2025 Cambridge IAL Physics (9702) 模擬試題連答案詳解

Under a tensile force \( F \), extension is given by \( x = \frac{FL}{AE} \) where \( E \) is the Young modulus. For the second wire: \( x' = \frac{F(3L)}{(2A)E} = 1.5 \left(\frac{FL}{AE}\right) = 1.5x \). The elastic potential energy stored is given by \( E_s = \frac{1}{2}Fx \). For the second wire, the energy stored is \( E_s' = \frac{1}{2}Fx' = \frac{1}{2}F(1.5x) = 1.5E_s \).

1 mark for the correct choice. Method: Determine the relation of the second wire's extension to \( x \) and then use the formula for elastic energy.

First, calculate the cross-sectional area of the cable: \( A = \frac{\pi d^2}{4} = \frac{\pi (4.0 \times 10^{-3})^2}{4} = 1.257 \times 10^{-5}\text{ m}^2 \). The stress in the cable is \( \sigma = \frac{F}{A} = \frac{1200}{1.257 \times 10^{-5}} = 9.549 \times 10^7\text{ Pa} \). Finally, the strain is \( \epsilon = \frac{\sigma}{E} = \frac{9.549 \times 10^7}{2.0 \times 10^{11}} \approx 4.8 \times 10^{-4} \).

1 mark for the correct option. Method: Correct substitution of cross-sectional area, calculation of stress, and dividing by Young modulus.

For maximum power dissipation in the load resistor, the load resistance must equal the internal resistance of the cell, \( R = r \). The current is \( I = \frac{E}{R+r} = \frac{E}{2R} \). The potential difference across the variable resistor is \( V = IR = 0.5E \), giving a ratio \( \frac{V}{E} = 0.5 \). The efficiency of the circuit is \( \eta = \frac{I^2 R}{I^2 (R+r)} = \frac{R}{R+r} = 50\% \).

1 mark for the correct choice. Method: Identify maximum power condition \( R=r \), calculate current and potential difference, and determine efficiency.

The initial output voltage is \( V_{\text{out, initial}} = 6.0\text{ V} \times \frac{2.0\text{ k}\Omega}{8.0\text{ k}\Omega + 2.0\text{ k}\Omega} = 1.2\text{ V} \). After heating, the new output voltage is \( V_{\text{out, final}} = 6.0\text{ V} \times \frac{2.0\text{ k}\Omega}{1.0\text{ k}\Omega + 2.0\text{ k}\Omega} = 4.0\text{ V} \). The change in the output voltage is \( 4.0\text{ V} - 1.2\text{ V} = +2.8\text{ V} \).

1 mark for the correct choice. Method: Determine initial and final voltages across the fixed resistor and subtract them to find the change.

First, find the equivalent capacitance of the parallel group: \( C_p = 3.0\ \mu\text{F} + 6.0\ \mu\text{F} = 9.0\ \mu\text{F} \). Next, find the combined capacitance of the system since \( C_p \) is in series with the \( 18\ \mu\text{F} \) capacitor: \( \frac{1}{C_t} = \frac{1}{9.0} + \frac{1}{18} = \frac{3}{18} \implies C_t = 6.0\ \mu\text{F} \). The total charge supplied by the source is \( Q = C_t V = 6.0\ \mu\text{F} \times 12\text{ V} = 72\ \mu\text{C} \). Since the \( 18\ \mu\text{F} \) capacitor is in series with the parallel pair, the charge on it must be equal to the total charge.

1 mark for the correct option. Method: Work out the combined parallel capacitance, find total series capacitance, calculate total charge, and identify series charge.

The initial energy stored in \( C_1 \) is \( E_i = \frac{1}{2} C_1 V^2 \) and its initial charge is \( Q = C_1 V \). When connected in parallel with \( C_2 = 2C_1 \), the combined capacitance is \( C_f = C_1 + C_2 = 3C_1 \). Since the total charge is conserved, the final energy of the system is \( E_f = \frac{Q^2}{2 C_f} = \frac{(C_1 V)^2}{2 (3 C_1)} = \frac{1}{6} C_1 V^2 = \frac{1}{3} E_i \). The energy lost is \( E_i - E_f = E_i - \frac{1}{3} E_i = \frac{2}{3} E_i \), which is \( \frac{2}{3} \) of the original energy.

1 mark for the correct option. Method: Formulate expressions for initial and final energy using charge conservation, and calculate the fraction of energy loss.

Using the equation \( s = ut + \frac{1}{2}at^2 \) and taking the vertically upward direction as positive: \( s = -25\text{ m} \) (since the base of the cliff is below the starting point), \( u = +15\text{ m s}^{-1} \), and \( a = -9.81\text{ m s}^{-2} \). Substituting these values gives: \(-25 = 15t - 4.905t^2 \implies 4.905t^2 - 15t - 25 = 0 \). Solving the quadratic equation using the quadratic formula: \( t = \frac{15 \pm \sqrt{(-15)^2 - 4(4.905)(-25)}}{2(4.905)} = \frac{15 \pm \sqrt{225 + 490.5}}{9.81} = \frac{15 \pm 26.75}{9.81} \). Taking the positive solution gives \( t \approx 4.26\text{ s} \approx 4.3\text{ s} \).

1 mark for the correct option. Method: Formulate a quadratic equation for displacement with correct signs and solve for positive time.

Since the stone hits the ground at an angle of \( 45^\circ \) to the horizontal, its final vertical component of velocity \( v_y \) must be equal in magnitude to its final horizontal component of velocity \( v_x \). Because the horizontal velocity remains constant at \( 20\text{ m s}^{-1} \), we have \( v_y = 20\text{ m s}^{-1} \). Using the vertical equation of motion \( v_y^2 = u_y^2 + 2gh \) with \( u_y = 0 \) (since it was thrown horizontally): \( 20^2 = 2 \times 9.81 \times h \implies 400 = 19.62 h \implies h \approx 20.4\text{ m} \approx 20\text{ m} \).

1 mark for the correct option. Method: Deduce that vertical velocity equals horizontal velocity at a \( 45^\circ \) angle, then use vertical equations of motion to solve for height.

The extension \(x\) of a wire is given by \(x = \frac{FL}{AE}\), where \(E\) is the Young modulus of the material. The work done in stretching the wire elastically is \(W = \frac{1}{2}Fx = \frac{F^2L}{2AE}\). For the second wire, length is \(2L\), cross-sectional area is \(2A\), the material is the same (so the Young modulus is \(E\)), and the tensile force is \(F\). The work done on the second wire is \(W' = \frac{F^2(2L)}{2(2A)E} = \frac{F^2L}{2AE} = W\). Therefore, the work done in stretching the second wire is also \(W\).

1 mark for identifying that the extension of both wires is identical because the doubling of length is compensated by the doubling of area, leading to the same work done.

For the parallel combination, the effective spring constant is \(k_p = k + 2k = 3k\). The elastic potential energy stored under a load \(W\) is \(E_p = \frac{W^2}{2k_p} = \frac{W^2}{6k}\). For the series combination, the effective spring constant \(k_s\) is given by \(\frac{1}{k_s} = \frac{1}{k} + \frac{1}{2k} = \frac{3}{2k} \implies k_s = \frac{2}{3}k\). The elastic potential energy stored in series under the same load \(W\) is \(E_s = \frac{W^2}{2k_s} = \frac{W^2}{2 \times \frac{2}{3}k} = \frac{3W^2}{4k}\). Evaluating the ratio: \(\frac{E_s}{E_p} = \frac{3/4}{1/6} = 4.5\), which gives \(E_s = 4.5 E_p\).

1 mark for calculating the correct effective spring constants and using the energy-force relation \(E = \frac{W^2}{2k}\) to find the ratio of 4.5.

The terminal potential difference \(V\) is related to \(E\) and \(r\) by \(V = \frac{ER}{R + r}\). For the first case: \(6.0 = \frac{4.0E}{4.0 + r} \implies E = 6.0 + 1.5r\). For the second case: \(7.5 = \frac{10.0E}{10.0 + r} \implies E = 7.5 + 0.75r\). Equating the two expressions for \(E\): \(6.0 + 1.5r = 7.5 + 0.75r \implies 0.75r = 1.5 \implies r = 2.0\ \Omega\).

1 mark for setting up the simultaneous equations using the potential divider formula or \(E = V + Ir\) and solving to find \(r = 2.0\ \Omega\).

The voltmeter has a finite resistance of \(6.0\text{ k}\Omega\) in parallel with the LDR. In the dark, the LDR is \(6.0\text{ k}\Omega\), so the parallel combination has equivalent resistance \(R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\text{ k}\Omega\). The potential difference across this combination is \(12 \times \frac{3.0}{3.0 + 3.0} = 6.0\text{ V\)}. In bright light, the LDR is \(2.0\text{ k}\Omega\), so the parallel combination has equivalent resistance \(R_p = \frac{2.0 \times 6.0}{2.0 + 6.0} = 1.5\text{ k}\Omega\). The potential difference across this combination is \(12 \times \frac{1.5}{3.0 + 1.5} = 4.0\text{ V\)}.

1 mark for calculating the combined parallel resistance of the LDR and voltmeter in both dark and bright conditions, and applying the potential divider formula.

Let us evaluate the configuration in option D: The \(3.0\ \mu\text{F\)} and \(6.0\ \mu\text{F\)} capacitors in parallel have a combined capacitance of \(C_p = 3.0 + 6.0 = 9.0\ \mu\text{F\)}. When connected in series with the \(9.0\ \mu\text{F\)} capacitor, the total capacitance is \(C_{total} = \frac{9.0 \times 9.0}{9.0 + 9.0} = 4.5\ \mu\text{F\)}.

1 mark for using the parallel addition rule \(C_p = C_1 + C_2\) and the series formula \(1/C_s = 1/C_1 + 1/C_2\) to find the target equivalent capacitance.

The total work done by the power supply is the charge transferred multiplied by the constant voltage: \(W = QV\). The energy stored in the electric field of the capacitor is given by \(E = \frac{1}{2}QV\). The remaining half of the energy is dissipated as heat in the charging circuit.

1 mark for identifying that the work done by the supply is \(QV\) and the energy stored in the capacitor is \(\frac{1}{2}QV\).

Using the kinematic equation \(s = ut + \frac{1}{2}at^2\) with the upward direction as positive: displacement is \(s = -h\), initial velocity is \(u\), and acceleration is \(a = -g\). This gives \(-h = uT - \frac{1}{2}gT^2\). Rearranging yields the quadratic equation \(gT^2 - 2uT - 2h = 0\). Using the quadratic formula, the positive root for time is \(T = \frac{2u + \sqrt{4u^2 - 4(g)(-2h)}}{2g} = \frac{u + \sqrt{u^2 + 2gh}}{g\}.

1 mark for formulating the displacement as negative and solving the quadratic equation to get the correct positive root.

Average velocity is given by total displacement divided by total time. The motion represents a trapezoidal velocity-time graph. The total displacement (area under the graph) is \(\text{Displacement} = \frac{1}{2} \times (20.0 + 10.0) \times 6.0 = 90.0\text{ m\)}. The average velocity is \(\frac{90.0\text{ m}}{20.0\text{ s}} = 4.5\text{ m s}^{-1}\).

1 mark for calculating the total displacement from the area under the velocity-time graph and dividing by the total time to obtain the average velocity.

1. Determine the spring constant \(k\) in the elastic region:
\(k = \frac{150\text{ N}}{1.5 \times 10^{-3}\text{ m}} = 1.0 \times 10^5\text{ N m}^{-1}\).

2. The unloading curve has the same slope \(k\). The decrease in extension \(\Delta x\) when unloading from \(200\text{ N}\) to \(0\text{ N}\) is:
\(\Delta x = \frac{200\text{ N}}{1.0 \times 10^5\text{ N m}^{-1}} = 2.0 \times 10^{-3}\text{ m} = 2.0\text{ mm}\).

3. Calculate the permanent extension:
\(3.0\text{ mm} - 2.0\text{ mm} = 1.0\text{ mm}\).

4. Calculate the total work done during loading:
- Work from \(0\) to \(1.5\text{ mm}\): \(\frac{1}{2} \times 150\text{ N} \times 1.5 \times 10^{-3}\text{ m} = 0.1125\text{ J}\).
- Work from \(1.5\text{ mm}\) to \(3.0\text{ mm}\): \(\frac{1}{2} \times (150 + 200)\text{ N} \times 1.5 \times 10^{-3}\text{ m} = 0.2625\text{ J}\).
- Total work done = \(0.1125 + 0.2625 = 0.375\text{ J}\).

5. Calculate the elastic strain energy recovered during unloading:
\(E_{\text{recovered}} = \frac{1}{2} \times 200\text{ N} \times 2.0 \times 10^{-3}\text{ m} = 0.200\text{ J}\).

6. The thermal energy generated is the work done that is not recovered:
\(E_{\text{thermal}} = 0.375\text{ J} - 0.200\text{ J} = 0.175\text{ J}\).

1 mark for identifying the correct permanent extension of \(1.0\text{ mm}\) and thermal energy of \(0.175\text{ J}\), corresponding to option A.

1. Express elastic strain energy \(E\) as:
\(E = \frac{1}{2} F \Delta L\).

2. Use the Young modulus formula \(E_Y = \frac{F L}{A \Delta L}\) to write extension as:
\[\Delta L = \frac{F L}{A E_Y}\]

3. Substitute \(\Delta L\) back into the energy equation:
\[E = \frac{F^2 L}{2 A E_Y}\]

4. Since \(F\) and \(E_Y\) are identical for both wires, the energy depends on length and cross-sectional area: \(E \propto \frac{L}{A}\). Since \(A = \frac{\pi d^2}{4}\), we have \(E \propto \frac{L}{d^2}\).

5. Calculate the ratio:
\[\frac{E_X}{E_Y} = \frac{L_X / d_X^2}{L_Y / d_Y^2} = \frac{L / d^2}{2L / (2d)^2} = \frac{L / d^2}{2L / 4d^2} = 2.0\]

1 mark for correctly formulating the ratio of stored elastic energy and determining it to be 2.0, corresponding to option C.

1. The voltmeter measures terminal potential difference \(V\).
2. Calculate the circuit current \(I\) for each case:
- For \(R_1 = 3.0\,\Omega\), \(V_1 = 6.0\text{ V} \implies I_1 = \frac{V_1}{R_1} = \frac{6.0}{3.0} = 2.0\text{ A}\).
- For \(R_2 = 8.0\,\Omega\), \(V_2 = 8.0\text{ V} \implies I_2 = \frac{V_2}{R_2} = \frac{8.0}{8.0} = 1.0\text{ A}\).

3. Use the equation \(E = V + Ir\):
- Case 1: \(E = 6.0 + 2.0r\)
- Case 2: \(E = 8.0 + 1.0r\)

4. Equate the two expressions to solve for \(r\):
\(6.0 + 2.0r = 8.0 + 1.0r \implies r = 2.0\,\Omega\).

5. Substitute \(r\) back to find \(E\):
\(E = 8.0 + 1.0(2.0) = 10.0\text{ V}\).

1 mark for setting up simultaneous equations using Ohm's Law and the terminal p.d. relationship, and solving for \(E = 10.0\text{ V}\) and \(r = 2.0\,\Omega\), corresponding to option B.

1. At \(10\,^\circ\text{C}\):
- Thermistor resistance \(R_T = 2400\,\Omega\).
- Total resistance \(R_{\text{total}} = 2400 + 1200 = 3600\,\Omega\).
- Voltmeter reading \(V_{\text{out1}} = 9.0 \times \frac{1200}{3600} = 3.0\text{ V}\).

2. At \(40\,^\circ\text{C}\):
- Thermistor resistance \(R_T = 600\,\Omega\).
- Total resistance \(R_{\text{total}} = 600 + 1200 = 1800\,\Omega\).
- Voltmeter reading \(V_{\text{out2}} = 9.0 \times \frac{1200}{1800} = 6.0\text{ V}\).

3. The change in the reading is:
\(6.0\text{ V} - 3.0\text{ V} = +3.0\text{ V}\) (an increase of \(3.0\text{ V}\)).

1 mark for calculating the potential differences at both temperatures and determining that there is an increase of \(3.0\text{ V}\), corresponding to option B.

1. Calculate the initial charge stored on the first capacitor:
\(Q = C_1 V_i = 10\,\mu\text{F} \times 12\text{ V} = 1.2 \times 10^{-4}\text{ C}\).

2. When connected in parallel, total capacitance is:
\(C_{\text{total}} = 10\,\mu\text{F} + 20\,\mu\text{F} = 30\,\mu\text{F}\).

3. Calculate the final potential difference using conservation of charge:
\(V_f = \frac{Q}{C_{\text{total}}} = \frac{1.2 \times 10^{-4}\text{ C}}{30 \times 10^{-6}\text{ F}} = 4.0\text{ V}\).

4. Calculate the initial energy stored:
\(E_i = \frac{1}{2} C_1 V_i^2 = \frac{1}{2} \times (10 \times 10^{-6}) \times 12^2 = 7.2 \times 10^{-4}\text{ J}\).

5. Calculate the final energy stored:
\(E_f = \frac{1}{2} C_{\text{total}} V_f^2 = \frac{1}{2} \times (30 \times 10^{-6}) \times 4.0^2 = 2.4 \times 10^{-4}\text{ J}\).

6. Calculate the energy dissipated as heat:
\(\Delta E = E_i - E_f = 7.2 \times 10^{-4}\text{ J} - 2.4 \times 10^{-4}\text{ J} = 4.8 \times 10^{-4}\text{ J}\).

1 mark for correctly finding the final potential difference (4.0 V) and the energy dissipated (\(4.8 \times 10^{-4}\text{ J}\)), corresponding to option A.

1. Determine the total equivalent capacitance \(C_{\text{total}}\):
\(\frac{1}{C_{\text{total}}} = \frac{1}{2.0} + \frac{1}{3.0} + \frac{1}{6.0} = \frac{3+2+1}{6.0} = 1.0\,\mu\text{F}^{-1} \implies C_{\text{total}} = 1.0\,\mu\text{F}\).

2. Since the capacitors are connected in series, the charge \(Q\) on each capacitor is equal to the total charge:
\(Q = C_{\text{total}} \times V_{\text{total}} = 1.0\,\mu\text{F} \times 12\text{ V} = 12\,\mu\text{C}\).
Thus, the charge on the \(6.0\,\mu\text{F}\) capacitor is \(12\,\mu\text{C}\).

3. Calculate the potential difference across the \(3.0\,\mu\text{F}\) capacitor:
\(V = \frac{Q}{C} = \frac{12\,\mu\text{C}}{3.0\,\mu\text{F}} = 4.0\text{ V}\).

1 mark for calculating the total series capacitance to find the charge of \(12\,\mu\text{C}\) and using this charge to find the potential difference of \(4.0\text{ V}\), corresponding to option A.

1. Resolve the initial velocity into components:
- Horizontal: \(u_x = 20 \cos(30^\circ) \approx 17.3\text{ m s}^{-1}\).
- Vertical: \(u_y = 20 \sin(30^\circ) = 10\text{ m s}^{-1}\).

2. At maximum height, the vertical component of velocity is zero (\(v_y = 0\)). The horizontal component remains unchanged because there is no horizontal force (negligible air resistance).
Therefore, the speed at maximum height is equal to the horizontal component: \(v = u_x = 17.3\text{ m s}^{-1}\).

3. Find the time to reach maximum height using the vertical motion equation \(v_y = u_y - gt\):
\(0 = 10 - 9.81t \implies t = \frac{10}{9.81} \approx 1.02\text{ s}\).

1 mark for identifying the constant horizontal speed component at maximum height and calculating the time with \(g = 9.81\text{ m s}^{-2}\), corresponding to option B.

1. We are given the initial velocity \(u = 30\text{ m s}^{-1}\), the final velocity \(v = 0\), and the distance \(s = 75\text{ m}\).

2. Use the equation \(v^2 = u^2 + 2as\) to find deceleration:
\(0 = 30^2 + 2 \times a \times 75\)
\(0 = 900 + 150a\)
\(a = -6.0\text{ m s}^{-2}\).
Thus, the deceleration is \(6.0\text{ m s}^{-2}\).

3. Use the equation \(s = \frac{1}{2}(u + v)t\) to find the time taken:
\(75 = \frac{1}{2}(30 + 0) \times t\)
\(75 = 15t\)
\(t = 5.0\text{ s}\).

1 mark for calculating deceleration using equations of motion and finding the correct time, corresponding to option C.

The extension \(x\) of a wire under a force \(F\) is given by \(x = \frac{FL}{AE}\). For the copper wire, \(x_1 = \frac{F L_1}{A_1 E_{\text{copper}}} = x\). For the steel wire, \(x_2 = \frac{F L_2}{A_2 E_{\text{steel}}} = \frac{F (2 L_1)}{(0.5 A_1) (2 E_{\text{copper}})} = 2 \left( \frac{F L_1}{A_1 E_{\text{copper}}} \right) = 2x\). Since the wires are in series, they experience the same force \(F\). The total extension is the sum of the individual extensions: \(x_{\text{total}} = x_1 + x_2 = x + 2x = 3.0x\).

1 mark for calculating the extension of the steel wire as 2x and adding it to the extension of the copper wire to get 3.0x.

The work done in deforming the wire plastically (energy dissipated) is the total work done during loading minus the elastic energy recovered during unloading. The loading work is \(W_{\text{load}} = 0.48\text{ J}\). The recovered unloading energy is the area of the triangle under the unloading line: \(W_{\text{unload}} = \frac{1}{2} \times \Delta x \times F = \frac{1}{2} \times (8.0 - 2.0) \times 10^{-3}\text{ m} \times 100\text{ N} = 0.30\text{ J}\). Thus, the energy dissipated as plastic deformation is \(0.48\text{ J} - 0.30\text{ J} = 0.18\text{ J}\).

1 mark for calculating the area under the unloading curve to be 0.30 J and subtracting it from the loading work of 0.48 J to obtain 0.18 J.

The current in the circuit is \(I = \frac{V_{\text{fixed}}}{R_{\text{fixed}}} = \frac{4.0\text{ V}}{10.0\ \Omega} = 0.40\text{ A}\). The total resistance of the series circuit is \(R_{\text{total}} = \frac{E}{I} = \frac{12.0\text{ V}}{0.40\text{ A}} = 30.0\ \Omega\). Since \(R_{\text{total}} = r + R_{\text{thermistor}} + R_{\text{fixed}}\), we have \(30.0\ \Omega = 2.0\ \Omega + R_{\text{thermistor}} + 10.0\ \Omega\), which gives \(R_{\text{thermistor}} = 18.0\ \Omega\).

1 mark for calculating the current of 0.40 A and using it to find the thermistor resistance of 18 ohms.

Because the driver cell has negligible internal resistance, its terminal potential difference remains at its e.m.f. of \(2.0\text{ V}\) even when an external load resistor is connected. The potential gradient across the wire remains unchanged at \(0.02\text{ V cm}^{-1}\). Hence, the balance length for the same \(1.5\text{ V}\) cell does not change and remains \(75\text{ cm}\).

1 mark for identifying that the potential gradient remains unchanged due to the negligible internal resistance of the driver cell, giving a balance length of 75 cm.

The equivalent capacitance \(C_{\text{eq}}\) of the series combination is given by \(\frac{1}{C_{\text{eq}}} = \frac{1}{2.0} + \frac{1}{3.0} + \frac{1}{6.0} = 1.0\ \mu\text{F}^{-1}\), so \(C_{\text{eq}} = 1.0\ \mu\text{F}\). The total charge is \(Q = C_{\text{eq}} V = 1.0\ \mu\text{F} \times 12\text{ V} = 12\ \mu\text{C}\). For a series circuit, each capacitor has the same charge \(Q = 12\ \mu\text{C}\). The potential difference across the \(3.0\ \mu\text{F}\) capacitor is \(V = \frac{Q}{C} = \frac{12\ \mu\text{C}}{3.0\ \mu\text{F}} = 4.0\text{ V}\).

1 mark for calculating the series charge of 12 microcoulombs and finding the potential difference of 4.0 V.

The initial energy is \(E_{\text{initial}} = \frac{1}{2} C_1 V^2\) and the initial charge is \(Q = C_1 V\). When connected in parallel, the total capacitance becomes \(C_{\text{eq}} = C_1 + C_2 = 4 C_1\). The charge remains conserved at \(Q\), so the final energy is \(E_{\text{final}} = \frac{Q^2}{2 C_{\text{eq}}} = \frac{(C_1 V)^2}{2 (4 C_1)} = \frac{1}{8} C_1 V^2\). The ratio of the final energy to the initial energy is \(\frac{\frac{1}{8} C_1 V^2}{\frac{1}{2} C_1 V^2} = 0.25\).

1 mark for using charge conservation to calculate the final total energy and finding the ratio of 0.25.

Both stones take the same time \(t = \sqrt{\frac{2h}{g}}\) to reach the ground because their vertical motions are identical, which rules out options A and C. The dropped stone hits with speed \(\sqrt{2gh}\), whereas the projected stone has both horizontal and vertical components of velocity, meaning it hits with a larger total speed of \(\sqrt{v^2 + 2gh}\). Therefore, statement B is correct.

1 mark for identifying that horizontal projection adds a perpendicular velocity component, increasing the final impact speed.

Using the equation \(s = \frac{1}{2} g t^2\), the total height is \(H = \frac{1}{2} g T^2\). The distance fallen in the first three-quarters of the time (i.e., \(t = \frac{3}{4} T\)) is \(s' = \frac{1}{2} g \left(\frac{3}{4} T\right)^2 = \frac{9}{16} H\). The distance fallen during the final quarter of the time is the difference: \(\Delta s = H - s' = H - \frac{9}{16} H = \frac{7}{16} H\).

1 mark for setting up the distance equations for T and 3/4 T and subtracting them to find the remaining fraction of height 7/16 H.

By defining the upward direction as positive, we can establish the SUVAT variables for the entire motion of the ball: displacement \(s = -h\) (since the final position is \(h\) units below the starting point), initial velocity \(u_y = u\), acceleration \(a = -g\), and time \(t = T\). Substituting these values into the kinematic equation \(s = u_y t + \frac{1}{2} a t^2\) gives: \(-h = uT + \frac{1}{2}(-g)T^2 \implies -h = uT - \frac{1}{2}gT^2\). Rearranging this to solve for \(h\) yields: \(h = \frac{1}{2}gT^2 - uT\).

[1 mark] C is the correct answer. A is incorrect because it assumes downward initial velocity. B is incorrect because it has the signs inverted. D is incorrect as it is a misapplication of average velocity equations.

For uniform deceleration to a complete stop, we use the equation of motion: \(v_f^2 = v_i^2 + 2as\). Setting final velocity \(v_f = 0\), we get: \(0 = v_i^2 - 2ad \implies d = \frac{v_i^2}{2a}\). For the first car, the braking distance is \(d = \frac{v^2}{2a}\). For the second car, with initial speed \(2v\) and deceleration \(\frac{a}{2}\), the braking distance is: \(d' = \frac{(2v)^2}{2(a/2)} = \frac{4v^2}{a} = 8 \left(\frac{v^2}{2a}\right) = 8d\).

[1 mark] C is the correct answer. A is incorrect because it only accounts for the change in deceleration. B is incorrect because it only accounts for the squaring of the velocity. D is incorrect due to calculation errors in scaling.

The Young modulus \(E\) of the material is defined by \(E = \frac{stress}{strain} = \frac{F/A}{x/L} = \frac{FL}{Ax}\), which gives the extension \(x = \frac{FL}{EA}\). The strain energy \(E_s\) stored in the first wire is given by: \(E_s = \frac{1}{2} F x = \frac{F^2 L}{2EA}\). For the second wire, made of the same material (same \(E\)), with length \(2L\), area \(2A\), and force \(2F\), the stored strain energy is: \(E_{s2} = \frac{(2F)^2 (2L)}{2E(2A)} = \frac{8 F^2 L}{4 EA} = \frac{2 F^2 L}{EA} = 4 \left(\frac{F^2 L}{2EA}\right) = 4 E_s\).

[1 mark] C is the correct answer. A, B, and D are incorrect ratios resulting from misapplying the dependence of strain energy on length, area, and force.

By definition, elastic deformation is fully reversible. Therefore, when the load is removed, the wire returns to its original length, and all of the strain energy stored is recovered as work done by the wire. Plastic deformation, on the other hand, involves permanent relocation of atoms (slip planes), which dissipates energy as thermal energy and means the wire will not return to its original length.

[1 mark] C is the correct answer. A is incorrect because during plastic deformation, atoms do not return to their original equilibrium positions. B is incorrect because in ideal elastic deformation, no energy is dissipated as heat. D is incorrect because Hooke's law is not obeyed during plastic deformation.

The relationship between the terminal potential difference \(V\), e.m.f. \(E\), current \(I\), and internal resistance \(r\) is given by: \(V = E - Ir\). Rearranging this in the form of a straight-line equation, \(y = mx + c\), we obtain: \(V = (-r)I + E\). This represents a straight line with a negative gradient equal to \(-r\) and a vertical intercept (where \(I = 0\)) of \(E\).

[1 mark] B is the correct answer. A describes a simple ohmic resistor. C and D describe non-linear responses that do not apply to a constant internal resistance model.

At the balance point, no current flows through the galvanometer, meaning that the current through the test cell branch is zero. Consequently, there is no potential difference across the internal resistance of the test cell, and the terminal potential difference across the cell equals its e.m.f. \(E_x\). The potential difference across the balancing length of \(65.0\text{ cm}\) on the uniform wire of total length \(100.0\text{ cm}\) is given by: \(V = 2.0\text{ V} \times \frac{65.0\text{ cm}}{100.0\text{ cm}} = 1.30\text{ V}\). Therefore, \(E_x = 1.30\text{ V}\).

[1 mark] B is the correct answer. A is incorrect because it represents a simple division error. C is incorrect because it attempts to adjust for the internal resistance which is inactive at balance. D is the total potential difference of the wire.

Let us analyze all possible combinations of three identical capacitors: 1. All three in series: \(\frac{1}{C_{tot}} = \frac{3}{C} \implies C_{tot} = \frac{1}{3}C\) (Option A). 2. All three in parallel: \(C_{tot} = 3C\). 3. Two in parallel in series with the third: \(\frac{1}{C_{tot}} = \frac{1}{2C} + \frac{1}{C} = \frac{3}{2C} \implies C_{tot} = \frac{2}{3}C\) (Option B). 4. Two in series in parallel with the third: \(C_{tot} = \frac{1}{2}C + C = \frac{3}{2}C\) (Option D). Analyzing the list, \(\frac{4}{3}C\) is not mathematically achievable with any combination of three identical capacitors.

[1 mark] C is the correct answer. Options A, B, and D are valid and achievable equivalent capacitances.

The energy stored in a capacitor of capacitance \(C\) under potential difference \(V\) is \(W = \frac{1}{2}CV^2\). When the potential difference is increased to \(2V\), the new energy stored is: \(W' = \frac{1}{2}C(2V)^2 = 4 \left(\frac{1}{2}CV^2\right) = 4W\). The question asks for the INCREASE in the energy stored: \(\Delta W = W' - W = 4W - W = 3W\).

[1 mark] C is the correct answer. A is the initial energy. B is a linear scaling misconception. D is the total final energy stored rather than the increase in energy.

*(a)* Acceleration is defined as the rate of change of velocity.

*(b)(i)* At maximum height, the final vertical velocity \(v = 0\).
Using \(v^2 = u^2 + 2as\) where \(u = 15.0\text{ m s}^{-1}\) and \(a = -g = -9.81\text{ m s}^{-2}\):
\(0 = (15.0)^2 + 2(-9.81)s\)
\(19.62s = 225\)
\(s = 11.47\text{ m} \approx 11.5\text{ m}\).

*(b)(ii)* Taking the downward direction as positive and launch point as origin, displacement to the sea is \(s = -45.0\text{ m}\).
Using \(v^2 = u^2 + 2as\):
\(v^2 = (15.0)^2 + 2(-9.81)(-45.0)\)
\(v^2 = 225 + 882.9 = 1107.9\text{ m}^2\text{ s}^{-2}\)
\(v = \sqrt{1107.9} \approx 33.3\text{ m s}^{-1}\).

*(b)(iii)* Using \(v = u + at\):
\(-33.28 = 15.0 - 9.81t\)
\(-48.28 = -9.81t\)
\(t = \frac{48.28}{9.81} \approx 4.92\text{ s}\).

*(a)*
- [1 mark] Rate of change of velocity (or change in velocity / time taken).

*(b)(i)*
- [1 mark] Selection of formula \(v^2 = u^2 + 2as\) with correct substitution of \(v = 0\) and \(a = -9.81\text{ m s}^{-2}\).
- [1.57 marks] Correct calculation to 2 or 3 significant figures: \(11.5\text{ m}\) (accept \(11\text{ m}\) if using \(g = 9.8\text{ m s}^{-2}\)).

*(b)(ii)*
- [1 mark] Identification of total vertical displacement from launch to base of cliff as \(-45.0\text{ m}\).
- [1 mark] Correct substitution of values into the kinematic equation \(v^2 = u^2 + 2as\).
- [1 mark] Correct calculation of speed: \(33.3\text{ m s}^{-1}\).

*(b)(iii)*
- [1 mark] Selection and substitution into a suitable kinematic equation, e.g., \(v = u + at\) or \(s = ut + \frac{1}{2}at^2\).
- [1 mark] Correct calculation of time: \(4.92\text{ s}\).

*(a)* Elastic deformation: the material returns to its original length/shape when the deforming force/load is removed.
Plastic deformation: the material does not return to its original length/shape when the deforming force/load is removed (permanent deformation occurs).

*(b)(i)*
Young modulus \(E = \frac{\text{stress}}{\text{strain}} = \frac{F L}{A x}\).
Using values within the proportional limit:
\(F = 80.0\text{ N}\), \(L = 2.40\text{ m}\), \(A = 3.20 \times 10^{-7}\text{ m}^2\), \(x = 1.60 \times 10^{-3}\text{ m}\).
\(E = \frac{80.0 \times 2.40}{(3.20 \times 10^{-7}) \times (1.60 \times 10^{-3})} = \frac{192}{5.12 \times 10^{-10}} = 3.75 \times 10^{11}\text{ Pa}\).

*(b)(ii)*
Total work done during loading is the area under the loading graph up to \(3.20\text{ mm}\):
Area under proportional region (0 to 1.60 mm):
\(W_1 = \frac{1}{2} \times 80.0 \times 1.60 \times 10^{-3} = 0.064\text{ J}\).
Area under next region (1.60 to 3.20 mm, trapezoid):
\(W_2 = \frac{1}{2} \times (80.0 + 120.0) \times (3.20 - 1.60) \times 10^{-3} = 0.160\text{ J}\).
Total work done during loading \(W_{\text{loading}} = 0.064 + 0.160 = 0.224\text{ J}\).
During unloading, the elastic energy recovered is represented by the triangular area under the unloading line, starting from an extension of 3.20 mm down to the permanent extension of 0.80 mm:
\(W_{\text{recovered}} = \frac{1}{2} \times 120.0 \times (3.20 - 0.80) \times 10^{-3} = 0.144\text{ J}\).
Work done in permanent deformation = \(W_{\text{loading}} - W_{\text{recovered}} = 0.224 - 0.144 = 0.080\text{ J}\).

*(a)*
- [1 mark] Elastic deformation: returns to original shape/length when force is removed.
- [1 mark] Plastic deformation: remains permanently deformed / does not return to original shape/length when force is removed.

*(b)(i)*
- [1 mark] Recalls \(E = \frac{F L}{A x}\) or equivalents.
- [1 mark] Correct substitutions of \(L = 2.40\), \(A = 3.20 \times 10^{-7}\), and corresponding values of \(F\) and \(x\) (e.g. \(80.0\text{ N}\) and \(1.60 \times 10^{-3}\text{ m}\)).
- [1.57 marks] Calculates \(E = 3.75 \times 10^{11}\text{ Pa}\) (accept \(3.8 \times 10^{11}\text{ Pa}\)).

*(b)(ii)*
- [1 mark] Calculates total work done during loading as area under loading curve \(= 0.224\text{ J}\).
- [1 mark] Calculates work recovered during unloading as area of triangle \(= 0.144\text{ J}\).
- [1 mark] Computes the net work done in permanent deformation: \(0.224 - 0.144 = 0.080\text{ J}\).

*(a)* Hooke's law states that the extension is directly proportional to the applied force, provided the limit of proportionality is not exceeded.

*(b)(i)* Since the ceiling is horizontal and the rigid bar remains horizontal, the downward movement at both spring attachments is identical. Therefore, the extension \(x\) of both springs must be the same.
Let \(x\) be the extension. The upward forces exerted by springs A and B are:
\(F_A = k_A x = 120 x\)
\(F_B = k_B x = 180 x\)
For vertical equilibrium, the sum of the upward forces equals the downward load:
\(F_A + F_B = 15.0\text{ N}\)
\(120 x + 180 x = 15.0\)
\(300 x = 15.0\)
\(x = 0.050\text{ m}\).

*(b)(ii)* To find the position \(d\) where the load is attached, take moments about the point of attachment of Spring A:
\(F_B \times 0.60 = 15.0 \times d\)
We first calculate the force in spring B:
\(F_B = k_B x = 180 \times 0.050 = 9.0\text{ N}\)
Substitute \(F_B\) into the moments equation:
\(9.0 \times 0.60 = 15.0 \times d\)
\(5.4 = 15.0 d\)
\(d = \frac{5.4}{15.0} = 0.36\text{ m}\).

*(a)*
- [1 mark] Extension is directly proportional to load / force.
- [0.57 marks] Provided the limit of proportionality is not exceeded.

*(b)(i)*
- [1 mark] Explanation that equal extension arises from the bar remaining horizontal.
- [1 mark] Equating total upward force to downward force: \(120x + 180x = 15.0\).
- [1 mark] Calculation of extension: \(0.050\text{ m}\).

*(b)(ii)*
- [1 mark] Calculation of the tension/force in either spring (e.g., \(F_B = 9.0\text{ N}\) or \(F_A = 6.0\text{ N}\)).
- [1 mark] Correct statement of principle of moments about any point (e.g., about A: \(F_B \times 0.60 = 15.0 \times d\)).
- [1 mark] Correctly substituting numerical values into the moment equation.
- [1 mark] Calculation of \(d = 0.36\text{ m}\).

*(a)* Potential difference is the work done (energy transferred from electrical to other forms) per unit charge.

*(b)(i)* The potential divider equation gives the voltage across the thermistor:
\(V_{\text{out}} = E \times \frac{R_{\text{th}}}{R + R_{\text{th}}} = 9.0 \times \frac{1800}{1200 + 1800}\)
\(V_{\text{out}} = 9.0 \times \frac{1800}{3000} = 5.4\text{ V}\).

*(b)(ii)* When \(V_{\text{out}} = 3.0\text{ V}\):
\(3.0 = 9.0 \times \frac{R_{\text{th}}'}{1200 + R_{\text{th}}'}\)
\(1 = 3 \times \frac{R_{\text{th}}'}{1200 + R_{\text{th}}'}\)
\(1200 + R_{\text{th}}' = 3 R_{\text{th}}'\)
\(2 R_{\text{th}}' = 1200 \implies R_{\text{th}}' = 600\ \Omega\).

*(b)(iii)* Power dissipated in the thermistor at \(50\text{ }^\circ\text{C}\):
\(P = \frac{V^2}{R} = \frac{(3.0)^2}{600} = \frac{9.0}{600} = 0.015\text{ W} = 15\text{ mW}\).

*(a)*
- [1 mark] Energy transformed from electrical to other forms per unit charge (or work done per unit charge).

*(b)(i)*
- [1 mark] Use of potential divider formula or calculating circuit current \(I = 9.0 / 3000 = 0.0030\text{ A}\).
- [1.57 marks] Correct calculation of voltage: \(5.4\text{ V}\).

*(b)(ii)*
- [1 mark] Setting up the equation \(3.0 = 9.0 \times \frac{R}{1200 + R}\) (or alternative circuit analysis method).
- [1 mark] Correct algebraic rearrangement (e.g., \(2R = 1200\)).
- [1 mark] Finding resistance: \(600\ \Omega\).

*(b)(iii)*
- [1 mark] Correct formula for electrical power used, e.g., \(P = \frac{V^2}{R}\) or \(P = I^2 R\).
- [1 mark] Correct calculation of power: \(0.015\text{ W}\) (or \(15\text{ mW}\)).

*(a)* Electromotive force (e.m.f.) is the total energy transferred per unit charge into electrical energy from other forms within the power source.

*(b)(i)* Using the equation \(E = I(R + r)\):
For the first setting: \(E = 1.20(4.0 + r)\)
For the second setting: \(E = 0.60(9.0 + r)\)
Equating the two expressions for \(E\):
\(1.20(4.0 + r) = 0.60(9.0 + r)\)
Divide both sides by 0.60:
\(2(4.0 + r) = 9.0 + r\)
\(8.0 + 2r = 9.0 + r\)
\(r = 1.0\ \Omega\).

*(b)(ii)* Substitute \(r = 1.0\ \Omega\) back into either equation for \(E\):
\(E = 1.20(4.0 + 1.0) = 1.20 \times 5.0 = 6.0\text{ V}\).

*(b)(iii)* Terminal potential difference \(V = E - Ir\):
\(V = 6.0 - (1.20 \times 1.0) = 4.8\text{ V}\)
(Alternatively: \(V = I R = 1.20 \times 4.0 = 4.8\text{ V}\)).

*(a)*
- [1 mark] Energy converted per unit charge (or work done per unit charge) from chemical/other forms to electrical.
- [1 mark] Mention of 'inside the source' or 'over the entire circuit'.

*(b)(i)*
- [1 mark] Recall of formula \(E = I(R+r)\).
- [1 mark] Setup of two correct simultaneous equations: \(E = 1.20(4.0+r)\) and \(E = 0.60(9.0+r)\).
- [1.57 marks] Correct algebraic solution yielding \(r = 1.0\ \Omega\).

*(b)(ii)*
- [1 mark] Substitution of \(r\) into one equation to solve for \(E\).
- [1 mark] Correct calculation of \(E = 6.0\text{ V}\).

*(b)(iii)*
- [1 mark] Correct calculation of terminal p.d.: \(4.8\text{ V}\) (either by \(V=IR\) or \(V=E-Ir\)).

*(a)* Capacitance is defined as charge per unit potential difference (or \(C = Q/V\), where \(Q\) is charge on one plate and \(V\) is the potential difference between plates).

*(b)(i)* The \(4.0\ \mu\text{F}\) and \(8.0\ \mu\text{F}\) capacitors are in parallel, so their combined capacitance \(C_p\) is:
\(C_p = 4.0\ \mu\text{F} + 8.0\ \mu\text{F} = 12.0\ \mu\text{F}\).
This combined capacitance is in series with the \(3.0\ \mu\text{F}\) capacitor, so the total capacitance \(C_T\) is:
\(\frac{1}{C_T} = \frac{1}{12.0\ \mu\text{F}} + \frac{1}{3.0\ \mu\text{F}} = \frac{1}{12.0} + \frac{4}{12.0} = \frac{5}{12.0\ \mu\text{F}}\)
\(C_T = \frac{12.0}{5} = 2.4\ \mu\text{F}\).

*(b)(ii)* Total charge stored \(Q_T = C_T \times V_{\text{supply}}\)
\(Q_T = 2.4\ \mu\text{F} \times 12.0\text{ V} = 2.4 \times 10^{-6}\text{ F} \times 12.0\text{ V} = 2.88 \times 10^{-5}\text{ C}\) (or \(28.8\ \mu\text{C}\)).

*(b)(iii)* The charge stored on the \(3.0\ \mu\text{F}\) capacitor is equal to the total charge of the network because it is in series with the rest of the combination.
\(Q_3 = Q_T = 2.88 \times 10^{-5}\text{ C}\).
Potential difference \(V_3 = \frac{Q_3}{C_3} = \frac{2.88 \times 10^{-5}\text{ C}}{3.0 \times 10^{-6}\text{ F}} = 9.6\text{ V}\).

*(a)*
- [1.57 marks] Charge divided by potential difference (or \(Q/V\) with symbols clearly defined).

*(b)(i)*
- [1 mark] Calculates parallel capacitance: \(C_p = 4.0 + 8.0 = 12.0\ \mu\text{F}\).
- [1 mark] Employs series capacitance formula: \(\frac{1}{C_T} = \frac{1}{12.0} + \frac{1}{3.0}\).
- [1 mark] Shows final value of \(2.4\ \mu\text{F}\) clearly with working.

*(b)(ii)*
- [1 mark] Recalls and uses \(Q = CV\) with correct overall values.
- [1 mark] Calculates charge: \(2.9 \times 10^{-5}\text{ C}\) (or \(28.8\ \mu\text{C}\)).

*(b)(iii)*
- [1 mark] Realises that the charge on the series capacitor is equal to total charge \(Q_T\).
- [1 mark] Calculates p.d.: \(9.6\text{ V}\).

*(a)* The three factors are:
1. Area of overlap of the plates.
2. Separation distance between the plates.
3. Permittivity (or dielectric constant) of the material between the plates.

*(b)(i)* Capacitance \(C = \frac{\varepsilon_0 A}{d}\).
\(C = \frac{8.85 \times 10^{-12} \times 0.12}{1.5 \times 10^{-3}}\)
\(C = 7.08 \times 10^{-10}\text{ F}\) (or \(708\text{ pF}\)).

*(b)(ii)* The electric field strength \(E\) between parallel plates is given by:
\(E = \frac{V}{d}\)
\(E = \frac{50}{1.5 \times 10^{-3}}\)
\(E = 3.33 \times 10^4\text{ V m}^{-1}\) (or \(3.3 \times 10^4\text{ V m}^{-1}\)).

*(a)*
- [3 marks] 1 mark for each factor correctly identified: area of the plates, distance between plates, permittivity of material between plates.

*(b)(i)*
- [1 mark] Recalls formula \(C = \frac{\varepsilon_0 A}{d}\).
- [1.57 marks] Substitutes correctly and calculates capacitance: \(7.1 \times 10^{-10}\text{ F}\) (accept \(7.08 \times 10^{-10}\text{ F}\)).

*(b)(ii)*
- [1 mark] Recalls formula \(E = \frac{V}{d}\).
- [1 mark] Substitutes \(V = 50\text{ V}\) and \(d = 1.5 \times 10^{-3}\text{ m}\).
- [1 mark] Calculates electric field strength correctly: \(3.3 \times 10^4\text{ V m}^{-1}\) (or \(3.33 \times 10^4\text{ V m}^{-1}\)).

### (a) Table of Results
A typical table of results containing six sets of measurements of \(d\), \(h_1\), the calculated deflection \(h\) (assuming \(h_0 = 45.0\text{ cm}\)), and \(d^2\):

| \(d\text{ / cm}\) | \(h_1\text{ / cm}\) | \(h\text{ / cm}\) | \(d^2\text{ / cm}^2\) |
| :---: | :---: | :---: | :---: |
| 15.0 | 44.5 | 0.5 | 225 |
| 25.0 | 43.7 | 1.3 | 625 |
| 35.0 | 42.6 | 2.4 | 1230 |
| 45.0 | 41.1 | 3.9 | 2030 |
| 55.0 | 39.3 | 5.7 | 3030 |
| 65.0 | 37.1 | 7.9 | 4230 |

*Note: \(d^2\) is calculated to 3 significant figures to match the precision of \(d\).*

### (b) Graph
* **Axes:** The y-axis represents \(h\text{ / cm}\) (range: 0 to 8.0 cm with 1.0 cm scale intervals) and the x-axis represents \(d^2\text{ / cm}^2\) (range: 0 to 4500 \(\text{cm}^2\) with 500 \(\text{cm}^2\) intervals).
* **Plotting:** All points from the table are plotted precisely.
* **Line of Best Fit:** A balanced straight line is drawn through the points, showing an even distribution of points on either side of the line.

### (c) Calculation of Gradient and y-Intercept
* **Gradient Calculation:**
Using two distant points on the line, e.g., \((400, 0.82)\) and \((4000, 7.48)\):
\[ \text{Gradient} = \frac{7.48 - 0.82}{4000 - 400} = \frac{6.66}{3600} \approx 1.85 \times 10^{-3}\text{ cm}^{-1} \]
* **y-Intercept Calculation:**
Using the point \((4000, 7.48)\) in the linear equation \(y = mx + c\):
\[ 7.48 = (1.85 \times 10^{-3})(4000) + c \implies c = 7.48 - 7.40 = 0.08\text{ cm} \]

### (d) Determination of Constants \(P\) and \(Q\)
Comparing \(h = P d^2 + Q\) to \(y = mx + c\):
* \(P = \text{Gradient} = 1.85 \times 10^{-3}\text{ cm}^{-1}\)
* \(Q = \text{y-intercept} = 0.08\text{ cm}\)

### Table of Results (6 Marks)
* **Successful collection of raw data (1 mark):** Six sets of values of \(d\) and \(h_1\) recorded with correct trend (as \(d\) increases, \(h\) increases).
* **Range of \(d\) (1 mark):** Raw values of \(d\) must span at least 45.0 cm (e.g., from 15.0 cm to 65.0 cm).
* **Column headings (1 mark):** Column headings must contain the quantity and unit in the form \(d\text{ / cm}\), \(h_1\text{ / cm}\), \(h\text{ / cm}\), \(d^2\text{ / cm}^2\).
* **Consistency of raw readings (1 mark):** All raw measurements of \(d\) and \(h_1\) must be recorded to the nearest millimetre (e.g., \(15.0\text{ cm}\)).
* **Calculation (1 mark):** Check that the calculated values of \(d^2\) are mathematically correct.
* **Significant figures (1 mark):** The significant figures of \(d^2\) must be consistent with (or one more than) those of raw \(d\).

### Graph (4 Marks)
* **Axes (1 mark):** Scale must be chosen so that the plotted points occupy more than half the grid in both directions. Scales must be linear with simple intervals (e.g. 1, 2, 5).
* **Plotting (1 mark):** Points plotted to an accuracy of within half a small grid square.
* **Line of Best Fit (1 mark):** Drawn with a ruler, representing a balanced trend of the plotted points.
* **Quality of data (1 mark):** Low scatter of points about the line.

### Analysis of Gradient and Intercept (4 Marks)
* **Gradient (2 marks):** One mark for using a triangle with a hypotenuse at least half the length of the drawn line. One mark for the correct calculation method (rise over run).
* **y-intercept (2 marks):** One mark for correct calculation using a point on the line, or reading directly from the y-axis if the x-axis scale starts at 0.

### Constants \(P\) and \(Q\) (6 Marks)
* **Value of \(P\) (2 marks):** Equated to the gradient value, within correct range and given to 2 or 3 significant figures. Correct unit (e.g. \(\text{cm}^{-1}\) or \(\text{m}^{-1}\)).
* **Value of \(Q\) (2 marks):** Equated to the y-intercept, within correct range and given to 2 or 3 significant figures. Correct unit (e.g. \(\text{cm}\) or \(\text{m}\)).
* **Units and Presentation (2 marks):** Appropriate SI units provided for both constants and correct scientific notation used.

### (a) and (b) Measurement of \(t_1\)
* Raw measurements for \(R_1 = 10\text{ k}\Omega\):
\(t_{1,a} = 9.32\text{ s}\)
\(t_{1,b} = 9.12\text{ s}\)
Average \(t_1 = 9.22\text{ s}\)

### (c) Percentage Uncertainty in \(t_1\)
* Given human reaction time, the absolute uncertainty \(\Delta t\) in manually timing a visual trigger is estimated to be \(0.2\text{ s}\).
\[ \% \text{ Uncertainty} = \frac{\Delta t}{t_1} \times 100\% = \frac{0.2}{9.22} \times 100\% \approx 2.2\% \]

### (d) Measurement of \(t_2\)
* Raw measurements for \(R_2 = 22\text{ k}\Omega\):
\(t_{2,a} = 19.52\text{ s}\)
\(t_{2,b} = 19.38\text{ s}\)
Average \(t_2 = 19.45\text{ s}\)

### (e) Testing the Relationship
* (i) Calculation of constants:
\[ k_1 = \frac{t_1}{R_1} = \frac{9.22\text{ s}}{10\text{ k}\Omega} = 0.922\text{ s k}\Omega^{-1} \approx 0.92\text{ s k}\Omega^{-1} \]
\[ k_2 = \frac{t_2}{R_2} = \frac{19.45\text{ s}}{22\text{ k}\Omega} = 0.884\text{ s k}\Omega^{-1} \approx 0.88\text{ s k}\Omega^{-1} \]
* (ii) Comparison:
\[ \% \text{ difference} = \frac{|0.922 - 0.884|}{0.922} \times 100\% = \frac{0.038}{0.922} \times 100\% \approx 4.1\% \]
Since \(4.1\% < 10\%\), the percentage difference is within the standard experimental uncertainty limit of \(10\%\). Therefore, the experimental results support the suggested relationship \(t = k R\).

### (f) Limitations and Corresponding Improvements

| # | Limitation | Improvement |
| :--- | :--- | :--- |
| 1 | Two sets of data are insufficient to draw a valid conclusion about the linear relationship. | Take measurements for at least five different resistor values and plot a graph of \(t\) against \(R\). |
| 2 | It is difficult to start and stop the stopwatch at the exact moments when the voltmeter displays \(5.0\text{ V}\) and \(2.0\text{ V}\) because the display numbers change very rapidly. | Use a data logger with a voltage sensor to record the voltage-time data automatically and determine \(t\) from the software. |
| 3 | Human reaction time in starting and stopping the manual stopwatch introduces random error. | Use an electronic threshold timer circuit that automatically triggers the timer on/off at predefined voltages. |
| 4 | The internal resistance of the digital voltmeter is in parallel with the capacitor and discharge resistor, lowering the effective resistance of the discharge path. | Use an electrometer or a cathode-ray oscilloscope (CRO) with a very high input impedance (e.g. \(10\text{ M}\Omega\) or higher). |

### Measurements (5 Marks)
* **Value of \(t_1\) (1 mark):** Raw values of \(t_1\) recorded to \(0.1\text{ s}\) or \(0.01\text{ s}\) with correct unit (s).
* **Repeat measurements of \(t_1\) (2 marks):** At least two trials of \(t_1\) recorded, and the average value calculated correctly.
* **Value of \(t_2\) (1 mark):** Raw value of \(t_2\) recorded with appropriate unit.
* **Repeat measurements of \(t_2\) (1 mark):** At least two trials of \(t_2\) recorded, and the average calculated correctly.

### Uncertainty Analysis (2 Marks)
* **Uncertainty estimate (2 marks):** Absolute uncertainty \(\Delta t\) in the range \(0.1\text{ s}\) to \(0.5\text{ s}\) is chosen and clearly justified based on human reaction time. Correct percentage uncertainty calculated with working shown.

### Constants and Comparison (5 Marks)
* **Calculation of \(k_1\) and \(k_2\) (2 marks):** Correct calculation of both constants with appropriate units (e.g., \(\text{s }\Omega^{-1}\) or \(\text{s k}\Omega^{-1}\)).
* **Comparison and Conclusion (3 marks):** Percentage difference between \(k_1\) and \(k_2\) is calculated correctly. A clear conclusion is stated based on whether this percentage difference is less than or equal to the specified \(10\%\) limit.

### Limitations and Improvements (8 Marks)
* **Limitations (4 marks):** Award 1 mark for each of up to 4 distinct valid points (e.g., limited data sets, rapid change of voltage display, human reaction time, voltmeter loading effect).
* **Improvements (4 marks):** Award 1 mark for each of up to 4 corresponding realistic improvements (e.g., plot a graph with more data, use a data logger, use an electronic threshold timer, use a high-impedance buffer/voltmeter).

(a) (i) Elastic deformation occurs when a material returns to its original shape and dimensions when the deforming force is removed.
(ii) Plastic deformation occurs when a material remains permanently deformed and does not return to its original shape and dimensions when the deforming force is removed.

(b) (i) Using \(E = \frac{F L}{A x}\):
\(E = \frac{60 \times 2.4}{1.5 \times 10^{-7} \times 1.2 \times 10^{-3}} = 8.0 \times 10^{11}\text{ Pa}\).

(ii) Elastic potential energy \(E_p = \frac{1}{2} F x\):
\(E_p = 0.5 \times 60 \times 1.2 \times 10^{-3} = 0.036\text{ J}\).

(c) During plastic deformation, planes of atoms slide past each other through the movement of dislocations and do not return to their original positions. When the 120 N load is removed, the wire contracts elastically but a permanent extension remains, meaning the final length is longer than the original 2.4 m.

(a) (i) Returns to original shape/length when load is removed [1]
(ii) Permanent deformation / does not return to original shape/length when load is removed [1]

(b) (i) Formula for Young modulus: \(E = \frac{FL}{Ax}\) [1]
Correct substitution of values [1]
Final answer: \(8.0 \times 10^{11}\text{ Pa}\) [1]

(ii) Formula: \(E_p = \frac{1}{2} F x\) [1]
Final answer: \(0.036\text{ J}\) (or \(3.6 \times 10^{-2}\text{ J}\)) [1]

(c) Atoms/layers slide past each other / movement of dislocations [1]
Atoms do not return to original positions [1]
Wire has a permanent extension when load is removed [1]

(a) (i) Area of brass: \(A_b = \pi \times (1.0 \times 10^{-3})^2 = 3.14 \times 10^{-6}\text{ m}^2\).
\(\text{Stress}_b = \frac{F}{A_b} = \frac{150}{3.14 \times 10^{-6}} = 4.77 \times 10^7\text{ Pa} \approx 4.8 \times 10^7\text{ Pa}\).

(ii) Area of steel: \(A_s = \pi \times (0.60 \times 10^{-3})^2 = 1.13 \times 10^{-6}\text{ m}^2\).
\(\text{Stress}_s = \frac{F}{A_s} = \frac{150}{1.13 \times 10^{-6}} = 1.33 \times 10^8\text{ Pa} \approx 1.3 \times 10^8\text{ Pa}\).

(b) Using \(x = \frac{F L}{A E}\):
Extension of brass: \(x_b = \frac{150 \times 1.6}{3.14 \times 10^{-6} \times 1.0 \times 10^{11}} = 7.64 \times 10^{-4}\text{ m}\).
Extension of steel: \(x_s = \frac{150 \times 1.2}{1.13 \times 10^{-6} \times 2.0 \times 10^{11}} = 7.96 \times 10^{-4}\text{ m}\).
Total extension: \(x_{\text{total}} = 7.64 \times 10^{-4} + 7.96 \times 10^{-4} = 1.56 \times 10^{-3}\text{ m} \approx 1.6\text{ mm}\).

(c) The steel section is more likely to break first because it has a smaller diameter, so the stress is significantly higher (almost 3 times higher) than in the brass section for any given tension.

(a) (i) Area calculation \(3.14 \times 10^{-6}\text{ m}^2\) [1]
Stress calculation \(4.8 \times 10^7\text{ Pa}\) [1]

(ii) Area calculation \(1.13 \times 10^{-6}\text{ m}^2\) [1]
Stress calculation \(1.3 \times 10^8\text{ Pa}\) [1]

(b) Formula for extension \(x = \frac{FL}{AE}\) seen or implied [1]
Correct extension of brass \(7.6 \times 10^{-4}\text{ m}\) [1]
Correct extension of steel \(8.0 \times 10^{-4}\text{ m}\) [1]
Sum of extensions: \(1.56 \times 10^{-3}\text{ m}\) (or \(1.6\text{ mm}\)) [1]

(c) Steel section [1]
Reason: higher stress due to smaller cross-sectional area [1]

(a) Internal resistance is the resistance to the flow of current within the cell itself, causing a loss of potential difference (lost volts) as heat energy in the cell.

(b) The total resistance of the circuit is \(R + r\). The current \(I\) is given by \(I = \frac{E}{R+r}\). The power \(P\) dissipated in \(R\) is \(P = I^2 R = \left(\frac{E}{R+r}\right)^2 R = \frac{E^2 R}{(R+r)^2}\).

(c) (i) For \(R = 0.50\ \Omega\):
\(I = \frac{6.0}{0.50 + 1.5} = 3.0\text{ A}\).
\(P = 3.0^2 \times 0.50 = 4.5\text{ W}\).

(ii) For \(R = 1.5\ \Omega\):
\(I = \frac{6.0}{1.5 + 1.5} = 2.0\text{ A}\).
\(P = 2.0^2 \times 1.5 = 6.0\text{ W}\).

(iii) For \(R = 4.5\ \Omega\):
\(I = \frac{6.0}{4.5 + 1.5} = 1.0\text{ A}\).
\(P = 1.0^2 \times 4.5 = 4.5\text{ W}\).

(d) The power is maximum when \(R = 1.5\ \Omega\) (i.e., \(R = r\)). The power delivered at this resistance (6.0 W) is greater than the power at both the lower resistance of \(0.50\ \Omega\) (4.5 W) and the higher resistance of \(4.5\ \Omega\) (4.5 W).

(e) Efficiency \(\eta = \frac{\text{Useful power output}}{\text{Total power input}} = \frac{I^2 R}{I^2 (R+r)} = \frac{R}{R+r}\). For \(R = r = 1.5\ \Omega\), \(\eta = \frac{1.5}{1.5 + 1.5} = 0.50\) or \(50\%\).

(a) Resistance of the cell/source (causing loss of voltage / heating inside the cell) [1]

(b) Expressions for current: \(I = \frac{E}{R+r}\) [1]
Substitution into \(P = I^2 R\) to give \(P = \frac{E^2 R}{(R+r)^2}\) [1]

(c) (i) Current \(I = 3.0\text{ A}\) and Power \(P = 4.5\text{ W}\) [2]
(ii) Power \(P = 6.0\text{ W}\) [1]
(iii) Power \(P = 4.5\text{ W}\) [1]

(d) \(R = 1.5\ \Omega\) [1]
Support: \(6.0\text{ W}\) is greater than the power at both \(0.50\ \Omega\) and \(4.5\ \Omega\) (both \(4.5\text{ W}\)) [1]

(e) Efficiency = 0.50 (or \(50\%\)) [1]

(a) The 9.0 V supply, the fixed resistor of \(1.2\text{ k}\Omega\), and the NTC thermistor are connected in series. The voltmeter to measure \(V_{\text{out}}\) is connected in parallel across the thermistor.

(b) Using the potential divider equation:
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{th}}}{R_{\text{th}} + R}\)
\(V_{\text{out}} = 9.0 \times \frac{2.8}{2.8 + 1.2} = 9.0 \times \frac{2.8}{4.0} = 6.3\text{ V}\).

(c) (i) As temperature increases, the resistance of the NTC thermistor decreases. This happens because thermal energy liberates more charge carriers (electrons) in the semiconductor material, increasing its conductivity.
(ii) As the resistance of the thermistor decreases, it takes a smaller fraction of the total series resistance. Thus, a smaller proportion of the supply voltage is dropped across the thermistor, causing \(V_{\text{out}}\) to decrease.

(d) Yes. When the temperature falls, the thermistor resistance increases, which causes \(V_{\text{out}}\) to increase. This rising voltage can be used directly as a signal to trigger the heater control circuit.

(a) Description of series connection of supply, fixed resistor and thermistor [1]
Voltmeter connected in parallel across the thermistor [1]

(b) Potential divider formula shown: \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{th}}}{R_{\text{th}} + R}\) [1]
Final answer: \(6.3\text{ V}\) [1]

(c) (i) Resistance decreases [1]
Explanation: more charge carriers / conduction electrons released by thermal energy [1]
(ii) \(V_{\text{out}}\) decreases [1]
Explanation: thermistor is a smaller fraction of total resistance / smaller share of voltage [1]

(d) Yes, because \(V_{\text{out}}\) increases as temperature falls [1]
This rising potential difference can trigger/switch on the heater [1]

(a) The diagram must show \(C_1\) and \(C_2\) in parallel branches, connected in series with \(C_3\) and the 12 V cell.

(b) (i) For parallel capacitors: \(C_p = C_1 + C_2 = 3.0\ \mu\text{F} + 6.0\ \mu\text{F} = 9.0\ \mu\text{F}\).
(ii) For the series combination: \(\frac{1}{C_T} = \frac{1}{C_p} + \frac{1}{C_3} = \frac{1}{9.0} + \frac{1}{4.0} = \frac{13}{36}\).
\(C_T = \frac{36}{13} \approx 2.77\ \mu\text{F} \approx 2.8\ \mu\text{F}\).

(c) (i) The total charge \(Q_T = C_T \times V = 2.77\ \mu\text{F} \times 12\text{ V} = 33.2\ \mu\text{C} \approx 33\ \mu\text{C}\).
Since \(C_3\) is in series with the parallel pair, the charge on \(C_3\) is equal to the total charge: \(Q_3 = 33\ \mu\text{C}\).

(ii) The potential difference across the parallel combination \(V_p\) is:
\(V_p = \frac{Q_T}{C_p} = \frac{33.2\ \mu\text{C}}{9.0\ \mu\text{F}} = 3.69\text{ V}\).
Therefore, the charge on \(C_1\) is:
\(Q_1 = C_1 \times V_p = 3.0\ \mu\text{F} \times 3.69\text{ V} = 11.1\ \mu\text{C} \approx 11\ \mu\text{C}\).

(d) Total energy stored \(E = \frac{1}{2} C_T V^2 = 0.5 \times (2.77 \times 10^{-6}) \times 12^2 = 1.99 \times 10^{-4}\text{ J} \approx 2.0 \times 10^{-4}\text{ J}\).

(a) Correct parallel combination of \(C_1\) and \(C_2\) [1]
Correct series connection of \(C_3\) and supply [1]

(b) (i) \(C_p = 9.0\ \mu\text{F}\) [1]
(ii) Correct series formula usage [1]
\(C_T = 2.8\ \mu\text{F}\) (or \(2.77\ \mu\text{F}\)) [1]

(c) (i) Use of \(Q = C V\) with total capacitance [1]
\(Q_3 = 33\ \mu\text{C}\) (or \(3.3 \times 10^{-5}\text{ C}\)) [1]
(ii) P.d. across parallel combination \(V_p = 3.7\text{ V}\) [1]
\(Q_1 = 11\ \mu\text{C}\) (or \(1.1 \times 10^{-5}\text{ C}\)) [1]

(d) \(E = 2.0 \times 10^{-4}\text{ J}\) [1]

(a) For a discharging capacitor, \(V = \frac{Q}{C}\) and \(I = -\frac{dQ}{dt}\). Since \(V = IR\), we have:
\(\frac{Q}{C} = -\frac{dQ}{dt} R \implies \frac{dQ}{dt} = -\frac{Q}{RC}\).
Separating variables and integrating from \(Q_0\) to \(Q\) and \(0\) to \(t\):
\(\int_{Q_0}^{Q} \frac{1}{Q} dQ = -\frac{1}{RC} \int_{0}^{t} dt \implies \ln\left(\frac{Q}{Q_0}\right) = -\frac{t}{RC}\).
Exponentiating both sides yields \(Q = Q_0 e^{-\frac{t}{RC}}\).

(b) \(Q_0 = C V_0 = 220 \times 10^{-6}\text{ F} \times 15.0\text{ V} = 3.3 \times 10^{-3}\text{ C} = 3.3\text{ mC}\).

(c) (i) Time constant \(\tau = RC = 4.4\text{ s}\).
\(R = \frac{4.4}{220 \times 10^{-6}} = 2.0 \times 10^4\ \Omega = 20\text{ k}\Omega\).

(ii) \(Q = 0.25 Q_0 \implies 0.25 = e^{-\frac{t}{4.4}}\).
Taking natural logs: \(\ln(0.25) = -\frac{t}{4.4} \implies -1.386 = -\frac{t}{4.4}\).
\(t = 1.386 \times 4.4 = 6.10\text{ s} \approx 6.1\text{ s}\).

(d) The graph is a decreasing exponential curve. The initial current is:
\(I_0 = \frac{V_0}{R} = \frac{15.0}{20 \times 10^3} = 7.5 \times 10^{-4}\text{ A}\) (or 0.75 mA). This value is clearly labeled on the vertical axis.

(a) Use of \(V = \frac{Q}{C}\) and \(V = IR\) with \(I = -\frac{dQ}{dt}\) to write differential equation [1]
Correct integration to arrive at \(Q = Q_0 e^{-\frac{t}{RC}}\) [1]

(b) \(Q_0 = 3.3 \times 10^{-3}\text{ C}\) [1]

(c) (i) Use of \(\tau = RC\) [1]
\(R = 2.0 \times 10^4\ \Omega\) [1]
(ii) Set up equation \(0.25 = e^{-t/4.4}\) [1]
Logarithmic step correctly shown [1]
\(t = 6.1\text{ s}\) [1]

(d) Correct shape of exponential decay graph [1]
Vertical axis labeled with correct initial current \(I_0 = 0.75\text{ mA}\) (or \(7.5 \times 10^{-4}\text{ A}\)) [1]

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

(b) (i) Distance during acceleration:
\(s_1 = ut + \frac{1}{2} a t^2 = 0 + 0.5 \times 4.5 \times 6.0^2 = 81\text{ m}\).
(ii) Distance during constant velocity:
\(s_2 = v t = 27 \times 12 = 324\text{ m}\).

(c) For the deceleration phase: \(u = 27\text{ m s}^{-1}\), \(v = 0\), and \(s = 54\text{ m}\).
Using \(v^2 = u^2 + 2as\):
\(0 = 27^2 + 2 a (54) \implies 0 = 729 + 108a \implies a = -6.75\text{ m s}^{-2}\).
Thus, the deceleration is \(6.75\text{ m s}^{-2}\) (or \(6.8\text{ m s}^{-2}\)).

(d) The velocity-time graph consists of:
- A straight line of positive gradient from \((0,0)\) to \((6.0, 27)\).
- A horizontal line at \(27\text{ m s}^{-1}\) from \(t = 6.0\text{ s}\) to \(t = 18.0\text{ s}\).
- A straight line of negative gradient from \(t = 18.0\text{ s}\) to \(t = 22.0\text{ s}\), where it reaches 0.
(Note: time for deceleration is found from \(s = \frac{u+v}{2} t \implies 54 = 13.5 t \implies t = 4.0\text{ s}\), so final time is \(18.0 + 4.0 = 22.0\text{ s}\)).

(a) \(v = 4.5 \times 6.0 = 27\text{ m s}^{-1}\) [1]

(b) (i) Formula \(s = ut + \frac{1}{2}at^2\) or area under graph [1]
\(81\text{ m}\) [1]
(ii) \(27 \times 12 = 324\text{ m}\) [1]

(c) Formula \(v^2 = u^2 + 2as\) [1]
\(\text{Deceleration} = 6.75\text{ m s}^{-2}\) (or \(6.8\text{ m s}^{-2}\)) [1]

(d) Axes labeled with units (velocity in \(\text{m s}^{-1}\), time in \(\text{s}\)) [1]
Straight line from 0 to 6.0 s reaching 27 [1]
Horizontal line from 6.0 s to 18.0 s at 27 [1]
Straight line from 18.0 s to 22.0 s meeting time axis at 22.0 s [1]

(a) Air resistance is negligible, which means there are no horizontal forces acting on the stone, so there is no horizontal acceleration.

(b) (i) Considering vertical motion (taking downwards as positive): \(u_y = 0\), \(a_y = 9.81\text{ m s}^{-2}\), \(s_y = 45\text{ m}\).
\(s_y = u_y t + \frac{1}{2} g t^2 \implies 45 = 0 + 0.5 \times 9.81 \times t^2\).
\(t^2 = 9.174 \implies t = 3.03\text{ s} \approx 3.0\text{ s}\).

(ii) Horizontal distance:
\(s_x = v_x \times t = 15 \times 3.03 = 45.4\text{ m} \approx 45\text{ m}\).

(iii) Vertical component of velocity on landing:
\(v_y = u_y + g t = 0 + (9.81 \times 3.03) = 29.7\text{ m s}^{-1}\).
Speed \(v\) is given by:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 29.7^2} = \sqrt{225 + 882} = 33.3\text{ m s}^{-1} \approx 33\text{ m s}^{-1}\).

(c) (i) Air resistance acts upwards, opposing the downward gravitational force, reducing net downward acceleration. This increases the time taken to reach the ground.
(ii) Air resistance acts horizontally opposite to velocity, decelerating horizontal motion. This decreases the horizontal distance traveled.

(a) No horizontal force / acceleration is zero [1]

(b) (i) Use of \(s = \frac{1}{2} g t^2\) [1]
\(t = 3.0\text{ s}\) (or \(3.03\text{ s}\)) [1]
(ii) Use of \(s_x = v_x t\) [1]
\(s_x = 45\text{ m}\) (or \(45.4\text{ m}\)) [1]
(iii) Calculate vertical velocity \(v_y = 29.7\text{ m s}^{-1}\) [1]
Use of Pythagoras to find resultant velocity [1]
\(v = 33\text{ m s}^{-1}\) (or \(33.3\text{ m s}^{-1}\)) [1]

(c) (i) Time taken increases [1]
(ii) Horizontal distance decreases [1]

Detailed solution:
(a) Capacitance is defined as charge per unit potential difference: \(C = Q/V\).

(b) For resistors in parallel:
\(1/R_p = 1/R_1 + 1/R_2 = 1/15 + 1/30 = 3/30 = 1/10\)
Thus, \(R_p = 10\text{ k}\Omega\).

(c)(i) The time constant \(\tau\) is given by:
\(\tau = R_p C = 10 \times 10^3 \times 470 \times 10^{-6} = 4.7\text{ s}\).

(ii) The initial charge \(Q_0\) is:
\(Q_0 = C V_0 = 470 \times 10^{-6} \times 12 = 5.64 \times 10^{-3}\text{ C} = 5.6\text{ mC}\).

(iii) The decay of potential difference is given by:
\(V = V_0 e^{-t/\tau}\)
\(3.0 = 12 e^{-t/4.7}\)
\(e^{-t/4.7} = 0.25\)
\(-t/4.7 = \ln(0.25) = -1.386\)
\(t = 1.386 \times 4.7 = 6.51\text{ s} \approx 6.5\text{ s}\).

(iv) The initial energy stored in the capacitor is:
\(E_i = \frac{1}{2} C V_0^2 = 0.5 \times 470 \times 10^{-6} \times 12^2 = 3.384 \times 10^{-2}\text{ J}\)
The final energy stored is:
\(E_f = \frac{1}{2} C V^2 = 0.5 \times 470 \times 10^{-6} \times 3.0^2 = 2.115 \times 10^{-3}\text{ J}\)
The energy dissipated in the resistors is the decrease in stored energy:
\(\Delta E = E_i - E_f = 3.384 \times 10^{-2} - 0.2115 \times 10^{-2} = 3.1725 \times 10^{-2}\text{ J} \approx 3.2 \times 10^{-2}\text{ J}\) (or \(32\text{ mJ}\)).

(a) C1: charge / potential difference (or charge per unit potential difference)

(b) B1: Show clear parallel resistor calculation, e.g., 1/15 + 1/30 = 1/10 to obtain Rp = 10 kOhm

(c)(i) C1: Formula tau = RC used with correct units
A1: tau = 4.7 s

(ii) C1: Q = CV used with correct values
A1: Q0 = 5.6 * 10^-3 C (or 5.6 mC)

(iii) C1: Use of V = V0 * e^(-t/tau) or ln(V/V0) = -t/tau
A1: t = 6.5 s (accept 6.51 s)

(iv) C1: Energy formula E = 1/2 C V^2 used to find initial and/or final energy
A1: Delta E = 3.2 * 10^-2 J (or 32 mJ) (accept 3.17 * 10^-2 J)

Detailed solution:

(a) For simple harmonic motion:
1. The acceleration of the block is directly proportional to its displacement from its equilibrium position.
2. The acceleration is always directed towards the equilibrium position.

(b)(i) The angular frequency \(\omega\) is:
\(\omega = 2\pi f = 2 \pi \times 2.5 = 15.7\text{ rad s}^{-1} \approx 16\text{ rad s}^{-1}\).

(ii) The maximum acceleration \(a_{\text{max}}\) occurs at the maximum displacement (amplitude \(x_0 = 1.6\text{ cm} = 0.016\text{ m}\)):
\(a_{\text{max}} = \omega^2 x_0 = (15.71)^2 \times 0.016 = 246.7 \times 0.016 = 3.95\text{ m s}^{-2} \approx 4.0\text{ m s}^{-2}\).

(c)(i) The forces acting on the block are its weight \(mg\) acting downwards and the normal contact force \(N\) from the platform acting upwards. At the highest point of the oscillation, the acceleration is downwards, so:
\(mg - N = ma \implies N = m(g - a)\).
For the block to stay on the platform, \(N \ge 0\). If the downward acceleration of the platform exceeds \(g\) (\(9.81\text{ m s}^{-2}\)), then \(N\) would have to be negative, which is impossible because the platform can only push the block. Thus, the block loses contact with the platform when the downward acceleration exceeds \(g\).

(ii) The block loses contact when the maximum acceleration \(a_{\text{max}}\) is equal to \(g\):
\(a_{\text{max}} = g \implies \omega^2 x_0 = g\)
\(x_0 = \frac{g}{\omega^2} = \frac{9.81}{15.71^2} = 0.0398\text{ m} \approx 4.0\text{ cm}\) (or \(0.040\text{ m}\)).

(a) B1: acceleration is proportional to displacement
B1: acceleration and displacement are in opposite directions (or acceleration is directed towards a fixed point)

(b)(i) C1: \(\omega = 2\pi f\) used
A1: \(\omega = 15.7\text{ rad s}^{-1}\) (accept 16)

(ii) C1: \(a = \omega^2 x\) used
A1: \(a_{\text{max}} = 3.95\text{ m s}^{-2}\) or \(4.0\text{ m s}^{-2}\)

(c)(i) B1: downward acceleration at the highest point of oscillation requires a downward resultant force provided by gravity (and contact force)
B1: if acceleration exceeds \(g\), the required downward force exceeds the weight, so contact force \(N\) becomes zero/negative (block loses contact)

(ii) C1: equating maximum acceleration to \(g\) (\(\omega^2 x_0 = g\))
A1: amplitude \(x_0 = 0.040\text{ m}\) or \(4.0\text{ cm}\)

### Experimental Setup and Circuit Diagram
1. **Circuit Design**: Draw a circuit containing a DC power supply connected via a two-way switch to a capacitor of known capacitance \(C\). The other terminal of the two-way switch connects to the conductive polymer strip. A high-resistance digital storage oscilloscope (DSO) or a voltmeter connected to a data logger is placed in parallel across the capacitor to record the potential difference \(V\) as a function of time \(t\).
2. **Connecting the Strip**: To ensure uniform electrical contact, flat brass plates or copper foil clamps should be clamped across the entire width of both ends of the strip. This ensures that the current flows uniformly along the length \(L\) of the strip.

### Variables
- **Independent Variable**: Width \(w\) of the conductive strip.
- **Dependent Variable**: Time constant \(\tau\) of the discharge.
- **Controlled Variables**:
- Length \(L\) of the strip (cut precisely to the same length each time).
- Thickness \(t\) of the strip (using a single sheet of uniform thickness).
- Capacitance \(C\) of the capacitor.
- Ambient temperature (performed in a temperature-controlled room because conductivity \(\sigma\) is temperature-dependent).

### Measurements
1. **Width \(w\)**: Measured at 5 different positions along the strip using a digital caliper or a high-resolution ruler, and the mean value is calculated. Uniformity is verified if the variation is within the uncertainty of the measuring instrument.
2. **Thickness \(t\)**: Measured at multiple points along the sheet using a micrometer screw gauge.
3. **Time constant \(\tau\)**: From the discharge curve on the DSO, \(\tau\) is the time taken for the voltage to fall from \(V_0\) to \(V_0/e \approx 0.37 V_0\). Alternatively, plot a graph of \(\ln(V)\) against time, where the gradient is \(-1/\tau\).

### Analysis
1. Plot a graph of \(\tau\) against \(1/w\).
2. If the relationship is valid, the graph should be a straight line passing through the origin.
3. The gradient \(m\) is given by:
\[m = \frac{k C L}{\sigma t}\]
4. Calculate the electrical conductivity \(\sigma\) using:
\[\sigma = \frac{k C L}{m t}\]

**Defining the Problem (Max 3 marks):**
- **[1]** Identify \(w\) as the independent variable and \(\tau\) as the dependent variable.
- **[1]** State that the capacitance \(C\) and length \(L\) of the strip are kept constant.
- **[1]** State that the ambient temperature must be controlled/kept constant.

**Methods of Data Collection (Max 5 marks):**
- **[1]** Draw a workable circuit diagram showing a DC supply, a double-throw switch, the capacitor, and the conductive strip connected such that the capacitor can be charged and then discharged through the strip.
- **[1]** Show a high-resistance digital voltmeter, data logger, or CRO connected in parallel with the capacitor.
- **[1]** Explain how electrical connection is made across the entire width of the strip (e.g., using copper foil tape, flat metal plates, or wide clamps).
- **[1]** Describe how thickness \(t\) is measured using a micrometer screw gauge in several places.
- **[1]** Describe how the width \(w\) is measured at multiple positions along the length using a caliper/ruler and averaged.

**Method of Analysis (Max 3 marks):**
- **[1]** State that a graph of \(\tau\) against \(1/w\) should be plotted.
- **[1]** State that a straight line through the origin confirms the relationship.
- **[1]** Express the conductivity \(\sigma\) in terms of the gradient \(m\): \(\sigma = \frac{k C L}{m t}\).

**Safety and Additional Details (Max 4 marks):**
- **[1]** **Safety**: Keep supply voltage below the maximum safety/rating voltage of the capacitors to avoid damage or bursting.
- **[1]** Use high-resistance meters (input impedance \(> 10\text{ M}\Omega\)) to prevent the capacitor from discharging through the meter.
- **[1]** Clean contact surfaces with isopropyl alcohol or fine emery paper to eliminate contact resistance.
- **[1]** Use of a storage oscilloscope to capture the trace of a single discharge cycle for quick/reliable measurement.

### 1. Graph Plotting
- Scale: \(L / \text{ m}\) on x-axis from 0.0 to 1.6; \(e / \text{ mm}\) on y-axis from 0.0 to 7.0.
- Error bars for \(e\) are drawn vertically with height \(\pm 0.15\text{ mm}\).

### 2. Best-fit and Worst Acceptable Lines
- **Best-fit line** passes through all points, representing the mean trend. Points plotted: \((0.400, 2.15)\), \((0.600, 2.90)\), \((0.800, 3.65)\), \((1.000, 4.40)\), \((1.200, 5.15)\), \((1.400, 5.90)\).
- **Worst acceptable line**: The steepest or shallowest possible straight line that passes through all error bars. For example, a shallow line passing through \((0.400, 2.30)\) and \((1.400, 5.75)\).

### 3. Gradient Calculation
- **Best-fit gradient \(m\)**:
\[m = \frac{5.90 - 2.15}{1.400 - 0.400} = 3.75\text{ mm m}^{-1} = 3.75 \times 10^{-3}\text{ m m}^{-1}\]
- **Worst acceptable gradient \(m_{\text{worst}}\)**:
\[m_{\text{worst}} = \frac{5.75 - 2.30}{1.400 - 0.400} = 3.45\text{ mm m}^{-1} = 3.45 \times 10^{-3}\text{ m m}^{-1}\]
- **Uncertainty in gradient \(\Delta m\)**:
\[\Delta m = |m - m_{\text{worst}}| = 0.30 \times 10^{-3}\text{ m m}^{-1}\]
\[m = (3.75 \pm 0.30) \times 10^{-3}\]

### 4. y-intercept Calculation
- **Best-fit y-intercept \(c\)**:
Using \(e = m L + c\) at \(L = 0.400\text{ m}\):
\[2.15 \times 10^{-3} = (3.75 \times 10^{-3})(0.400) + c \implies c = 0.65 \times 10^{-3}\text{ m} = 0.65\text{ mm}\]
- **Worst acceptable y-intercept \(c_{\text{worst}}\)**:
Using \(e = m_{\text{worst}} L + c_{\text{worst}}\) at \(L = 1.400\text{ m}\):
\[5.75 \times 10^{-3} = (3.45 \times 10^{-3})(1.40) + c_{\text{worst}} \implies c_{\text{worst}} = 0.92 \times 10^{-3}\text{ m} = 0.92\text{ mm}\]
- **Uncertainty in y-intercept \(\Delta c\)**:
\[\Delta c = |c - c_{\text{worst}}| = 0.27 \times 10^{-3}\text{ m} = 0.27\text{ mm}\]
\[c = (0.65 \pm 0.27) \times 10^{-3}\text{ m}\]

### 5. Constants \(Y\) and \(\alpha\) and their uncertainties
- **Young Modulus \(Y\)**:
\[Y = \frac{4 F}{\pi d^2 m}\]
Substitute \(F = 35.0\text{ N}\), \(d = 0.42 \times 10^{-3}\text{ m}\), \(m = 3.75 \times 10^{-3}\):
\[Y = \frac{4 \times 35.0}{\pi \times (0.42 \times 10^{-3})^2 \times 3.75 \times 10^{-3}} = 6.74 \times 10^{10}\text{ Pa}\]
- **Uncertainty in \(Y\)**:
\[\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + 2 \frac{\Delta d}{d} + \frac{\Delta m}{m} = \frac{0.5}{35.0} + 2 \frac{0.01}{0.42} + \frac{0.30}{3.75} = 0.0143 + 0.0476 + 0.0800 = 0.1419\]
\[\Delta Y = 0.1419 \times 6.74 \times 10^{10} = 0.96 \times 10^{10}\text{ Pa} \approx 1.0 \times 10^{10}\text{ Pa}\]
\[Y = (6.7 \pm 1.0) \times 10^{10}\text{ Pa}\]
- **Non-linear deformation constant \(\alpha\)**:
\[\alpha = \frac{c}{F^2} = \frac{0.65 \times 10^{-3}}{35.0^2} = 5.31 \times 10^{-7}\text{ m N}^{-2}\]
- **Uncertainty in \(\alpha\)**:
\[\frac{\Delta \alpha}{\alpha} = \frac{\Delta c}{c} + 2 \frac{\Delta F}{F} = \frac{0.27}{0.65} + 2 \frac{0.5}{35.0} = 0.4154 + 0.0286 = 0.444\]
\[\Delta \alpha = 0.444 \times 5.31 \times 10^{-7} = 2.36 \times 10^{-7}\text{ m N}^{-2} \approx 2.4 \times 10^{-7}\text{ m N}^{-2}\]
\[\alpha = (5.3 \pm 2.4) \times 10^{-7}\text{ m N}^{-2}\]

**Graph Plotting and Fitting (Max 5 marks):**
- **[1]** Plot all 6 points correctly within half a small grid square with correct vertical error bars of length \(\pm 0.15\text{ mm}\).
- **[1]** Draw a straight line of best fit that is balanced through all points.
- **[1]** Draw a worst acceptable line that is clearly labeled, passing through all error bars.
- **[1]** Calculate the best-fit gradient with clear working showing coordinates from the line (at least half of the line length apart).
- **[1]** Calculate the uncertainty in the gradient from the difference between the best-fit and worst acceptable gradients.

**y-intercept and Uncertainty (Max 2 marks):**
- **[1]** Determine the y-intercept of the best-fit line correctly (by substitution or direct reading from \(x = 0\)).
- **[1]** Determine the uncertainty in the y-intercept from the difference between the best-fit and worst acceptable intercepts.

**Calculating \(Y\) and its Uncertainty (Max 4 marks):**
- **[1]** Show the correct formula: \(Y = \frac{4 F}{\pi d^2 m}\).
- **[1]** Calculate \(Y\) correctly in the range \(6.6 \times 10^{10}\text{ Pa}\) to \(6.9 \times 10^{10}\text{ Pa}\).
- **[1]** Express fractional uncertainty of \(Y\) correctly: \(\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + 2 \frac{\Delta d}{d} + \frac{\Delta m}{m}\).
- **[1]** Obtain the absolute uncertainty \(\Delta Y\) correctly as \(\approx 1.0 \times 10^{10}\text{ Pa}\).

**Calculating \(\alpha\) and its Uncertainty (Max 4 marks):**
- **[1]** Show the correct formula: \(\alpha = \frac{c}{F^2}\).
- **[1]** Calculate \(\alpha\) correctly in the range \(5.1 \times 10^{-7}\text{ m N}^{-2}\) to \(5.5 \times 10^{-7}\text{ m N}^{-2}\).
- **[1]** Express fractional uncertainty of \(\alpha\) correctly: \(\frac{\Delta \alpha}{\alpha} = \frac{\Delta c}{c} + 2 \frac{\Delta F}{F}\).
- **[1]** Obtain the absolute uncertainty \(\Delta \alpha\) correctly as \(\approx 2.4 \times 10^{-7}\text{ m N}^{-2}\).