2026 Cambridge IAS-Level Physics (9702) 模拟试题及答案详解

Rearrange the formula for dynamic viscosity \(\eta\):

\[\eta = \frac{F}{6 \pi r v}\]

The factor \(6\pi\) is dimensionless. Substitute the base units for force \(F\), radius \(r\), and velocity \(v\):
- Base units of force \(F = ma\): \(\text{kg m s}^{-2}\)
- Base units of radius \(r\): \(\text{m}\)
- Base units of velocity \(v\): \(\text{m s}^{-1}\)

\[[\eta] = \frac{\text{kg m s}^{-2}}{\text{m} \times \text{m s}^{-1}} = \frac{\text{kg m s}^{-2}}{\text{m}^2\text{ s}^{-1}} = \text{kg m}^{-1}\text{s}^{-1}\]

Thus, the correct option is A.

A [1 mark] for identifying the correct SI base units \(\text{kg m}^{-1}\text{s}^{-1}\).

Let \(A\) be the uniform cross-sectional area of the cylinder, \(h = 12.0\text{ cm}\) be its total height, \(\sigma = 800\text{ kg m}^{-3}\) be its density, and \(x\) be the height submerged in the oil of density \(\rho_{\text{oil}} = 700\text{ kg m}^{-3}\).

The height submerged in water is \((h - x)\) where \(\rho_{\text{water}} = 1000\text{ kg m}^{-3}\).

For floating equilibrium, total weight = total upthrust:

\[\sigma A h g = \rho_{\text{oil}} A x g + \rho_{\text{water}} A (h - x) g\]

Dividing both sides by \(A g\):

\[\sigma h = \rho_{\text{oil}} x + \rho_{\text{water}} h - \rho_{\text{water}} x\]
\[x(\rho_{\text{water}} - \rho_{\text{oil}}) = h(\rho_{\text{water}} - \sigma)\]
\[x = h \left(\frac{\rho_{\text{water}} - \sigma}{\rho_{\text{water}} - \rho_{\text{oil}}}\right) = 12.0 \times \left(\frac{1000 - 800}{1000 - 700}\right) = 12.0 \times \frac{200}{300} = 8.0\text{ cm}\]

Thus, the height of the block in the oil is \(8.0\text{ cm}\).

C [1 mark] for calculating \(8.0\text{ cm}\) using Archimedes' principle in two liquids.

Let \(v_1\) be the velocity of mass \(m\) and \(v_2\) be the velocity of mass \(2m\) after the collision.

From conservation of linear momentum:
\[m u = m v_1 + 2m v_2 \implies u = v_1 + 2v_2\]

For a perfectly elastic collision, relative speed of approach equals relative speed of separation:
\[u = v_2 - v_1\]

Adding these two equations:
\[2u = 3v_2 \implies v_2 = \frac{2}{3}u\]

Initial kinetic energy \(E_{k,\text{init}} = \frac{1}{2} m u^2\).

Final kinetic energy of mass \(2m\):
\[E_{k, 2m} = \frac{1}{2}(2m) v_2^2 = m\left(\frac{2}{3}u\right)^2 = \frac{4}{9} m u^2 = \frac{8}{9}\left(\frac{1}{2}m u^2\right)\]

Therefore, the fraction of kinetic energy transferred is \(\frac{8}{9}\).

D [1 mark] for determining the fractional transfer of kinetic energy as \(\frac{8}{9}\).

First calculate the grating spacing \(d\):

\[d = \frac{1}{N} = \frac{1}{4.00 \times 10^5\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m} = 2500\text{ nm}\]

Using the grating equation \(d \sin\theta = n \lambda\), the maximum possible order \(n_{\text{max}}\) occurs when \(\sin\theta \le 1\):

\[n_{\text{max}} \le \frac{d}{\lambda} = \frac{2500\text{ nm}}{589\text{ nm}} \approx 4.24\]

Since the order \(n\) must be an integer, the maximum observable order is \(n = 4\).

The total number of bright maxima includes the central zeroth-order maximum \((n = 0)\) and \(4\) orders on each side of the centre:

\[\text{Total maxima} = 2n + 1 = 2(4) + 1 = 9\]

Thus, 9 maxima are observed in total.

D [1 mark] for correctly calculating the maximum order \(n = 4\) and finding the total number of maxima \(2n + 1 = 9\).

Elastic strain energy stored is given by \(U = \frac{1}{2} F x\).

From the definition of Young modulus \(E = \frac{F/A}{x/L}\), the extension is \(x = \frac{F L}{A E}\).

Substituting \(x\) into the strain energy formula:

\[U = \frac{F^2 L}{2 A E}\]

The cross-sectional area of a wire of diameter \(d\) is \(A = \frac{\pi d^2}{4} \propto d^2\).

Therefore, \(U \propto \frac{L}{d^2}\).

For wire \(Y\):
- Length \(L_Y = 2L\)
- Diameter \(d_Y = 2d \implies d_Y^2 = 4d^2\)

\[U_Y = U_X \times \frac{2L}{L} \times \frac{d^2}{(2d)^2} = U \times 2 \times \frac{1}{4} = \frac{1}{2}U\]

Hence, the energy stored in \(Y\) is \(\frac{1}{2}U\).

B [1 mark] for applying \(U \propto L/d^2\) to obtain \(\frac{1}{2}U\).

Initial state (\(R_1 = 2.0\ \Omega\)):
- Total circuit resistance \(R_{\text{total}, 1} = r + R_{\text{fixed}} + R_1 = 2.0 + 6.0 + 2.0 = 10.0\ \Omega\)
- Circuit current \(I_1 = \frac{E}{R_{\text{total}, 1}} = \frac{12.0}{10.0} = 1.20\text{ A}\)
- Initial voltmeter reading \(V_1 = I_1 R_1 = 1.20 \times 2.0 = 2.40\text{ V}\)

Final state (\(R_2 = 10.0\ \Omega\)):
- Total circuit resistance \(R_{\text{total}, 2} = r + R_{\text{fixed}} + R_2 = 2.0 + 6.0 + 10.0 = 18.0\ \Omega\)
- Circuit current \(I_2 = \frac{E}{R_{\text{total}, 2}} = \frac{12.0}{18.0} = \frac{2}{3}\text{ A}\)
- Final voltmeter reading \(V_2 = I_2 R_2 = \frac{2}{3} \times 10.0 = \frac{20}{3}\text{ V} \approx 6.67\text{ V}\)

Change in voltmeter reading:
\[\Delta V = V_2 - V_1 = 6.67 - 2.40 = 4.27\text{ V}\]

B [1 mark] for correctly finding \(V_1 = 2.40\text{ V}\), \(V_2 = 6.67\text{ V}\), and their difference \(\Delta V = 4.27\text{ V}\).

In \(\beta^+\) decay, a proton transforms into a neutron:
\[p \rightarrow n + e^+ + \nu_e\]

The quark composition of a proton is \(uud\) and that of a neutron is \(udd\).
Therefore, an up quark (\(u\)) changes into a down quark (\(d\)):
\[u \rightarrow d + e^+ + \nu_e\]

To conserve lepton number (\(L = 0\) initially), the emission of a positron (antilepton, \(L = -1\)) must be accompanied by an electron neutrino (lepton, \(L = +1\)), not an antineutrino.

Hence, an up quark changes into a down quark, and an electron neutrino is emitted (Option D).

D [1 mark] for stating that an up quark changes to a down quark and an electron neutrino is emitted.

Using the Doppler effect formula for a moving source approaching a stationary observer:

\[f_o = f_s \left(\frac{v}{v - v_s}\right) = 850 \times \left(\frac{340}{340 - 25.0}\right) = 850 \times \frac{340}{315} \approx 917.46\text{ Hz} \approx 917\text{ Hz}\]

The wavelength \(\lambda\) of the wavefronts compressed in front of the moving source is:

\[\lambda = \frac{v - v_s}{f_s} = \frac{340 - 25.0}{850} = \frac{315}{850} \approx 0.3706\text{ m} \approx 0.371\text{ m}\]

(Alternatively, using \(\lambda = \frac{v}{f_o} = \frac{340}{917.46} = 0.371\text{ m}\).)

Thus, row C is correct.

C [1 mark] for correctly calculating observed frequency \(917\text{ Hz}\) and observed wavelength \(0.371\text{ m}\).

Rearranging the equation for viscosity gives \(\eta = \frac{F}{6\pi r v}\). The numerical constant \(6\pi\) is dimensionless. In SI base units: force \(F\) has units \(\text{kg m s}^{-2}\), radius \(r\) has units \(\text{m}\), and speed \(v\) has units \(\text{m s}^{-1}\). Substituting these units gives \([\eta] = \frac{\text{kg m s}^{-2}}{\text{m} \times \text{m s}^{-1}} = \text{kg m}^{-1}\text{s}^{-1}\).

A is correct [1]. Base units derived by correct dimensional substitution: force is kg m s^-2, dividing by m and m s^-1 gives kg m^-1 s^-1.

Let \(m_1 = 0.40\text{ kg}\), \(u_1 = 1.5\text{ m s}^{-1}\), \(m_2 = 0.20\text{ kg}\), and \(u_2 = 0\). In an elastic collision, total linear momentum is conserved and the relative speed of approach equals the relative speed of separation: \(u_1 - u_2 = v_2 - v_1 \Rightarrow 1.5 = v_2 - v_1 \Rightarrow v_2 = v_1 + 1.5\). By conservation of linear momentum: \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \Rightarrow (0.40)(1.5) + 0 = 0.40 v_1 + 0.20 (v_1 + 1.5) \Rightarrow 0.60 = 0.60 v_1 + 0.30 \Rightarrow 0.60 v_1 = 0.30 \Rightarrow v_1 = +0.50\text{ m s}^{-1}\). The positive sign indicates motion in the original direction.

A is correct [1]. Correct simultaneous solution of momentum conservation and relative speed equation yields v1 = +0.50 m s^-1 in the original direction.

For a floating body, the upthrust equals the weight of the body. Let \(A\) be the uniform cross-sectional area of the cylinder. In liquid X: \(W = \rho_{\text{X}} g A h\). In liquid Y: \(W = \rho_{\text{Y}} g A h_{\text{Y}} = 1.25 \rho_{\text{X}} g A h_{\text{Y}}\). Equating the two expressions for the weight \(W\): \(\rho_{\text{X}} g A h = 1.25 \rho_{\text{X}} g A h_{\text{Y}} \Rightarrow h_{\text{Y}} = \frac{h}{1.25} = 0.80 h\). Therefore, the fraction of the total length submerged in liquid Y is \(\frac{h_{\text{Y}}}{L} = 0.80 \frac{h}{L}\).

A is correct [1]. Equating upthrusts for both floating states gives h_Y = h / 1.25 = 0.80 h, so fraction submerged is 0.80 (h / L).

The elastic potential energy stored in a wire obeying Hooke's law is \(E_{\text{p}} = \frac{1}{2} T x\), where \(x\) is the extension. From the definition of the Young modulus: \(E = \frac{\text{stress}}{\text{strain}} = \frac{T / A}{x / L} = \frac{T L}{A x} \Rightarrow x = \frac{T L}{E A}\). Substituting \(x\) into the energy formula gives \(E_{\text{p}} = \frac{1}{2} T \left(\frac{T L}{E A}\right) = \frac{T^2 L}{2 E A}\).

A is correct [1]. Using E_p = (1/2) T x and substituting extension x = T L / (E A) gives T^2 L / (2 E A).

The double-slit interference fringe spacing is given by \(x = \frac{\lambda D}{a}\). For the new configuration: \(\lambda' = 450\text{ nm} = 0.75 \times 600\text{ nm} = 0.75 \lambda\) and \(a' = 0.75 a\). The new fringe separation is \(x' = \frac{\lambda' D}{a'} = \frac{0.75 \lambda D}{0.75 a} = \frac{\lambda D}{a} = x\). Thus, the fringe separation remains unchanged.

C is correct [1]. The ratio lambda / a is (0.75 lambda) / (0.75 a) = lambda / a, so the fringe separation remains x.

Using the Doppler effect formula for a moving source approaching a stationary observer: \(f_{\text{o}} = f_{\text{s}} \left(\frac{v}{v - v_{\text{s}}}\right)\), where \(f_{\text{s}} = 680\text{ Hz}\), \(v = 340\text{ m s}^{-1}\), and \(v_{\text{s}} = 25\text{ m s}^{-1}\). Substituting the values: \(f_{\text{o}} = 680 \times \left(\frac{340}{340 - 25}\right) = 680 \times \left(\frac{340}{315}\right) \approx 734\text{ Hz}\).

C is correct [1]. Correct substitution into f_o = f_s * v / (v - v_s) gives 680 * (340 / 315) = 734 Hz.

The potential difference \(V_{\text{v}}\) across the variable resistor \(R_{\text{v}}\) is given by \(V_{\text{v}} = E \times \frac{R_{\text{v}}}{r + R_{\text{fixed}} + R_{\text{v}}}\), where \(E = 12.0\text{ V}\), \(r = 2.0\,\Omega\), and \(R_{\text{fixed}} = 10.0\,\Omega\). When \(R_{\text{v}} = 4.0\,\Omega\): \(V_{\text{v}} = 12.0 \times \frac{4.0}{2.0 + 10.0 + 4.0} = 12.0 \times \frac{4.0}{16.0} = 3.0\text{ V}\). When \(R_{\text{v}} = 20.0\,\Omega\): \(V_{\text{v}} = 12.0 \times \frac{20.0}{2.0 + 10.0 + 20.0} = 12.0 \times \frac{20.0}{32.0} = 7.5\text{ V}\). The potential difference ranges from \(3.0\text{ V}\) to \(7.5\text{ V}\).

B is correct [1]. Minimum p.d. = 12.0 * (4.0 / 16.0) = 3.0 V; maximum p.d. = 12.0 * (20.0 / 32.0) = 7.5 V.

A proton has quark composition \(\text{uud}\) and a neutron has quark composition \(\text{udd}\). In \(\beta^+\) decay, an up quark (charge \(+\frac{2}{3}e\)) changes into a down quark (charge \(-\frac{1}{3}e\)): \(\text{u} \rightarrow \text{d} + \text{e}^+ + \nu_{\text{e}}\). The emitted leptons are a positron (\(\beta^+\)) and an electron neutrino (\(\nu_{\text{e}}\)).

A is correct [1]. Up quark changes to down quark, emitting a positron and an electron neutrino to conserve charge and lepton number.

Electrical resistivity \(\rho\) is given by \(\rho = \frac{RA}{L}\).
Resistance \(R = \frac{V}{I} = \frac{W}{Q I} = \frac{W}{I^2 t}\).
In base units:
- Work \(W\) has units \(\text{kg m}^2 \text{s}^{-2}\)
- Current squared times time \(I^2 t\) has units \(\text{A}^2 \text{s}\)
Thus, the base units of resistance \(R\) are \(\frac{\text{kg m}^2 \text{s}^{-2}}{\text{A}^2 \text{s}} = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\).
Substituting into the formula for resistivity:
\[ [\rho] = \text{kg m}^2 \text{s}^{-3} \text{A}^{-2} \times \frac{\text{m}^2}{\text{m}} = \text{kg m}^3 \text{s}^{-3} \text{A}^{-2} \]

A is correct [1]. Base units for power or energy derived correctly to find resistance in SI base units \(\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\) and multiplied by metres to yield \(\text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\).

For horizontal motion, the velocity remains constant: \(v_x = 12\text{ m s}^{-1}\).
For vertical motion, using \(v_y^2 = u_y^2 + 2gs\):
\[ v_y^2 = 0 + 2(9.81)(45) = 882.9\text{ m}^2\text{ s}^{-2} \]
\[ v_y = \sqrt{882.9} \approx 29.71\text{ m s}^{-1} \]
The resultant speed \(v\) is:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 882.9} = \sqrt{144 + 882.9} = \sqrt{1026.9} \approx 32\text{ m s}^{-1} \]

C is correct [1]. Vertical velocity calculated using equations of motion (\(v_y \approx 29.7\text{ m s}^{-1}\)) and combined with horizontal component (\(12\text{ m s}^{-1}\)) via Pythagoras theorem to give \(32\text{ m s}^{-1}\).

By conservation of linear momentum:
\[ m_1 u_1 = (m_1 + m_2) v \]
\[ (0.40)(1.5) = (0.40 + 0.60) v \implies 0.60 = 1.00 v \implies v = 0.60\text{ m s}^{-1} \]
Initial kinetic energy:
\[ E_{k,\text{initial}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2}(0.40)(1.5)^2 = 0.45\text{ J} \]
Final kinetic energy:
\[ E_{k,\text{final}} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2}(1.00)(0.60)^2 = 0.18\text{ J} \]
Loss in kinetic energy:
\[ \Delta E_k = 0.45 - 0.18 = 0.27\text{ J} \]

B is correct [1]. Correct conservation of momentum gives \(v = 0.60\text{ m s}^{-1}\); subtracting final kinetic energy (\(0.18\text{ J}\)) from initial kinetic energy (\(0.45\text{ J}\)) gives \(0.27\text{ J}\).

For a floating body in equilibrium, the weight of the body equals the upthrust (weight of displaced liquid):
\[ \text{Weight} = \rho_{\text{wood}} A h g \]
\[ \text{Upthrust} = \rho_{\text{liquid}} A h_{\text{sub}} g \]
Equating the two expressions:
\[ \rho_{\text{wood}} A h g = \rho_{\text{liquid}} A h_{\text{sub}} g \]
\[ \rho_{\text{wood}} = \rho_{\text{liquid}} \left(\frac{h_{\text{sub}}}{h}\right) = 1200 \times \left(\frac{0.090}{0.15}\right) = 1200 \times 0.60 = 720\text{ kg m}^{-3} \]

A is correct [1]. Archimedes' principle gives \(\rho_{\text{wood}} / \rho_{\text{liquid}} = h_{\text{sub}} / h = 0.090 / 0.15 = 0.60\), leading to \(720\text{ kg m}^{-3}\).

The Young modulus is defined as:
\[ E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{x / L} = \frac{F L}{A x} \implies F = \frac{E A x}{L} \]
The elastic potential energy \(E_p\) stored within the limit of proportionality is the area under the force-extension graph:
\[ E_p = \frac{1}{2} F x = \frac{1}{2} \left(\frac{E A x}{L}\right) x = \frac{E A x^2}{2L} \]

D is correct [1]. Using \(E_p = \frac{1}{2}Fx\) and substituting \(F = \frac{EAx}{L}\) gives \(\frac{EAx^2}{2L}\).

The slit spacing \(d\) is:
\[ d = \frac{1}{N} = \frac{1}{5.00 \times 10^5\text{ m}^{-1}} = 2.00 \times 10^{-6}\text{ m} \]
The condition for principal maxima is \(d \sin \theta = n \lambda\).
Since the maximum possible value of \(\sin \theta\) is \(1\):
\[ n \le \frac{d}{\lambda} = \frac{2.00 \times 10^{-6}\text{ m}}{589 \times 10^{-9}\text{ m}} \approx 3.396 \]
Thus, the maximum order number is \(n = 3\).
The total number of observed maxima is:
\[ N_{\text{max}} = 2n + 1 = 2(3) + 1 = 7 \] (corresponding to \(n = 0, \pm 1, \pm 2, \pm 3\)).

C is correct [1]. Calculating grating spacing \(d = 2.00 \times 10^{-6}\text{ m}\), maximum order \(n_{\text{max}} = 3\), and total maxima = \(2(3) + 1 = 7\).

The total resistance in the series circuit is:
\[ R_{\text{total}} = r + R_{\text{fixed}} + R = 2.0 + 10.0 + 12.0 = 24.0\,\Omega \]
The circuit current is:
\[ I = \frac{E}{R_{\text{total}}} = \frac{12.0\text{ V}}{24.0\,\Omega} = 0.50\text{ A} \]
The potential difference across the \(10.0\,\Omega\) resistor is:
\[ V = I R_{\text{fixed}} = 0.50\text{ A} \times 10.0\,\Omega = 5.0\text{ V} \]

B is correct [1]. Circuit current \(I = 12.0 / 24.0 = 0.50\text{ A}\) and \(V = 0.50 \times 10.0 = 5.0\text{ V}\).

In \(\beta^-\) decay, a neutron converts into a proton, an electron (\(\beta^-\)), and an electron antineutrino:
\[ n \rightarrow p + e^- + \bar{\nu}_e \]
In terms of quark composition:
- A neutron is \(udd\)
- A proton is \(uud\)
Therefore, a down quark changes into an up quark (\(d \rightarrow u\)).
Beta decay is mediated by the weak interaction (weak nuclear force).

D is correct [1]. In \(\beta^-\) decay, \(d \rightarrow u\) and the weak interaction is the responsible force.

Rearranging the equation for viscosity \( \eta \) gives:
\[ \eta = \frac{F_{\text{D}}}{6\pi r v} \]

The constant \( 6\pi \) is dimensionless. Substituting the SI base units of each quantity:
- Force \( F \): \( \text{kg}\text{ m}\text{ s}^{-2} \)
- Radius \( r \): \( \text{m} \)
- Speed \( v \): \( \text{m}\text{ s}^{-1} \)

\[ [\eta] = \frac{\text{kg}\text{ m}\text{ s}^{-2}}{\text{m} \times \text{m}\text{ s}^{-1}} = \frac{\text{kg}\text{ m}\text{ s}^{-2}}{\text{m}^2\text{ s}^{-1}} = \text{kg}\text{ m}^{-1}\text{ s}^{-1} \]

B [1 mark]: Correctly equates base units of force to \( \text{kg}\text{ m}\text{ s}^{-2} \) and divides by \( \text{m}^2\text{ s}^{-1} \) to obtain \( \text{kg}\text{ m}^{-1}\text{ s}^{-1} \).

Taking motion to the right as positive:
Initial total momentum \( p_{\text{initial}} = m_{\text{X}} u_{\text{X}} + m_{\text{Y}} u_{\text{Y}} = (0.40)(+3.0) + (0.60)(-2.0) = +1.2 - 1.2 = 0\text{ kg m s}^{-1} \).

For a perfectly elastic collision, relative speed of approach equals relative speed of separation:
\( u_{\text{X}} - u_{\text{Y}} = v_{\text{Y}} - v_{\text{X}} \implies 3.0 - (-2.0) = 5.0\text{ m s}^{-1} \).
So \( v_{\text{Y}} - v_{\text{X}} = 5.0 \implies v_{\text{Y}} = v_{\text{X}} + 5.0 \).

By conservation of linear momentum:
\( m_{\text{X}} v_{\text{X}} + m_{\text{Y}} v_{\text{Y}} = 0 \)
\( 0.40 v_{\text{X}} + 0.60(v_{\text{X}} + 5.0) = 0 \)
\( 1.00 v_{\text{X}} + 3.0 = 0 \implies v_{\text{X}} = -3.0\text{ m s}^{-1} \).

Thus, trolley X moves at \( 3.0\text{ m s}^{-1} \) to the left.

A [1 mark]: Correct application of conservation of linear momentum and conservation of kinetic energy (or relative speed relation) to find \( v_{\text{X}} = 3.0\text{ m s}^{-1} \text{ to the left} \).

When the object of mass \( m \) is added, the extra downward weight is balanced by the increase in upthrust (Archimedes' principle):

Extra weight \( = m g \)
Extra upthrust \( = \rho g \Delta V = \rho g (A h) \)

Equating the forces:
\[ m g = \rho g A h \implies m = \rho A h \]

B [1 mark]: Correctly identifies that the additional weight \( mg \) equals the additional upthrust \( \rho g A h \), giving \( m = \rho A h \).

Using the diffraction grating equation \( d \sin \theta = n \lambda \):
For \( \lambda = 600\text{ nm} \) in the first order (\( n = 1 \)):
\[ d \sin(18.0^\circ) = 1 \times 600\text{ nm} \implies d = \frac{600\text{ nm}}{\sin(18.0^\circ)} \]

For maxima of \( \lambda_1 = 450\text{ nm} \) and \( \lambda_2 = 675\text{ nm} \) to coincide:
\[ d \sin \theta = n_1 \lambda_1 = n_2 \lambda_2 \]
\[ n_1 (450) = n_2 (675) \implies \frac{n_1}{n_2} = \frac{675}{450} = \frac{3}{2} \]

The smallest non-zero integers are \( n_1 = 3 \) and \( n_2 = 2 \).

Therefore:
\[ d \sin \theta = 3 \times 450\text{ nm} = 1350\text{ nm} \]
\[ \sin \theta = \frac{1350\text{ nm}}{d} = \frac{1350\text{ nm}}{600\text{ nm}} \times \sin(18.0^\circ) = 2.25 \times \sin(18.0^\circ) \approx 2.25 \times 0.30902 = 0.6953 \]
\[ \theta = \arcsin(0.6953) = 44.1^\circ \]

C [1 mark]: Correctly identifies order ratio \( n_1/n_2 = 3/2 \) and uses grating equation to calculate \( \theta = 44.1^\circ \).

The Young modulus is \( E_{\text{Y}} = \frac{F/A}{e/L} = \frac{F L}{A e} \implies e = \frac{F L}{A E_{\text{Y}}} \).

The elastic potential energy stored is given by:
\[ E = \frac{1}{2} F e = \frac{F^2 L}{2 A E_{\text{Y}}} \]

For the second wire:
- Force \( = 2F \)
- Length \( = 2L \)
- Area \( = 2A \)
- Same material, so Young modulus is \( E_{\text{Y}} \)

\[ E_2 = \frac{(2F)^2 (2L)}{2 (2A) E_{\text{Y}}} = \frac{4 F^2 \times 2L}{4 A E_{\text{Y}}} = \frac{2 F^2 L}{A E_{\text{Y}}} = 4 \left(\frac{F^2 L}{2 A E_{\text{Y}}}\right) = 4E \]

C [1 mark]: Deduces that the extension doubles, so energy \( E_2 = \frac{1}{2}(2F)(2e) = 4E \).

The terminal potential difference is given by \( V = E \frac{R}{R + r} \).
- When \( R = r \), \( V = \frac{1}{2}E = 0.50 E \).
- When \( R = 3r \), \( V = \frac{3}{4}E = 0.75 E \).
Thus, the terminal potential difference increases.

The current in the circuit is \( I = \frac{E}{R + r} \).
As \( R \) increases from \( r \) to \( 3r \), the current decreases from \( \frac{E}{2r} \) to \( \frac{E}{4r} \).
The rate of heat dissipation in the internal resistance is \( P = I^2 r \).
Since current \( I \) decreases, the rate of heat dissipation in \( r \) decreases.

A [1 mark]: Correctly identifies that terminal p.d. increases as external resistance increases, and power in internal resistance decreases due to reduced current.

A proton has quark composition \( uud \) and a neutron has quark composition \( udd \).
When a proton turns into a neutron, an up quark (\( +\frac{2}{3}e \)) changes into a down quark (\( -\frac{1}{3}e \)):
\[ u \rightarrow d + e^+ + \nu_{\text{e}} \]

To conserve lepton number (or by conservation rules in \( \beta^+ \) decay), an electron neutrino \( \nu_{\text{e}} \) must be emitted alongside the positron (anti-electron).

D [1 mark]: Identifies quark transition as up to down and the emitted particle as an electron neutrino.

Using the Doppler effect formula for a moving source and stationary observer:

When approaching the observer:
\[ f_1 = f_{\text{s}} \left( \frac{v}{v - v_{\text{s}}} \right) \]

When moving away from the observer:
\[ f_2 = f_{\text{s}} \left( \frac{v}{v + v_{\text{s}}} \right) \]

Dividing \( f_1 \) by \( f_2 \):
\[ \frac{f_1}{f_2} = \frac{f_{\text{s}} \left( \frac{v}{v - v_{\text{s}}} \right)}{f_{\text{s}} \left( \frac{v}{v + v_{\text{s}}} \right)} = \frac{v + v_{\text{s}}}{v - v_{\text{s}}} \]

A [1 mark]: Applies Doppler effect formula for moving source approaching and receding, then simplifies the ratio to \( \frac{v + v_{\text{s}}}{v - v_{\text{s}}} \).

Resistivity \(\rho\) is given by \(\rho = \frac{RA}{L}\). Resistance \(R = \frac{V}{I}\), and potential difference \(V = \frac{W}{Q} = \frac{W}{It}\). The base units of work \(W\) are \(\text{kg m}^2\text{s}^{-2}\), so the base units of \(V\) are \(\frac{\text{kg m}^2\text{s}^{-2}}{\text{A s}} = \text{kg m}^2\text{s}^{-3}\text{A}^{-1}\). The base units of \(R\) are \(\frac{\text{kg m}^2\text{s}^{-3}\text{A}^{-1}}{\text{A}} = \text{kg m}^2\text{s}^{-3}\text{A}^{-2}\). Therefore, the base units of \(\rho\) are \(\text{kg m}^2\text{s}^{-3}\text{A}^{-2} \times \frac{\text{m}^2}{\text{m}} = \text{kg m}^3\text{s}^{-3}\text{A}^{-2}\).

A is correct [1 mark]. B is the unit of resistance or potential difference derivative, C has incorrect time power, D has incorrect length power.

The horizontal component of velocity remains constant: \(v_x = 12\text{ m s}^{-1}\). For the vertical motion, using \(v_y^2 = u_y^2 + 2gs\) with \(u_y = 0\), \(g = 9.81\text{ m s}^{-2}\), and \(s = 45\text{ m}\): \(v_y^2 = 2 \times 9.81 \times 45 = 882.9\text{ m}^2\text{s}^{-2}\). The resultant speed \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 882.9} = \sqrt{144 + 882.9} = \sqrt{1026.9} \approx 32\text{ m s}^{-1}\).

C is correct [1 mark]. A is from subtracting components, B is vertical component only (\(29.7\text{ m s}^{-1}\)), D is scalar sum (\(12 + 29.7\)).

Taking the direction to the right as positive: Total initial momentum \(p_i = (0.40)(+3.0) + (0.20)(-2.0) = 1.20 - 0.40 = +0.80\text{ kg m s}^{-1}\). Total final momentum \(p_f = (0.40)(+1.0) + (0.20)(v_Y) = 0.40 + 0.20v_Y\). By conservation of linear momentum: \(0.40 + 0.20v_Y = 0.80 \implies 0.20v_Y = 0.40 \implies v_Y = +2.0\text{ m s}^{-1}\), which is \(2.0\text{ m s}^{-1}\) to the right.

B is correct [1 mark]. A sign error leads to option A, neglecting mass ratio leads to C or D.

By Archimedes' principle, weight of block equals upthrust: \(\rho_{\text{block}} g A H = \rho_{\text{liquid}} g A h_{\text{sub}}\), where \(h_{\text{sub}}\) is the submerged depth. Thus, \(h_{\text{sub}} = \left(\frac{\rho_{\text{block}}}{\rho_{\text{liquid}}}\right) H = \left(\frac{640}{800}\right) \times 0.15 = 0.80 \times 0.15 = 0.12\text{ m}\). The height above the surface is \(H - h_{\text{sub}} = 0.15\text{ m} - 0.12\text{ m} = 0.030\text{ m}\).

A is correct [1 mark]. D is the submerged height \(0.12\text{ m}\), B and C result from arithmetic or ratio inversions.

Elastic potential energy is \(E = \frac{1}{2} F x\). Extension \(x = \frac{F L}{A E_{\text{Young}}}\), so \(E = \frac{F^2 L}{2 A E_{\text{Young}}}\). For the second wire: \(E_2 = \frac{(2F)^2 (2L)}{2 (2A) E_{\text{Young}}} = \frac{4 F^2 \times 2L}{4 A E_{\text{Young}}} = \frac{2 F^2 L}{A E_{\text{Young}}} = 4 \left(\frac{F^2 L}{2 A E_{\text{Young}}}\right) = 4E\).

C is correct [1 mark]. A incorrectly assumes scaling cancels out, B is linear scaling without squaring force, D is cubing factor.

Grating spacing \(d = \frac{1}{N} = \frac{1}{4.0 \times 10^5\text{ m}^{-1}} = 2.5 \times 10^{-6}\text{ m}\). The condition for maxima is \(d \sin\theta = n \lambda\). Since \(\sin\theta \le 1\), the maximum order is \(n_{\text{max}} \le \frac{d}{\lambda} = \frac{2.5 \times 10^{-6}}{580 \times 10^{-9}} \approx 4.31\). The highest integer order is \(n = 4\). The total number of maxima is \(2 n_{\text{max}} + 1 = 2(4) + 1 = 9\).

D is correct [1 mark]. A is \(n_{\text{max}}\) only, B is one side plus central maximum, C omits the central maximum (\(2n\)).

The relationship between terminal potential difference and current is \(V = E - Ir = -r I + E\). Comparing this with the equation of a straight line \(y = mx + c\), a plot of \(V\) against \(I\) gives a straight line with a constant negative gradient equal to \(-r\) and a vertical intercept on the \(V\)-axis equal to the e.m.f. \(E\).

B is correct [1 mark]. A describes an ohmic resistor without a source, C describes a non-linear component, D describes an ideal cell with zero internal resistance.

A proton has quark composition \(uud\) and a neutron has quark composition \(udd\). Therefore, in \(\beta^+\) decay, an up quark (\(u\)) changes into a down quark (\(d\)): \(u \rightarrow d + e^+ + \nu_e\). The other particle emitted alongside the positron is an electron neutrino (\(\nu_e\)) to conserve lepton number.

C is correct [1 mark]. A and B confuse \(\beta^-\)-decay quark transition (\(d \rightarrow u\)), D incorrectly states antineutrino.

Linear momentum is defined as the product of an object's mass and its velocity (\(p = mv\)).

B1: product of mass and velocity (allow: mass \(\times\) velocity). [Do not allow: mass \(\times\) speed].

The torque of a couple is defined as the product of the magnitude of one of the forces and the perpendicular distance between the lines of action of the forces.

B1: product of (one of the) force(s) and the perpendicular distance between the (lines of action of the) forces.

Power is defined as the work done per unit time (or the rate at which energy is transferred).

B1: work done per unit time OR rate of doing work OR rate of energy transfer (allow: \(\Delta W / \Delta t\) with terms defined).

Coherent waves are waves that have a constant phase difference between them.

B1: waves having a constant phase difference (allow: same frequency and constant phase difference). [Do not allow: just 'same frequency' or 'in phase'].

The electromotive force (e.m.f.) of a source is the energy transferred from other forms (e.g. chemical) to electrical energy per unit charge that passes through the source (or in driving charge around a complete circuit).

M1: energy transferred (from other forms to electrical) per unit charge OR work done per unit charge
A1: from other forms of energy to electrical energy (in driving unit charge around a complete circuit).

1. Express the weight \( W \) (force \( \Delta F \)) exerted by the liquid column on its base in terms of mass \( m \) and density \( \rho \):
\[ \Delta F = W = mg = (\rho V)g = \rho (A \Delta h) g \]

2. Use the definition of pressure \( p = \frac{F}{A} \):
\[ \Delta p = \frac{\Delta F}{A} = \frac{\rho A \Delta h g}{A} = \rho g \Delta h \]

[M1] Expresses the weight/force of the column as \( W = mg = \rho V g = \rho A \Delta h g \) (or mass \( m = \rho A \Delta h \))
[A1] States \( \Delta p = \frac{F}{A} \) and correctly cancels \( A \) with all intermediate steps shown clearly to yield \( \Delta p = \rho g \Delta h \)

1. Express current in terms of drift speed and in terms of Ohm's law:
\[ I = n A v e \quad \text{and} \quad I = \frac{V}{R} \]

2. Substitute the resistance formula \( R = \frac{\rho L}{A} \) into Ohm's law:
\[ I = \frac{V}{\frac{\rho L}{A}} = \frac{V A}{\rho L} \]

3. Equate the two expressions for current:
\[ n A v e = \frac{V A}{\rho L} \]
Cancelling \( A \) from both sides and rearranging for \( v \):
\[ v = \frac{V}{n e \rho L} \]

[M1] Equates \( I = n A v e \) with \( I = \frac{V}{R} \) and uses \( R = \frac{\rho L}{A} \) (or substitutes \( v = \frac{I}{nAe} \) and \( I = \frac{VA}{\rho L} \))
[A1] Shows clear cancellation of cross-sectional area \( A \) and complete algebraic rearrangement to arrive at \( v = \frac{V}{n e \rho L} \)

1. Calculate the extension \( x \) using the definition of Young modulus \( E = \frac{\sigma}{\varepsilon} = \frac{F L}{A x} \):
\[ x = \frac{F L}{A E} = \frac{48.0 \times 2.40}{(3.20 \times 10^{-7}) \times (2.00 \times 10^{11})} = \frac{115.2}{6.40 \times 10^4} = 1.80 \times 10^{-3}\text{ m} \]

2. Calculate the strain energy (elastic potential energy) \( E_{\text{p}} \):
\[ E_{\text{p}} = \frac{1}{2} F x = \frac{1}{2} \times 48.0 \times 1.80 \times 10^{-3} = 4.32 \times 10^{-3}\text{ J} \]
Alternatively, combine into a single formula:
\[ E_{\text{p}} = \frac{F^2 L}{2 A E} = \frac{48.0^2 \times 2.40}{2 \times (3.20 \times 10^{-7}) \times (2.00 \times 10^{11})} = 4.32 \times 10^{-3}\text{ J} \]

[C1] Correct calculation of extension \( x = 1.80 \times 10^{-3}\text{ m} \) OR algebraic combination \( E_{\text{p}} = \frac{F^2 L}{2AE} \) with full substitution shown
[A1] Fully evaluated working shown leading explicitly to \( 4.32 \times 10^{-3}\text{ J} \) (at least 3 significant figures shown without intermediate rounding errors)

1. Volume of the cylinder:
$$V = \pi \left(\frac{d}{2}\right)^2 L = \frac{\pi d^2 L}{4}$$
$$V = \frac{\pi \times (0.0260)^2 \times 0.0580}{4} = 3.0795 \times 10^{-5}\text{ m}^3$$

2. Density \(\rho\):
$$\rho = \frac{m}{V} = \frac{0.245}{3.0795 \times 10^{-5}} = 7.9559 \times 10^3 \approx 7.97 \times 10^3\text{ kg m}^{-3}$$

3. Percentage uncertainty in \(\rho\):
$$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta L}{L}$$
$$\frac{\Delta \rho}{\rho} = \frac{0.002}{0.245} + 2\left(\frac{0.04}{2.60}\right) + \frac{0.05}{5.80} = 0.00816 + 0.03077 + 0.00862 = 0.04755 \text{ (or } 4.76\%\text{)}$$

4. Absolute uncertainty \(\Delta \rho\):
$$\Delta \rho = 0.04755 \times 7956 = 378\text{ kg m}^{-3} \approx 0.38 \times 10^3\text{ kg m}^{-3}$$
(or to 1–2 s.f., \(\pm 0.38 \times 10^3\text{ kg m}^{-3}\), giving \((7.96 \pm 0.38) \times 10^3\text{ kg m}^{-3}\) or \((8.0 \pm 0.4) \times 10^3\text{ kg m}^{-3}\)).

[C1] For correct calculation of volume \(V = 3.08 \times 10^{-5}\text{ m}^3\) and density \(\rho = 7.96 \times 10^3\text{ kg m}^{-3}\) (or \(7.97\text{ g cm}^{-3}\)).
[C1] For summing fractional / percentage uncertainties: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L} = 4.76\%\) (must include factor of 2 for diameter).
[A1] For absolute uncertainty \(\Delta \rho = \pm 0.38 \times 10^3\text{ kg m}^{-3}\) (accept \(\pm 380\text{ kg m}^{-3}\) or \(\pm 400\text{ kg m}^{-3}\)) and final value quoted consistently.

1. Calculate the volume \(V\) of the block:
$$V = A \times h = 8.0 \times 10^{-3} \times 0.15 = 1.20 \times 10^{-3}\text{ m}^3$$

2. Calculate the weight \(W\) of the block:
$$W = m g = \rho_{\text{wood}} V g = 650 \times 1.20 \times 10^{-3} \times 9.81 = 7.652\text{ N}$$

3. Calculate the upthrust \(U\) acting on the block:
$$U = \rho_{\text{liquid}} V g = 1.05 \times 10^3 \times 1.20 \times 10^{-3} \times 9.81 = 12.361\text{ N}$$

4. Apply equilibrium conditions:
$$U = W + T \implies T = U - W$$
$$T = 12.361 - 7.652 = 4.709\text{ N} \approx 4.7\text{ N}$$

[C1] For calculating weight \(W = 7.65\text{ N}\) or upthrust \(U = 12.4\text{ N}\).
[M1] For equating forces in vertical equilibrium: \(T = U - W\) or \(T = (\rho_{\text{liq}} - \rho_{\text{wood}}) V g\).
[A1] For final answer \(T = 4.7\text{ N}\) (accept \(4.71\text{ N}\)).

1. Apply conservation of linear momentum to find the velocity \(v_A\) of trolley \(A\) after the collision:
$$m_A u_A + m_B u_B = m_A v_A + m_B v_B$$
$$(0.80 \times 2.4) + 0 = (0.80 \times v_A) + (1.20 \times 1.6)$$
$$1.92 = 0.80 v_A + 1.92$$
$$0.80 v_A = 0 \implies v_A = 0\text{ m s}^{-1}$$

2. Initial kinetic energy \(E_{k\text{, initial}}\):
$$E_{k\text{, initial}} = \frac{1}{2} m_A u_A^2 = \frac{1}{2} \times 0.80 \times (2.4)^2 = 2.304\text{ J}$$

3. Final kinetic energy \(E_{k\text{, final}}\):
$$E_{k\text{, final}} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = 0 + \frac{1}{2} \times 1.20 \times (1.6)^2 = 1.536\text{ J}$$

4. Loss in kinetic energy \(\Delta E_k\):
$$\Delta E_k = E_{k\text{, initial}} - E_{k\text{, final}} = 2.304 - 1.536 = 0.768\text{ J} \approx 0.77\text{ J}$$

[C1] For using conservation of momentum to determine \(v_A = 0\text{ m s}^{-1}\).
[C1] For calculating initial \(E_k = 2.30\text{ J}\) and final \(E_k = 1.54\text{ J}\).
[A1] For correct difference \(\Delta E_k = 0.77\text{ J}\) (accept \(0.768\text{ J}\)).

1. Determine grating spacing \(d\):
$$d = \frac{1 \times 10^{-3}\text{ m}}{450} = 2.222 \times 10^{-6}\text{ m}$$

2. Use the diffraction grating formula \(d \sin\theta = n\lambda\) with \(\theta \le 90^\circ\) (so \(\sin\theta \le 1\)):
$$n \le \frac{d}{\lambda} = \frac{2.222 \times 10^{-6}}{589 \times 10^{-9}} = 3.77$$

3. The maximum integer order is \(n_{\text{max}} = 3\).

4. Total number of maxima observed is:
$$\text{Total} = 2 n_{\text{max}} + 1 = (2 \times 3) + 1 = 7$$

[C1] For calculating grating spacing \(d = 2.22 \times 10^{-6}\text{ m}\).
[M1] For substituting into \(d \sin\theta = n\lambda\) with \(\sin\theta = 1\) to get \(n = 3.77\) and deducing \(n_{\text{max}} = 3\).
[A1] For calculating total number of maxima \(N = 7\) (must include central maximum).

1. Calculate the tensile force \(F\):
$$F = mg = 6.50 \times 9.81 = 63.765\text{ N}$$

2. Use the Young modulus formula \(E = \frac{\sigma}{\varepsilon} = \frac{F L}{A x}\) to calculate the extension \(x\):
$$x = \frac{F L}{A E} = \frac{63.765 \times 2.40}{(3.80 \times 10^{-7}) \times (1.20 \times 10^{11})} = \frac{153.036}{4.56 \times 10^4} = 3.356 \times 10^{-3}\text{ m}$$

3. Calculate the elastic potential energy \(E_p\):
$$E_p = \frac{1}{2} F x = \frac{1}{2} \times 63.765 \times 3.356 \times 10^{-3} = 1.070 \times 10^{-2}\text{ J} \approx 1.07 \times 10^{-2}\text{ J}$$

[C1] For calculating tension \(F = 63.8\text{ N}\) and correct rearrangement for extension \(x = 3.36 \times 10^{-3}\text{ m}\).
[M1] For using \(E_p = \frac{1}{2} F x\) or \(E_p = \frac{F^2 L}{2 A E}\).
[A1] For final value \(E_p = 1.07 \times 10^{-2}\text{ J}\) (accept \(0.0107\text{ J}\)).

1. Current in the circuit \(I\) is determined from the fixed resistor:
$$I = \frac{V_R}{R} = \frac{3.60\text{ V}}{4.50\,\Omega} = 0.800\text{ A}$$

2. Using the e.m.f. equation for the series circuit:
$$E = I(R + R_T + r)$$
$$9.00 = 0.800 \times (4.50 + R_T + 1.50)$$
$$9.00 = 0.800 \times (6.00 + R_T)$$

3. Solve for \(R_T\):
$$6.00 + R_T = \frac{9.00}{0.800} = 11.25\,\Omega$$
$$R_T = 11.25 - 6.00 = 5.25\,\Omega$$

[C1] For calculating the circuit current \(I = \frac{3.60}{4.50} = 0.800\text{ A}\).
[M1] For applying Kirchhoff's second law / e.m.f. formula: \(E = I(R + r + R_T)\) or equating total resistance \(R_{\text{total}} = \frac{9.00}{0.800} = 11.25\,\Omega\).
[A1] For calculating \(R_T = 5.25\,\Omega\).

1. Convert the energy from \(\text{MeV}\) to joules (\(\text{J}\)):
$$E = 0.156 \times 10^6 \times 1.60 \times 10^{-19}\text{ J} = 2.496 \times 10^{-14}\text{ J}$$

2. Use the mass-energy equivalence relation \(\Delta E = \Delta m \, c^2\):
$$\Delta m = \frac{\Delta E}{c^2}$$
$$\Delta m = \frac{2.496 \times 10^{-14}}{(3.00 \times 10^8)^2} = \frac{2.496 \times 10^{-14}}{9.00 \times 10^{16}} = 2.7733 \times 10^{-31}\text{ kg} \approx 2.77 \times 10^{-31}\text{ kg}$$
(or \(2.78 \times 10^{-31}\text{ kg}\) using standard rounding).

[C1] For converting energy to joules: \(E = 2.50 \times 10^{-14}\text{ J}\).
[M1] For using \(\Delta m = \frac{E}{c^2}\) with \(c = 3.00 \times 10^8\text{ m s}^{-1}\).
[A1] For answer \(\Delta m = 2.77 \times 10^{-31}\text{ kg}\) (or \(2.78 \times 10^{-31}\text{ kg}\)).

1. Calculate observed frequency \(f_o\) using the Doppler effect formula for a moving source approaching a stationary observer:
$$f_o = f_s \left( \frac{v}{v - v_s} \right)$$
$$f_o = 850 \times \left( \frac{340}{340 - 28.0} \right) = 850 \times \frac{340}{312} = 926.28\text{ Hz}$$

2. Calculate the observed wavelength \(\lambda_{\text{obs}}\) using the wave equation:
$$\lambda_{\text{obs}} = \frac{v}{f_o} = \frac{340}{926.28} = 0.3671\text{ m} \approx 0.367\text{ m}$$

Alternatively, using the compressed wavefront formula directly:
$$\lambda_{\text{obs}} = \frac{v - v_s}{f_s} = \frac{340 - 28.0}{850} = \frac{312}{850} = 0.3671\text{ m} \approx 0.367\text{ m}$$

[C1] For calculating the observed frequency \(f_o = 926\text{ Hz}\) OR writing \(\lambda = \frac{v - v_s}{f_s}\).
[M1] For relating wave speed, frequency, and wavelength \(\lambda = \frac{v}{f_o}\) or substitution \(\frac{312}{850}\).
[A1] For final answer \(\lambda_{\text{obs}} = 0.367\text{ m}\) (3 s.f.).

(a) At \( t = 0 \), the ball is at rest so \( v = 0 \). Initially, drag is zero, so acceleration is maximum (finite positive initial gradient). As the ball accelerates, resistive drag increases, causing the net downward force and hence the acceleration (gradient) to decrease continuously until the net force is zero and velocity reaches a constant terminal value \( v_{\text{T}} \) (gradient = 0).

(b) The gradient of a velocity--time graph represents the acceleration \( a = \frac{\Delta v}{\Delta t} \).

(c) The forces acting are the constant downward weight \( W \), the constant upward upthrust \( U \), and the upward viscous drag \( D \). Drag \( D \) increases with speed \( v \). The resultant force is \( F_{\text{net}} = W - U - D \). Since \( D \) increases with \( v \), \( F_{\text{net}} \) decreases. From Newton's second law, \( a = \frac{F_{\text{net}}}{m} \), so acceleration decreases.

(a) [B1] Graph starts at origin \( (0, 0) \) with positive initial gradient and smoothly curves to become horizontal at terminal velocity \( v_{\text{T}} \).
(b) [B1] (Instantaneous) acceleration.
(c) [B1] Drag / viscous resistance increases as velocity increases AND net / resultant force decreases (so acceleration decreases via \( F = ma \)).

(a)
1. Tensile stress \( \sigma = \frac{F}{A} = \frac{60.0\text{ N}}{4.00 \times 10^{-7}\text{ m}^2} = 1.50 \times 10^8\text{ Pa} \).
Thus, value for [ A ] is \( 1.50 \).

2. Tensile strain \( \varepsilon = \frac{\Delta L}{L_0} = \frac{1.50 \times 10^{-3}\text{ m}}{2.50\text{ m}} = 6.00 \times 10^{-4} \).
Thus, value for [ B ] is \( 6.00 \).

(b) Young modulus \( E = \frac{\sigma}{\varepsilon} = \frac{1.50 \times 10^8\text{ Pa}}{6.00 \times 10^{-4}} = 2.50 \times 10^{11}\text{ Pa} = 250\text{ GPa} \).

(a) [B1] [ A ] = \( 1.50 \) (allow \( 1.5 \)) AND [ B ] = \( 6.00 \) (allow \( 6.0 \) or \( 6 \)).
(b) [C1] \( E = \frac{\text{stress}}{\text{strain}} \) or \( E = \frac{FL_0}{A\Delta L} \) with values substituted.
[A1] \( E = 250\text{ GPa} \) (or \( 2.50 \times 10^{11}\text{ Pa} \) with unit clearly stated if not in GPa; allow 2 or 3 s.f.).

(a)
(i) A neutron consists of one up quark and two down quarks: \( udd \).
(ii) An antiproton is the antiparticle of a proton (\( uud \)), so its quark composition is \( \bar{u}\bar{u}\bar{d} \).
(iii) Charge of an antiproton: \( 2(-\frac{2}{3}e) + (+\frac{1}{3}e) = -1e \).
(iv) A pion consists of a quark-antiquark pair (\( u\bar{d} \)), which classifies it as a meson.

(b) In \( \beta^- \) decay, a neutron changes into a proton (\( n \rightarrow p + e^- + \bar{\nu}_e \)). At the fundamental level, one down quark transforms into an up quark: \( d \rightarrow u + e^- + \bar{\nu}_e \).

(a) [B1] (i) \( udd \) AND (iv) Meson.
[B1] (ii) \( \bar{u}\bar{u}\bar{d} \) AND (iii) \( -1 \) (or \( -e \)).
(b) [B1] \( d \rightarrow u \) / a down quark changes into an up quark.

1. Increasing the width of one slit allows more light through, increasing the amplitude of the wave emerging from that slit compared to the other.
2. At the positions of interference minima, the two waves still meet with a phase difference of \(\pi\text{ rad}\) (or path difference of \((n + \frac{1}{2})\lambda\)), but because their amplitudes are unequal, destructive interference is incomplete. The resultant amplitude, and therefore the intensity (since \(I \propto A^2\)), is greater than zero.

B1: (Increasing slit width causes) greater amplitude / intensity of light from that slit than from the other slit.
B1: Waves at the minima still arrive in antiphase / path difference of \((n + \frac{1}{2})\lambda\), but unequal amplitudes prevent complete cancellation / destructive interference is incomplete.

1. As the external resistance \(R\) decreases, the total resistance \((R + r)\) of the circuit decreases, which causes the current \(I = \frac{E}{R + r}\) in the circuit to increase.
2. The potential difference across the internal resistance (the 'lost volts'), given by \(V_{\text{lost}} = Ir\), increases as \(I\) increases.
3. Since the terminal potential difference is \(V = E - Ir\) and \(E\) is constant, an increase in \(Ir\) causes \(V\) to decrease.

B1: (Decreasing \(R\) reduces total circuit resistance so) current \(I\) in the circuit increases.
B1: Larger current results in a greater potential drop across the internal resistance / larger 'lost volts' (\(Ir\)), so terminal p.d. \(V = E - Ir\) decreases.

1. At release, the downward force (weight) exceeds the upward forces (upthrust and drag), producing an initial acceleration.
2. As the sphere's speed increases, the upward viscous drag force increases with speed.
3. Consequently, the downward resultant force (\(F_{\text{net}} = W - U - D\)) decreases, causing the acceleration (\(a = \frac{F_{\text{net}}}{m}\)) to decrease until the upward drag plus upthrust equals the downward weight. At this point, the resultant force and acceleration become zero (terminal velocity).

B1: (As speed increases,) drag / resistive force increases (acting upwards).
B1: Resultant (downward) force decreases (hence acceleration decreases) until upward forces equal weight / net force is zero.

1. The area under the loading curve represents the work done on the rubber to stretch it, whereas the area under the unloading curve represents the elastic potential energy recovered as it contracts.
2. Because the unloading curve lies below the loading curve, the work done in stretching is greater than the energy released. The net energy transferred (represented by the area enclosed by the hysteresis loop) is dissipated internally as thermal energy (internal energy), raising the temperature of the rubber.

B1: Work done to stretch / load the rubber is greater than energy returned / work done by the rubber during unloading (or area under loading curve > area under unloading curve).
B1: The net work done / energy difference (area enclosed by hysteresis loop) is transformed into thermal / internal energy within the rubber.

1. During \(\beta^-\ decay, an electron antineutrino (\)\bar{\nu}_e\)) is emitted along with the \(\beta^-\ particle (electron) and the daughter nucleus. 2. The discrete energy released in the decay is shared between the emitted \)\beta^-\ particle and the antineutrino. Because the energy can be distributed between these two particles in continuously variable proportions, the kinetic energy of the \(\beta^-\ particle forms a continuous spectrum up to the maximum possible energy (which occurs when the antineutrino carries negligible kinetic energy).

B1: An (electron) antineutrino is emitted along with the \(\beta^-\) particle / electron (three-body decay).
B1: The fixed decay energy is shared / divided between the \(\beta^-\) particle and the antineutrino in varying proportions.

(a)
For \( L = 0.600\text{ m} \):
\( t_{\text{mean}} = \frac{31.0 + 31.2}{2} = 31.1\text{ s} \)
\( T = \frac{31.1}{20} = 1.555\text{ s} \) (or \( 1.56\text{ s} \))
\( T^2 = (1.555)^2 = 2.418 \approx 2.42\text{ s}^2 \)

For \( L = 0.700\text{ m} \):
\( t_{\text{mean}} = \frac{33.6 + 33.4}{2} = 33.5\text{ s} \)
\( T = \frac{33.5}{20} = 1.675\text{ s} \) (or \( 1.68\text{ s} \))
\( T^2 = (1.675)^2 = 2.806 \approx 2.81\text{ s}^2 \)

(b)
Raw time measurements \( t_1 \) and \( t_2 \) have 3 significant figures (recorded to \( 0.1\text{ s} \)), therefore \( T \) is given to the same number of significant figures (3 s.f.) or one more (4 s.f.).

(c)
Calculate the ratio \( \frac{T^2}{L} \) for the data sets:
- \( \frac{1.20}{0.300} = 4.00\text{ s}^2\text{ m}^{-1} \)
- \( \frac{1.61}{0.400} = 4.03\text{ s}^2\text{ m}^{-1} \)
- \( \frac{2.02}{0.500} = 4.04\text{ s}^2\text{ m}^{-1} \)
- \( \frac{2.42}{0.600} = 4.03\text{ s}^2\text{ m}^{-1} \)
- \( \frac{2.81}{0.700} = 4.01\text{ s}^2\text{ m}^{-1} \)

Since \( \frac{T^2}{L} \approx 4.02\text{ s}^2\text{ m}^{-1} \) (constant within experimental error), \( T^2 \) is directly proportional to \( L \).

(d)
Place a fiducial marker at the equilibrium position (centre of the swing) and start/stop timing when the pendulum passes this marker.

(a)
[1] B1: Both values of \( t_{\text{mean}} \) correct (\( 31.1 \) and \( 33.5 \)).
[1] B1: Both values of \( T \) correct (\( 1.555 \) / \( 1.56 \) and \( 1.675 \) / \( 1.68 \)).
[1] B1: Both values of \( T^2 \) calculated correctly to 3 (or 4) s.f. consistent with previous column (\( 2.42 \) and \( 2.81 \)).

(b)
[1] B1: Reference to raw time readings (\( t \) values) having 3 significant figures / 1 decimal place, so \( T \) is given to 3 (or 3/4) significant figures.

(c)
[1] M1: Evidence of calculating \( \frac{T^2}{L} \) (or \( \frac{L}{T^2} \)) for at least two pairs of values (e.g. \( \approx 4.0\text{ s}^2\text{ m}^{-1} \)).
[1] A1: Conclusion that \( T^2 \) is directly proportional to \( L \) (or linear relationship through origin) supported by the constant ratio.

(d)
[1] B1: Use a fiducial mark / pin / reference line at the centre of oscillation / equilibrium position OR use a light gate and datalogger / timer.

(a) Graph plotting:
1. Axes: Linear scales chosen such that the plotted points occupy at least half the grid in both the \(x\)-direction and the \(y\)-direction. Scales must not be awkward (e.g. multiples of 3, 7, etc.). Axes are labelled clearly with quantity and unit: \(x / \text{cm}\) on the horizontal axis, \(y / \text{mm}\) on the vertical axis.
2. Plotting: All six pairs of coordinates are plotted accurately to within half a small grid square.
3. Line of best fit: A single, continuous straight line drawn with a transparent ruler, balancing points evenly on either side of the line over its entire length.

(b) Gradient and intercept determination:
1. Gradient calculation:
Choose two points \((x_1, y_1)\) and \((x_2, y_2)\) on the drawn line of best fit separated by a distance whose hypotenuse is greater than half the length of the drawn line.
For example, using points on the line \((15.0, 4.2)\) and \((40.0, 21.5)\):
$$\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{21.5 - 4.2}{40.0 - 15.0} = \frac{17.3}{25.0} = 0.692\text{ mm cm}^{-1}$$

2. \(y\)-intercept calculation:
Using \(y = mx + c\) with a point on the line, e.g., \((40.0, 21.5)\):
$$c = y - mx = 21.5 - (0.692 \times 40.0) = 21.5 - 27.68 = -6.18\text{ mm} \approx -6.2\text{ mm}$$

(a)
- Axes [1]: Scales must be linear and non-awkward. Plotted points occupy \(\ge 50\%\) of the grid in each direction. Both axes labelled with quantity and unit (\(x / \text{cm}\) and \(y / \text{mm}\)).
- Plotting [1]: All points plotted correctly to within half a small square diameter. Sharp pencil and distinct crosses/dots in circles required.
- Line of best fit [1]: Straight line with balanced distribution of points along the whole length; line must not be kinked, thick, or feathery.
- Trend/Quality [1]: All data points lie close to the drawn trend line.

(b)
- Gradient [1]: Triangle hypotenuse \(> 50\%\) of line drawn. Coordinates read correctly from line to within half a small square and substituted into \(\Delta y / \Delta x\). Value in range \(0.68\text{ to }0.70\text{ mm cm}^{-1}\).
- Intercept [1]: Correctly calculated from \(c = y_1 - m x_1\) using a point on the line, or read directly from the \(y\)-axis if \(x = 0\) is on the grid. Value in range \(-6.5\text{ mm to }-5.8\text{ mm}\).

(a)(i)
Rearranging the expression for Young modulus \( E \):
\[ E = \frac{4 g L^3}{G b d^3} \]

Substitute the values in consistent SI units:
- \( L = 0.800 \text{ m} \implies L^3 = 0.512 \text{ m}^3 \)
- \( b = 28.4 \times 10^{-3} \text{ m} \)
- \( d = 4.20 \times 10^{-3} \text{ m} \implies d^3 = 7.4088 \times 10^{-8} \text{ m}^3 \)
- \( G = 0.0542 \text{ m kg}^{-1} \)
- \( g = 9.81 \text{ m s}^{-2} \)

\[ E = \frac{4 \times 9.81 \times 0.512}{0.0542 \times (28.4 \times 10^{-3}) \times (7.4088 \times 10^{-8})} \]
\[ E = \frac{20.09088}{1.14042 \times 10^{-10}} = 1.7617 \times 10^{11} \approx 1.76 \times 10^{11} \text{ Pa} \text{ (or N m}^{-2}\text{)} \]

(a)(ii)
From \( E = \frac{\text{Force}/\text{Area}}{\text{Strain}} \):
\[ [E] = \frac{\text{N}}{\text{m}^2} = \frac{\text{kg m s}^{-2}}{\text{m}^2} = \text{kg m}^{-1} \text{s}^{-2} \]

(b)(i)
Percentage uncertainty in \( d^3 = 3 \times \left( \frac{\Delta d}{d} \right) \times 100\% \):
\[ \text{Percentage uncertainty} = 3 \times \left( \frac{0.05}{4.20} \right) \times 100\% = 3.57\% \text{ (or } 3.6\%\text{)} \]

(b)(ii)
Sum the fractional/percentage uncertainties in \( L \), \( b \), and \( d \):
\[ \frac{\Delta E}{E} = 3\left(\frac{\Delta L}{L}\right) + \frac{\Delta b}{b} + 3\left(\frac{\Delta d}{d}\right) \]
\[ 3\left(\frac{\Delta L}{L}\right) = 3 \times \frac{0.002}{0.800} = 0.00750 \text{ (0.75%)} \]
\[ \frac{\Delta b}{b} = \frac{0.1}{28.4} \approx 0.00352 \text{ (0.352%)} \]
\[ 3\left(\frac{\Delta d}{d}\right) = 0.03571 \text{ (3.571%)} \]
\[ \frac{\Delta E}{E} = 0.00750 + 0.00352 + 0.03571 = 0.04673 \text{ (4.67%)} \]

Absolute uncertainty:
\[ \Delta E = 0.04673 \times 1.762 \times 10^{11} \approx 8.2 \times 10^9 \text{ Pa} = 0.08 \times 10^{11} \text{ Pa} \]

(c)
Since \( G \propto \frac{1}{b} \), doubling \( b \) causes the gradient \( G \) to decrease (halve).

(a)(i)
- C1: Correct rearrangement of formula and substitution of all values with appropriate power of ten conversions: \( E = \frac{4 \times 9.81 \times (0.800)^3}{0.0542 \times (28.4 \times 10^{-3}) \times (4.20 \times 10^{-3})^3} \)
- A1: Value of \( E = 1.76 \times 10^{11} \text{ Pa} \) (accept \( 1.76 \times 10^{11} \text{ N m}^{-2} \), range \( 1.75 \times 10^{11} \) to \( 1.77 \times 10^{11} \)) given to 2 or 3 s.f. [2]

(a)(ii)
- B1: \( \text{kg m}^{-1} \text{s}^{-2} \) only. [1]

(b)(i)
- B1: Correct percentage uncertainty in \( d^3 = 3.6\% \) or \( 3.57\% \). [1]

(b)(ii)
- C1: Correct method to determine total fractional or percentage uncertainty in \( E \) by summing individual terms: \( 3(\Delta L/L) + \Delta b/b + 3(\Delta d/d) \) \( (= 0.0467 \text{ or } 4.7\%) \)
- A1: Absolute uncertainty \( \Delta E = \pm 0.08 \times 10^{11} \text{ Pa} \) or \( \pm 8 \times 10^9 \text{ Pa} \) (accept \( 0.082 \times 10^{11} \text{ Pa} \), allow ECF from (a)(i)). [2]

(c)
- B1: Gradient decreases AND refers to \( G \propto 1/b \) or inverse relationship between gradient and width. [1]

(a)(i)
- Estimated absolute uncertainty in stopwatch timing \(\Delta t = 0.2\text{ s}\) (acceptable range \(0.1\text{ s} - 0.4\text{ s}\) due to human reaction time / spread of readings).
\[\text{Percentage uncertainty} = \frac{\Delta t}{t_1} \times 100\% = \frac{0.2\text{ s}}{7.43\text{ s}} \times 100\% = 2.7\%\]

(a)(ii)
- \(L_1 = 28.0\text{ cm} = 0.280\text{ m}\), \(T_1 = 0.743\text{ s}\)
\[k_1 = \frac{T_1^2}{L_1^3} = \frac{(0.743)^2}{(0.280)^3} = \frac{0.552049}{0.021952} = 25.15 \approx 25.2\text{ s}^2\text{ m}^{-3}\]

(b)
- \(L_2 = 18.5\text{ cm} = 0.185\text{ m}\), \(T_2 = 0.382\text{ s}\)
\[k_2 = \frac{T_2^2}{L_2^3} = \frac{(0.382)^2}{(0.185)^3} = \frac{0.145924}{0.0063316} = 23.046 \approx 23.0\text{ s}^2\text{ m}^{-3}\]

(c)
- The raw values of length \(L\) (\(28.0\text{ cm}\) and \(18.5\text{ cm}\)) and time \(t\) (\(7.43\text{ s}\) and \(3.82\text{ s}\)) are recorded to 3 significant figures. Therefore, \(k\) should be quoted to the same number of significant figures (3 s.f., or optionally 2 s.f.).

(d)(i)
\[\text{Percentage difference} = \frac{|k_1 - k_2|}{\frac{k_1 + k_2}{2}} \times 100\% = \frac{|25.15 - 23.05|}{24.10} \times 100\% = \frac{2.10}{24.10} \times 100\% = 8.7\% \text{ to } 9.1\%\]

(d)(ii)
- Since the percentage difference (\(8.7\% - 9.1\%\)) is less than the \(10\%\) limit specified, the experimental evidence supports the proposed relationship.

(e) Limitations & Improvements:
1. Limitation: Two sets of data are insufficient to draw a valid conclusion.
Improvement: Take readings for at least 5 or 6 different lengths \(L\) and plot a graph of \(T^2\) against \(L^3\).
2. Limitation: The cantilever undergoes lateral twisting/wobbling, making it difficult to judge complete vertical oscillations.
Improvement: Use a fiducial marker / position indicator placed at the equilibrium position.
3. Limitation: Human reaction time creates significant uncertainty over small total oscillation times.
Improvement: Use a light gate connected to an electronic datalogger / timing system to record periods automatically.

(a)(i) [1 mark]
- B1: Reasonable absolute uncertainty in \(t_1\) stated (\(0.1\text{ s} \le \Delta t \le 0.4\text{ s}\)) AND correct calculation of percentage uncertainty to 2 or 3 s.f. with \(\%\) sign.

(a)(ii) [1 mark]
- A1: Correct calculation of \(k_1\) with correct unit (\(25.1\) to \(25.2\text{ s}^2\text{ m}^{-3}\) or \(2.51 \times 10^{-5} - 2.52 \times 10^{-5}\text{ s}^2\text{ cm}^{-3}\)).

(b) [1 mark]
- A1: Correct calculation of \(k_2\) to 2 or 3 s.f. (\(23.0\text{ s}^2\text{ m}^{-3}\) or \(2.30 \times 10^{-5}\text{ s}^2\text{ cm}^{-3}\)).

(c) [1 mark]
- B1: Clear statement that \(k\) is given to 3 s.f. (or 2 s.f.) because the raw measurements of \(L\) and \(t\) are both known to 3 s.f. (or the least s.f. of raw data is 3).

(d)(i) [1 mark]
- A1: Correct calculation of percentage difference between \(k_1\) and \(k_2\) (e.g., \(8.7\%\) to \(9.1\%\)).

(d)(ii) [1 mark]
- B1: Valid conclusion stating whether the relationship is supported based on comparing calculated percentage difference with the stated \(10\%\) criterion.

(e) [6 marks] (1 mark for each valid limitation, 1 mark for each matched improvement, max 6):
- L1: Two readings are not enough to establish a relationship / draw a firm conclusion.
I1: Take more values of \(L\) (at least 5-6 values) and plot a graph of \(T^2\) against \(L^3\).
- L2: Difficult to judge when an oscillation is completed / oscillations damp out quickly.
I2: Use a fiducial marker placed at the centre/equilibrium position to time oscillations.
- L3: Human reaction time error significant / small total time recorded.
I3: Use an optical pin/light gate connected to a digital timer/datalogger to measure \(T\).

Limitations and Matched Improvements:

1. Data sufficiency:
- Limitation: Two values of \( x_0 \) (initial displacement) are not sufficient to draw a valid conclusion or verify the relationship.
- Improvement: Take measurements for at least five or six different values of \( x_0 \) and plot a graph of \( t \) against \( x_0 \) (or test a specific relationship graphically).

2. Timing / Measurement of decaying amplitude:
- Limitation: It is difficult to determine the precise instant the bob passes through the \( \frac{1}{2} x_0 \) position because the amplitude decays continuously and rapidly, leading to human reaction time errors.
- Improvement: Record the motion using a high-frame-rate video camera placed perpendicular to the scale with a digital clock in view, and analyse frame-by-frame / use position sensors (e.g. motion sensor or light gates).

3. Parallax and visual distortion:
- Limitation: Parallax error occurs when reading the position of the sphere because there is a physical gap between the sphere inside the tank and the scale mounted outside / refraction through the glass and fluid distorts the apparent position.
- Improvement: Place the scale inside the fluid directly behind/alongside the path of the bob, or place a plane mirror behind the external scale and align the line of sight with the bob and its reflected image.

4. Motion stability / Hydrodynamic effects:
- Limitation: The sphere may drift, oscillate out of plane, or experience wall-proximity effects (boundary layer drag) in a narrow tank.
- Improvement: Use a much wider tank to eliminate wall proximity drag / use a bifilar (two-thread) suspension to constrain the oscillation to a single vertical plane.

Marking Scheme (8 marks maximum: 1 mark for each valid limitation [L1–L4] and 1 mark for each correctly matched improvement [I1–I4]):

- A (Data points):
- L1 [1 mark]: Two sets of readings are not enough / insufficient to draw a valid conclusion / establish a relationship.
- I1 [1 mark]: Take readings for more / at least 5 sets of \( x_0 \) and plot a graph.
- Reject: Just "repeat readings" or "repeat and average" for I1.

- B (Judging amplitude / timing):
- L2 [1 mark]: Difficult to identify / judge when amplitude reaches \( \frac{1}{2}x_0 \) / large uncertainty in stopwatch timing due to rapid damping / reaction time.
- I2 [1 mark]: Use video recording with a timer / frame-by-frame playback / motion sensor connected to a data logger.
- Reject: Just "use a motion sensor" without stating data logger / computer recording.

- C (Scale alignment & parallax):
- L3 [1 mark]: Parallax error due to gap / distance between sphere and external scale (or refraction through tank walls/fluid).
- I3 [1 mark]: Position scale inside the fluid adjacent to the bob / use a mirror behind the scale to eliminate parallax / view using a travelling microscope.
- Reject: "Look straight at it" / "eye level".

- D (Motion out of plane / Wall effects):
- L4 [1 mark]: Sphere wobbles / swings out of a straight line / experiences wall effects from the narrow tank.
- I4 [1 mark]: Use a wider/larger container / use a bifilar suspension to constrain motion to a single plane.
- Reject: Irrelevant suggestions such as "do experiment in a vacuum" or "change the liquid".

Refer to the evaluation table in the marking scheme.

Refer to marking scheme for question evaluation.