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

Pressure is defined as force per unit area: \(p = \frac{F}{A}\).

The SI base units of force are \(\text{kg}\text{ m}\text{ s}^{-2}\).

The SI base units of area are \(\text{m}^2\).

Therefore, the SI base units of pressure are:
\[ [p] = \frac{\text{kg}\text{ m}\text{ s}^{-2}}{\text{m}^2} = \text{kg}\text{ m}^{-1}\text{ s}^{-2} \]

B is correct [1 mark].
A: Energy has base units \(\text{kg}\text{ m}^{2}\text{ s}^{-2}\).
C: Power has base units \(\text{kg}\text{ m}^{2}\text{ s}^{-3}\).
D: Momentum has base units \(\text{kg}\text{ m}\text{ s}^{-1}\).

As the sphere falls, its downward speed \(v\) increases, so the upward drag force \(D\) increases.

The resultant downward force is \(F_{\text{net}} = mg - U - D\), where \(U\) is the constant upthrust.

Since acceleration \(a = \frac{F_{\text{net}}}{m}\), as drag \(D\) increases, the acceleration \(a\) decreases continuously from an initial non-zero value towards zero when terminal velocity is achieved.

Graph C shows \(a\) starting at a positive value and decreasing non-linearly to zero.

C is correct [1 mark].
A incorrectly shows constant acceleration.
B incorrectly shows acceleration increasing.
D incorrectly shows acceleration decreasing linearly to a negative value.

By conservation of linear momentum:
\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \]
\[ (0.80)(1.5) + (1.2)(0) = (0.80 + 1.2) v \]
\[ 1.2 = 2.0 v \implies v = 0.60\text{ m}\text{ s}^{-1} \]

Initial total kinetic energy:
\[ E_{\text{k, initial}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2}(0.80)(1.5)^2 = 0.90\text{ J} \]

Final total kinetic energy:
\[ E_{\text{k, final}} = \frac{1}{2}(m_1 + m_2) v^2 = \frac{1}{2}(2.0)(0.60)^2 = 0.36\text{ J} \]

Loss in kinetic energy:
\[ \Delta E_{\text{k}} = 0.90 - 0.36 = 0.54\text{ J} \]

A is correct [1 mark].
B is the final kinetic energy \(0.36\text{ J}\).
C is the initial kinetic energy \(0.90\text{ J}\).
D is the sum of initial and final kinetic energy \(1.26\text{ J}\).

For a uniform plank, the centre of gravity is at the midpoint, which is \(1.20\text{ m}\) from the left end \(P\).

Taking moments about support \(P\):
\[ \text{Clockwise moment} = \text{Anticlockwise moment} \]
\[ W \times 1.20 = F_Q \times 1.80 \]
\[ 180 \times 1.20 = F_Q \times 1.80 \]
\[ 216 = 1.80 F_Q \]
\[ F_Q = \frac{216}{1.80} = 120\text{ N} \]

D is correct [1 mark].
A is the force at support \(P\) (\(60\text{ N}\)).
B is half the total weight (\(90\text{ N}\)).
C is an incorrect calculation using distance to the right edge (\(100\text{ N}\)).

Useful work done:
\[ W = mgh = 45 \times 9.81 \times 12 = 5297.4\text{ J} \]

Useful output power:
\[ P_{\text{out}} = \frac{W}{t} = \frac{5297.4}{15} = 353.16\text{ W} \]

Efficiency:
\[ \text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{353.16}{480} \times 100\% \approx 73.6\% \approx 74\% \]

B is correct [1 mark].
A uses \(g = 10\text{ m}\text{ s}^{-2}\) and inverts or miscalculates power ratio.
C uses \(P_{\text{out}} / 480\) with incorrect time factor.
D omits the time division.

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 given values:
\[ f_o = 850 \times \left( \frac{340}{340 - 28.0} \right) = 850 \times \left( \frac{340}{312.0} \right) \approx 926.3\text{ Hz} \approx 926\text{ Hz} \]

C is correct [1 mark].
A is obtained using the formula for a receding source \(850 \times \frac{340}{368} \approx 785\text{ Hz}\).
B is obtained using \(850 \times \left(1 + \frac{28}{340}\right) = 920\text{ Hz}\).
D is obtained by adding simple velocity offset.

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

Using the diffraction grating formula \(d \sin \theta = n \lambda\):
Since the maximum possible angle is \(\theta = 90^\circ\) (where \(\sin 90^\circ = 1\)):
\[ n_{\text{max}} \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}}{589 \times 10^{-9}} \approx 4.24 \]

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

The total number of maxima observed includes the central zeroth-order maximum and the positive and negative orders up to \(n = 4\):
\[ \text{Total number} = 2n_{\text{max}} + 1 = 2(4) + 1 = 9 \]

B is correct [1 mark].
A only counts the highest order \(n = 4\).
C calculates \(2n = 8\) forgetting the central maximum.
D mistakenly uses \(n = 5\).

The circuit current is given by:
\[ I = \frac{E}{R + r} \]
As \(R\) increases, the denominator \((R + r)\) increases, so current \(I\) decreases.

The terminal potential difference is given by:
\[ V = E - Ir = E \left( \frac{R}{R + r} \right) \]
As \(R\) increases, \(Ir\) (the lost volts) decreases, which means \(V\) increases towards \(E\).

A is correct [1 mark].
B incorrectly states that \(V\) decreases.
C incorrectly states that \(I\) increases.
D incorrectly states that \(V\) decreases and \(I\) increases.

Potential difference \(V\) is defined as work done per unit charge, \(V = \frac{W}{Q}\).
In terms of base units:
Work / Energy \(W = \text{force} \times \text{distance} = (\text{kg}\;\text{m}\;\text{s}^{-2}) \times \text{m} = \text{kg}\;\text{m}^2\;\text{s}^{-2}\).
Charge \(Q = I \times t = \text{A}\;\text{s}\).
Therefore, the unit of potential difference (volt) in base units is:
\[ [V] = \frac{\text{kg}\;\text{m}^2\;\text{s}^{-2}}{\text{A}\;\text{s}} = \text{kg}\;\text{m}^2\;\text{s}^{-3}\;\text{A}^{-1} \]

B is correct (1 mark). A is energy per second (\(\text{kg}\;\text{m}^2\;\text{s}^{-3}\)); C is resistance (\(\text{kg}\;\text{m}^2\;\text{s}^{-3}\;\text{A}^{-2}\)); D is electric charge (\(\text{A}\;\text{s}\)).

The initial vertical component of velocity is \(u_y = u\sin\theta\).
At the maximum height \(H\), the vertical velocity \(v_y = 0\).
Using the equation of motion \(v_y^2 = u_y^2 - 2gH\):
\[ 0 = (u\sin\theta)^2 - 2gH \implies H = \frac{u^2\sin^2\theta}{2g} \]

C is correct (1 mark). Using \(v_y^2 = u_y^2 - 2gh\) with \(u_y = u\sin\theta\) yields \(H = \frac{u^2\sin^2\theta}{2g}\).

1. Apply conservation of linear momentum to find the final velocity \(v\):
\[ m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \]
\[ (0.60)(2.5) + (0.40)(0) = (0.60 + 0.40) v \]
\[ 1.50 = 1.00 v \implies v = 1.50\;\text{m}\;\text{s}^{-1} \]

2. Initial kinetic energy \(E_{\text{k, initial}}\):
\[ E_{\text{k, initial}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2}(0.60)(2.5)^2 = 1.875\;\text{J} \]

3. Final kinetic energy \(E_{\text{k, final}}\):
\[ E_{\text{k, final}} = \frac{1}{2}(m_1 + m_2) v^2 = \frac{1}{2}(1.00)(1.50)^2 = 1.125\;\text{J} \]

4. Loss in kinetic energy \(\Delta E_{\text{k}}\):
\[ \Delta E_{\text{k}} = 1.875 - 1.125 = 0.75\;\text{J} \]

A is correct (1 mark). Momentum conservation gives \(v = 1.5\;\text{m}\;\text{s}^{-1}\); \(\Delta E_{\text{k}} = 1.875 - 1.125 = 0.75\;\text{J}\).

For equilibrium, the upward forces must balance the downward forces:
\[ T + U = W \]
where:
- Tension \(T\)
- Upthrust \(U = \rho_{\text{L}} V g = \rho_{\text{L}} A h g\)
- Weight \(W = m g = \rho_{\text{s}} V g = \rho_{\text{s}} A h g\)

Rearranging for tension:
\[ T = W - U = \rho_{\text{s}} A h g - \rho_{\text{L}} A h g = A h g (\rho_{\text{s}} - \rho_{\text{L}}) \]

D is correct (1 mark). \(T = W - U = Ahg(\rho_{\text{s}} - \rho_{\text{L}})\).

The elastic potential energy stored is given by:
\[ E_{\text{p}} = \frac{1}{2} F x \]

From the definition of the Young modulus:
\[ E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{x / L} = \frac{F L}{A x} \implies F = \frac{E A x}{L} \]

Substitute \(F\) into the energy expression:
\[ E_{\text{p}} = \frac{1}{2} \left(\frac{E A x}{L}\right) x = \frac{E A x^2}{2 L} \]

B is correct (1 mark). \(E_{\text{p}} = \frac{1}{2}Fx = \frac{1}{2}\left(\frac{EAx}{L}\right)x = \frac{EAx^2}{2L}\).

For a source moving towards a stationary observer, the Doppler-shifted observed frequency \(f_{\text{o}}\) is given by:
\[ f_{\text{o}} = f_{\text{s}} \left( \frac{v}{v - v_{\text{s}}} \right) \]

Substitute the given values:
- \(f_{\text{s}} = 720\;\text{Hz}\)
- \(v = 336\;\text{m}\;\text{s}^{-1}\)
- \(v_{\text{s}} = 28.0\;\text{m}\;\text{s}^{-1}\)

\[ f_{\text{o}} = 720 \times \left( \frac{336}{336 - 28.0} \right) = 720 \times \left( \frac{336}{308} \right) = 720 \times 1.0909 = 785.45\;\text{Hz} \approx 785\;\text{Hz} \]

C is correct (1 mark). Direct substitution into the Doppler formula: \(f_{\text{o}} = 720 \times (336 / (336 - 28)) = 785\;\text{Hz}\).

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

2. The diffraction grating formula is \(d \sin\theta = n \lambda\).
For the maximum order \(n_{\text{max}}\), \(\sin\theta \le 1\):
\[ n_{\text{max}} \le \frac{d}{\lambda} = \frac{2.50 \times 10^{-6}}{540 \times 10^{-9}} = 4.63 \]
Since \(n\) must be an integer, \(n_{\text{max}} = 4\).

3. The total number of maxima observed includes the zero-order maximum \((n = 0)\) and maxima on both sides \((n = 1, 2, 3, 4)\):
\[ \text{Total number} = 2 n_{\text{max}} + 1 = 2(4) + 1 = 9 \]

B is correct (1 mark). \(d = 2.50 \times 10^{-6}\;\text{m}\); \(n_{\text{max}} = 4\); total number \(= 2(4) + 1 = 9\).

1. Current in the circuit is \(I = \frac{E}{R + r}\).
As \(R\) increases, the total resistance \((R + r)\) increases, so the current \(I\) decreases.

2. The terminal potential difference \(V\) is given by:
\[ V = E - I r \]
As \(I\) decreases, the lost volts \(I r\) decrease, which causes the terminal potential difference \(V\) to increase (approaching \(E\) as \(R \to \infty\)).

C is correct (1 mark). \(I = E / (R+r)\) decreases, so \(V = E - Ir\) increases.

Pressure is defined as force per unit area: \(p = \frac{F}{A}\).

The SI unit of force is the newton (\(\text{N}\)), where \(F = ma\), so \(\text{N} = \text{kg}\;\text{m}\;\text{s}^{-2}\).

Area has the base units of \(\text{m}^2\).

Therefore, the base units of the pascal are:
\[\text{Pa} = \frac{\text{kg}\;\text{m}\;\text{s}^{-2}}{\text{m}^2} = \text{kg}\;\text{m}^{-1}\;\text{s}^{-2}\]

A is correct [1 mark].
B is the unit of force (\(\text{N}\)).
C is the unit of energy or work (\(\text{J}\)).
D represents an incorrect combination of powers.

The density \(\rho\) of a cylinder is given by:
\[ \rho = \frac{m}{V} = \frac{m}{\pi \left(\frac{d}{2}\right)^2 L} = \frac{4m}{\pi d^2 L} \]
The percentage uncertainty in density is the sum of the percentage uncertainties of the measured quantities, with the power of \(d\) multiplying its uncertainty:
\[ \frac{\Delta \rho}{\rho} \times 100\% = \left( \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L} \right) \times 100\% \]
Calculating each individual percentage uncertainty:
\[ \frac{\Delta m}{m} \times 100\% = \frac{0.9}{45.0} \times 100\% = 2.0\% \]
\[ \frac{\Delta d}{d} \times 100\% = \frac{0.2}{20.0} \times 100\% = 1.0\% \]
\[ \frac{\Delta L}{L} \times 100\% = \frac{0.5}{50.0} \times 100\% = 1.0\% \]
Combining these:
\[ \text{Total percentage uncertainty} = 2.0\% + 2(1.0\%) + 1.0\% = 5.0\% \]

C is correct (1 mark). Summing individual fractional uncertainties with factor of 2 for diameter gives 5.0%.

Let the launch speed be \(v_0\). The initial kinetic energy is:
\[ E_{\text{k0}} = \frac{1}{2}m v_0^2 \]
At the highest point, the vertical component of velocity is zero, so the speed is equal to the horizontal component:
\[ v_x = v_0 \cos\theta \]
The kinetic energy at the highest point is:
\[ E_{\text{k}} = \frac{1}{2}m v_x^2 = \frac{1}{2}m (v_0 \cos\theta)^2 = E_{\text{k0}} \cos^2\theta \]
Given that \(E_{\text{k}} = \frac{1}{3}E_{\text{k0}}\):
\[ \cos^2\theta = \frac{1}{3} \implies \cos\theta = \frac{1}{\sqrt{3}} \approx 0.5774 \]
\[ \theta = \arccos(0.5774) \approx 54.7^\circ \approx 55^\circ \]

C is correct (1 mark). Using horizontal velocity conservation: cos^2(theta) = 1/3, giving theta = 55 degrees.

According to Newton's second law, force is the rate of change of momentum:
\[ F = \frac{\Delta p}{\Delta t} = \frac{\Delta (m v)}{\Delta t} \]
Since the velocity \(u\) is constant:
\[ F = u \frac{\Delta m}{\Delta t} \]
Given that the sand enters at rate \(\frac{\Delta m}{\Delta t} = R\), the sand must be accelerated from zero horizontal speed to horizontal speed \(u\).
\[ F = R u \]

C is correct (1 mark). Rate of momentum increase of added mass is (dm/dt) * u = Ru.

The submerged block experiences three vertical forces in equilibrium:
1. Upthrust \(U = \rho_{\text{L}} V g\) acting upwards.
2. Weight \(W = \rho_{\text{b}} V g\) acting downwards.
3. Tension \(T\) in the string acting downwards (preventing the block from floating upwards).

For equilibrium:
\[ U = W + T \]
\[ T = U - W = \rho_{\text{L}} V g - \rho_{\text{b}} V g = (\rho_{\text{L}} - \rho_{\text{b}}) V g \]

B is correct (1 mark). Tension = Upthrust - Weight = (rho_L - rho_b) * V * g.

The forward driving force \(F\) provided by the engine must balance the component of the car's weight acting down the slope and the resistive force:
\[ F = m g \sin\theta + F_{\text{resistive}} \]
Substitute the values:
\[ F = (1200 \times 9.81 \times \sin 5.7^\circ) + 400 \]
\[ \sin 5.7^\circ \approx 0.09932 \]
\[ m g \sin\theta = 1200 \times 9.81 \times 0.09932 \approx 1169\text{ N} \]
\[ F = 1169 + 400 = 1569\text{ N} \]
The power output is given by \(P = F v\):
\[ P = 1569\text{ N} \times 20\text{ m s}^{-1} = 31380\text{ W} \approx 31\text{ kW} \]

C is correct (1 mark). Total force = mg sin(theta) + 400 N = 1569 N; Power = Fv = 1569 * 20 = 31 kW.

The elastic potential energy \(E_{\text{p}}\) stored in a wire of cross-sectional area \(A\), length \(L\), and Young modulus \(E\) stretched by force \(F\) is:
\[ E_{\text{p}} = \frac{1}{2} F e = \frac{1}{2} F \left(\frac{F L}{A E}\right) = \frac{F^2 L}{2 A E} \]
Since \(A = \frac{\pi d^2}{4}\), we have:
\[ E_{\text{p}} \propto \frac{L}{d^2} \]
For wire X:
\[ E_{\text{p, X}} \propto \frac{L}{d^2} \]
For wire Y:
\[ E_{\text{p, Y}} \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \left(\frac{L}{d^2}\right) \]
Therefore, the ratio is:
\[ \frac{E_{\text{p, Y dogma}}}{E_{\text{p, X}}} = \frac{1}{2} = 0.50 \]

B is correct (1 mark). Energy is proportional to L / d^2. Ratio = (2) / (2^2) = 2/4 = 0.50.

The observed frequency \(f_{\text{o}}\) for a source moving towards a stationary observer is given by the Doppler equation:
\[ f_{\text{o}} = f_{\text{s}} \left( \frac{v}{v - v_{\text{s}}} \right) \]
where:
• \(f_{\text{s}} = 680\text{ Hz}\)
• \(v = 340\text{ m s}^{-1}\)
• \(v_{\text{s}} = 24.0\text{ m s}^{-1}\)

Substitute the values:
\[ f_{\text{o}} = 680 \times \left( \frac{340}{340 - 24.0} \right) = 680 \times \frac{340}{316} \approx 731.6\text{ Hz} \approx 732\text{ Hz} \]

D is correct (1 mark). Correct Doppler formula substitution: 680 * (340 / (340 - 24.0)) = 732 Hz.

Maximum power is transferred to a load resistor when its resistance \(R\) is equal to the internal resistance \(r\) of the power source:
\[ R = r = 2.0\,\Omega \]
When \(R = 2.0\,\Omega\), the current in the circuit is:
\[ I = \frac{E}{R + r} = \frac{12.0}{2.0 + 2.0} = 3.0\text{ A} \]
The power delivered to the variable resistor is:
\[ P = I^2 R = (3.0)^2 \times 2.0 = 18\text{ W} \]

A is correct (1 mark). Maximum power occurs when R = r = 2.0 ohms; P = (E / (2r))^2 * r = (12/4)^2 * 2 = 18 W.

The density \(\rho\) of a cylinder is given by:
$$\rho = \frac{m}{V} = \frac{m}{\pi \left(\frac{d}{2}\right)^2 L} = \frac{4m}{\pi d^2 L}$$

The fractional uncertainty in \(\rho\) is determined by summing the fractional uncertainties of each independent measurement, taking powers into account:
$$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta L}{L}$$

Substitute the given values:
$$\frac{\Delta m}{m} = \frac{3}{120} = 0.025 = 2.5\%$$
$$\frac{\Delta d}{d} = \frac{0.02}{2.00} = 0.010 = 1.0\%$$
$$\frac{\Delta L}{L} = \frac{0.05}{5.00} = 0.010 = 1.0\%$$

Therefore:
$$\frac{\Delta \rho}{\rho} = 2.5\% + 2(1.0\%) + 1.0\% = 5.5\%$$

Hence, the percentage uncertainty is \(5.5\%\).

[1] C: 5.5%
- Correctly adds percentage uncertainties taking the power of 2 for diameter into account.

Consider the vertical motion of the ball:
$$v_y^2 = u_y^2 + 2gh = 0 + 2gh = 2gh \implies v_y = \sqrt{2gh}$$

The horizontal velocity remains constant:
$$v_x = u$$

At the point of impact, the velocity makes an angle of \(60^\circ\) to the horizontal:
$$\tan(60^\circ) = \frac{v_y}{v_x} = \frac{\sqrt{2gh}}{u}$$

Since \(\tan(60^\circ) = \sqrt{3}\):
$$\sqrt{3} = \frac{\sqrt{2gh}}{u} \implies u = \frac{\sqrt{2gh}}{\sqrt{3}} = \sqrt{\frac{2gh}{3}}$$

[1] A: \(\sqrt{\frac{2gh}{3}}\)
- Uses \(v_y = \sqrt{2gh}\) and \(\tan 60^\circ = v_y / u\).

Let motion to the right be positive.

Initial linear momentum of the system:
$$p_i = (3m)(+2u) + m(-4u) = +6mu - 4mu = +2mu$$

Total mass after collision \(M = 3m + m = 4m\).

By conservation of momentum, the common final velocity \(v\) is:
$$v = \frac{p_i}{M} = \frac{2mu}{4m} = +\frac{1}{2}u$$

Initial total kinetic energy:
$$E_{\text{k}, i} = \frac{1}{2}(3m)(2u)^2 + \frac{1}{2}m(4u)^2 = \frac{1}{2}(3m)(4u^2) + \frac{1}{2}m(16u^2) = 6mu^2 + 8mu^2 = 14mu^2$$

Final total kinetic energy:
$$E_{\text{k}, f} = \frac{1}{2}(4m)\left(\frac{u}{2}\right)^2 = 2m \left(\frac{u^2}{4}\right) = \frac{1}{2}mu^2$$

Loss in kinetic energy:
$$\Delta E_\text{k} = E_{\text{k}, i} - E_{\text{k}, f} = 14mu^2 - \frac{1}{2}mu^2 = \frac{27}{2}mu^2$$

[1] D: \(\frac{27}{2}mu^2\)
- Correct application of conservation of momentum and calculation of kinetic energy change.

Let \(V\) be the volume of the submerged object. Its weight is:
$$w = (3\rho) V g \implies V = \frac{w}{3\rho g}$$

The upthrust \(U\) exerted by the liquid on the submerged object is:
$$U = \rho V g = \rho \left(\frac{w}{3\rho g}\right) g = \frac{1}{3}w$$

The apparent weight (downward force exerted on the beam at distance \(L/2\)) is:
$$F_\text{down} = w - U = w - \frac{1}{3}w = \frac{2}{3}w$$

Taking moments about the hinge at the wall:
$$\text{Clockwise moments} = W \left(\frac{L}{2}\right) + \left(\frac{2}{3}w\right)\left(\frac{L}{2}\right) = \frac{1}{2}WL + \frac{1}{3}wL$$
$$\text{Anticlockwise moment} = T \times L$$

Equating clockwise and anticlockwise moments:
$$T L = \frac{1}{2}WL + \frac{1}{3}wL \implies T = \frac{1}{2}W + \frac{1}{3}w$$

[1] A: \(\frac{1}{2}W + \frac{1}{3}w\)
- Determines upthrust as \(w/3\), net weight as \(2w/3\), and takes moments about the hinge.

The elastic potential energy stored in a wire within its limit of proportionality is:
$$U = \frac{1}{2} F x$$

From Hooke's law and the definition of the Young modulus \(E\):
$$x = \frac{F L}{A E} = \frac{F L}{\pi r^2 E}$$

Substituting \(x\) into the energy expression:
$$U = \frac{F^2 L}{2 \pi r^2 E}$$

For the second wire, \(L_2 = 2L\), \(r_2 = 2r\), and \(E_2 = E\):
$$U_2 = \frac{F^2 (2L)}{2 \pi (2r)^2 E} = \frac{2 F^2 L}{2 \pi (4r^2) E} = \frac{1}{2} \left(\frac{F^2 L}{2 \pi r^2 E}\right) = \frac{1}{2} U$$

Therefore, the energy stored is \(\frac{1}{2}U\).

[1] B: \(\frac{1}{2} U\)
- Applies \(U = \frac{F^2 L}{2 A E}\) with \(L \to 2L\) and \(A \to 4A\).

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

When approaching:
$$f_1 = f_\text{s} \left(\frac{v}{v - v_\text{s}}\right)$$

When receding:
$$f_2 = f_\text{s} \left(\frac{v}{v + v_\text{s}}\right)$$

Taking the ratio \(\frac{f_1}{f_2}\):
$$\frac{f_1}{f_2} = \frac{v + v_\text{s}}{v - v_\text{s}}$$

Cross-multiplying and rearranging for \(v_\text{s}\):
$$f_1(v - v_\text{s}) = f_2(v + v_\text{s})$$
$$f_1 v - f_1 v_\text{s} = f_2 v + f_2 v_\text{s}$$
$$v(f_1 - f_2) = v_\text{s}(f_1 + f_2)$$
$$v_\text{s} = v\left(\frac{f_1 - f_2}{f_1 + f_2}\right)$$

[1] A: \(v\left(\frac{f_1 - f_2}{f_1 + f_2}\right)\)
- Correct application and algebraic combination of the Doppler effect equations for approaching and receding source.

Let \(R_\text{ext}\) be the resistance of the variable resistor and \(R_\text{tot} = R + r\).

In the first case (\(R_\text{ext} = 4.0\ \Omega\), \(V_1 = 4.0\text{ V}\)):
$$I_1 = \frac{V_1}{R_\text{ext}} = \frac{4.0}{4.0} = 1.0\text{ A}$$
$$E = I_1(R_\text{tot} + R_\text{ext}) = 1.0(R_\text{tot} + 4.0) = R_\text{tot} + 4.0$$

In the second case (\(R_\text{ext} = 10.0\ \Omega\), \(V_2 = 6.0\text{ V}\)):
$$I_2 = \frac{V_2}{R_\text{ext}} = \frac{6.0}{10.0} = 0.60\text{ A}$$
$$E = I_2(R_\text{tot} + R_\text{ext}) = 0.60(R_\text{tot} + 10.0) = 0.60R_\text{tot} + 6.0$$

Equating the two expressions for \(E\):
$$R_\text{tot} + 4.0 = 0.60R_\text{tot} + 6.0$$
$$0.40 R_\text{tot} = 2.0$$
$$R_\text{tot} = \frac{2.0}{0.40} = 5.0\ \Omega$$

[1] B: \(5.0\ \Omega\)
- Determines currents as 1.0 A and 0.60 A, sets up simultaneous equations with e.m.f., and solves for \((R + r)\).

A neutron has quark composition \(\text{u}\text{d}\text{d}\) and a proton has quark composition \(\text{u}\text{u}\text{d}\).

In \(\beta^-\)-decay (neutron decay):
$$\text{n} \to \text{p} + \text{e}^- + \bar{\nu}_\text{e}$$

One down quark (\(\text{d}\)) transforms into an up quark (\(\text{u}\)):
$$\text{d} \to \text{u} + \text{e}^- + \bar{\nu}_\text{e}$$

This flavour-changing process is governed exclusively by the weak interaction (weak nuclear force).

[1] D: down to up, weak
- Identifies \(\text{d} \to \text{u}\) quark transformation and weak interaction.

(a) Acceleration is defined as the rate of change of velocity, or \(\Delta v / \Delta t\).

(b) (i) Using \(v = u + at\):
\(v = 0 + 4.5 \times 6.0 = 27\text{ m s}^{-1}\).

(ii) Height reached during the powered stage (\(s_1\)):
\(s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(4.5)(6.0)^2 = 81.0\text{ m}\)

Height gained during the free-flight upward stage until \(v = 0\) (\(s_2\)):
\(v^2 = u^2 + 2as_2 \implies 0^2 = 27^2 - 2(9.81)s_2\)
\(s_2 = \frac{729}{19.62} = 37.16\text{ m}\)

Total maximum height \(H = s_1 + s_2 = 81.0 + 37.16 = 118.16\text{ m} \approx 118\text{ m}\) (or \(1.2 \times 10^2\text{ m}\) to 2 s.f.).

(iii) Time from engine shut-off to the peak (\(t_2\)):
\(v = u + at_2 \implies 0 = 27 - 9.81 t_2 \implies t_2 = \frac{27}{9.81} \approx 2.75\text{ s}\)
Time at apex = \(6.0 + 2.75 = 8.75\text{ s}\).

Time to fall from apex to ground (\(t_3\)):
\(H = \frac{1}{2}gt_3^2 \implies 118.16 = \frac{1}{2}(9.81)t_3^2\)
\(t_3 = \sqrt{\frac{2 \times 118.16}{9.81}} = \sqrt{24.09} \approx 4.91\text{ s}\)

Total time \(T = 6.0 + 2.75 + 4.91 = 13.66\text{ s} \approx 13.7\text{ s}\).

(c) (i) Velocity before impact:
\(v^2 = u^2 + 2gH = 0 + 2(9.81)(118.16) = 2318.3\)
\(v = -\sqrt{2318.3} = -48.1\text{ m s}^{-1}\) (downwards).

(ii) The sketch consists of:
1. A straight line from \((0, 0)\) to \((6.0, 27)\) with constant positive gradient \(+4.5\text{ m s}^{-2}\).
2. A straight line with constant negative gradient \(-9.81\text{ m s}^{-2}\) from \((6.0, 27)\), crossing the time axis at \(t = 8.75\text{ s}\) (\(v = 0\)), and ending at \(t = 13.7\text{ s}\) with \(v = -48.1\text{ m s}^{-1}\).

(a) B1: rate of change of velocity OR change in velocity / time taken [Allow \(\Delta v / \Delta t\) with terms defined]. Reject 'change in speed / time'.

(b)(i) B1: \(v = 0 + 4.5 \times 6.0 = 27\text{ m s}^{-1}\) [Must show substitution and formula or clear working].

(b)(ii) C1: \(s_1 = \frac{1}{2} \times 4.5 \times 6.0^2 = 81\text{ m}\) (or \(\text{average speed} \times t = 13.5 \times 6.0 = 81\text{ m}\))
C1: \(s_2 = \frac{27^2}{2 \times 9.81} = 37.2\text{ m}\) (or \(37.16\text{ m}\))
A1: \(H = 81 + 37.2 = 118\text{ m}\) (accept 118 to 118.2 m, or \(1.2 \times 10^2\text{ m}\)).

(b)(iii) C1: \(t_2 = 27 / 9.81 = 2.75\text{ s}\) OR setting up displacement equation \(-81 = 27t - \frac{1}{2}(9.81)t^2\)
C1: \(t_3 = \sqrt{(2 \times 118.2) / 9.81} = 4.91\text{ s}\) OR solving quadratic to find \(t = 7.66\text{ s}\)
A1: \(\text{Total time} = 6.0 + 2.75 + 4.91 = 13.7\text{ s}\) (accept 13.6 to 13.7 s).

(c)(i) A1: \((-48.1\text{ m s}^{-1}\) or speed \(48.1\text{ m s}^{-1}\) (accept 48 or 48.1 m s\(^{-1}\)).

(c)(ii) B1: straight line with positive gradient starting at origin \((0, 0)\) and ending at \((6.0, 27)\).
B1: continuous single straight line from \((6.0, 27)\) with steeper negative gradient crossing \(v = 0\) at \(t \approx 8.8\text{ s}\) and terminating at \((13.7, -48.1)\).

(a)
1. The resultant force (in any direction) must be zero (\(\sum F = 0\)).
2. The resultant torque / resultant moment about any point must be zero (\(\sum \tau = 0\)).

(b)(i)
Weight of the uniform beam acts at its midpoint, at a distance of \(0.60\text{ m}\) from A.
Taking moments about hinge A:
Clockwise moments:
\[\text{Clockwise moment} = (48 \times 0.60) + (75 \times 0.90) = 28.8 + 67.5 = 96.3\text{ N m}\]
Anticlockwise moment from the wire tension \(T\):
\[\text{Anticlockwise moment} = (T \sin 35^\circ) \times 1.20\]
For equilibrium, clockwise moment = anticlockwise moment:
\[T \times 1.20 \sin 35^\circ = 96.3\]
\[T = \frac{96.3}{1.20 \sin 35^\circ} = \frac{96.3}{0.6883} = 139.9\text{ N} \approx 140\text{ N}\]

(b)(ii)
Using the definition of the Young modulus \(E\):
\[E = \frac{\text{stress}}{\text{strain}} = \frac{T / A}{\Delta L / L_0} = \frac{T L_0}{A \Delta L}\]
Rearranging for extension \(\Delta L\):
\[\Delta L = \frac{T L_0}{A E}\]
\[\Delta L = \frac{140 \times 0.85}{1.5 \times 10^{-6} \times 2.0 \times 10^{11}} = \frac{119}{3.0 \times 10^5} = 3.97 \times 10^{-4}\text{ m} = 4.0 \times 10^{-4}\text{ m}\]

(b)(iii)
Elastic potential energy \(E_p\):
\[E_p = \frac{1}{2} T \Delta L\]
\[E_p = \frac{1}{2} \times 140 \times 3.97 \times 10^{-4} = 0.0278\text{ J} = 2.8 \times 10^{-2}\text{ J}\]

(a)
- Resultant force is zero / sum of forces in any direction is zero [B1]
- Resultant torque is zero / sum of clockwise moments = sum of anticlockwise moments about any point [B1]

(b)(i)
- Sum of moments of weights evaluated: \((48 \times 0.60) + (75 \times 0.90) = 96.3\text{ N m}\) [C1]
- Equating to anticlockwise moment: \(T \sin(35^\circ) \times 1.20 = 96.3\) [M1]
- Calculation shown with intermediate value to at least 3 s.f. before rounding: \(T = 139.9\text{ N} = 140\text{ N}\) [A1]

(b)(ii)
- \(\Delta L = \frac{F L}{A E}\) with correct substitutions shown [C1]
- \(\Delta L = 4.0 \times 10^{-4}\text{ m}\) (accept \(3.96 \times 10^{-4}\text{ m}\) to \(4.0 \times 10^{-4}\text{ m}\)) [A1]

(b)(iii)
- \(E_p = \frac{1}{2} F x = 0.5 \times 140 \times 3.97 \times 10^{-4} = 2.8 \times 10^{-2}\text{ J}\) (or \(0.028\text{ J}\)) [A1] (allow ECF from (b)(ii))

(a)(i)
According to the principle of superposition, when two or more waves meet at a point, the resultant displacement is equal to the vector (or algebraic) sum of the individual displacements of the waves.

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

(b)(i)
Using the double-slit interference relation:
\[\lambda = \frac{ax}{D} \implies x = \frac{\lambda D}{a}\]

Substitute the given values:
\[x = \frac{630 \times 10^{-9} \times 1.50}{0.28 \times 10^{-3}} = 3.375 \times 10^{-3}\text{ m} \approx 3.4 \times 10^{-3}\text{ m}\]

(b)(ii) 1.
Using Malus's law:
\[I = I_0 \cos^2 \theta\]
\[I = I_0 \cos^2(55^\circ) = I_0 \times (0.5736)^2 = 0.329\,I_0 \approx 0.33\,I_0\]

(b)(ii) 2.
Since intensity \(I \propto A^2\), amplitude \(A \propto \sqrt{I}\):
\[\frac{A_2}{A_1} = \sqrt{\frac{I_2}{I_1}} = \sqrt{\frac{0.329\,I_0}{I_0}} = \sqrt{0.329} = 0.5736 \approx 0.57\]
(Alternatively: \(\frac{A_2}{A_1} = \cos 55^\circ = 0.57\))

(b)(ii) 3.
At a minimum, the two waves meet in antiphase (phase difference is \(180^\circ\) or \(\pi\text{ rad}\)). However, since the amplitude of the wave from \(S_2\) is now smaller than the amplitude of the wave from \(S_1\) (\(A_2 \neq A_1\)), the waves do not completely cancel each other out, resulting in a non-zero resultant amplitude and non-zero intensity.

(a)(i)
- When two (or more) waves meet/overlap (at a point), resultant displacement is the (vector / algebraic) sum of individual displacements. `[B1]`

(a)(ii)
- Constant / fixed phase difference (between the waves). `[B1]`
(Do not accept 'constant frequency' or 'constant wavelength' alone without constant phase difference)

(b)(i)
- \(x = \frac{\lambda D}{a}\) or substitution \(630 \times 10^{-9} = \frac{0.28 \times 10^{-3} \times x}{1.50}\) `[C1]`
- \(x = 3.4 \times 10^{-3}\text{ m}\) (or \(3.38 \times 10^{-3}\text{ m}\)) `[A1]`

(b)(ii) 1.
- Correct use of Malus's law \(I = I_0 \cos^2(55^\circ)\) `[C1]`
- \(0.33\) (or \(0.329\)) `[A1]`

(b)(ii) 2.
- \(\text{ratio} = \sqrt{0.329}\) or \(\cos 55^\circ\) `[C1]`
- \(0.57\) (allow \(0.574\)) `[A1]`

(b)(ii) 3.
- Waves meet in antiphase / with a phase difference of \(180^\circ\) / \(\pi\text{ rad}\) / path difference of \((n + \frac{1}{2})\lambda\) `[B1]`
- Amplitudes of the two waves are unequal / do not completely cancel `[B1]`

(a) Work done is the product of the force and the distance (or displacement) moved in the direction of the force.

(b) (i) Component of weight parallel to the slope \(W_\parallel = mg \sin\theta\)
\(W_\parallel = 450 \times 9.81 \times \sin(14^\circ)\)
\(W_\parallel = 4414.5 \times 0.24192 = 1068\text{ N} \approx 1070\text{ N}\)

(ii) 1. At constant speed, resultant force parallel to track is zero:
\(F_{\text{motor}} = W_\parallel + F_{\text{resistive}} = 1068 + 380 = 1448\text{ N}\) (or \(1070 + 380 = 1450\text{ N}\))
Power \(P = Fv\)
\(P = 1448 \times 4.2 = 6082\text{ W} = 6.08\text{ kW}\) (or \(1450 \times 4.2 = 6090\text{ W} = 6.09\text{ kW}\))

2. \(\text{Efficiency} = \frac{P_{\text{useful}}}{P_{\text{input}}} \times 100\%\)
\(P_{\text{input}} = \frac{6082}{0.68} = 8944\text{ W} = 8.9\text{ kW}\) (or \(\frac{6090}{0.68} = 8956\text{ W} = 9.0\text{ kW}\))

(c) By conservation of energy, work done against resistive force = initial kinetic energy:
\(F_{\text{resistive}} \times d = \frac{1}{2}mv^2\)
\(380 \times d = \frac{1}{2} \times 450 \times (4.2)^2\)
\(380 \times d = 3969\text{ J}\)
\(d = \frac{3969}{380} = 10.44\text{ m} \approx 10\text{ m}\) (or \(10.4\text{ m}\))

Alternatively, using \(F = ma \Rightarrow a = -\frac{380}{450} = -0.8444\text{ m s}^{-2}\) and \(v^2 = u^2 + 2as \Rightarrow 0 = 4.2^2 + 2(-0.8444)s \Rightarrow s = 10.4\text{ m}\).

(a) force \(\times\) distance / displacement moved in the direction of the force [B1]

(b)(i) \(W = mg \sin\theta\) or \(450 \times 9.81 \times \sin(14^\circ)\) [C1]
\(= 1068\text{ N}\) shown to at least 3 significant figures leading to \(1070\text{ N}\) [A1]

(b)(ii) 1. \(F = 1068 + 380 = 1448\text{ N}\) (or \(1070 + 380 = 1450\text{ N}\)) [C1]
\(P = Fv\) (or substitution \(1448 \times 4.2\) or \(1450 \times 4.2\)) [C1]
\(P = 6.1\text{ kW}\) or \(6.08\text{ kW}\) or \(6080\text{ W}\) (allow \(6.09\text{ kW}\) or \(6090\text{ W}\)) [A1]

2. \(\text{efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\) or \(P_{\text{in}} = \frac{6082}{0.68}\) (allow ECF from (b)(ii)1.) [C1]
\(P_{\text{in}} = 8.9\text{ kW}\) or \(8940\text{ W}\) (allow \(9.0\text{ kW}\) or \(8960\text{ W}\)) [A1]

(c) \(W = Fd = \frac{1}{2}mv^2\) or \(F = ma\) and \(v^2 = u^2 + 2as\) [C1]
\(d = 10\text{ m}\) or \(10.4\text{ m}\) [A1]

(a) (i) Electromotive force (e.m.f.) is the energy transferred from chemical (or other forms) to electrical energy per unit charge, or work done per unit charge in driving charge around a complete circuit.

(ii) Kirchhoff's second law states that the algebraic sum of the e.m.f.s is equal to the algebraic sum of the potential differences (p.d.s) around any closed loop in a circuit. It is based on the conservation of energy.

(b) (i) Total resistance of the primary circuit:
\(R_{\text{total}} = R_{\text{XY}} + R + r = 8.40 + 3.60 + 0.50 = 12.50\,\Omega\)

Current in the circuit:
\(I = \frac{E_1}{R_{\text{total}}} = \frac{4.50}{12.50} = 0.360\text{ A} P.d. across wire XY: \)V_{\text{XY}} = I R_{\text{XY}} = 0.360 \times 8.40 = 3.024\text{ V} \approx 3.02\text{ V}

(b) (ii) Using the drift speed equation:
\(I = n A v q \implies v = \frac{I}{n A e} \)v = \frac{0.360}{(8.50 \times 10^{28})(1.40 \times 10^{-7})(1.60 \times 10^{-19})}
\(v = \frac{0.360}{1.904 \times 10^3} = 1.89 \times 10^{-4}\text{ m s}^{-1} (c) (i) At null deflection (zero current in the galvanometer branch), no current flows through the second cell, so there is no potential difference across its internal resistance (no lost volts). (c) (ii) Potential gradient along wire XY: \)k = \frac{V_{\text{XY}}}{L} = \frac{3.024\text{ V}}{1.20\text{ m}} = 2.52\text{ V m}^{-1}

Length \(l = 78.0\text{ cm} = 0.780\text{ m} \)E_2 = k \times l = 2.52 \times 0.780 = 1.9656\text{ V} \approx 1.97\text{ V}

(a) (i)
B1: Energy transferred from other forms to electrical energy (or work done)
B1: per unit charge (driving charge round a complete circuit)

(a) (ii)
B1: (algebraic) sum of e.m.f.s = (algebraic) sum of p.d.s around a closed loop / circuit
B1: (conservation of) energy

(b) (i)
C1: Total resistance \(R = 8.40 + 3.60 + 0.50 = 12.50\,\Omega\)
C1: Current \(I = \frac{4.50}{12.50} = 0.360\text{ A}\) OR potential divider ratio expression \(V_{\text{XY}} = 4.50 \times \frac{8.40}{12.50}\)
A1: \(V_{\text{XY}} = 3.02\text{ V}\) (allow 3.0 V or 3.024 V)

(b) (ii)
C1: \(v = \frac{I}{n A e}\) with substitution of \(e = 1.60 \times 10^{-19}\text{ C}\) and candidate's \(I\)
A1: \(v = 1.89 \times 10^{-4}\text{ m s}^{-1}\) (allow ECF from (b)(i))

(c) (i)
B1: (At balance point) current in the cell/galvanometer branch is zero, so no p.d. / lost volts across internal resistance

(c) (ii)
C1: \(E_2 = V_{\text{XY}} \times \frac{0.780}{1.20}\) OR \(\text{potential gradient} = 2.52\text{ V m}^{-1}\)
A1: \(E_2 = 1.97\text{ V}\) (allow 1.96 V or 1.966 V, allow ECF from (b)(i))

(a) A hadron is a non-fundamental particle made of quarks (or a particle that experiences the strong nuclear force).
A baryon consists of three quarks (or three antiquarks), whereas a meson consists of a quark and an antiquark pair.

(b)(i) In \(\beta^+\) decay, a proton changes to a neutron, emitting a positron and an electron neutrino.
Nucleon number is conserved: \(18 = 18 + 0 + 0\).
Proton number is conserved: \(9 = 8 + (+1) + 0\).
Hence: \(^{18}_{9}\text{F} \rightarrow \,^{18}_{8}\text{O} + \,^{0}_{+1}\beta^+ + \nu_{\text{e}}\).

(b)(ii) A proton (uud) changes into a neutron (udd), so an up (u) quark changes into a down (d) quark: \(\text{u} \rightarrow \text{d}\).

(c) The total decay energy available is fixed (equal to the mass-energy difference between parent and daughter nuclei). This total energy is shared between the positron and the electron neutrino (as well as a small recoil energy of the daughter nucleus). Because the neutrino can take any fraction of this energy, the kinetic energy of the positron can take any value between zero and the maximum.

(d) A fluorine-18 nucleus contains 9 protons and \(18 - 9 = 9\) neutrons.
Each proton consists of uud: charge = \(+1e\).
Each neutron consists of udd: charge = \(0\).
Total quark charge = \(9 \times (+1e) + 9 \times (0) = 9e\).
Alternatively, total up quarks = \(9(2) + 9(1) = 27\), total down quarks = \(9(1) + 9(2) = 27\).
Total charge = \(27 \times (+\frac{2}{3}e) + 27 \times (-\frac{1}{3}e) = 27 \times (+\frac{1}{3}e) = +9e\).
\[Q = 9 \times 1.60 \times 10^{-19}\text{ C} = +1.44 \times 10^{-18}\text{ C}\]

(a)
- Hadron: particle composed of quarks / experiences the strong force [B1]
- Baryon: (combination of) 3 quarks AND Meson: quark and antiquark (pair) [B1]

(b)(i)
- Nucleon and proton numbers correct for oxygen (18 and 8) and positron (0 and +1) [B1]
- Electron neutrino (\(\nu_{\text{e}}\) or \(\nu\)) correctly included [B1]

(b)(ii)
- Up (quark) changes to down (quark) / \(\text{u} \rightarrow \text{d}\) [B1]

(c)
- Fixed / total energy released is shared between the positron and the neutrino [B1]
- Neutrino can carry away variable / any fraction of the energy (leading to a range of positron energies) [B1]

(d)
- Clear method shown to determine total charge, e.g. \(9 \times 1.60 \times 10^{-19}\) or calculation using individual up and down quarks [C1]
- \((+)1.44 \times 10^{-18}\text{ C}\) (allow 2 or 3 s.f.) [A1]

(a)
(i) Raw length of rule \( L = (1.000 \pm 0.001)\text{ m} = 100.0\text{ cm} \).
Raw mass of rule \( M = (0.120 \pm 0.001)\text{ kg} = 120.0\text{ g} \).

(ii) Measured cord spacing \( s = 0.200\text{ m} = 20.0\text{ cm} \).
Absolute uncertainty in \( s \), \( \Delta s = 1\text{ mm} = 0.001\text{ m} \) (or \( 2\text{ mm} \) accounting for positioning both cord clamps).
\[ \text{Percentage uncertainty} = \frac{\Delta s}{s} \times 100\% = \frac{0.002}{0.200} \times 100\% = 1.0\% \]

(b)
Timing at least 20 oscillations twice to reduce random uncertainty:
- Run 1: \( 20 T = 26.4\text{ s} \implies T_1 = 1.32\text{ s} \)
- Run 2: \( 20 T = 26.6\text{ s} \implies T_2 = 1.33\text{ s} \)
Mean period \( T = 1.325\text{ s} \).

(c) Data Table:
Six distinct sets of readings covering the required range:

| \( x / \text{m} \) | \( x^2 / \text{m}^2 \) | \( t_1 / \text{s} \text{ (20 osc)} \) | \( t_2 / \text{s} \text{ (20 osc)} \) | \( T / \text{s} \) | \( T^2 / \text{s}^2 \) |
|---|---|---|---|---|---|
| 0.100 | 0.0100 | 16.4 | 16.6 | 0.825 | 0.681 |
| 0.180 | 0.0324 | 18.2 | 18.0 | 0.905 | 0.819 |
| 0.250 | 0.0625 | 20.2 | 20.4 | 1.015 | 1.030 |
| 0.320 | 0.1024 | 22.8 | 23.0 | 1.145 | 1.311 |
| 0.390 | 0.1521 | 25.8 | 25.6 | 1.285 | 1.651 |
| 0.450 | 0.2025 | 28.4 | 28.6 | 1.425 | 2.031 |

All columns have proper quantity/unit headings; raw times given to 0.1 s; calculated \( T^2 \) given to consistent significant figures (3 or 4 s.f.).

(d) Graph:
- Scales chosen so points occupy \( > 50\% \) of grid in both directions (e.g. \( x^2 \)-axis: \( 0 \text{ to } 0.25\text{ m}^2 \); \( y \)-axis: \( 0.5 \text{ to } 2.2\text{ s}^2 \)).
- Points plotted accurately to within \( \pm 0.5 \) small square.
- Line of best fit drawn with balanced point distribution.

(e) Gradient and Intercept:
Using two well-separated points on the line, e.g., \( (x_1^2, y_1) = (0.020, 0.750) \) and \( (x_2^2, y_2) = (0.200, 2.010) \):
\[ \text{Gradient } m = \frac{2.010 - 0.750}{0.200 - 0.020} = \frac{1.260}{0.180} = 7.00\text{ s}^2\text{ m}^{-2} \]
\[ y\text{-intercept } c = y_1 - m x_1^2 = 0.750 - 7.00(0.020) = 0.610\text{ s}^2 \]

(f)
(i) Comparing \( T^2 = P x^2 + Q \) with \( y = m x + c \):
\[ P = \text{gradient} = 7.00\text{ s}^2\text{ m}^{-2} \]
\[ Q = y\text{-intercept} = 0.610\text{ s}^2 \]

(ii) From the theoretical expression:
\[ P = \frac{32 \pi^2 h m}{g s^2 (M + 2m)} \]
Rearranging for \( g \):
\[ g = \frac{32 \pi^2 h m}{P s^2 (M + 2m)} \]
Substituting values (\( h = 0.400\text{ m} \), \( m = 0.100\text{ kg} \), \( M = 0.120\text{ kg} \), \( s = 0.200\text{ m} \), \( P = 7.00\text{ s}^2\text{ m}^{-2} \)):
\[ g = \frac{32 \pi^2 (0.400)(0.100)}{7.00 \times (0.200)^2 \times (0.120 + 2 \times 0.100)} = \frac{12.633}{7.00 \times 0.0400 \times 0.320} = \frac{12.633}{0.0896} = 9.81\text{ m s}^{-2} \]

Part (a) [2 marks]
- B1: Value of \( L \) recorded to nearest mm with unit (e.g. 1.000 m or 100.0 cm) AND value of \( M \) recorded with unit.
- B1: Correct percentage uncertainty calculation in \( s \) using absolute uncertainty \( 1\text{ mm} \le \Delta s \le 3\text{ mm} \).

Part (b) [2 marks]
- M1: Evidence of repeated timings for a minimum of 10 oscillations (or \( n \ge 10 \)).
- A1: \( T \) calculated correctly from raw times, falling within realistic experimental range (\( 0.6\text{ s} \le T \le 2.0\text{ s} \)) with unit.

Part (c) [7 marks]
- Table - Range (B1): At least 6 sets of readings with \( x \) values spanning at least \( 0.300\text{ m} \).
- Table - Headings (B1): All column headings include quantity and SI unit separating symbol (e.g., \( x^2 / \text{m}^2 \), \( T^2 / \text{s}^2 \)).
- Table - Raw data precision (B1): All raw time measurements recorded to consistent precision (0.1 s or 0.01 s).
- Table - Derived quantities (B1): Values of \( x^2 \) and \( T^2 \) calculated correctly.
- Table - Significant figures (B1): Number of significant figures for \( T^2 \) matches or is one more than the significant figures of raw timing data.
- Quality of data (B2): Scatter of plotted points around a straight line trend is minimal (2 marks for very tight fit, 1 mark for moderate scatter).

Part (d) [4 marks]
- Axes (B1): Sensible linear scales where points occupy at least half the grid in both dimensions; no awkward scale factors (e.g. 3, 7, 9).
- Plotting (B2): All points plotted accurately to within \( \pm 0.5 \) small grid square. Deduct 1 mark for any single plotting error.
- Line of best fit (B1): Straight line drawn with balanced number of points on either side along the entire length; no thick or kinked lines.

Part (e) [2 marks]
- Gradient (B1): Triangle hypotenuse used is at least half the length of the drawn line; coordinates read correctly to within \( 0.5 \) small square.
- Intercept (B1): Correctly read from vertical axis where \( x^2 = 0 \) or correctly computed via \( y - m x \) using a point on the line.

Part (f) [3 marks]
- Values and units (B1): \( P \) equated to gradient and \( Q \) equated to intercept, with correct units (\( \text{s}^2\text{ m}^{-2} \) or \( \text{s}^2 / \text{m}^2 \) for \( P \), and \( \text{s}^2 \) for \( Q \)).
- **Method for \( g \) (M1)**: Correct algebraic transposition linking \( P \) to \( g \): \( g = \frac{32 \pi^2 h m}{P s^2 (M + 2m)} \) with candidate's values substituted.
- **Final \( g \) value (A1)**: Value of \( g \) calculated to 2 or 3 significant figures with standard unit (\( \text{m s}^{-2} \) or \( \text{N kg}^{-1} \)).

(a) (i) A radian is the angle subtended at the centre of a circle by an arc of length equal to the radius of the circle.

(ii) Velocity is a vector quantity that possesses both magnitude (speed) and direction. As the object travels around the circular path, its direction of motion is constantly changing, even though its speed is constant. Since acceleration is defined as the rate of change of velocity, the changing direction implies a non-zero acceleration directed towards the centre of the circle.

(b) (i) Resolving forces vertically for the ball in vertical equilibrium:
\[T \cos 32^\circ = mg\]
\[T = \frac{0.24 \times 9.81}{\cos 32^\circ} = \frac{2.3544}{0.84805} = 2.78\text{ N} \approx 2.8\text{ N}\]

(ii) The radius of the horizontal circular path is given by:
\[r = L \sin 32^\circ = 0.75 \times \sin 32^\circ = 0.397\text{ m} \approx 0.40\text{ m}\]

(iii) The horizontal component of tension provides the centripetal force:
\[T \sin 32^\circ = m r \omega^2\]
Alternatively, using \(\tan 32^\circ = \frac{r \omega^2}{g}\):
\[\omega^2 = \frac{g \tan 32^\circ}{r} = \frac{9.81 \times \tan 32^\circ}{0.3974} = \frac{6.130}{0.3974} = 15.42\text{ rad}^2\text{ s}^{-2}\]
\[\omega = \sqrt{15.42} = 3.93\text{ rad s}^{-1} \approx 3.9\text{ rad s}^{-1}\]

(iv) In vertical equilibrium at the new angle \(\theta = 55^\circ\):
\[T_{\text{new}} \cos 55^\circ = mg\]
\[T_{\text{new}} = \frac{0.24 \times 9.81}{\cos 55^\circ} = \frac{2.3544}{0.5736} = 4.10\text{ N} \approx 4.1\text{ N}\]

(a) (i) [B1] Angle subtended at the centre of a circle by an arc equal in length to the radius.

(ii) [M1] Velocity is a vector / has direction, AND direction is continuously changing.
[A1] Acceleration is the rate of change of velocity (so changing velocity implies acceleration).

(b) (i) [C1] Vertical resolution equation stated: \(T \cos 32^\circ = mg\) or \(T \cos 32^\circ = 0.24 \times 9.81\).
[A1] Calculation shown to at least 3 s.f.: \(T = \frac{2.3544}{0.84805} = 2.78\text{ N}\) (hence \(2.8\text{ N}\)).

(ii) [A1] \(r = 0.75 \sin 32^\circ = 0.40\text{ m}\) (accept \(0.397\text{ m}\)).

(iii) [C1] Correct centripetal force equation, e.g. \(T \sin 32^\circ = m r \omega^2\) or \(\tan 32^\circ = \frac{r \omega^2}{g}\).
[A1] \(\omega = 3.9\text{ rad s}^{-1}\) (or \(3.93\text{ rad s}^{-1}\)).

(iv) [C1] \(T_{\text{new}} = \frac{mg}{\cos 55^\circ}\) or \(T_{\text{new}} \cos 55^\circ = 0.24 \times 9.81\).
[A1] \(T_{\text{new}} = 4.1\text{ N}\) (or \(4.10\text{ N}\)).

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

(b)(i)
The gravitational force between the planet and the satellite provides the required centripetal force for circular motion:
\[ F_g = F_c \]
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]
Dividing both sides by \(m\) and multiplying by \(r\):
\[ v^2 = \frac{GM}{r} \]
Taking the square root gives:
\[ v = \sqrt{\frac{GM}{r}} \]

(b)(ii)
The gravitational potential \(\phi\) at distance \(R\) from the centre of mass is:
\[ \phi = -\frac{GM}{R} \]
Substituting the given values:
\[ \phi = -\frac{(6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2})(6.40 \times 10^{23} \text{ kg})}{4.20 \times 10^6 \text{ m}} \]
\[ \phi = -1.016 \times 10^7 \text{ J kg}^{-1} \approx -1.02 \times 10^7 \text{ J kg}^{-1} \]

(b)(iii)
At the maximum height \(h = R\), the distance of the probe from the centre of the planet is \(r_2 = R + h = 2R\).
At maximum height, the kinetic energy of the probe is zero, so all initial kinetic energy has been converted into gravitational potential energy:
\[ E_k = \Delta E_p = m(\phi_2 - \phi_1) = GMm \left( \frac{1}{R} - \frac{1}{2R} \right) = \frac{GMm}{2R} \]
\[ E_k = \frac{1}{2} m |\phi_1| = \frac{1}{2} \times 850 \text{ kg} \times 1.0164 \times 10^7 \text{ J kg}^{-1} \]
\[ E_k = 4.32 \times 10^9 \text{ J} \]

(a)
- work done per unit mass [B1]
- (in bringing a small test mass) from infinity (to the point) [B1]

(b)(i)
- Equates gravitational force to centripetal force: \(\frac{GMm}{r^2} = \frac{mv^2}{r}\) [M1]
- Simplifies algebraically to obtain \(v = \sqrt{\frac{GM}{r}}\) with at least one intermediate step shown [A1]

(b)(ii)
- Uses \(\phi = -\frac{GM}{R}\) and substitutes values: \(\frac{6.67 \times 10^{-11} \times 6.40 \times 10^{23}}{4.20 \times 10^6}\) [C1]
- \(\phi = -1.02 \times 10^7 \text{ J kg}^{-1}\) (must include negative sign and correct unit) [A1]

(b)(iii)
- Recognises that final distance from centre is \(2R\) or calculates \(\Delta \phi = \frac{GM}{2R} = 5.08 \times 10^6 \text{ J kg}^{-1}\) [C1]
- Uses \(E_k = \Delta E_p = m \Delta \phi\) or \(E_k = GMm\left(\frac{1}{R} - \frac{1}{2R}\right)\) [C1]
- \(E_k = 4.32 \times 10^9 \text{ J}\) (allow \(4.3 \times 10^9 \text{ J}\)) [A1]

(a) Internal energy is defined as the sum of the random distribution of kinetic and potential energies associated with the molecules (or atoms/particles) of a system.

(b) The first law of thermodynamics is given by $\Delta U = q + W$ (or equivalent word statement), where:
$\Delta U =$ increase in internal energy of the system,
$q =$ thermal energy (heat) supplied to the system,
$W =$ work done on the system.

(c) In an ideal gas, there are no intermolecular forces between particles (except during collisions), which means that the intermolecular potential energy is zero. Consequently, the internal energy consists solely of the random kinetic energy of the particles. Because the average translational kinetic energy of gas molecules is directly proportional to the thermodynamic temperature ($E_k = \frac{3}{2}kT$), the internal energy depends only on temperature.

(d)(i) Using the ideal gas equation $pV = nRT$ at state $P$:
$$T_P = \frac{p_P V_P}{nR} = \frac{(1.20 \times 10^5\text{ Pa}) \times (1.80 \times 10^{-3}\text{ m}^3)}{(0.250\text{ mol}) \times (8.31\text{ J K}^{-1}\text{ mol}^{-1})} = \frac{216}{2.0775} = 103.97\text{ K} \approx 104\text{ K}$$

(ii) At constant volume $V_P = V_Q$, pressure is directly proportional to temperature ($p \propto T$):
$$T_Q = T_P \times \frac{p_Q}{p_P} = 103.97\text{ K} \times \frac{3.60 \times 10^5}{1.20 \times 10^5} = 103.97 \times 3 = 311.9\text{ K} \approx 312\text{ K}$$

(iii) The work done on the gas at constant pressure is $W = -p \Delta V$ (or $W = p \Delta V$ with sign convention for work done on the gas):
$$\Delta V = V_P - V_R = 1.80 \times 10^{-3} - 5.40 \times 10^{-3} = -3.60 \times 10^{-3}\text{ m}^3$$
$$W = -p \Delta V = -(1.20 \times 10^5\text{ Pa}) \times (-3.60 \times 10^{-3}\text{ m}^3) = +432\text{ J}$$
(Work is positive because the gas is compressed, i.e. work is done on the gas).

(iv) Values for each process:
- Process $P \to Q$: Constant volume means no displacement, so $W = 0$. From $\Delta U = q + W \implies q = \Delta U - W = +648 - 0 = +648\text{ J}$.
- Process $Q \to R$: Isothermal means $\Delta T = 0$, so $\Delta U = 0\text{ J}$. From $\Delta U = q + W \implies 0 = +712 + W \implies W = -712\text{ J}$.
- Process $R \to P$: Since $T_R = T_Q$, the internal energy at state $R$ is equal to that at state $Q$. Thus, $\Delta U_{R \to P} = U_P - U_R = U_P - U_Q = -\Delta U_{P \to Q} = -648\text{ J}$. Using $W = +432\text{ J}$, $q = \Delta U - W = -648 - 432 = -1080\text{ J}$.
- Complete cycle: For a complete cycle, internal energy returns to its starting value, so $\Delta U = 0\text{ J}$. Sum of $W = 0 + (-712) + 432 = -280\text{ J}$. Sum of $q = +648 + 712 + (-1080) = +280\text{ J}$.

(a)
- Sum of (random distribution of) kinetic and potential energies [B1]
- of atoms / molecules / particles (in a system) [B1]

(b)
- $\Delta U = q + W$ (or statement in words representing this relationship) [B1]
- Correct identification of symbols with signs/directions: $\Delta U$ = increase in internal energy, $q$ = thermal energy supplied to the system, $W$ = work done on the system [B1]

(c)
- Ideal gas has no intermolecular forces / potential energy is zero [B1]
- (Mean) kinetic energy of molecules is proportional to thermodynamic temperature (hence internal energy depends only on $T$) [B1]

(d)(i)
- $pV = nRT \implies T = \frac{pV}{nR}$ with correct substitution: $T_P = \frac{1.20 \times 10^5 \times 1.80 \times 10^{-3}}{0.250 \times 8.31}$ [C1]
- $T_P = 104\text{ K}$ (accept $103.97\text{ K}$ or $100\text{ K}$) [A1]

(d)(ii)
- $T_Q = 312\text{ K}$ (allow ECF from (d)(i), $T_Q = 3 \times T_P$) [A1]

(d)(iii)
- $W = p \Delta V = 1.20 \times 10^5 \times (5.40 \times 10^{-3} - 1.80 \times 10^{-3})$ [C1]
- $W = (+)432\text{ J}$ [A1]

(d)(iv)
- Process $P \to Q$: $q = +648\text{ J}$ and $W = 0$ [B1]
- Process $Q \to R$: $W = -712\text{ J}$ and $\Delta U = 0$ [B1]
- Process $R \to P$ and Complete cycle: all four entries correct ($R \to P$: $q = -1080\text{ J}$, $\Delta U = -648\text{ J}$; cycle: $q = +280\text{ J}$, $W = -280\text{ J}$, $\Delta U = 0$) [B1]

(a)
Any two valid assumptions of the kinetic theory of gases:
- Volume of the gas molecules/particles is negligible compared to the volume occupied by the gas (or volume of the container).
- No intermolecular forces between molecules except during collisions (potential energy is negligible / zero).
- Collisions between molecules and with the walls of the container are perfectly elastic.
- The duration of a collision is negligible compared to the time between collisions.
- Molecules move in continuous, rapid, random motion.

(b)(i)
Using the ideal gas equation:
\[ pV = NkT \]
Convert temperature to Kelvin:
\[ T_1 = 27.0 + 273.15 = 300.15\text{ K} \approx 300\text{ K} \]
Substitute values:
\[ N = \frac{pV}{kT} = \frac{2.50 \times 10^5 \times 4.20 \times 10^{-2}}{1.38 \times 10^{-23} \times 300.15} = 2.535 \times 10^{24} \approx 2.54 \times 10^{24} \]

(b)(ii)
The mean translational kinetic energy per molecule is given by:
\[ \langle E_{\text{k}} \rangle = \frac{3}{2}kT \]
Total kinetic energy:
\[ E_{\text{k}} = \frac{3}{2}NkT \]
Increase in temperature:
\[ \Delta T = 147.0 - 27.0 = 120.0\text{ K} \]
\[ \Delta E_{\text{k}} = \frac{3}{2} N k \Delta T = \frac{3}{2} (2.535 \times 10^{24})(1.38 \times 10^{-23})(120.0) = 6.30 \times 10^3\text{ J} \]
*(Alternatively: \(\Delta E_{\text{k}} = \frac{3}{2} V \Delta p = \frac{3}{2} (4.20 \times 10^{-2}) (3.50 \times 10^5 - 2.50 \times 10^5) = 6.30 \times 10^3\text{ J}\))*

(c)
At \(T_2 = 147.0^\circ\text{C} = 420.15\text{ K}\) and \(p = 2.50 \times 10^5\text{ Pa}\), the number \(N_2\) of molecules remaining is:
\[ N_2 = \frac{pV}{kT_2} = \frac{2.50 \times 10^5 \times 4.20 \times 10^{-2}}{1.38 \times 10^{-23} \times 420.15} = 1.810 \times 10^{24} \]
*(Alternatively, \(N_2 = N_1 \times \frac{T_1}{T_2} = 2.535 \times 10^{24} \times \frac{300.15}{420.15} = 1.810 \times 10^{24}\))*

Number of molecules that escape:
\[ \Delta N = N_1 - N_2 = 2.535 \times 10^{24} - 1.810 \times 10^{24} = 7.25 \times 10^{23} \]

(a)
- Any two valid assumptions: [B1] each, max [2]
- Volume of molecules negligible compared to volume of container / gas
- No forces between molecules except during collisions
- Collisions are (perfectly) elastic
- Time / duration of collisions is negligible compared to time between collisions
- Molecules in random / rapid / continuous motion

(b)(i)
- \(T = 300\text{ K}\) and \(pV = NkT\) or \(pV = nRT\) with \(N = n N_{\text{A}}\) [C1]
- \(N = 2.54 \times 10^{24}\) (allow \(2.53 \times 10^{24}\) to \(2.54 \times 10^{24}\)) [A1]

(b)(ii)
- \(\Delta T = 120\text{ K}\) seen or used [C1]
- \(\Delta E_{\text{k}} = \frac{3}{2} N k \Delta T\) or \(\Delta E_{\text{k}} = \frac{3}{2} V \Delta p\) [C1]
- \(\Delta E_{\text{k}} = 6.30 \times 10^3\text{ J}\) (or \(6300\text{ J}\)) [A1]

(c)
- Calculation of remaining molecules \(N_2 = \frac{pV}{kT_2}\) or ratio \(N_2 = N_1 \left(\frac{T_1}{T_2}\right)\) [C1]
- \(N_2 = 1.81 \times 10^{24}\) [C1]
- \(\Delta N = N_1 - N_2 = 7.25 \times 10^{23}\) (allow \(7.2 \times 10^{23}\) to \(7.3 \times 10^{23}\)) [A1]

(a) Simple harmonic motion is defined as motion where the acceleration of a body is directly proportional to its displacement from a fixed/equilibrium position, and the acceleration is always directed towards that fixed/equilibrium position (or in the opposite direction to the displacement).

(b)(i) The angular frequency \(\omega\) is:
\[\omega = 2\pi f = 2\pi \times 2.40 = 15.08\text{ rad s}^{-1}\]
The total energy \(E\) is given by:
\[E = \frac{1}{2} m \omega^2 x_0^2\]
Converting amplitude to metres: \(x_0 = 4.5\text{ cm} = 0.045\text{ m}\).
\[E = \frac{1}{2} \times 0.350 \times (15.08)^2 \times (0.045)^2\]
\[E = 0.5 \times 0.350 \times 227.4 \times 0.002025 = 8.06 \times 10^{-2}\text{ J} = 0.081\text{ J}\]

(b)(ii) The speed \(v\) at displacement \(x = 2.8\text{ cm} = 0.028\text{ m}\) is given by:
\[v = \omega \sqrt{x_0^2 - x^2}\]
\[v = 15.08 \times \sqrt{0.045^2 - 0.028^2} = 15.08 \times \sqrt{0.002025 - 0.000784}\]
\[v = 15.08 \times \sqrt{0.001241} = 15.08 \times 0.03523 = 0.531\text{ m s}^{-1}\]

(c) The oscillations are damped (or subject to light damping). The total mechanical energy decreases continuously because work is done against viscous forces (drag), transforming mechanical energy into internal/thermal energy in the fluid.

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

(b)(i)
- \(\omega = 2\pi \times 2.40 = 15.1\text{ rad s}^{-1}\) and \(E = \frac{1}{2} m \omega^2 x_0^2\) or \(E = \frac{1}{2} k x_0^2\) with \(k = m\omega^2\) [C1]
- \(E = 0.081\text{ J}\) or \(8.06 \times 10^{-2}\text{ J}\) (allow 2 or 3 s.f.) [A1]

(b)(ii)
- \(v = \omega \sqrt{x_0^2 - x^2}\) with correct substitutions \(x_0 = 0.045\) and \(x = 0.028\) [C1]
- \(v = 0.53\text{ m s}^{-1}\) or \(0.531\text{ m s}^{-1}\) [A1]

(c)
- Damped (oscillation) / damping (allow light damping) [B1]
- Mechanical energy decreases / is dissipated as thermal energy (or heat) in the liquid / surroundings [B1]

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

(b)(i)
On the outer surface of the sphere.

(b)(ii)
For an isolated sphere:
\[V = \frac{Q}{4\pi\varepsilon_0 R}\]
Since \(C = \frac{Q}{V}\):
\[C = 4\pi\varepsilon_0 R\]
Substitute the values:
\[C = 4\pi \times 8.85\times 10^{-12}\text{ F m}^{-1} \times 0.15\text{ m} = 1.668\times 10^{-11}\text{ F} = 16.7\text{ pF} \approx 17\text{ pF}\]

(b)(iii)
\[V_0 = \frac{Q}{C} = \frac{3.6\times 10^{-8}\text{ C}}{1.668\times 10^{-11}\text{ F}} = 2.158\times 10^3\text{ V} \approx 2.2\times 10^3\text{ V}\]

(c)(i)
The electric potential at infinity is \(V_\infty = 0\).
Work done against the repulsive electric field to move from the surface to infinity:
\[E_{\text{k}} = q(V_0 - V_\infty) = qV_0\]
\[E_{\text{k}} = 1.2\times 10^{-9}\text{ C} \times 2.158\times 10^3\text{ V} = 2.59\times 10^{-6}\text{ J} \approx 2.6\times 10^{-6}\text{ J}\]

(c)(ii)
\[E_{\text{k}} = \frac{1}{2}mv^2\]
\[v = \sqrt{\frac{2E_{\text{k}}}{m}} = \sqrt{\frac{2 \times 2.59\times 10^{-6}\text{ J}}{4.5\times 10^{-5}\text{ kg}}} = \sqrt{0.1151} = 0.339\text{ m s}^{-1} \approx 0.34\text{ m s}^{-1}\]

(d)(i)
Total charge is conserved:
\[Q = C_1 V_1 = 47\times 10^{-6}\text{ F} \times 24\text{ V} = 1.128\times 10^{-3}\text{ C}\]
When connected in parallel:
\[Q = (C_1 + C) V_2\]
\[C_1 + C = \frac{1.128\times 10^{-3}\text{ C}}{8.0\text{ V}} = 1.41\times 10^{-4}\text{ F} = 141\text{ }\mu\text{F}\]
\[C = 141\text{ }\mu\text{F} - 47\text{ }\mu\text{F} = 94\text{ }\mu\text{F}\]

(d)(ii)
Initial energy stored:
\[E_{\text{initial}} = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 47\times 10^{-6} \times 24^2 = 0.01354\text{ J}\]
Final energy stored:
\[E_{\text{final}} = \frac{1}{2} (C_1 + C) V_2^2 = \frac{1}{2} \times 1.41\times 10^{-4} \times 8.0^2 = 0.00451\text{ J}\]
Loss in energy:
\[\Delta E = 0.01354\text{ J} - 0.00451\text{ J} = 0.00903\text{ J} \approx 9.0\times 10^{-3}\text{ J}\]

(a)
- Work done per unit charge [B1]
- (work done) in bringing a positive test charge from infinity (to the point) [B1]

(b)(i)
- On the (outer) surface / exterior of the sphere [B1]

(b)(ii)
- Uses \(C = 4\pi\varepsilon_0 R\) or \(V = \frac{Q}{4\pi\varepsilon_0 R}\) and \(C = \frac{Q}{V}\) [C1]
- Correct substitution shown: \(4\pi \times 8.85\times 10^{-12} \times 0.15 = 1.67\times 10^{-11}\text{ F}\) leading to \(17\text{ pF}\) [A1]

(b)(iii)
- \(V_0 = \frac{3.6\times 10^{-8}}{4\pi \times 8.85\times 10^{-12} \times 0.15}\) or \(\frac{3.6\times 10^{-8}}{1.67\times 10^{-11}}\) [C1]
- \(V_0 = 2.2\times 10^3\text{ V}\) (allow \(2.16\times 10^3\text{ V}\)) [A1]

(c)(i)
- \(E_{\text{k}} = q V_0\) or \(E_{\text{k}} = \frac{Q q}{4\pi\varepsilon_0 R}\) [C1]
- \(E_{\text{k}} = 1.2\times 10^{-9} \times 2160 = 2.6\times 10^{-6}\text{ J}\) [A1]

(c)(ii)
- \(\frac{1}{2} m v^2 = 2.59\times 10^{-6}\) [C1]
- \(v = 0.34\text{ m s}^{-1}\) (allow \(0.339\text{ m s}^{-1}\)) [A1]

(d)(i)
- \(C = 94\text{ }\mu\text{F}\) [A1]

(d)(ii)
- \(\text{energy loss} = 9.0\times 10^{-3}\text{ J}\) (or \(9.03\times 10^{-3}\text{ J}\)) [A1]

(a) Magnetic flux density is the force per unit length acting on a straight conductor carrying unit current placed at right angles to the magnetic field.

(b) When current flows, the charge carriers move through the magnetic field and experience a magnetic force \(F = Bvq\) perpendicular to both their velocity and the magnetic field. This force deflects the charge carriers towards one side of the slice, causing an accumulation of charge on opposite faces. This charge separation sets up a transverse electric field, which exerts an opposing electric force on subsequent carriers until the electric force balances the magnetic force (\(qE = Bvq\)), resulting in a steady potential difference across the slice known as the Hall voltage.

(c)(i) Using the Hall voltage formula:
\(V_{\text{H}} = \frac{BI}{ntq}\)

Substitute the given values:
\(B = 0.32\text{ T}\)
\(I = 45 \times 10^{-3}\text{ A}\)
\(n = 1.4 \times 10^{21}\text{ m}^{-3}\)
\(t = 0.18 \times 10^{-3}\text{ m}\)
\(q = 1.60 \times 10^{-19}\text{ C}\)

\(V_{\text{H}} = \frac{0.32 \times 45 \times 10^{-3}}{1.4 \times 10^{21} \times 0.18 \times 10^{-3} \times 1.60 \times 10^{-19}}\)
\(V_{\text{H}} = \frac{0.0144}{0.04032} = 0.357\text{ V} \approx 0.36\text{ V}\) (or \(360\text{ mV}\))

(c)(ii) The Hall voltage is inversely proportional to the number density of charge carriers (\(V_{\text{H}} \propto 1/n\)). Since metals have a much higher charge carrier density \(n\) than semiconductors, \(V_{\text{H}}\) in the metal slice is much smaller.

(a)
- Force per unit length [1]
- on a straight conductor carrying unit current normal to the magnetic field [1]

(b)
- Charge carriers moving perpendicular to the magnetic field experience a magnetic force (given by \(F = Bvq\)) [1]
- Charge carriers are deflected to one face/side, creating a separation of opposite charges on opposite faces [1]
- This creates a transverse electric field / p.d. that exerts an electric force until electric force balances magnetic force (\(qE = Bvq\)) [1]

(c)(i)
- Formula quoted or correct substitution: \(V_{\text{H}} = \frac{0.32 \times 45 \times 10^{-3}}{1.4 \times 10^{21} \times 0.18 \times 10^{-3} \times 1.60 \times 10^{-19}}\) [1] (C1)
- Final answer: \(V_{\text{H}} = 0.36\text{ V}\) (allow 0.357 V or 360 mV) [1] (A1)

(c)(ii)
- (Voltage is much) smaller because the number density \(n\) of charge carriers in a metal is much greater (and \(V_{\text{H}} \propto 1/n\)) [1] (B1)

(a) The root-mean-square (r.m.s.) value of an alternating current is that value of steady direct current which produces the same rate of dissipation of thermal energy (or mean power) in a given resistor.

(b)(i)
1. Angular frequency \(\omega = 2\pi f\)
\(f = \frac{\omega}{2\pi} = \frac{120\pi}{2\pi} = 60\text{ Hz} 2. \)V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{18}{\sqrt{2}} \approx 12.7\text{ V}

(b)(ii)
\(\langle P \rangle = \frac{V_{\text{rms}}^2}{R} = \frac{V_0^2}{2R} = \frac{18^2}{2 \times 450} = \frac{324}{900} = 0.36\text{ W} (c)(i) The ripple is reduced (or output is smoother) because a larger capacitance increases the time constant (\)\tau = RC\)), causing the capacitor to discharge more slowly between peaks.

(c)(ii) The ripple increases (fluctuates more) because a smaller load resistance decreases the time constant (\(\tau = RC\)), causing the capacitor to discharge more rapidly between peaks.

(a)
- The value of direct / steady current [1]
- that produces the same (rate of) heat dissipation / power in a resistor (accept 'in the same load') [1]

(b)(i)
- \(f = 60\text{ Hz}\) [B1]
- \(V_{\text{rms}} = 13\text{ V}\) or \(12.7\text{ V}\) [A1]

(b)(ii)
- \(\langle P \rangle = 0.36\text{ W}\) (allow calculation using \(V_{\text{rms}}^2 / R\) with ecf from (b)(i)) [A1]

(c)(i)
- Ripple decreases / smaller fluctuation because time constant / \(RC\) increases OR capacitor discharges more slowly [B1]

(c)(ii)
- Ripple increases / larger fluctuation because time constant / \(RC\) decreases OR capacitor discharges more quickly [B1]

(a) A photon is a discrete packet (or quantum) of energy of electromagnetic radiation.

(b) (i) The work function energy is the minimum amount of energy required to remove an electron from the surface of a metal.

(b) (ii) 1.
First, calculate the energy \(E\) of an incident photon:
\[ E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34} \text{ J s})(3.00 \times 10^8 \text{ m s}^{-1})}{240 \times 10^{-9} \text{ m}} = 8.288 \times 10^{-19} \text{ J} \]

Convert the work function energy \(\Phi\) into joules:
\[ \Phi = 3.68 \times 1.60 \times 10^{-19} \text{ J} = 5.888 \times 10^{-19} \text{ J} \]

Using Einstein's photoelectric equation:
\[ E_{\text{k,max}} = E - \Phi = 8.288 \times 10^{-19} - 5.888 \times 10^{-19} = 2.40 \times 10^{-19} \text{ J} \]

2.
The momentum \(p\) of the electron with maximum kinetic energy is:
\[ p = \sqrt{2 m_e E_{\text{k,max}}} = \sqrt{2 \times (9.11 \times 10^{-31} \text{ kg}) \times (2.40 \times 10^{-19} \text{ J})} = 6.613 \times 10^{-25} \text{ kg m s}^{-1} \]

The de Broglie wavelength is:
\[ \lambda_{\text{dB}} = \frac{h}{p} = \frac{6.63 \times 10^{-34} \text{ J s}}{6.613 \times 10^{-25} \text{ kg m s}^{-1}} = 1.00 \times 10^{-9} \text{ m} \text{ (or } 1.00 \text{ nm)} \]

(c) (i) The maximum kinetic energy remains unchanged. The energy of an individual photon depends only on the frequency (or wavelength) of the radiation (\(E = hf\)), which is constant.

(c) (ii) The rate of emission of photoelectrons doubles (increases by a factor of 2). Doubling the intensity doubles the rate of arrival of photons at the metal surface. Since each emitted electron results from a one-to-one interaction with a single photon, the rate of emission is directly proportional to the intensity.

(a) B1: quantum / packet of energy of electromagnetic radiation [1]

(b)(i) B1: minimum energy required to release / emit an electron from the (metal) surface [1]

(b)(ii)1.
C1: calculation of photon energy \(E = hc/\lambda = 8.29 \times 10^{-19}\text{ J}\) (or \(5.18\text{ eV}\))
C1: conversion of work function \(\Phi = 3.68 \times 1.60 \times 10^{-19} = 5.89 \times 10^{-19}\text{ J}\) (or subtraction in eV: \(5.18 - 3.68 = 1.50\text{ eV}\))
A1: \(E_{\text{k,max}} = 2.40 \times 10^{-19}\text{ J}\) (allow \(2.39 \times 10^{-19}\text{ J}\) to \(2.41 \times 10^{-19}\text{ J}\)) [3]

(b)(ii)2.
C1: \(p = \sqrt{2mE_k}\) or \(v = \sqrt{2E_k/m} = 7.26 \times 10^5\text{ m s}^{-1}\) and \(\lambda = h/mv\) / \(\lambda = h/p\)
A1: \(\lambda_{\text{dB}} = 1.00 \times 10^{-9}\text{ m}\) (allow \(1.0 \times 10^{-9}\text{ m}\) or \(1.00\text{ nm}\); allow ecf from (b)(ii)1) [2]

(c)(i) B1: no change / remains constant AND photon energy depends only on frequency / wavelength / is unchanged [1]

(c)(ii)
B1: (rate of emission / number of electrons per second) doubles / increases by a factor of 2
B1: (doubling intensity) doubles the number of photons incident per unit time / photon arrival rate AND one photon absorbed by one electron (one-to-one interaction) [2]

(a) An astronomical object whose luminosity is known.

(b) (i) The total energy emitted per unit time (or total power radiated) by the star across all wavelengths.

(ii) Using the Stefan-Boltzmann law:
\[L = 4\pi \sigma r^2 T^4\]
\[L = 4\pi \times (5.67 \times 10^{-8}) \times (3.20 \times 10^9)^2 \times (6400)^4\]
\[L = 4\pi \times 5.67 \times 10^{-8} \times 1.024 \times 10^{19} \times 1.6777 \times 10^{15}\]
\[L = 1.2244 \times 10^{28} \text{ W} \approx 1.22 \times 10^{28} \text{ W}\]

(c) Radiant flux intensity:
\[F = \frac{L}{4\pi d^2}\]
\[d = \sqrt{\frac{L}{4\pi F}} = \sqrt{\frac{1.2244 \times 10^{28}}{4\pi \times 2.70 \times 10^{-22}}}\]
\[d = \sqrt{3.6088 \times 10^{48}} = 1.90 \times 10^{24} \text{ m}\]

(d) (i) Redshift calculation:
\[\Delta \lambda = 493.1 - 486.1 = 7.0 \text{ nm}\]
\[\frac{\Delta \lambda}{\lambda} = \frac{v}{c}\]
\[v = c \times \frac{\Delta \lambda}{\lambda} = (3.00 \times 10^8) \times \frac{7.0}{486.1} = 4.320 \times 10^6 \text{ m s}^{-1} \approx 4.32 \times 10^6 \text{ m s}^{-1}\]

(ii) Hubble's law:
\[v = H_0 d \implies H_0 = \frac{v}{d}\]
\[H_0 = \frac{4.320 \times 10^6 \text{ m s}^{-1}}{1.900 \times 10^{24} \text{ m}} = 2.27 \times 10^{-18} \text{ s}^{-1}\]

(a) An astronomical object of known luminosity [B1]

(b) (i) Total power radiated / energy emitted per unit time by the star [B1]

(ii) \(L = 4\pi \sigma r^2 T^4\) AND substitution of values: \(4\pi \times 5.67 \times 10^{-8} \times (3.20 \times 10^9)^2 \times 6400^4\) [C1]
Calculation evaluated to at least 3 significant figures: \(1.224 \times 10^{28} \text{ W}\) leading to \(1.22 \times 10^{28} \text{ W}\) [A1]

(c) \(F = \frac{L}{4\pi d^2}\) rearranged to \(d = \sqrt{\frac{L}{4\pi F}}\) or correct numerical substitution [C1]
\(d = 1.90 \times 10^{24} \text{ m}\) (allow \(1.89 \times 10^{24} \text{ m}\) to \(1.90 \times 10^{24} \text{ m}\)) [A1]

(d) (i) \(\frac{\Delta \lambda}{\lambda} = \frac{v}{c}\) with \(\Delta \lambda = 7.0 \text{ nm}\) [C1]
\(v = 4.32 \times 10^6 \text{ m s}^{-1}\) (allow \(4.3 \times 10^6 \text{ m s}^{-1}\)) [A1]

(ii) \(v = H_0 d\) or \(H_0 = \frac{v}{d}\) [C1]
\(H_0 = 2.27 \times 10^{-18}\) (allow ecf from (c) and (d)(i)) [A1]
Unit: \(\text{s}^{-1}\) [B1]

### 1. Variables
- Independent variable: Potential difference \(V\) across the capacitor.
- Dependent variable: Maximum vertical height \(h\) reached by the swinging rod.
- Controlled variables: Mass of the rod \(M\), length of the rod within the magnetic field \(L\), capacitance \(C\), length of suspension threads, and the magnetic flux density \(B\).

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### 2. Diagram and Apparatus
- Diagram elements:
- A circuit containing a variable d.c. power supply, a protective charging resistor, a two-way switch (SPDT), a capacitor of known capacitance \(C\), and a digital voltmeter connected across the capacitor.
- The switch connects the capacitor either to the charging supply or to the suspended conducting rod.
- The conducting rod is suspended horizontally by two identical light insulating threads attached to a rigid clamp stand.
- The rod is positioned symmetrically within the pole pieces of a large U-shaped magnet (or Helmholtz coil pair) producing a uniform horizontal magnetic field perpendicular to the rod.
- Connecting wires attached to the ends of the rod are thin, flexible, and have sufficient slack so as not to resist the swinging motion.
- A vertical scale / grid and a high-frame-rate video camera mounted on a tripod perpendicular to the plane of oscillation.

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### 3. Procedure and Measurements
1. Measure the mass \(M\) of the conducting rod using an electronic top-pan balance.
2. Measure the effective length \(L\) of the rod inside the magnetic field using a vernier caliper or ruler.
3. Connect the circuit with the two-way switch in the charging position. Adjust the variable power supply until the digital voltmeter reads a chosen potential difference \(V\).
4. Ensure the rod is completely stationary in its equilibrium position.
5. Set the video camera to record. Flip the two-way switch to discharge the capacitor through the rod.
6. Play back the video recording frame-by-frame (or in slow motion) against the vertical scale to record the highest vertical position of the rod. Subtract the initial equilibrium height to find the maximum vertical displacement \(h\).
7. Repeat the discharge for the same value of \(V\) at least three times to calculate a mean value of \(h\).
8. Change the power supply voltage to obtain at least six different values of \(V\) over a wide range and measure the corresponding mean values of \(h\).

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### 4. Data Analysis
- Plot a graph of \(h\) on the \(y\)-axis against \(V^2\) on the \(x\)-axis (or \(\sqrt{h}\) against \(V\)).
- If the suggested relationship holds, the graph of \(h\) against \(V^2\) will be a straight line passing through the origin.
- Determine the gradient \(m\) of the line of best fit.
- From the equation \(h = \left( \frac{B^2 L^2 C^2}{2 M^2 g} \right) V^2\), the gradient is:
\[ m = \frac{B^2 L^2 C^2}{2 M^2 g} \]
- Calculate the magnetic flux density \(B\) using:
\[ B = \frac{M}{L C} \sqrt{2 g m} \]
where \(g = 9.81\text{ m s}^{-2}\).

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### 5. Safety Considerations
- High voltages or large discharge currents may present an electric shock hazard; ensure supply voltages remain within safe laboratory limits and avoid touching exposed terminals during discharge.
- Strong permanent magnets can cause pinching injuries or dislodge nearby ferromagnetic items; clamp magnets firmly in place and keep steel tools away.

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### 6. Additional Detail / Quality of Practical Technique
- Use thin, highly flexible copper flying leads so that mechanical resistance from the wires does not hinder the swing of the rod.
- Position the camera with its optical axis exactly horizontal and level with the rod's path to eliminate parallax error.
- Verify the capacitance \(C\) independently with a digital capacitance meter.
- Use pole pieces that are significantly wider than the displacement of the rod to ensure the magnetic field remains uniform throughout the initial impulse.

Defining the Problem (2 marks maximum):
- B1: Identifies \(V\) as the independent variable and \(h\) as the dependent variable (or specifies varying \(V\) and measuring \(h\)).
- B1: Identifies at least two relevant controlled variables (e.g. mass of rod \(M\), active length of rod \(L\), capacitance \(C\), length of suspension threads).

Methods of Data Collection (4 marks maximum):
- M1: Clear, labelled diagram of workable apparatus showing horizontal rod suspended by threads within a magnetic field, connected to capacitor discharge circuit, and a height-measuring scale.
- M1: Circuit diagram correctly showing d.c. power supply, voltmeter across capacitor, and two-way switch (SPDT) to switch from charging to discharging through the rod.
- M1: Method of measuring \(h\) dynamically (e.g. video camera with slow-motion / frame-by-frame playback positioned perpendicular to scale/grid, or motion sensor).
- M1: Method of measuring \(M\) using a balance / top-pan balance and \(L\) using a ruler / vernier caliper.

Method of Analysis (3 marks maximum):
- A1: States appropriate plot, e.g. \(h\) against \(V^2\) (or \(\sqrt{h}\) against \(V\), or \(\ln h\) against \(\ln V\)).
- A1: States that the relationship is confirmed if the graph is a straight line through the origin (for \(h\) vs \(V^2\) or \(\sqrt{h}\) vs \(V\)) [or straight line of gradient 2 for \(\ln h\) vs \(\ln V\)].
- A1: Correct expression for determining \(B\) from the gradient, e.g. \(B = \frac{M}{LC}\sqrt{2gm}\) where \(m\) is the gradient of \(h\) vs \(V^2\).

Safety Considerations (1 mark maximum):
- S1: Valid safety precaution linked to the experiment with specific reason (e.g. avoid touching bare contacts/switch off supply before making adjustments to prevent electric shock; clamp heavy magnets to prevent clamping/pinching hazard or tipping).

Additional Detail (up to 5 marks from the following points):
- D1: Repeat measurements of \(h\) for each voltage \(V\) and calculate a mean.
- D2: Ensure the rod is stationary in equilibrium before each discharge.
- D3: Use fine, flexible, lightweight leads with slack to minimize resistance to rod motion.
- D4: Align video camera optical axis perpendicular to the plane of motion to avoid parallax errors.
- D5: Measure capacitance \(C\) directly using a digital capacitance meter.
- D6: Use large/wide pole pieces to ensure the magnetic field is uniform across the region of swing.
- D7: Include a resistor in the charging path to limit charging current.

(Total marks: 15)