2026 Cambridge IAL Physics (9702) 模拟试题及答案详解

Rearranging the equation for dynamic viscosity gives:
\[\eta = \frac{F}{6 \pi r v}\]

Substitute the SI base units of each quantity:
- Force \(F = m a\), so units are \(\text{kg m s}^{-2}\)
- Radius \(r\) has units \(\text{m}\)
- Speed \(v\) has units \(\text{m s}^{-1}\)
- \(6\pi\) is dimensionless.

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

A is correct [1 mark].
B is the base unit of force.
C and D have incorrect powers of \(\text{m}\) resulting from algebraic errors in rearranging.

The final velocity \(v\) is composed of a horizontal component \(v_x = u\) (since horizontal velocity remains constant) and a vertical component \(v_y\).

By Pythagoras' theorem:
\[v^2 = v_x^2 + v_y^2 = u^2 + v_y^2 \implies v_y^2 = v^2 - u^2\]

Using the kinematic equation for vertical motion with initial vertical velocity \(u_y = 0\):
\[v_y^2 = u_y^2 + 2gh = 2gh\]

Equating the expressions for \(v_y^2\):
\[2gh = v^2 - u^2 \implies h = \frac{v^2 - u^2}{2g}\]

A is correct [1 mark].
B incorrectly adds the squares instead of subtracting.
C uses \((v-u)^2\) rather than \(v^2 - u^2\).
D confuses dimensions of velocity and height.

Taking the initial direction of motion as positive:
- Initial velocity \(u = +20\text{ m s}^{-1}\)
- Final velocity \(v = -14\text{ m s}^{-1}\)

Change in momentum \(\Delta p\) of the ball:
\[\Delta p = m(v - u) = 0.15 \times (-14 - 20) = 0.15 \times (-34) = -5.10\text{ N s}\]

Magnitude of the average force \(F\):
\[F = \frac{|\Delta p|}{\Delta t} = \frac{5.10}{0.025} = 204\text{ N}\]

C is correct [1 mark].
A results from subtracting speeds without considering direction: \(0.15 \times (20 - 14) / 0.025 = 36\text{ N}\).
B and D are arithmetic or misread factor errors.

Let \(A\) be the cross-sectional area of the block, \(H = 0.20\text{ m}\) be the total height, and \(h\) be the submerged depth.

For floating equilibrium:
\[\text{Weight of block} = \text{Upthrust}\]
\[\rho_{\text{block}} A H g = \rho_{\text{liquid}} A h g\]

Cancelling \(A\) and \(g\):
\[h = H \times \frac{\rho_{\text{block}}}{\rho_{\text{liquid}}} = 0.20 \times \frac{650}{850} \approx 0.153\text{ m} = 0.15\text{ m}\]

B is correct [1 mark].
A is the height of the block remaining above the liquid surface: \(0.20 - 0.153 = 0.047\text{ m}\).
C and D arise from inverted density ratios or incorrect scaling.

In each second (mass \(m = 4.5\text{ kg}\)):
1. Gravitational potential energy gained per second:
\[P_{\text{p}} = \frac{mgh}{t} = 4.5 \times 9.81 \times 18 = 794.61\text{ W}\]

2. Kinetic energy given to water per second:
\[P_{\text{k}} = \frac{\frac{1}{2}mv^2}{t} = 0.5 \times 4.5 \times 6.0^2 = 81.0\text{ W}\]

Total useful output power:
\[P_{\text{out}} = 794.61 + 81.0 = 875.61\text{ W}\]

Input power required:
\[P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{875.61}{0.75} \approx 1170\text{ W} = 1.17\text{ kW}\]

D is correct [1 mark].
A multiplies output power by efficiency rather than dividing: \(876 \times 0.75 = 660\text{ W}\).
B is the useful output power without accounting for efficiency: \(880\text{ W}\).
C neglects the kinetic energy of the emerging water: \(795 / 0.75 = 1060\text{ W}\).

The Young modulus \(E\) is given by:
\[E = \frac{\sigma}{\varepsilon} = \frac{F / A}{e / L} = \frac{F L}{\pi r^2 e}\]

Rearranging for force \(F\):
\[F = \frac{E \pi r^2 e}{L}\]

For the second wire with length \(L' = 2L\), radius \(r' = 2r\), and extension \(e' = e\):
\[F' = \frac{E \pi (2r)^2 e}{2L} = \frac{E \pi (4r^2) e}{2L} = 2 \left(\frac{E \pi r^2 e}{L}\right) = 2F\]

C is correct [1 mark].
A incorrectly assumes force is inversely proportional to radius.
B incorrectly assumes the changes in length and radius cancel out linearly without squaring the radius.
D fails to divide by the doubled length.

The diffraction grating equation is:
\[d \sin\theta = n \lambda\]

Since the angles \(\theta\) and grating spacing \(d\) are the same for both maxima:
\[d \sin\theta = 3 \lambda_1 = 2 \lambda_2\]

Rearranging to find \(\lambda_2\):
\[\lambda_2 = \frac{3}{2} \lambda_1 = 1.5 \times 480\text{ nm} = 720\text{ nm}\]

D is correct [1 mark].
A is calculated using \(\frac{2}{3} \times 480 = 320\text{ nm}\) (inverted ratio).
B assumes \(\lambda_2 = \lambda_1\).
C corresponds to \(\frac{4}{3} \times 480 = 640\text{ nm}\).

1. Current in the circuit is \(I = \frac{E}{R + r}\). As \(R\) increases, \(I\) decreases.
2. Terminal potential difference is \(V = E - Ir\). Since \(I\) decreases, the lost volts \(Ir\) decreases, so \(V\) increases towards \(E\).
3. Power dissipated in the internal resistance is \(P_{\text{int}} = I^2 r\). Since \(I\) decreases and \(r\) is constant, \(P_{\text{int}}\) decreases.

Therefore, the terminal potential difference increases and the power dissipated inside the cell decreases.

A is correct [1 mark].
B incorrectly predicts that internal power dissipation increases despite decreasing current.
C and D incorrectly predict that terminal potential difference decreases as load resistance increases.

Intensity \(I = \frac{\text{Power}}{\text{Area}}\).
In SI base units:
Power \(= \text{J s}^{-1} = (\text{kg m}^2 \text{s}^{-2}) \text{s}^{-1} = \text{kg m}^2 \text{s}^{-3}\).
Area \(= \text{m}^2\).
Therefore, \(I = \frac{\text{kg m}^2 \text{s}^{-3}}{\text{m}^2} = \text{kg s}^{-3}\).

From the equation \(\sigma = \frac{I}{T^4}\):
Units of \(\sigma = \frac{\text{kg s}^{-3}}{\text{K}^4} = \text{kg s}^{-3} \text{K}^{-4}\).

A is correct [1 mark].
B incorrectly includes \(\text{m}^2\) from power without dividing by area.
C incorrectly uses units of pressure divided by \(\text{K}^4\).
D has an incorrect power of seconds (\(\text{s}^{-2}\)).

Taking the upwards direction as positive:
Initial velocity \(= +u\)
Acceleration \(= -g\)
Total displacement \(s = -h\)
Time \(= T\)

Using the equation of uniformly accelerated motion \(s = ut + \frac{1}{2}at^2\):
\(-h = uT - \frac{1}{2}gT^2\)

Multiplying both sides by \(-1\):
\(h = \frac{1}{2}gT^2 - uT\).

A is correct [1 mark].
B incorrectly uses \(+uT\) for upward projection while assigning positive displacement downwards.
C gives \(-h\) rather than \(h\).
D confuses the time-dependent kinematics with maximum height calculations.

Initial kinetic energy:
\(E_{\text{k, initial}} = \frac{1}{2}m(3v)^2 + \frac{1}{2}(2m)(-v)^2 = \frac{9}{2}mv^2 + mv^2 = \frac{11}{2}mv^2\).

Conservation of linear momentum:
\(p_{\text{initial}} = m(3v) + 2m(-v) = 3mv - 2mv = mv\).

Total combined mass \(= 3m\).
Final speed \(V = \frac{mv}{3m} = \frac{1}{3}v\).

Final kinetic energy:
\(E_{\text{k, final}} = \frac{1}{2}(3m)V^2 = \frac{3}{2}m\left(\frac{v}{3}\right)^2 = \frac{1}{6}mv^2\).

Kinetic energy dissipated:
\(\Delta E_{\text{k}} = E_{\text{k, initial}} - E_{\text{k, final}} = \frac{11}{2}mv^2 - \frac{1}{6}mv^2 = \frac{32}{6}mv^2 = \frac{16}{3}mv^2\).

Fraction dissipated:
\(\frac{\Delta E_{\text{k}}}{E_{\text{k, initial}}} = \frac{\frac{16}{3}mv^2}{\frac{11}{2}mv^2} = \frac{16}{3} \times \frac{2}{11} = \frac{32}{33}\).

D is correct [1 mark].
A is the fraction of kinetic energy remaining (\(1/33\)).
B and C arise from arithmetic mistakes in evaluating the momentum or initial kinetic energy sum.

For a floating body, Upthrust = Weight of object.

In the first liquid:
\(W = \rho g A x\), where \(A\) is the cross-sectional area of the cylinder.

In the second liquid:
\(W = (1.25\rho) g A x_{\text{new}}\).

Equating the two expressions for the weight \(W\):
\(\rho g A x = 1.25\rho g A x_{\text{new}}\)
\(x_{\text{new}} = \frac{x}{1.25} = 0.80x\).

A is correct [1 mark].
B incorrectly uses the square root of the ratio.
C multiplies by 1.25 rather than dividing.
D squares the density factor.

The Young modulus \(E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{e / L} = \frac{F L}{A e}\).
Since the cross-section is circular, \(A = \frac{\pi d^2}{4}\), so:
\(e = \frac{4FL}{\pi d^2 E}\).

For the second wire:
\(e_2 = \frac{4 F_2 (2L)}{\pi (2d)^2 E} = \frac{8 F_2 L}{4 \pi d^2 E} = \frac{2 F_2 L}{\pi d^2 E}\).

We require \(e_2 = e\):
\(\frac{2 F_2 L}{\pi d^2 E} = \frac{4 F L}{\pi d^2 E} \implies 2F_2 = 4F \implies F_2 = 2F\).

C is correct [1 mark].
A and B fail to account for the \(d^2\) dependence of area.
D incorrectly assumes extension is inversely proportional to length.

Using the Doppler effect formula for a source moving towards a stationary observer:
\(f_o = f_s \left(\frac{v}{v - v_s}\right)\)
\(f_o = 680 \times \left(\frac{340}{340 - 30.0}\right) = 680 \times \left(\frac{340}{310}\right) = 746\text{ Hz}\).

The observed wavelength is the distance between consecutive wavefronts approaching the observer:
\(\lambda_o = \frac{v - v_s}{f_s} = \frac{340 - 30.0}{680} = \frac{310}{680} = 0.456\text{ m}\)
(or \(\lambda_o = \frac{v}{f_o} = \frac{340}{745.8} = 0.456\text{ m}\)).

B is correct [1 mark].
A uses \(v + v_s\) (source moving away).
C calculates the correct observed frequency but uses the unshifted source wavelength \(\lambda = \frac{340}{680} = 0.500\text{ m}\).
D assumes frequency does not change.

At temperature \(T_1\), the potential difference across the thermistor is:
\(V_1 = 12.0 \times \frac{300}{180 + 300} = 12.0 \times \frac{300}{480} = 7.50\text{ V}\).

At temperature \(T_2\), the potential difference decreases by \(2.50\text{ V}\):
\(V_2 = 7.50 - 2.50 = 5.00\text{ V}\).

Let the new resistance of the thermistor be \(R_T\):
\(V_2 = 12.0 \times \frac{R_T}{180 + R_T} = 5.00\text{ V}\)
\(5.00(180 + R_T) = 12.0 R_T\)
\(900 + 5.00 R_T = 12.0 R_T\)
\(7.00 R_T = 900\)
\(R_T = \frac{900}{7.00} \approx 129\,\Omega\).

B is correct [1 mark].
A incorrectly sets \(V_2 = 2.50\text{ V}\) instead of subtracting 2.50 V from 7.50 V.
C and D arise from applying the voltage divider across the fixed resistor rather than the thermistor.

A proton has quark composition \(\text{uud}\) and a neutron has quark composition \(\text{udd}\).
Therefore, the decay involves an up quark transforming into a down quark:
\(\text{u} \rightarrow \text{d} + \text{W}^+\).

The exchange particle is the \(\text{W}^+\) boson, which carries a charge of \(+e\) to conserve charge at the vertex (\(+\frac{2}{3}e \rightarrow -\frac{1}{3}e + 1e\)).

A is correct [1 mark].
B incorrectly identifies the charge of the exchange particle as \(-e\).
C and D describe \(\beta^-\) decay where a down quark changes to an up quark.

Base units of each term:
- Intensity \(I = \frac{\text{Power}}{\text{Area}} = \frac{\text{kg}\text{ m}^2\text{ s}^{-3}}{\text{m}^2} = \text{kg}\text{ s}^{-3}\)
- Density \(\rho = \text{kg}\text{ m}^{-3}\)
- Speed \(c = \text{m}\text{ s}^{-1}\)
- Angular frequency \(\omega = \text{s}^{-1}\), so \(\omega^n = \text{s}^{-n}\)
- Amplitude squared \(x_0^2 = \text{m}^2\)

Equating base units on both sides:
\[\text{kg}\text{ s}^{-3} = (\text{kg}\text{ m}^{-3})(\text{m}\text{ s}^{-1})(\text{s}^{-n})(\text{m}^2) = \text{kg}\text{ s}^{-(1+n)}\]

Comparing exponents of \(\text{s}\):
\[-3 = -(1+n) \implies n = 2\]

B [1 mark]: Correctly equates base units of intensity \(\text{kg}\text{ s}^{-3}\) to the product of base units on RHS to find \(n = 2\).

Horizontal component of velocity remains constant:
\[v_x = 15\text{ m s}^{-1}\]

Vertical component of velocity just before impact:
\[v_y^2 = u_y^2 + 2gs = 0 + 2(9.81)(45) = 882.9\text{ m}^2\text{ s}^{-2}\]
\[v_y = \sqrt{882.9} \approx 29.71\text{ m s}^{-1}\]

The angle \(\theta\) to the horizontal is given by:
\[\tan \theta = \frac{v_y}{v_x} = \frac{29.71}{15} \approx 1.981\]
\[\theta = \arctan(1.981) \approx 63^\circ\]

D [1 mark]: Correctly calculates vertical velocity \(v_y \approx 29.7\text{ m s}^{-1}\) and angle to horizontal \(\theta = \arctan(v_y/v_x) \approx 63^\circ\).

Take right as positive.

Initial momentum of the system:
\[p_i = (0.40)(+3.0) + (0.60)(-2.0) = 1.2 - 1.2 = 0\text{ kg m s}^{-1}\]

By conservation of linear momentum, since the total initial momentum is zero, the combined mass is stationary after the collision (\(v = 0\)).

Initial kinetic energy:
\[E_{k,i} = \frac{1}{2}(0.40)(3.0)^2 + \frac{1}{2}(0.60)(2.0)^2 = 1.8 + 1.2 = 3.0\text{ J}\]

Final kinetic energy:
\[E_{k,f} = 0\text{ J}\]

Total loss in kinetic energy:
\[\Delta E_k = 3.0\text{ J} - 0 = 3.0\text{ J}\]

D [1 mark]: Recognises final momentum and KE are zero, yielding a total kinetic energy loss of \(3.0\text{ J}\).

For a floating object, the upthrust equals the weight \(W\) of the cylinder.

In the first liquid:
\[W = \rho_1 g A h\]

In the second liquid:
\[W = \rho_2 g A (1.25 h)\]

Equating the two expressions for \(W\):
\[\rho_1 g A h = \rho_2 g A (1.25 h)\]
\[\rho_1 = 1.25 \rho_2 \implies \frac{\rho_1}{\rho_2} = 1.25\]

C [1 mark]: Equates upthrust in both cases to weight and solves for \(\frac{\rho_1}{\rho_2} = 1.25\).

Young modulus \(E = \frac{F/A}{x/L} \implies x = \frac{F L}{A E}\).

The elastic strain energy stored in a wire obeying Hooke's law is:
\[E_{\text{s}} = \frac{1}{2} F x = \frac{F^2 L}{2 A E}\]

For the second wire, force is \(2F\), length is \(2L\), and area is \(2A\):
\[E_{\text{s}, 2} = \frac{(2F)^2 (2L)}{2 (2A) E} = \frac{4 F^2 \times 2L}{4 A E} = 2 \left(\frac{F^2 L}{A E}\right) = 4 \left(\frac{F^2 L}{2 A E}\right) = 4 E_{\text{s}}\]

C [1 mark]: Correctly determines that strain energy scales as \(\frac{F^2 L}{A}\), giving a factor of \(\frac{2^2 \times 2}{2} = 4\).

The grating spacing \(d\) is:
\[d = \frac{1}{N} = \frac{1}{4.00 \times 10^5\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m}\]

Using the diffraction grating equation \(d \sin\theta = n \lambda\):
For the maximum order \(n_{\text{max}}\), \(\sin\theta \le 1\):
\[n \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}\text{ m}}{600 \times 10^{-9}\text{ m}} \approx 4.17\]

Since \(n\) must be an integer, the highest order is \(n_{\text{max}} = 4\).

The total number of observable maxima is:
\[N_{\text{total}} = 2 n_{\text{max}} + 1 = 2(4) + 1 = 9\]

D [1 mark]: Identifies maximum order \(n=4\) and calculates total maxima as \(2n+1 = 9\).

When the variable resistor \(R_{\text{var}} = 0\text{ }\Omega\):
- Total circuit resistance \(R_{\text{total}, 1} = 1.0 + 5.0 + 0 = 6.0\text{ }\Omega\)
- Circuit current \(I_1 = \frac{12}{6.0} = 2.0\text{ A}\)
- Potential difference across the fixed resistor \(V_1 = I_1 R_{\text{fixed}} = 2.0 \times 5.0 = 10.0\text{ V}\)

When the variable resistor \(R_{\text{var}} = 6.0\text{ }\Omega\):
- Total circuit resistance \(R_{\text{total}, 2} = 1.0 + 5.0 + 6.0 = 12.0\text{ }\Omega\)
- Circuit current \(I_2 = \frac{12}{12.0} = 1.0\text{ A}\)
- Potential difference across the fixed resistor \(V_2 = I_2 R_{\text{fixed}} = 1.0 \times 5.0 = 5.0\text{ V}\)

Change in potential difference:
\[\Delta V = 10.0\text{ V} - 5.0\text{ V} = 5.0\text{ V}\]

C [1 mark]: Correctly calculates initial p.d. (\(10.0\text{ V}\)) and final p.d. (\(5.0\text{ V}\)), giving a change of \(5.0\text{ V}\).

In \(\beta^-\) decay, a neutron converts into a proton:
\[\text{n} \rightarrow \text{p} + \text{e}^- + \bar{\nu}_\text{e}\]

In terms of fundamental quarks:
\[\text{udd} \rightarrow \text{uud} + \text{e}^- + \bar{\nu}_\text{e}\]
Hence, a down quark transforms into an up quark (\(\text{d} \rightarrow \text{u}\)).

The emitted leptons are an electron (\(\text{e}^-\)) and an electron antineutrino (\(\bar{\nu}_\text{e}\)).

D [1 mark]: Correctly identifies the quark change as \(\text{d} \rightarrow \text{u}\) and the emitted leptons as an electron and an electron antineutrino.

Resistivity is defined by \(R = \frac{\rho L}{A}\), so \(\rho = \frac{R A}{L}\). Resistance \(R = \frac{V}{I} = \frac{P}{I^2}\). Power \(P\) in base units is \(\text{kg m}^2 \text{s}^{-3}\), so \(R\) has base units \(\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\). Therefore, \(\rho = \frac{R A}{L}\) has base units \(\frac{(\text{kg m}^2 \text{s}^{-3} \text{A}^{-2})(\text{m}^2)}{\text{m}} = \text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\).

A is correct [1 mark]. B confuses resistance with resistivity. C omits power of length in area/length. D has incorrect time base exponent.

Vertical component of velocity just before impact: \(v_y^2 = u_y^2 + 2 g h = 0 + 2 \times 9.81 \times 45 = 882.9\text{ m}^2\text{ s}^{-2}\). Horizontal component remains constant: \(v_x = 12\text{ m s}^{-1}\). 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 only the horizontal speed. B is only the vertical speed \(\approx 29.7\text{ m s}^{-1}\). D adds the speeds directly.

In a perfectly elastic collision, relative speed of approach equals relative speed of separation: \(u_P - u_Q = v_Q - v_P \implies 3.0 - (-1.5) = 4.5\text{ m s}^{-1}\), so \(v_Q = v_P + 4.5\). By conservation of linear momentum: \(m_P u_P + m_Q u_Q = m_P v_P + m_Q v_Q \implies (0.40)(3.0) + (0.20)(-1.5) = 0.40 v_P + 0.20(v_P + 4.5)\). This gives \(1.20 - 0.30 = 0.60 v_P + 0.90 \implies 0.90 = 0.60 v_P + 0.90 \implies v_P = 0\text{ m s}^{-1}\).

C is correct [1 mark]. A assumes velocities are simply reversed. B and D arise from arithmetic errors in signs.

When completely submerged, the volume of liquid displaced is \(V = A H\). The upward upthrust is \(U = \rho_L A H g\). The downward weight of the cylinder is \(W = \rho_s A H g\). For equilibrium when held submerged: \(F + W = U \implies F = U - W = (\rho_L - \rho_s) A H g\).

B is correct [1 mark]. A ignores the weight of the cylinder. C inverts the density difference. D adds the weight instead of subtracting.

Elastic strain energy is given by \(W = \frac{1}{2} F e\). By definition of Young modulus, \(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{e/L} \implies F = \frac{E A e}{L}\). Substituting for \(F\) gives \(W = \frac{1}{2} \left(\frac{E A e}{L}\right) e = \frac{E A e^2}{2 L}\).

A is correct [1 mark]. B misses the factor of 1/2. C and D have incorrect dimensional forms.

The observed frequency is \(f_o = f_s \left(\frac{v}{v - v_s}\right) = 680 \times \left(\frac{340}{340 - 20}\right) = 680 \times \frac{340}{320} = 722.5\text{ Hz} \approx 723\text{ Hz}\). The observed wavelength is \(\lambda_o = \frac{v - v_s}{f_s} = \frac{340 - 20}{680} = \frac{320}{680} = 0.471\text{ m}\).

B is correct [1 mark]. A uses \(v + v_s\) instead of \(v - v_s\). C and D use unshifted wavelengths or incorrect formulas.

Initial total resistance \(R_1 = 1.5 + 4.5 + 6.0 = 12.0\ \Omega\). Initial current \(I_1 = \frac{12.0\text{ V}}{12.0\ \Omega} = 1.0\text{ A}\). Initial terminal p.d. \(V_1 = E - I_1 r = 12.0 - (1.0 \times 1.5) = 10.5\text{ V}\).

Final total resistance \(R_2 = 1.5 + 4.5 + 2.0 = 8.0\ \Omega\). Final current \(I_2 = \frac{12.0\text{ V}}{8.0\ \Omega} = 1.5\text{ A}\). Final terminal p.d. \(V_2 = E - I_2 r = 12.0 - (1.5 \times 1.5) = 9.75\text{ V}\).

Change in terminal p.d. \(= 9.75 - 10.5 = -0.75\text{ V}\) (a decrease of \(0.75\text{ V}\)).

A is correct [1 mark]. B identifies the magnitude correctly but states an increase. C and D confuse lost volts with terminal p.d.

A neutron has quark composition \(\text{udd}\) and a proton has quark composition \(\text{uud}\). In \(\beta^-\)-decay, a down quark transforms into an up quark (\(\text{d} \rightarrow \text{u}\)). Beta decay is mediated by the weak interaction.

D is correct [1 mark]. A names the strong interaction instead of weak. B and C reverse the quark flavour change.

The ohm is the unit of electrical resistance, defined by \(R = \frac{V}{I}\).

Potential difference is work done per unit charge:
\[V = \frac{W}{Q} = \frac{\text{N}\cdot\text{m}}{\text{A}\cdot\text{s}} = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{m}}{\text{A}\cdot\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}\]

Therefore, the resistance in base units is:
\[R = \frac{V}{I} = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-1}}{\text{A}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}\cdot\text{A}^{-2}\]

A is correct [1 mark].
B is incorrect (has units of \(\text{J A}^{-2}\)).
C is missing the factor of \(\text{m}^2\).
D has incorrect powers for time.

The horizontal component of velocity remains constant:
\[v_x = 15\text{ m s}^{-1}\]

The vertical component of velocity immediately before impact is given by:
\[v_y^2 = u_y^2 + 2gs_y = 0 + 2(9.81)(45) = 882.9\text{ m}^2\text{s}^{-2}\]

The resultant speed \(v\) is:
\[v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 882.9} = \sqrt{225 + 882.9} = \sqrt{1107.9} \approx 33\text{ m s}^{-1}\]

C is correct [1 mark].
A is only the horizontal component \(v_x\).
B is only the vertical component \(v_y \approx 29.7\text{ m s}^{-1} \approx 30\text{ m s}^{-1}\).
D is the height of the cliff.

For an elastic collision between a moving object of mass \(m_1\) and a stationary object of mass \(m_2\):
\[v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) u_1\]

Substituting the given values:
\[v_1 = \left(\frac{0.40 - 0.20}{0.40 + 0.20}\right) \times 3.0 = \left(\frac{0.20}{0.60}\right) \times 3.0 = 1.0\text{ m s}^{-1}\]
Since the sign is positive, the velocity is in the original direction of motion.

A is correct [1 mark].
B has the wrong direction.
C and D represent the velocity of the second block or incorrect algebraic signs.

The volume \(V\) of the cylinder is:
\[V = A \times h = (2.5 \times 10^{-3}\text{ m}^2) \times 0.12\text{ m} = 3.0 \times 10^{-4}\text{ m}^3\]

By Archimedes' principle, the upthrust \(U\) is the weight of the displaced liquid:
\[U = \rho g V = 850 \times 9.81 \times (3.0 \times 10^{-4}) \approx 2.50\text{ N}\]

B is correct [1 mark].
A, C, and D result from powers of 10 or forgetting \(g\).

The extension \(x\) is given by the Young modulus formula:
\[E = \frac{F/A}{x/L} \implies x = \frac{FL}{AE}\]

The elastic potential energy stored within the limit of proportionality is:
\[E_p = \frac{1}{2} F x = \frac{1}{2} F \left(\frac{FL}{AE}\right) = \frac{F^2 L}{2AE}\]

A is correct [1 mark].
B has incorrect power of \(L\).
C has inverted area and length.
D omits the factor of \(1/2\).

Using the Doppler effect formula for a moving source approaching a stationary observer:
\[f_o = f_s \left(\frac{v}{v - v_s}\right)\]

Substitute the values:
\[f_o = 680 \times \left(\frac{340}{340 - 24}\right) = 680 \times \frac{340}{316} \approx 732\text{ Hz}\]

D is correct [1 mark].
A is obtained using \(v + v_s\) in the denominator (source moving away).
B assumes no Doppler shift.
C is an arithmetic slip.

The grating spacing \(d\) is:
\[d = \frac{1}{4.0 \times 10^5\text{ m}^{-1}} = 2.5 \times 10^{-6}\text{ m}\]

Using the diffraction grating equation \(d \sin\theta = n\lambda\) with \(\sin\theta \le 1\):
\[n \le \frac{d}{\lambda} = \frac{2.5 \times 10^{-6}}{5.4 \times 10^{-7}} \approx 4.63\]

Since \(n\) must be an integer, the highest observable order on either side is \(n = 4\).

The total number of maxima is:
\[N = 2n + 1 = 2(4) + 1 = 9\]

D is correct [1 mark].
A gives only the maximum order \(n\).
B gives the number of maxima on one side including the central maximum \(n+1\).
C forgets the central maximum \(2n\).

A neutron has quark composition \(\text{udd}\) and a proton has quark composition \(\text{uud}\). In \(\beta^-\text{ decay}\), one down quark changes into an up quark (\(\text{d} \rightarrow \text{u} + \text{e}^- + \bar{\nu}_e\)).

Quark flavour changes are mediated solely by the weak interaction.

A is correct [1 mark].
B gives the quark change for \(\beta^+\text{ decay}\).
C and D assign incorrect fundamental forces (strong and electromagnetic interactions cannot change quark flavour).

(a)
- Systematic error: Error that causes readings to be consistently higher or lower than the true value by a fixed amount (or predictable ratio).
- Random error: Error that causes readings to vary unpredictably in magnitude and sign about the true value / mean.

(b)
Rearrange the equation for \(\eta\):
\[ \eta = \frac{F_{\text{D}}}{6\pi r v} \]
Base units of \(F_{\text{D}}\) (force): \([F] = \text{kg m s}^{-2}\)
Base units of \(r\) (length): \([r] = \text{m}\)
Base units of \(v\) (speed): \([v] = \text{m s}^{-1}\)
Constant \(6\pi\) is dimensionless.
\[ [\eta] = \frac{\text{kg m s}^{-2}}{\text{m} \times \text{m s}^{-1}} = \frac{\text{kg m s}^{-2}}{\text{m}^2 \text{s}^{-1}} = \text{kg m}^{-1}\text{s}^{-1} \]

(c)(i)
Convert values to SI units:
\(m = 145.0 \times 10^{-3}\text{ kg} = 0.1450\text{ kg}\)
\(r = \frac{d}{2} = \frac{18.4 \times 10^{-3}}{2} = 9.20 \times 10^{-3}\text{ m}\)
\(L = 72.0 \times 10^{-3}\text{ m}\)
Volume \(V = \pi r^2 L = \pi \left(\frac{d}{2}\right)^2 L = \frac{\pi d^2 L}{4}\)
\[ V = \frac{\pi (18.4 \times 10^{-3})^2 (72.0 \times 10^{-3})}{4} = 1.9144 \times 10^{-5}\text{ m}^3 \]
Density \(\rho = \frac{m}{V}\):
\[ \rho = \frac{0.1450}{1.9144 \times 10^{-5}} = 7.574 \times 10^3\text{ kg m}^{-3} \approx 7.6 \times 10^3\text{ kg m}^{-3} \]

(c)(ii)
Formula for density: \(\rho = \frac{4m}{\pi d^2 L}\)
Fractional uncertainty in \(\rho\):
\[ \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta L}{L} \]
Percentage uncertainties:
- \(\frac{\Delta m}{m} = \frac{0.5}{145.0} \times 100\% = 0.345\%\)
- \(2 \times \frac{\Delta d}{d} = 2 \times \frac{0.2}{18.4} \times 100\% = 2.174\%\)
- \(\frac{\Delta L}{L} = \frac{0.5}{72.0} \times 100\% = 0.694\%\)
Total percentage uncertainty:
\[ \frac{\Delta \rho}{\rho} \times 100\% = 0.345\% + 2.174\% + 0.694\% = 3.213\% \approx 3.2\% \]

(c)(iii)
Absolute uncertainty \(\Delta \rho\):
\[ \Delta \rho = 0.03213 \times 7574\text{ kg m}^{-3} = 243\text{ kg m}^{-3} \approx 200\text{ kg m}^{-3} = 0.2 \times 10^3\text{ kg m}^{-3} \]
Density with uncertainty:
\[ \rho = (7.6 \pm 0.2) \times 10^3\text{ kg m}^{-3} \]

(d)
Weight \(W = mg = 0.1450 \times 9.81 = 1.422\text{ N}\)
Let \(T_1\) be the tension in the string at \(35^\circ\) and \(T_2\) be the tension in the string at \(55^\circ\).
Horizontal equilibrium:
\[ T_1 \cos 35^\circ = T_2 \cos 55^\circ \implies T_1 = T_2 \frac{\cos 55^\circ}{\cos 35^\circ} = 0.7002 T_2 \]
Vertical equilibrium:
\[ T_1 \sin 35^\circ + T_2 \sin 55^\circ = W \]
Substitute \(T_1\):
\[ (0.7002 T_2) \sin 35^\circ + T_2 \sin 55^\circ = 1.422 \]
\[ T_2 (0.4016 + 0.8192) = 1.422 \implies 1.2208 T_2 = 1.422 \]
\[ T_2 = 1.165\text{ N} \approx 1.17\text{ N} \]
\[ T_1 = 0.7002 \times 1.165 = 0.8158\text{ N} \approx 0.816\text{ N} \]

(a) [2 marks]
- B1: Systematic error: reading differs from the true value by a constant amount / error in one direction (e.g. zero error).
- B1: Random error: readings scatter unpredictably about the true/mean value in both directions.

(b) [3 marks]
- C1: Base units of force = kg m s^-2
- C1: Base units of r = m, v = m s^-1
- A1: Correctly rearranged and combined: kg m^-1 s^-1

(c)(i) [1 mark]
- A1: Correct substitution of values into V = pi d^2 L / 4 and rho = m/V to give 7.57 x 10^3 kg m^-3 (minimum 3 s.f. shown).

(c)(ii) [2 marks]
- C1: Recognition that % uncertainty in d must be multiplied by 2: % uncertainty = (0.5/145.0) + 2*(0.2/18.4) + (0.5/72.0)
- A1: 3.2% (or 3.21%)

(c)(iii) [1 mark]
- A1: (7.6 +/- 0.2) x 10^3 kg m^-3 (both value and uncertainty given to 2 s.f. and matching decimal place).

(d) [3 marks]
- C1: W = 0.1450 x 9.81 = 1.42 N (or 1.4 N)
- C1: Correct resolution of forces horizontally and vertically (e.g. T1 cos 35 = T2 cos 55 and T1 sin 35 + T2 sin 55 = W)
- A1: T1 = 0.816 N (or 0.82 N) and T2 = 1.17 N (or 1.2 N)

(a)
The total linear momentum of a system of interacting bodies remains constant provided no external resultant force acts on the system.

(b)(i)
Apply the principle of conservation of momentum:
\[ m_{\text{A}} u_{\text{A}} + m_{\text{B}} u_{\text{B}} = (m_{\text{A}} + m_{\text{B}}) V \]
\[ (0.450)(1.60) + (0.750)(0) = (0.450 + 0.750) V \]
\[ 0.720 = 1.200 V \implies V = \frac{0.720}{1.200} = 0.600\text{ m s}^{-1} \]

(b)(ii)
Initial total kinetic energy \(E_{\text{k, initial}}\):
\[ E_{\text{k, initial}} = \frac{1}{2} m_{\text{A}} u_{\text{A}}^2 = \frac{1}{2}(0.450)(1.60)^2 = 0.576\text{ J} \]
Final total kinetic energy \(E_{\text{k, final}}\):
\[ E_{\text{k, final}} = \frac{1}{2} (m_{\text{A}} + m_{\text{B}}) V^2 = \frac{1}{2}(1.200)(0.600)^2 = 0.216\text{ J} \]
Loss in kinetic energy:
\[ \Delta E_{\text{k}} = 0.576 - 0.216 = 0.360\text{ J} \]
Since kinetic energy is not conserved (\(\Delta E_{\text{k}} > 0\)), the collision is inelastic.

(b)(iii)
The lost kinetic energy is converted into internal thermal energy / heat (warming the trucks) and sound energy / work done in permanent deformation.

(c)(i)
Change in momentum of truck \(\text{B}\):
\[ \Delta p_{\text{B}} = m_{\text{B}} V - 0 = (0.750)(0.600) = 0.450\text{ kg m s}^{-1} \]
Average force:
\[ F_{\text{avg}} = \frac{\Delta p_{\text{B}}}{\Delta t} = \frac{0.450}{48.0 \times 10^{-3}} = 9.375\text{ N} \approx 9.38\text{ N} \]

(c)(ii)
By Newton's third law, the force on truck \(\text{A}\) has the same magnitude: \(9.38\text{ N}\).
Its direction is opposite to the motion of truck \(\text{A}\) (backwards / in the opposite direction to the force on truck \(\text{B}\)).

(a) [2 marks]
- B1: Total momentum remains constant / is conserved
- B1: In an isolated / closed system / when no external force acts

(b)(i) [2 marks]
- C1: (0.450)(1.60) = (0.450 + 0.750)V
- A1: V = 0.600 m s^-1

(b)(ii) [3 marks]
- C1: Initial KE = 0.5 x 0.450 x 1.60^2 = 0.576 J
- C1: Final KE = 0.5 x 1.200 x 0.600^2 = 0.216 J
- A1: Loss in KE = 0.360 J and statement that KE is not conserved so collision is inelastic

(b)(iii) [1 mark]
- B1: Converted to thermal energy (heat) and/or sound energy (or work done in deformation)

(c)(i) [2 marks]
- C1: F = Delta p / Delta t = (0.750 x 0.600) / (48.0 x 10^-3)
- A1: F = 9.38 N (accept 9.4 N)

(c)(ii) [2 marks]
- B1: Magnitude = 9.38 N (allow ECF from (c)(i))
- B1: Direction = Opposite to initial direction of truck A / backwards

(a)
Young modulus is defined as the ratio of tensile stress to tensile strain (or \(\text{stress} / \text{strain}\)).

(b)(i)
Radius \(r = \frac{d}{2} = \frac{0.800 \times 10^{-3}}{2} = 4.00 \times 10^{-4}\text{ m}\)
Cross-sectional area \(A = \pi r^2 = \pi (4.00 \times 10^{-4})^2 = 5.0265 \times 10^{-7}\text{ m}^2 \approx 5.03 \times 10^{-7}\text{ m}^2\).

(b)(ii)
Tensile stress \(\sigma = \frac{F}{A}\):
\[ \sigma = \frac{65.0}{5.0265 \times 10^{-7}} = 1.293 \times 10^8\text{ Pa} \approx 1.29 \times 10^8\text{ Pa} \]

(b)(iii)
Young modulus \(E = \frac{\sigma}{\varepsilon} = \frac{F L}{A x}\)
\[ x = \frac{F L}{A E} = \frac{(65.0)(2.40)}{(5.0265 \times 10^{-7})(2.10 \times 10^{11})} = \frac{156.0}{1.0556 \times 10^5} = 1.478 \times 10^{-3}\text{ m} \approx 1.48 \times 10^{-3}\text{ m} \text{ (or } 1.48\text{ mm)} \]

(b)(iv)
Elastic potential energy \(E_{\text{p}} = \frac{1}{2} F x\):
\[ E_{\text{p}} = \frac{1}{2}(65.0)(1.478 \times 10^{-3}) = 4.8035 \times 10^{-2}\text{ J} \approx 4.80 \times 10^{-2}\text{ J} \]

(c)(i)
Mass of the block \(m = \frac{W}{g} = \frac{65.0}{9.81} = 6.626\text{ kg}\)
Volume of the block \(V = \frac{m}{\rho_{\text{metal}}} = \frac{6.626}{7.85 \times 10^3} = 8.441 \times 10^{-4}\text{ m}^3\)
Upthrust \(U = \rho_{\text{water}} g V\):
\[ U = (1.00 \times 10^3)(9.81)(8.441 \times 10^{-4}) = 8.281\text{ N} \approx 8.28\text{ N} \]
(Alternatively: \(U = W \times \frac{\rho_{\text{water}}}{\rho_{\text{metal}}} = 65.0 \times \frac{1.00 \times 10^3}{7.85 \times 10^3} = 8.28\text{ N}\))

(c)(ii)
New tension in the wire \(T = W - U = 65.0 - 8.281 = 56.72\text{ N}\)
Since extension is directly proportional to tension:
\[ x_{\text{new}} = x \times \frac{T}{W} = 1.478 \times 10^{-3} \times \frac{56.72}{65.0} = 1.290 \times 10^{-3}\text{ m} \approx 1.29 \times 10^{-3}\text{ m} \text{ (or } 1.29\text{ mm)} \]

(a) [1 mark]
- B1: (Tensile) stress divided by (tensile) strain

(b)(i) [1 mark]
- A1: A = pi * (0.400 x 10^-3)^2 = 5.03 x 10^-7 m^2 (showing working with values)

(b)(ii) [2 marks]
- C1: stress = F / A = 65.0 / (5.03 x 10^-7)
- A1: stress = 1.29 x 10^8 Pa

(b)(iii) [2 marks]
- C1: x = FL / (AE) or strain = stress / E = 1.293 x 10^8 / 2.10 x 10^11 = 6.16 x 10^-4
- A1: x = 1.48 x 10^-3 m (or 1.48 mm)

(b)(iv) [2 marks]
- C1: Ep = 1/2 * F * x = 0.5 * 65.0 * 1.48 x 10^-3
- A1: Ep = 4.80 x 10^-2 J (or 0.0480 J)

(c)(i) [2 marks]
- C1: V = 65.0 / (7850 x 9.81) = 8.44 x 10^-4 m^3 (or ratio rho_water / rho_metal = 1.00/7.85)
- A1: Upthrust = rho_water * g * V = 8.28 N

(c)(ii) [2 marks]
- C1: New tension = 65.0 - 8.28 = 56.7 N
- A1: x_new = 1.29 x 10^-3 m (or 1.29 mm)

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

(b)(i)
Using the Doppler effect formula for a moving source approaching a stationary observer:
\[ f_{\text{o}} = f_{\text{s}} \left(\frac{v}{v - v_{\text{s}}}\right) \]
Substitute given values:
\[ f_{\text{o}} = 850 \left(\frac{340}{340 - 28.0}\right) = 850 \left(\frac{340}{312}\right) = 926.28\text{ Hz} \approx 926\text{ Hz} \]

(b)(ii)
As the ambulance passes the observer, the frequency drops abruptly from \(926\text{ Hz}\) to a value less than the source frequency:
\[ f_{\text{o, away}} = 850 \left(\frac{340}{340 + 28.0}\right) = 850 \left(\frac{340}{368}\right) = 785\text{ Hz} \]
The frequency remains constant at this lower value as it continues to move away at constant speed.

(c)(i)
Grating spacing \(d\):
\[ d = \frac{1}{N} = \frac{1}{4.00 \times 10^5\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m} \]

(c)(ii)
Diffraction grating equation:
\[ d \sin \theta = n \lambda \]
For \(n = 2\):
\[ \sin \theta = \frac{2 \lambda}{d} = \frac{2(532 \times 10^{-9})}{2.50 \times 10^{-6}} = \frac{1.064 \times 10^{-6}}{2.50 \times 10^{-6}} = 0.4256 \]
\[ \theta = \arcsin(0.4256) = 25.187^\circ \approx 25.2^\circ \]

(c)(iii)
For maximum observable order \(n_{\text{max}}\), \(\sin \theta \le 1\):
\[ n_{\text{max}} \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}}{532 \times 10^{-9}} = 4.699 \]
Since \(n\) must be an integer, \(n_{\text{max}} = 4\).
Total number of visible maxima:
\[ N_{\text{total}} = 2 n_{\text{max}} + 1 = 2(4) + 1 = 9 \]
(Orders: \(n = 0, \pm 1, \pm 2, \pm 3, \pm 4\)).

(d)
Using the double-slit formula \(\lambda = \frac{a x}{D}\):
\[ x = \frac{\lambda D}{a} = \frac{(532 \times 10^{-9})(1.80)}{0.350 \times 10^{-3}} = \frac{9.576 \times 10^{-7}}{3.50 \times 10^{-4}} = 2.736 \times 10^{-3}\text{ m} \approx 2.74\text{ mm} \]

(a) [1 mark]
- B1: Constant phase difference (do not accept 'same phase')

(b)(i) [2 marks]
- C1: f_o = f_s * v / (v - v_s) = 850 * 340 / (340 - 28.0)
- A1: f_o = 926 Hz

(b)(ii) [2 marks]
- B1: Observed frequency decreases / drops below 850 Hz
- B1: Frequency becomes constant at 785 Hz (or lower pitch remains steady)

(c)(i) [1 mark]
- A1: d = 1 / 4.00 x 10^5 = 2.50 x 10^-6 m

(c)(ii) [2 marks]
- C1: sin(theta) = 2 * (532 x 10^-9) / (2.50 x 10^-6) = 0.4256
- A1: theta = 25.2° (or 0.440 rad)

(c)(iii) [2 marks]
- C1: n_max = d / lambda = 2.50 x 10^-6 / 532 x 10^-9 = 4.70 -> n_max = 4
- A1: Total number = 2(4) + 1 = 9

(d) [2 marks]
- C1: x = lambda * D / a = (532 x 10^-9 * 1.80) / (0.350 x 10^-3)
- A1: x = 2.74 x 10^-3 m (or 2.74 mm)

(a)
- Kirchhoff's second law: The algebraic sum of e.m.f.s around any closed loop in a circuit is equal to the algebraic sum of potential differences (p.d.s) around that loop (
\(\sum E = \sum V\)).
- Physical law: Conservation of energy.

(b)(i)
For the parallel combination \(R_2\) and \(R_3\):
\[ \frac{1}{R_{\text{p}}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{12.0} + \frac{1}{24.0} = \frac{2 + 1}{24.0} = \frac{3}{24.0} \implies R_{\text{p}} = 8.00\ \Omega \]
Total external resistance:
\[ R_{\text{ext}} = R_1 + R_{\text{p}} = 6.80 + 8.00 = 14.80\ \Omega \]

(b)(ii)
Total circuit resistance:
\[ R_{\text{total}} = R_{\text{ext}} + r = 14.80 + 1.20 = 16.00\ \Omega \]
Total current supplied by battery:
\[ I = \frac{E}{R_{\text{total}}} = \frac{9.00}{16.00} = 0.5625\text{ A} \approx 0.563\text{ A} \]

(b)(iii)
Terminal p.d. \(V = E - I r\):
\[ V = 9.00 - (0.5625)(1.20) = 9.00 - 0.675 = 8.325\text{ V} \approx 8.33\text{ V} \]
(Or \(V = I R_{\text{ext}} = 0.5625 \times 14.80 = 8.325\text{ V}\)).

(b)(iv)
P.d. across the parallel combination \(V_{\text{p}} = I R_{\text{p}} = (0.5625)(8.00) = 4.50\text{ V}\).
Power dissipated in \(R_2\):
\[ P_2 = \frac{V_{\text{p}}^2}{R_2} = \frac{(4.50)^2}{12.0} = \frac{20.25}{12.0} = 1.6875\text{ W} \approx 1.69\text{ W} \]

(c)(i)
The nuclear decay equation is:
\[ ^{14}_{6}\text{C} \longrightarrow {^{14}_{7}\text{N}} + {^{0}_{-1}\text{e}} + \bar{\nu}_{\text{e}} \]
where \(^{0}_{-1}\text{e}\) (or \(\beta^-\)) is an electron and \(\bar{\nu}_{\text{e}}\) is an electron antineutrino.

(c)(ii)
- Quark composition of proton: \(\text{uud}\)
- Quark composition of neutron: \(\text{udd}\)
- During \(\beta^-\)-decay, a down quark transforms into an up quark, emitting an electron (\(\beta^-\)) and an electron antineutrino:
\[ \text{d} \longrightarrow \text{u} + \text{e}^- + \bar{\nu}_{\text{e}} \]

(a) [2 marks]
- B1: Sum of e.m.f.s = sum of p.d.s around a closed loop
- B1: Conservation of energy

(b)(i) [1 mark]
- A1: 1/Rp = 1/12 + 1/24 => Rp = 8.0 Omega; R_ext = 6.8 + 8.0 = 14.8 Omega

(b)(ii) [1 mark]
- A1: I = E / (R_ext + r) = 9.00 / 16.0 = 0.563 A (or 0.5625 A)

(b)(iii) [2 marks]
- C1: V = E - Ir or V = I * R_ext
- A1: V = 8.33 V (or 8.325 V)

(b)(iv) [2 marks]
- C1: V_parallel = 0.5625 * 8.00 = 4.50 V (or I_2 = 0.5625 * 24/36 = 0.375 A)
- A1: P = (4.50)^2 / 12.0 = 1.69 W (or 1.688 W)

(c)(i) [2 marks]
- B1: Correct symbols and nucleon/proton numbers for ^14_7N and ^0_-1e (or beta^-)
- B1: Inclusion of electron antineutrino (bar{nu}_e or bar{nu})

(c)(ii) [2 marks]
- B1: Proton = uud and Neutron = udd
- B1: Down quark changes to up quark (d -> u) (with emission of beta^- and bar{nu}_e)

Worked solution outline:

(a) (i) Measure \(d\) with a metre rule to the nearest millimetre, e.g., \(d = 0.400\text{ m}\) (or \(40.0\text{ cm}\)).
(ii) Time at least 10 or 20 complete oscillations (e.g., \(t_1 = 28.42\text{ s}\), \(t_2 = 28.50\text{ s}\), mean \(t = 28.46\text{ s}\) for 20 oscillations) to find \(T = 1.423\text{ s}\).

(b) Collect 6 sets of readings over a wide range of \(d\) (e.g. \(0.15\text{ m} \le d \le 0.70\text{ m}\)).
Record raw times with repeats, raw \(d\) to \(1\text{ mm}\), calculated \(T\), \(d^{-2}\) and \(T^2\).
Ensure correct column headings with units: \(d / \text{m}\), \(t_1 / \text{s}\), \(t_2 / \text{s}\), \(t_{\text{mean}} / \text{s}\), \(T / \text{s}\), \(d^{-2} / \text{m}^{-2}\), \(T^2 / \text{s}^2\).

(c) (i) Plot \(T^2\) against \(d^{-2}\). Scales must be linear, points occupy more than half the grid in both dimensions, plots accurate to within half a small square.
(ii) Line of best fit drawn straight with balanced scatter.
(iii) Gradient \(= \frac{\Delta (T^2)}{\Delta (d^{-2})}\) using points on the line where \(\Delta x\) is greater than half the length of the drawn line.
\(y\)-intercept determined either by direct reading from the vertical axis where \(d^{-2} = 0\) or calculated via \(y - mx\).

(d) Comparing \(T^2 = P(d^{-2}) + Q\) with \(y = mx + c\):
\(P = \text{gradient}\) with units \(\text{s}^2\,\text{m}^2\).
\(Q = y\text{-intercept}\) with units \(\text{s}^2\).

Marking breakdown (Total: 20 marks):

(a) [3 marks]
- Value of \(d\) recorded to the nearest millimetre with correct unit (B1)
- Measure time for at least 10 oscillations, repeated and averaged, raw time to at least 0.1 s (B1)
- Period \(T\) correctly calculated from raw time and number of oscillations \(N\) (B1)

(b) [7 marks]
- Successful collection of at least 6 sets of data without assistance (B1)
- Range: highest and lowest values of \(d\) differ by at least a factor of 2 or spanning at least 40 cm (B1)
- Table headings: each column clearly labelled with quantity and correct unit separated by a solidus (e.g., \(d^{-2} / \text{m}^{-2}\), \(T^2 / \text{s}^2\)) (B1)
- Raw data consistency: all values of \(d\) to same number of decimal places (nearest mm), raw times to 0.1 s or 0.01 s (B1)
- Significant figures: values of \(d^{-2}\) and \(T^2\) given to the same number of (or one more than) the s.f. of the raw data (B1)
- Calculation of \(d^{-2}\) and \(T^2\) correct across all table rows (B2)

(c) [6 marks]
- Axes: correctly labelled with quantity and unit, sensible scale with no awkward multiples (e.g., 3, 7, 9) and points occupy \(> 50\%\) of grid in both directions (B1)
- Plotting: all points plotted correctly within \(\pm 0.5\) small square (B1)
- Line of best fit: drawn with thin, continuous straight line, well-balanced point distribution (B1)
- Gradient calculation: hypotenuse of read-off triangle \(\ge 50\%\) of plotted point range, coordinates read correctly within 0.5 small square, correct arithmetic (B1)
- \(y\)-intercept: read directly from \(x = 0\) axis or calculated using \(c = y - mx\) using a point on the line (B2)

(d) [4 marks]
- Equates \(P\) to gradient and \(Q\) to \(y\)-intercept (B1)
- Value of \(P\) calculated with correct s.f. (2 or 3 s.f.) (B1)
- Value of \(Q\) calculated with correct s.f. (B1)
- Correct units for both constants: \(P\) in \(\text{s}^2\,\text{m}^2\) and \(Q\) in \(\text{s}^2\) (B1)

(a) Gravitational potential at a point is defined as the work done per unit mass in bringing a small test mass / point mass from infinity to that point.

(b)(i) For an object of mass $m$ escaping from the surface to infinity:
$\frac{1}{2} m v_{\text{esc}}^2 = \frac{G M m}{R}$
$v_{\text{esc}} = \sqrt{\frac{2 G M}{R}}$
$v_{\text{esc}} = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (4.80 \times 10^{18})}{8.50 \times 10^4}} = \sqrt{\frac{6.4032 \times 10^8}{8.50 \times 10^4}} = \sqrt{7533.18} = 86.79\text{ m s}^{-1} \approx 86.8\text{ m s}^{-1}$.

(b)(ii) By conservation of energy:
Total energy at surface = Total energy at maximum distance $(R + h)$:
$\frac{1}{2} m u^2 - \frac{G M m}{R} = 0 - \frac{G M m}{R + h}$
$\frac{1}{2} u^2 = G M \left( \frac{1}{R} - \frac{1}{R + h} \right)$
$\frac{1}{2} (60.0)^2 = (6.67 \times 10^{-11} \times 4.80 \times 10^{18}) \left( \frac{1}{8.50 \times 10^4} - \frac{1}{R + h} \right)$
$1800 = (3.2016 \times 10^8) \left( 1.17647 \times 10^{-5} - \frac{1}{R + h} \right)$
$5.6222 \times 10^{-6} = 1.17647 \times 10^{-5} - \frac{1}{R + h}$
$\frac{1}{R + h} = 6.1425 \times 10^{-6}\text{ m}^{-1}$
$R + h = 1.6280 \times 10^5\text{ m}$
$h = (1.6280 \times 10^5) - (8.50 \times 10^4) = 7.780 \times 10^4\text{ m} = 7.78 \times 10^4\text{ m}$ (or $7.79 \times 10^4\text{ m}$ depending on rounding).

(c) As the projectile moves through the atmosphere, work is done against drag / air resistance. This converts some of the projectile's kinetic energy into thermal energy, so a higher initial kinetic energy / launch speed is required.

(a)
- Work done per unit mass [M1]
- In bringing a point mass / test mass from infinity (to the point) [A1]

(b)(i)
- Equates initial kinetic energy to gravitational potential energy at surface: $\frac{1}{2} m v^2 = \frac{GMm}{R}$ OR states $v = \sqrt{2GM/R}$ [C1]
- Correct substitution with unrounded/intermediate values: $\sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 4.80 \times 10^{18}}{8.50 \times 10^4}}$ [M1]
- Correct calculation leading explicitly to $86.79\text{ m s}^{-1}$ shown to at least 3 s.f. before rounding to $86.8\text{ m s}^{-1}$ [A1]

(b)(ii)
- Equates loss in kinetic energy to gain in gravitational potential energy [C1]
- Correct substitution: $\frac{1}{2}(60.0)^2 = (6.67 \times 10^{-11})(4.80 \times 10^{18})(\frac{1}{8.50 \times 10^4} - \frac{1}{r})$ [C1]
- $h = 7.78 \times 10^4\text{ m}$ or $7.79 \times 10^4\text{ m}$ (allow $7.8 \times 10^4\text{ m}$) [A1]

(c)
- Work is done against atmospheric friction / drag [B1]
- Kinetic energy is dissipated / transferred to thermal energy (so more initial energy/speed is needed) [B1]

(a) Any two of:
1. Molecules are in continuous, random motion with a distribution of speeds and directions.
2. Molecules move in straight lines between collisions.
3. The time of a collision is negligible compared to the time between collisions.

(b)(i) $E_{\text{k}} = \frac{3}{2} k T$, so mean translational kinetic energy is directly proportional to thermodynamic temperature $T$.

(b)(ii) 1. Using $p V = n R T$:
$p = \frac{n R T}{V} = \frac{1.80 \times 8.31 \times 295}{0.0450} = 9.8058 \times 10^4\text{ Pa} = 9.81 \times 10^4\text{ Pa}$.

2. For a monatomic ideal gas, internal energy is entirely the kinetic energy of the molecules:
$U = N E_{\text{k}} = n N_{\text{A}} \left( \frac{3}{2} k T \right) = \frac{3}{2} n R T$
$U = \frac{3}{2} \times 1.80 \times 8.31 \times 295 = 6.6189 \times 10^3\text{ J} = 6.62 \times 10^3\text{ J}$.

(c) As temperature increases, the root-mean-square speed / mean kinetic energy of the molecules increases. Consequently, the molecules collide with the container walls more frequently (greater rate of collision) and with a greater change in momentum per collision. Therefore, the average force exerted per unit area (pressure) increases.

(a)
- Random motion (with a range of speeds and directions) [B1]
- Negligible collision time compared to time between collisions / motion in straight lines between collisions [B1]

(b)(i)
- $E_{\text{k}} = \frac{3}{2} k T$ or $E_{\text{k}} \propto T$ where $T$ is in kelvin / thermodynamic temperature [B1]

(b)(ii) 1.
- $p = \frac{nRT}{V}$ formula used [C1]
- $p = \frac{1.80 \times 8.31 \times 295}{0.0450} = 9.81 \times 10^4\text{ Pa}$ (allow $9.8 \times 10^4\text{ Pa}$) [A1]

(b)(ii) 2.
- $U = \frac{3}{2} nRT$ or $U = \frac{3}{2} pV$ [C1]
- Substitution: $\frac{3}{2} \times 1.80 \times 8.31 \times 295$ or $\frac{3}{2} \times (9.81 \times 10^4) \times 0.0450$ [C1]
- $U = 6.62 \times 10^3\text{ J}$ (allow $6.61 \times 10^3\text{ J}$ or $6.6 \times 10^3\text{ J}$) [A1]

(c)
- (Mean) speed / kinetic energy of molecules increases [B1]
- Greater change in momentum per collision AND greater frequency of collisions with walls (resulting in greater force per unit area) [B1]

(a) 1. Acceleration is directly proportional to displacement from the equilibrium position.
2. Acceleration is always directed towards a fixed point / equilibrium position (in the opposite direction to displacement).

(b)(i) $\omega = 2 \pi f = 2 \pi \times 2.40 = 15.08\text{ rad s}^{-1} \approx 15.1\text{ rad s}^{-1}$.

(b)(ii) $a_0 = \omega^2 x_0$
$a_0 = (15.08)^2 \times (3.20 \times 10^{-2}) = 227.4 \times 0.0320 = 7.277\text{ m s}^{-2} \approx 7.28\text{ m s}^{-2}$.

(b)(iii) $E_{\text{total}} = \frac{1}{2} m \omega^2 x_0^2 = \frac{1}{2} m v_0^2$
$E_{\text{total}} = \frac{1}{2} \times 0.350 \times (15.08)^2 \times (0.0320)^2 = \frac{1}{2} \times 0.350 \times 227.4 \times 0.001024 = 0.04075\text{ J} \approx 0.0408\text{ J}$ (or $0.0407\text{ J}$).

(c)(i) Critical damping is the minimum damping required for an oscillating system to return to its equilibrium position in the minimum possible time without overshooting / oscillating.

(c)(ii) Graph starts at $(0, +3.20\text{ cm})$ with zero initial gradient, drops rapidly and asymptotically approaches $x = 0$ without crossing the $t$-axis.

(a)
- Acceleration is directly proportional to displacement (from equilibrium) [B1]
- Acceleration is directed towards the equilibrium position / fixed point (or in opposite direction to displacement) [B1]

(b)(i)
- $\omega = 2 \pi f = 2 \pi \times 2.40 = 15.1\text{ rad s}^{-1}$ [A1]

(b)(ii)
- Uses $a_0 = \omega^2 x_0$ [C1]
- $a_0 = (15.08)^2 \times 0.0320 = 7.28\text{ m s}^{-2}$ (allow $7.3\text{ m s}^{-2}$) [A1]

(b)(iii)
- Uses $E = \frac{1}{2} m \omega^2 x_0^2$ or $E = \frac{1}{2} k x_0^2$ [C1]
- $E_{\text{total}} = 0.0407\text{ J}$ or $0.0408\text{ J}$ [A1]

(c)(i)
- System returns to equilibrium in the shortest/minimum possible time [B1]
- Without any oscillation / without overshooting [B1]

(c)(ii)
- Curve starts at $x = +3.20\text{ cm}$ at $t = 0$ with zero gradient and approaches $x = 0$ smoothly without crossing the time axis [B1]

(a) Electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point.

(b)(i) Electric potential is a scalar quantity:
$V = \frac{1}{4 \pi \varepsilon_0} \left( \frac{Q_A}{r_A} + \frac{Q_B}{r_B} \right)$
$V = (8.99 \times 10^9) \left( \frac{+6.40 \times 10^{-9}}{0.100} + \frac{-3.20 \times 10^{-9}}{0.050} \right)$
$V = (8.99 \times 10^9) (6.40 \times 10^{-8} - 6.40 \times 10^{-8}) = 0\text{ V}$.

(b)(ii) Work done by field $W = q (V_{\text{initial}} - V_{\text{final}}) = q (V_{\text{P}} - V_{\infty})$.
Since $V_{\text{P}} = 0\text{ V}$ and $V_{\infty} = 0\text{ V}$:
$W = (+3.20 \times 10^{-19}) \times (0 - 0) = 0\text{ J}$.

(c)(i) Electric field strength is a vector quantity. At point P, the field $E_A$ due to positive charge $Q_A$ points away from $Q_A$ (towards $Q_B$). The field $E_B$ due to negative charge $Q_B$ points towards $Q_B$. Both fields act in the same direction (from $Q_A$ towards $Q_B$):
$E_A = \frac{Q_A}{4 \pi \varepsilon_0 r_A^2} = \frac{(8.99 \times 10^9) \times (6.40 \times 10^{-9})}{(0.100)^2} = \frac{57.536}{0.0100} = 5.754 \times 10^3\text{ N C}^{-1}$
$E_B = \frac{|Q_B|}{4 \pi \varepsilon_0 r_B^2} = \frac{(8.99 \times 10^9) \times (3.20 \times 10^{-9})}{(0.050)^2} = \frac{28.768}{0.0025} = 1.1507 \times 10^4\text{ N C}^{-1}$
$E_{\text{resultant}} = E_A + E_B = 5.754 \times 10^3 + 1.1507 \times 10^4 = 1.726 \times 10^4\text{ N C}^{-1} \approx 1.73 \times 10^4\text{ N C}^{-1}$.

(c)(ii) Directed along the line from $Q_A$ towards $Q_B$ (or towards $Q_B$).

(a)
- Work done per unit positive charge [M1]
- In bringing a (small test) charge from infinity (to the point) [A1]

(b)(i)
- Uses $V = \frac{Q}{4 \pi \varepsilon_0 r}$ for both charges [C1]
- Correct substitution: $(8.99 \times 10^9) \times (\frac{6.40 \times 10^{-9}}{0.100} - \frac{3.20 \times 10^{-9}}{0.050})$ [M1]
- $V = 0\text{ V}$ [A1]

(b)(ii)
- $W = q \Delta V$ or $W = q(V_P - V_\infty)$ with $V_P = 0$ and $V_\infty = 0$ [M1]
- $W = 0\text{ J}$ [A1]

(c)(i)
- Calculates either $E_A = 5.75 \times 10^3\text{ N C}^{-1}$ or $E_B = 1.15 \times 10^4\text{ N C}^{-1}$ [C1]
- Adds the two field strengths together: $E = E_A + E_B$ [M1]
- $E = 1.73 \times 10^4\text{ N C}^{-1}$ (or $\text{V m}^{-1}$) [A1]

(c)(ii)
- From $Q_A$ to $Q_B$ / towards $Q_B$ / to the right [B1]

(a) The magnitude of the induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage.

(b)(i) Area of the coil $A = \pi r^2 = \pi \times (3.50 \times 10^{-2})^2 = 3.8485 \times 10^{-3}\text{ m}^2$.
Flux linkage $\Phi_{\text{total}} = N B A = 140 \times 0.0650 \times (3.8485 \times 10^{-3}) = 3.502 \times 10^{-2}\text{ Wb} \approx 3.50 \times 10^{-2}\text{ Wb}$.

(b)(ii) Using $E = \frac{\Delta (N \Phi)}{\Delta t}$:
$E = \frac{3.502 \times 10^{-2}}{0.0500} = 0.7004\text{ V} \approx 0.700\text{ V}$.

(c)(i) Hall voltage formula: $V_{\text{H}} = \frac{B I}{n t q}$
Rearranging for $n$ (with $q = e = 1.60 \times 10^{-19}\text{ C}$):
$n = \frac{B I}{V_{\text{H}} t e}$
$n = \frac{0.420 \times (18.0 \times 10^{-3})}{(1.45 \times 10^{-3}) \times (0.250 \times 10^{-3}) \times (1.60 \times 10^{-19})}$
$n = \frac{7.56 \times 10^{-3}}{5.80 \times 10^{-26}} = 1.303 \times 10^{21}\text{ m}^{-3} \approx 1.30 \times 10^{21}\text{ m}^{-3}$.

(c)(ii) Metals / copper have a much larger number density of charge carriers $n$ (free electrons) compared to semiconductors. Since $V_{\text{H}} \propto \frac{1}{n}$, a much larger value of $n$ produces a significantly smaller Hall voltage.

(a)
- Induced e.m.f. is directly proportional to [M1]
- Rate of change of magnetic flux linkage [A1]

(b)(i)
- Calculates area: $A = \pi \times (0.0350)^2 = 3.848 \times 10^{-3}\text{ m}^2$ [C1]
- Correct substitution leading to $3.502 \times 10^{-2}\text{ Wb}$ explicitly before rounding to $3.50 \times 10^{-2}\text{ Wb}$ [A1]

(b)(ii)
- Uses $E = \frac{\Delta (N\Phi)}{\Delta t}$ [C1]
- $E = \frac{3.50 \times 10^{-2}}{0.0500} = 0.700\text{ V}$ [A1]

(c)(i)
- Uses $n = \frac{BI}{V_H t q}$ [C1]
- Correct substitution of values with $e = 1.60 \times 10^{-19}\text{ C}$ [M1]
- $n = 1.30 \times 10^{21}\text{ m}^{-3}$ [A1]

(c)(ii)
- In a metal/copper, charge carrier density $n$ is much greater than in a semiconductor [B1]
- $V_{\text{H}} \propto \frac{1}{n}$ (for constant $B, I, t$), so $V_{\text{H}}$ is much smaller [B1]

(a) The work function energy is the minimum amount of energy needed to release/emit an electron from the surface of a metal.

(b)(i) $E = \frac{h c}{\lambda}$
$E = \frac{(6.63 \times 10^{-34}) \times (3.00 \times 10^8)}{260 \times 10^{-9}} = \frac{1.989 \times 10^{-25}}{2.60 \times 10^{-7}} = 7.650 \times 10^{-19}\text{ J}$
Converting to $\text{eV}$:
$E = \frac{7.650 \times 10^{-19}}{1.60 \times 10^{-19}} = 4.781\text{ eV} \approx 4.78\text{ eV}$.

(b)(ii) Using Einstein's photoelectric equation:
$h f = \Phi + E_{\text{k,max}}$
$E_{\text{k,max}} = E - \Phi = 4.781\text{ eV} - 3.68\text{ eV} = 1.101\text{ eV}$
In joules:
$E_{\text{k,max}} = 1.101 \times (1.60 \times 10^{-19}) = 1.762 \times 10^{-19}\text{ J} \approx 1.76 \times 10^{-19}\text{ J}$.

(b)(iii) $E_{\text{k}} = \frac{p^2}{2 m_{\text{e}}} \implies p = \sqrt{2 m_{\text{e}} E_{\text{k}}}$
$p = \sqrt{2 \times (9.11 \times 10^{-31}) \times (1.762 \times 10^{-19})} = \sqrt{3.210 \times 10^{-49}} = 5.666 \times 10^{-25}\text{ kg m s}^{-1}$
$\lambda_{\text{dB}} = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.666 \times 10^{-25}} = 1.170 \times 10^{-9}\text{ m} = 1.17\text{ nm}$.

(c) There is no change to the maximum kinetic energy. Doubling the intensity increases the number of photons incident per unit time, but does not change the energy of individual photons ($E = hf$), which is determined only by the frequency/wavelength.

(a)
- Minimum energy required [M1]
- To remove/emit an electron from the surface of a metal [A1]

(b)(i)
- Uses $E = \frac{hc}{\lambda}$ [C1]
- $E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{260 \times 10^{-9} \times 1.60 \times 10^{-19}} = 4.781\text{ eV}$ shown to at least 3 s.f. [A1]

(b)(ii)
- Uses $E_{\text{k,max}} = E - \Phi = 4.78 - 3.68 = 1.10\text{ eV}$ [C1]
- Converts to joules: $E_{\text{k,max}} = 1.76 \times 10^{-19}\text{ J}$ [A1]

(b)(iii)
- Uses $p = \sqrt{2 m_e E_k}$ or $v = \sqrt{2E_k/m}$ and $\lambda = h/p$ [C1]
- $\lambda_{\text{dB}} = 1.17 \times 10^{-9}\text{ m}$ (or $1.17\text{ nm}$) [A1]

(c)
- No change / no effect on maximum kinetic energy [B1]
- Photon energy depends only on frequency/wavelength (or intensity only affects the rate/number of photons incident per second) [B1]

(a) The decay constant is the probability per unit time of the decay of a nucleus.

(b)(i) Half-life $t_{1/2} = 5.01\text{ days} = 5.01 \times 24 \times 3600\text{ s} = 4.3286 \times 10^5\text{ s}$.
$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.69315}{4.3286 \times 10^5} = 1.6013 \times 10^{-6}\text{ s}^{-1} \approx 1.60 \times 10^{-6}\text{ s}^{-1}$.

(b)(ii) Activity $A_0 = \lambda N_0$
$A_0 = (1.6013 \times 10^{-6}\text{ s}^{-1}) \times (3.60 \times 10^{15}) = 5.765 \times 10^9\text{ Bq} \approx 5.76 \times 10^9\text{ Bq}$.

(b)(iii) $t = 12.0\text{ days} = 12.0 \times 86400\text{ s} = 1.0368 \times 10^6\text{ s}$.
$A = A_0 \text{e}^{-\lambda t}$
$A = (5.765 \times 10^9) \times \text{e}^{-(1.6013 \times 10^{-6} \times 1.0368 \times 10^6)} = (5.765 \times 10^9) \times \text{e}^{-1.6602} = (5.765 \times 10^9) \times 0.1901 = 1.096 \times 10^9\text{ Bq} \approx 1.10 \times 10^9\text{ Bq}$.
Alternatively, $A = A_0 \left( \frac{1}{2} \right)^{t / t_{1/2}} = (5.765 \times 10^9) \times (0.5)^{12.0 / 5.01} = 1.10 \times 10^9\text{ Bq}$.

(c)(i) Decay is random because it is impossible to predict which specific nucleus will decay next or when a particular nucleus will decay.

(c)(ii) Decay is spontaneous because the process is not affected by external environmental factors such as temperature, pressure, or chemical environment.

(a)
- Probability of decay of a nucleus per unit time [B1]

(b)(i)
- Converts days to seconds: $5.01 \times 24 \times 3600 = 4.329 \times 10^5\text{ s}$ [C1]
- Uses $\lambda = \frac{\ln 2}{t_{1/2}}$ leading to $1.601 \times 10^{-6}\text{ s}^{-1}$ shown before rounding [A1]

(b)(ii)
- Uses $A_0 = \lambda N_0$ [C1]
- $A_0 = (1.60 \times 10^{-6}) \times (3.60 \times 10^{15}) = 5.76 \times 10^9\text{ Bq}$ (or $\text{s}^{-1}$) [A1]

(b)(iii)
- Uses $A = A_0 \text{e}^{-\lambda t}$ or $A = A_0 (0.5)^{t / t_{1/2}}$ [C1]
- Correct calculation: $A = 1.10 \times 10^9\text{ Bq}$ (allow $1.09 \times 10^9\text{ Bq}$ or $1.1 \times 10^9\text{ Bq}$) [A1]

(c)(i)
- Impossible to predict which nucleus will decay next / when an individual nucleus will decay [B1]

(c)(ii)
- Unaffected by external environmental conditions (e.g. temperature, pressure, chemical combination) [B1]

(a) Acoustic impedance is defined as the product of the density of the medium and the speed of sound/ultrasound in that medium ($Z = \rho c$).

(b)(i) $\alpha = \frac{(Z_2 - Z_1)^2}{(Z_2 + Z_1)^2}$
$\alpha = \frac{(7.80 \times 10^6 - 1.70 \times 10^6)^2}{(7.80 \times 10^6 + 1.70 \times 10^6)^2} = \frac{(6.10 \times 10^6)^2}{(9.50 \times 10^6)^2} = \left( \frac{6.10}{9.50} \right)^2 = (0.6421)^2 = 0.4123 \approx 0.412$ (or $41.2\%$).

(b)(ii) Air has a very low acoustic impedance compared to skin/tissue. Without coupling gel, air trapped between the transducer and skin would cause an extremely large difference in acoustic impedance, reflecting almost all the ultrasound intensity at the skin surface. The coupling gel has an acoustic impedance very similar to skin, which drastically reduces reflection and ensures maximum transmission of ultrasound into the body.

(c)(i) A positron and an electron collide and undergo annihilation, converting their combined mass into energy. This produces two identical gamma-ray photons moving in opposite directions to conserve momentum.

(c)(ii) By conservation of mass-energy, the mass of one electron/positron converts into the energy of one gamma photon:
$E = m_{\text{e}} c^2$
$E = (9.11 \times 10^{-31}\text{ kg}) \times (3.00 \times 10^8\text{ m s}^{-1})^2 = 8.199 \times 10^{-14}\text{ J}$
Converting to $\text{MeV}$:
$E = \frac{8.199 \times 10^{-14}}{1.60 \times 10^{-13}} = 0.5124\text{ MeV} \approx 0.512\text{ MeV}$ (or $0.511\text{ MeV}$ using $1\text{ u}$ / standard data values).

(a)
- Product of density (of the medium) and speed of sound/wave (in the medium) / $Z = \rho c$ [B1]

(b)(i)
- Uses $\frac{I_R}{I_0} = \frac{(Z_1 - Z_2)^2}{(Z_1 + Z_2)^2}$ [C1]
- $\alpha = \left(\frac{7.80 - 1.70}{7.80 + 1.70}\right)^2 = 0.412$ or $41.2\%$ [A1]

(b)(ii)
- Air has very different acoustic impedance from skin, causing almost total reflection (at the air-skin boundary) [B1]
- Gel has acoustic impedance matched to/similar to skin, so (almost) all ultrasound is transmitted into the body [B1]

(c)(i)
- Positron and electron annihilate each other (mass converted into energy) [B1]
- Producing two gamma-ray (photons) emitted in opposite directions [B1]

(c)(ii)
- Uses $E = m_e c^2$ [C1]
- $E = 9.11 \times 10^{-31} \times (3.00 \times 10^8)^2 = 8.20 \times 10^{-14}\text{ J}$ [C1]
- Converts to $\text{MeV}$: $E = 0.512\text{ MeV}$ or $0.511\text{ MeV}$ [A1]

(a) Gravitational potential is the work done per unit mass in bringing a small test mass from infinity to that point in the gravitational field.

(b)(i) At escape speed, the probe reaches infinity with zero kinetic energy.
At infinity, gravitational potential energy \(E_{\text{p}} = 0\) and kinetic energy \(E_{\text{k}} = 0\), so total mechanical energy is 0.
At the surface of the planet:
\[E_{\text{total}} = \frac{1}{2}m v_{\text{esc}}^2 - \frac{GMm}{R} = 0\]
\[\frac{1}{2}m v_{\text{esc}}^2 = \frac{GMm}{R}\]
\[v_{\text{esc}}^2 = \frac{2GM}{R} \implies v_{\text{esc}} = \sqrt{\frac{2GM}{R}}\]

(b)(ii)
\[v_{\text{esc}} = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (6.42 \times 10^{23})}{3.40 \times 10^6}}\]
\[v_{\text{esc}} = \sqrt{\frac{8.5643 \times 10^{13}}{3.40 \times 10^6}} = \sqrt{2.5189 \times 10^7} = 5.0189 \times 10^3\text{ m s}^{-1} \approx 5.02 \times 10^3\text{ m s}^{-1}\]

(c)(i) Let \(r_{\text{max}}\) be the maximum distance from the centre of the planet.
At the maximum height, velocity is zero (\(E_{\text{k}} = 0\)).
By conservation of energy:
\[\frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{r_{\text{max}}}\]
Divide by \(m\):
\[\frac{1}{2}v^2 - \frac{GM}{R} = -\frac{GM}{r_{\text{max}}}\]
Substitute values:
\[\frac{1}{2}(3.20 \times 10^3)^2 - \frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{3.40 \times 10^6} = -\frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{r_{\text{max}}}\]
\[5.12 \times 10^6 - 1.2595 \times 10^7 = -\frac{4.2821 \times 10^{13}}{r_{\text{max}}}\]
\[-7.4745 \times 10^6 = -\frac{4.2821 \times 10^{13}}{r_{\text{max}}}\]
\[r_{\text{max}} = \frac{4.2821 \times 10^{13}}{7.4745 \times 10^6} = 5.729 \times 10^6\text{ m}\]
Height above the surface:
\[h = r_{\text{max}} - R = 5.729 \times 10^6 - 3.40 \times 10^6 = 2.33 \times 10^6\text{ m}\]

(c)(ii) Earth has an atmosphere, so work is done against air resistance / aerodynamic drag, dissipating kinetic energy as thermal energy.

(a)
- work done per unit mass [M1]
- in bringing a (small/point) mass from infinity (to the point) [A1]

(b)(i)
- Equating initial kinetic energy to the work done against gravitational field / change in potential energy, or \(\frac{1}{2}mv^2 - \frac{GMm}{R} = 0\) [M1]
- Clear algebraic rearrangement showing \(v_{\text{esc}} = \sqrt{\frac{2GM}{R}}\) [A1]

(b)(ii)
- Correct substitution into formula: \(\sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{3.40 \times 10^6}}\) [C1]
- \(v = 5.02 \times 10^3\text{ m s}^{-1}\) (allow \(5.0 \times 10^3\text{ m s}^{-1}\)) [A1]

(c)(i)
- Recognition of conservation of energy: \(\frac{1}{2}v^2 - \frac{GM}{R} = -\frac{GM}{r_{\text{max}}}\) [C1]
- Correct calculation of \(r_{\text{max}} = 5.73 \times 10^6\text{ m}\) [C1]
- Calculation of \(h = r_{\text{max}} - R = 2.33 \times 10^6\text{ m}\) (accept \(2.3 \times 10^6\text{ m}\)) [A1]

(c)(ii)
- (Work must be done against) air resistance / friction with the atmosphere (dissipating energy as heat) [B1]

(a) Acoustic impedance is the product of the density of the substance and the speed of the ultrasound wave in the substance (\(Z = \rho c\)).

(b)(i) The intensity reflection coefficient is:
\[\frac{I_{\text{R}}}{I_0} = \frac{(Z_2 - Z_1)^2}{(Z_2 + Z_1)^2}\]
\[\frac{I_{\text{R}}}{I_0} = \frac{(6.30 \times 10^6 - 1.70 \times 10^6)^2}{(6.30 \times 10^6 + 1.70 \times 10^6)^2} = \frac{(4.60)^2}{(8.00)^2} = \frac{21.16}{64.00} = 0.330625 \approx 0.331\]

(b)(ii) Transmitted fraction \(= 1 - 0.3306 = 0.6694\)
Percentage transmitted \(= 66.9\%\) (or \(67\%\)).

(c) Attenuation of ultrasound in a medium follows:
\[I = I_{\text{T}} e^{-\mu x}\]
Where \(x = 3.20\text{ cm} = 0.0320\text{ m}\) and \(\mu = 130\text{ m}^{-1}\).
\[\frac{I}{I_{\text{T}}} = e^{-130 \times 0.0320} = e^{-4.16} = 0.015607 \approx 0.0156\]

(d)(i) A positron (emitted by the radioactive tracer) collides/interacts with an electron in the patient's tissue.
The two particles annihilate, converting their mass-energy into two gamma-ray photons emitted in opposite directions.

(d)(ii) By conservation of mass-energy, the energy of each photon equals the rest energy of one electron (or positron):
\[E = m_{\text{e}} c^2\]
\[E = (9.11 \times 10^{-31}\text{ kg}) \times (3.00 \times 10^8\text{ m s}^{-1})^2 = 8.199 \times 10^{-14}\text{ J}\]
Convert to MeV:
\[E = \frac{8.199 \times 10^{-14}}{1.60 \times 10^{-19} \times 10^6} = 0.5124\text{ MeV} \approx 0.512\text{ MeV}\]

(a)
- Product of density (of medium) and speed of sound/ultrasound (in the medium) [B1]

(b)(i)
- Correct formula used with substitution: \(\frac{(6.30 - 1.70)^2}{(6.30 + 1.70)^2}\) [C1]
- \(\frac{I_{\text{R}}}{I_0} = 0.331\) (accept 0.33) [A1]

(b)(ii)
- \(66.9\%\) or \(67\%\) (allow ECF: \((1 - \text{value from (b)(i)}) \times 100\)) [B1]

(c)
- Correct substitution into \(I = I_0 e^{-\mu x}\): \(e^{-130 \times 0.0320}\) [C1]
- \(\text{fraction} = 0.0156\) (or \(1.56 \times 10^{-2}\)) [A1]

(d)(i)
- Positron and electron collide / interact / annihilate [B1]
- Mass is converted to energy producing two gamma-ray (\(\gamma\)) photons travelling in opposite directions [B1]

(d)(ii)
- Use of \(E = mc^2\) with \(m = 9.11 \times 10^{-31}\text{ kg}\) to find energy in joules (\(8.20 \times 10^{-14}\text{ J}\)) [C1]
- Division by \(1.60 \times 10^{-13}\) to give \(0.512\text{ MeV}\) (or \(0.51\text{ MeV}\)) [A1]

1. Defining the Problem:
- Independent variable: Magnetic flux density \(B\).
- Dependent variable: Damping half-time \(\tau\) (time for amplitude to decrease to half of \(\theta_0\)).
- Variables to be kept constant: Initial angular displacement \(\theta_0\), dimensions/mass of the metal disc, length and tension of the torsion suspension wire, temperature of the disc.

2. Methods of Data Collection:
- Apparatus setup: A circular copper or aluminium disc suspended horizontally by a thin vertical wire clamped firmly at its centre from a rigid stand. The disc is positioned symmetrically in the air gap between the pole pieces of a variable-current electromagnet powered by a low-voltage, high-current d.c. power supply with a rheostat and ammeter.
- **Measurement of \(B\):** Place the Hall probe flat in the region between the pole pieces where the disc oscillates. Align the probe perpendicular to the magnetic field lines. Connect the Hall probe to a calibrated millivoltmeter / digital teslameter to read \(B\).
- **Varying \(B\):** Adjust the current from the d.c. power supply using the variable resistor / variable power supply.
- **Measurement of \(\tau\):** Fix a circular protractor scale below/above the disc or place an angular pointer on the disc. Displace the disc by a fixed initial angle \(\theta_0\) (e.g. \(30^\circ\)) using the scale, release smoothly from rest, and simultaneously start a digital stopwatch (or use high-speed video recording with on-screen timecode). Note the time \(\tau\) when the amplitude decays to \(\frac{1}{2}\theta_0\) (e.g. \(15^\circ\)). Repeat and find the mean \(\tau\) for each setting of \(B\).

3. Method of Analysis:
- Plot a graph of \(\tau\) on the \(y\)-axis against \(\frac{1}{B^2}\) on the \(x\)-axis.
- The relationship is valid if the graph yields a straight line passing through the origin.
- Determine the gradient of the line of best fit. The constant \(K\) is given by:
$$K = \text{gradient}$$
- Alternative analysis: Plot \(\lg \tau\) against \(\lg B\). The relationship is valid if the graph is a straight line with a gradient of \(-2\). The vertical intercept equals \(\lg K\), so \(K = 10^{\text{intercept}}\).

4. Additional Detail & Safety Considerations:
- Safety: Electromagnet coils can become very hot when high currents pass through them for prolonged periods. Use heat-resistant gloves or switch off current between trials to prevent burns / overheating. Secure heavy electromagnet clamps so that the heavy iron core cannot fall and cause injury.
- Additional practical details:
- Ensure the disc does not touch the pole pieces during oscillation.
- Use a non-magnetic disc (e.g., pure copper or aluminium) to avoid magnetic attraction to the pole pieces.
- Ensure draught shields / transparent enclosure are placed around the apparatus to prevent air currents affecting the damping.
- Check that the magnetic field is uniform across the volume swept by the oscillating disc.
- Calibrate the Hall probe beforehand in a known magnetic field or zero the Hall probe in zero field before taking measurements.

Defining the Problem (2 marks max):
- P1 [B1]: \(B\) is the independent variable and \(\tau\) is the dependent variable.
- P2 [B1]: Keep initial angular displacement \(\theta_0\) constant (or keep torsion wire length / disc material and dimensions constant).

Methods of Data Collection (4 marks max):
- M1 [B1]: Labelled diagram showing the disc suspended between the pole pieces of an electromagnet connected in a complete circuit with a d.c. power source.
- M2 [B1]: Method of measuring magnetic flux density \(B\) using a Hall probe / search coil / teslameter positioned in the gap.
- M3 [B1]: Method to measure the initial angle \(\theta_0\) and subsequent amplitude using an angular scale / protractor / pointer.
- M4 [B1]: Method to determine \(\tau\) using a stopwatch / video camera with frame timer / motion sensor to find time for amplitude to halve.

Method of Analysis (2 marks max):
- A1 [B1]: Plot a graph of \(\tau\) against \(1/B^2\) (or \(\lg \tau\) against \(\lg B\) or \(\ln \tau\) against \(\ln B\)).
- A2 [B1]: Relationship is valid if the graph of \(\tau\) against \(1/B^2\) is a straight line through the origin, and \(K = \text{gradient}\) (or for \(\lg \tau\) against \(\lg B\), a straight line of gradient \(-2\) with \(K = 10^{\text{intercept}}\)).

Safety Considerations (1 mark max):
- S1 [B1]: Specific precaution relevant to high current in electromagnet coils causing heating (e.g. switch off between readings / use protective gloves) OR securing heavy electromagnet/pole pieces to prevent crushing.

Additional Detail (6 marks max from the following):
- D1: Repeat timing measurements for each \(B\) and calculate average \(\tau\).
- D2: Calibrate / zero the Hall probe prior to measurements (or ensure Hall slice is perpendicular to field lines).
- D3: Use of non-ferromagnetic disc (copper/aluminium) to avoid ferromagnetic attraction to pole faces.
- D4: Enclose apparatus in a draught box / shield to reduce air turbulence damping.
- D5: Method of ensuring release from rest without imparting initial angular impulse/transverse swing.
- D6: Detail on varying \(B\) by adjusting rheostat / variable d.c. voltage supply.
- D7: Measure background damping (damping without magnetic field, \(B=0\)) to ensure magnetic damping dominates.