2} = 2\frac{hc}{\lambda}\). The new maximum kinetic energy \(E'_{\text{max}}\) is therefore: \(E'_{\text{max}} = 2\frac{hc}{\lambda} - \phi = 2(E_{\text{max}} + \phi) - \phi = 2E_{\text{max}} + \phi\).

Correct answer is B (1 mark). Reject A, C, and D because they fail to correctly partition the work function subtraction when the photon energy is doubled.

Brittle materials are characterised by the fact that they undergo little or no plastic deformation before fracturing. They exhibit elastic behaviour up to their breaking point, where they fail suddenly. In contrast, ductile materials undergo significant plastic deformation.

Correct answer is B (1 mark). Reject options describing ductile behavior (A and C) or incorrect material properties (D).

The grating equation for the first-order maximum is \(d \sin\theta = \lambda\), which gives \(\sin\theta = \frac{\lambda}{d}\). For the new setup, the wavelength is \(\lambda' = 1.2\lambda\) and the slit spacing is \(d' = 0.5d\). Thus, the sine of the new angle is \(\sin\theta' = \frac{1.2\lambda}{0.5d} = 2.4 \frac{\lambda}{d} = 2.4 \sin\theta\).

Correct answer is C (1 mark). Reject other options due to incorrect multiplication or division of the scale factors.

(a) Coherent sources have a constant phase relationship/difference and the same frequency (or wavelength).

(b) Light passing through the slits undergoes diffraction (spreads out). The diffracted waves overlap in the space beyond the grating. Where they meet in phase (path difference of an integer number of wavelengths, \(n\lambda\)), constructive interference occurs, producing bright fringes (maxima). Where they meet in anti-phase (path difference of \((n + \frac{1}{2})\lambda\)), destructive interference occurs, producing dark regions.

(c) First, calculate the grating spacing \(d\):
\(d = \frac{1 \times 10^{-3}\text{ m}}{500} = 2.00 \times 10^{-6}\text{ m}\)

Using the diffraction grating equation for the second-order maximum (\(n = 2\)):
\(d \sin \theta = n \lambda\)
\(\sin \theta_2 = \frac{2 \times 633 \times 10^{-9}\text{ m}}{2.00 \times 10^{-6}\text{ m}} = 0.633\)
\(\theta_2 = \sin^{-1}(0.633) = 39.27^\circ\)

Using trigonometry to find the distance \(x\) on the screen:
\(x = D \tan \theta_2 = 1.80\text{ m} \times \tan(39.27^\circ) = 1.80 \times 0.8176 = 1.47\text{ m}\)

Part (a):
- Constant phase relationship/difference (1)
- Same frequency / wavelength (1)

Part (b):
- Mention of diffraction / spreading out of light as it passes through the slits (1)
- Overlapping / superposition of waves (1)
- Constructive interference where waves arrive in phase / path difference of \(n\lambda\) to form maxima (1)

Part (c):
- Calculation of \(d = 2.00 \times 10^{-6}\text{ m}\) (1)
- Calculation of \(\theta_2 = 39.3^\circ\) using \(d \sin \theta = 2\lambda\) (1)
- Calculation of \(x = 1.47\text{ m}\) using \(x = D \tan \theta\) (1)

(a) Tensile stress is defined as the applied force per unit cross-sectional area (\(\sigma = F/A\)). Tensile strain is defined as the extension per unit original length (\(\epsilon = \Delta L/L\)).

(b) Calculate cross-sectional area:
\(r = 0.40\text{ mm} = 4.0 \times 10^{-4}\text{ m}\)
\(A = \pi r^2 = \pi (4.0 \times 10^{-4}\text{ m})^2 = 5.03 \times 10^{-7}\text{ m}^2\)

Young Modulus definition:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}\)

Rearranging for extension \(\Delta L\):
\(\Delta L = \frac{FL}{AE} = \frac{35\text{ N} \times 2.40\text{ m}}{5.03 \times 10^{-7}\text{ m}^2 \times 1.2 \times 10^{11}\text{ Pa}}\)
\(\Delta L = \frac{84}{60360} = 1.39 \times 10^{-3}\text{ m} \approx 1.4\text{ mm}\) (or \(1.4 \times 10^{-3}\text{ m}\))

(c) When the wire is stretched within its elastic limit, the atoms are pulled slightly further apart but remain within their potential wells. The attractive interatomic forces act to pull the atoms back to their equilibrium positions once the external stretching force is removed.

Part (a):
- Stress = Force / Area AND Strain = Extension / Original Length (2 marks, 1 for each correct definition)

Part (b):
- Calculation of area \(A = 5.03 \times 10^{-7}\text{ m}^2\) (1)
- Recall of \(E = \frac{FL}{A\Delta L}\) or equivalent step-by-step stress and strain calculations (1)
- Correct algebraic rearrangement for \(\Delta L\) (1)
- Final value of \(1.4\text{ mm}\) or \(1.4 \times 10^{-3}\text{ m}\) (allow 1.39 to 1.40) (1)

Part (c):
- Atoms are displaced from equilibrium positions / interatomic spacing increases (1)
- Attractive forces pull atoms back to equilibrium positions when load is removed (1)

(a) Sound waves are generated at the open end and travel down the pipe. They reflect at the closed end. The incident and reflected waves, which have the same frequency and amplitude, travel in opposite directions and superpose/interfere. This superposition forms a stationary wave with nodes and antinodes.

(b) In the fundamental mode:
- A displacement node is located at the closed end (where air molecules cannot vibrate freely).
- A displacement antinode is located at the open end (where air molecules vibrate with maximum amplitude).

(c) For a pipe closed at one end, the boundary conditions only allow odd harmonics.
- Fundamental: \(L = \frac{\lambda_1}{4} \implies \lambda_1 = 4L\)
- First overtone (3rd harmonic): \(L = \frac{3\lambda_3}{4} \implies \lambda_3 = \frac{4L}{3}\)

Calculate wavelength \(\lambda_3\):
\(\lambda_3 = \frac{4 \times 0.85\text{ m}}{3} = 1.133\text{ m}\)

Calculate frequency \(f_3\):
\(f_3 = \frac{v}{\lambda_3} = \frac{340\text{ m s}^{-1}}{1.133\text{ m}} = 300\text{ Hz}\)

Part (a):
- Reflection of wave at the closed end (1)
- Superposition/interference of incident and reflected waves (1)
- Moving in opposite directions with same frequency / amplitude (1)

Part (b):
- Node at the closed end (1)
- Antinode at the open end (1)

Part (c):
- Identification that first overtone corresponds to \(L = \frac{3}{4}\lambda\) (1)
- Calculation of wavelength \(\lambda = 1.13\text{ m}\) (1)
- Calculation of frequency \(f = 300\text{ Hz}\) (1)

(a) Stokes' Law states that the viscous drag force \(F\) on a small spherical object moving through a fluid is given by \(F = 6\pi \eta r v\).
Conditions:
1. The flow of the fluid must be laminar (no turbulence).
2. The object must be spherical.

(b) The diagram should show a circle representing the sphere with three vertical forces acting on it:
- Weight (\(W\)) acting vertically downwards.
- Upthrust (\(U\)) acting vertically upwards.
- Viscous drag (\(F\)) acting vertically upwards.
At terminal velocity, \(W = U + F\).

(c) Set up the force balance equation:
\(W = U + F\)
\(\rho_s V g = \rho_f V g + 6\pi \eta r v\)
\(\frac{4}{3}\pi r^3 g (\rho_s - \rho_f) = 6\pi \eta r v\)

Rearranging to solve for terminal velocity \(v\):
\(v = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}\)

Substitute values:
\(v = \frac{2 \times (2.0 \times 10^{-3}\text{ m})^2 \times (7800 - 900)\text{ kg m}^{-3} \times 9.81\text{ m s}^{-2}}{9 \times 0.250\text{ Pa s}}\)
\(v = \frac{2 \times 4.0 \times 10^{-6} \times 6900 \times 9.81}{2.25}\)
\(v = \frac{0.5415}{2.25} = 0.241\text{ m s}^{-1}\)

Part (a):
- Correct definition \(F = 6\pi\eta r v\) with terms defined (1)
- Spherical object AND Laminar flow / low speed (1 mark for both conditions) (1)
- Fluid is infinite in extent / not close to walls (or equivalent valid condition) (1)

Part (b):
- Downward arrow for Weight/Gravity (1)
- Upward arrows for BOTH Upthrust and Drag/Viscous force (1)

Part (c):
- Identification of force balance: \(W = U + F\) (1)
- Substitution of correct densities and dimensions into \(v = \frac{2r^2(\rho_s - \rho_f)g}{9\eta}\) (1)
- Correct calculation of \(v = 0.24\text{ m s}^{-1}\) (or \(0.241\text{ m s}^{-1}\)) (1)

(a) According to classical wave theory, light waves continuously deliver energy to the metal surface. Therefore, given enough time, electrons should absorb enough energy to escape, regardless of the frequency of the light. Wave theory cannot explain why emission is instantaneous or why no emission occurs at all below a specific threshold frequency.

(b) Calculate photon energy \(E\):
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{320 \times 10^{-9}\text{ m}} = 6.216 \times 10^{-19}\text{ J}\)

Convert work function \(\phi\) to Joules:
\(\phi = 2.3\text{ eV} \times 1.60 \times 10^{-19}\text{ J/eV} = 3.68 \times 10^{-19}\text{ J}\)

Apply Einstein's photoelectric equation:
\(E_k = hf - \phi\)
\(E_k = 6.216 \times 10^{-19}\text{ J} - 3.68 \times 10^{-19}\text{ J} = 2.536 \times 10^{-19}\text{ J} \approx 2.5 \times 10^{-19}\text{ J}\)

(c) The stopping potential \(V_s\) is related to the maximum kinetic energy by:
\(e V_s = E_k\)
\(V_s = \frac{2.536 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ C}} = 1.585\text{ V} \approx 1.6\text{ V}\)

Part (a):
- Wave theory predicts continuous energy absorption over time (1)
- Hence, even low-frequency light should eventually cause emission (1)
- This contradicts the observation of an immediate cut-off threshold frequency (1)

Part (b):
- Correct calculation of photon energy \(E = 6.22 \times 10^{-19}\text{ J}\) (1)
- Correct conversion of \(\phi\) to Joules (\(3.68 \times 10^{-19}\text{ J}\)) (1)
- Correct subtraction to get \(E_k = 2.5 \times 10^{-19}\text{ J}\) (allow \(2.54 \times 10^{-19}\text{ J}\)) (1)

Part (c):
- Recall of \(e V_s = E_k\) (1)
- Correct calculation of \(V_s = 1.6\text{ V}\) (allow \(1.58\text{ V}\) to \(1.60\text{ V}\)) (1)

(a) Stiffness \(k\) is the force per unit extension in the linear region:
\(k = \frac{F}{x} = \frac{4500\text{ N}}{1.20\text{ m}} = 3750\text{ N m}^{-1}\)

(b) Work done is the area under the force-extension graph in the linear region:
\(W = \frac{1}{2} F x = 0.5 \times 4500\text{ N} \times 1.20\text{ m} = 2700\text{ J}\)

(c) Plastic deformation is a permanent deformation; the material does not return to its original length when the applied force is removed because atomic planes have slid past each other.

During unloading, the elastic strain is recovered. Because the unloading line is parallel to the loading line, the stiffness during recovery remains \(k = 3750\text{ N m}^{-1}\).
Elastic extension recovered:
\(\Delta x_{\text{recovered}} = \frac{F_{\text{max}}}{k} = \frac{5500\text{ N}}{3750\text{ N m}^{-1}} = 1.47\text{ m}\)

Permanent extension:
\(x_{\text{permanent}} = x_{\text{max}} - \Delta x_{\text{recovered}} = 1.80\text{ m} - 1.47\text{ m} = 0.33\text{ m}\)

Part (a):
- Recall of \(F = kx\) (1)
- Calculation of \(k = 3750\text{ N m}^{-1}\) (1)

Part (b):
- Use of \(W = \frac{1}{2} F x\) (1)
- Calculation of \(W = 2700\text{ J}\) (1)

Part (c):
- Explanation: Permanent deformation / does not return to original length because atoms/layers slide past each other (1)
- Recognition that the slope of unloading is equal to \(k = 3750\text{ N m}^{-1}\) (1)
- Calculation of recovered elastic extension \(\Delta x = 1.47\text{ m}\) (1)
- Correct calculation of permanent extension \(1.80 - 1.47 = 0.33\text{ m}\) (1)

(a) The work function is the minimum energy required to liberate a conduction electron from the surface of a metal. (b) Using \(E = \frac{hc}{\lambda}\), we have \(E = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{380 \times 10^{-9}\text{ m}} = 5.23 \times 10^{-19}\text{ J}\), which is approximately \(5.2 \times 10^{-19}\text{ J}\). (c) The stopping potential energy is \(E_{k,\text{max}} = e V_s = 1.60 \times 10^{-19}\text{ C} \times 1.05\text{ V} = 1.68 \times 10^{-19}\text{ J}\). From Einstein's equation, \(h f = \phi + E_{k,\text{max}}\), so \(\phi = 5.234 \times 10^{-19}\text{ J} - 1.68 \times 10^{-19}\text{ J} = 3.554 \times 10^{-19}\text{ J}\). In eV, this is \(\frac{3.554 \times 10^{-19}}{1.60 \times 10^{-19}} = 2.22\text{ eV}\). (d) According to wave theory, electromagnetic wave energy is delivered continuously over the wavefront. Any frequency of light could eventually deliver enough energy to emit electrons if shone for long enough or with enough intensity. In the photon model, light is quantized into packets (photons). One photon interacts with only one electron. If the energy of a single photon \(E = hf\) is less than the work function \(\phi\), no emission occurs, defining a threshold frequency \(f_0 = \frac{\phi}{h}\) below which no emission is possible, regardless of intensity.

(a) [2 marks] M1: Minimum energy needed. A1: to release/emit an electron from the surface of a metal. (b) [2 marks] M1: Recalls and substitutes into \(E = hf\) or \(E = hc/\lambda\). A1: Correct calculation showing at least 3 sig figs (\(5.23 \times 10^{-19}\text{ J}\)). (c) [4 marks] M1: Uses \(E_{k,\text{max}} = e V_s\) to find max KE (\(1.68 \times 10^{-19}\text{ J}\)). M1: Uses Einstein's photoelectric equation \(\phi = E - E_{k,\text{max}}\). A1: Calculates work function in Joules (\(3.55 \times 10^{-19}\text{ J}\)). A1: Converts correctly to eV (\(2.22\text{ eV}\), accept range 2.20 to 2.24 eV). (d) [4 marks] Wave Theory: M1: Wave energy is delivered continuously. A1: Predicts that any frequency would eventually cause emission (no threshold frequency). Photon Model: M1: One-to-one interaction between photon and electron. A1: If photon energy \(hf < \phi\), no emission can occur, hence threshold frequency exists.

(a) For constructive interference, the path difference between the interfering waves must be an integer multiple of the wavelength, \(n\lambda\) (or the phase difference must be an even multiple of \(\pi\) radians). (b) First-order maximum corresponds to \(n=1\). The angle of diffraction \(\theta\) is given by \(\tan\theta = \frac{x}{D} = \frac{0.58\text{ m}}{1.50\text{ m}} = 0.3867\). This gives \(\theta = 21.14^\circ\). Using the grating equation \(d \sin\theta = n \lambda\): \(d \sin(21.14^\circ) = 1 \times 632.8 \times 10^{-9}\text{ m}\). Since \(\sin(21.14^\circ) = 0.3606\), we get \(d = \frac{632.8 \times 10^{-9}}{0.3606} = 1.75 \times 10^{-6}\text{ m}\) (or \(1.75\ \mu\text{m}\)). (c) The energy of a single photon is \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{632.8 \times 10^{-9}\text{ m}} = 3.143 \times 10^{-19}\text{ J}\). The total power is \(1.50 \times 10^{-3}\text{ W}\). The number of photons emitted per second is \(N = \frac{\text{Power}}{\text{Energy of one photon}} = \frac{1.50 \times 10^{-3}\text{ W}}{3.143 \times 10^{-19}\text{ J}} = 4.77 \times 10^{15}\text{ s}^{-1}\). (d) Blue light has a shorter wavelength (\(405\text{ nm} < 632.8\text{ nm}\)). Since \(d\sin\theta = n\lambda\), a smaller wavelength \(\lambda\) for a constant spacing \(d\) results in a smaller angle of diffraction \(\theta\). Therefore, the maxima on the screen will be closer together (the distance \(x\) decreases).

(a) [1 mark] A1: Path difference is an integer multiple of wavelength (\(n\lambda\)) or phase difference is \(2\pi n\) radians. (b) [4 marks] M1: Uses \(\tan\theta = x/D\) to calculate angle \(\theta = 21.14^\circ\). M1: Recalls and uses \(d \sin\theta = n\lambda\). A1: Correct substitution of values with \(n=1\). A1: Correct calculation of \(d = 1.75 \times 10^{-6}\text{ m}\) (or \(1.75\ \mu\text{m}\); accept range 1.74 to 1.76). Note: Award max 2 marks if the small-angle approximation \(\sin\theta \approx \tan\theta\) was used, yielding \(1.64 \times 10^{-6}\text{ m}\). (c) [4 marks] M1: Recalls and uses \(E = hc/\lambda\). A1: Calculates energy of a photon correctly as \(3.14 \times 10^{-19}\text{ J}\). M1: Uses \(\text{Power} / E\) to find rate. A1: Correct calculation of \(4.77 \times 10^{15}\text{ s}^{-1}\) (accept range 4.75 to 4.80 \times 10^{15}). (d) [3 marks] B1: Identifies that blue light has a shorter wavelength than red light. B1: States that the angle of diffraction \(\theta\) decreases (referencing \(d\sin\theta = n\lambda\)). B1: Concludes that the maxima/spots on the screen will be closer together.