2025 Edexcel IAL Physics (YPH11) 模擬試題連答案詳解

The initial component of momentum perpendicular to the wall is \(mv \cos\theta\) (towards the wall). The final component of momentum perpendicular to the wall is \(-mv \cos\theta\) (away from the wall).

The change in momentum perpendicular to the wall is:
\(\Delta p = -mv \cos\theta - mv \cos\theta = -2mv \cos\theta\)

The magnitude of this change is \(2mv \cos\theta\).

Since the component of momentum parallel to the wall remains unchanged (\(mv \sin\theta\)), the total magnitude of the change in momentum of the ball is \(2mv \cos\theta\).

[1 mark] C is the correct response. Correct resolution of the perpendicular component of velocity and calculation of the change in momentum.

The extension of a wire is given by \(\Delta x = \frac{FL}{AE}\).

For the second wire:
- The material is the same, so the Young modulus is \(E\).
- The length is \(2L\).
- The diameter is halved, which means the radius is halved. Since cross-sectional area is proportional to the square of the radius, the new area is \(A' = \frac{A}{4}\).

Substituting these into the extension formula:
\(\Delta x' = \frac{F(2L)}{\left(\frac{A}{4}\right)E} = 8 \times \frac{FL}{AE} = 8\Delta x\)

[1 mark] C is the correct response. Deduces that halving the diameter reduces the area by a factor of 4, leading to an 8-fold increase in extension.

At the maximum height of its trajectory, the vertical component of the projectile's velocity is zero (\(v_y = 0\)).

However, because air resistance is negligible, there is no horizontal force acting on the projectile, so its horizontal velocity remains constant throughout the flight: \(v_x = u \cos\theta\).

Therefore, the velocity of the projectile at its maximum height is \(u \cos\theta\).

The only acceleration acting on the projectile is gravity, which remains constant at \(g\) directed vertically downwards throughout the entire flight.

[1 mark] B is the correct response. Identifies that horizontal velocity is constant at \(u \cos\theta\) and vertical acceleration is \(g\) downwards.

At terminal velocity, the sphere is in translational equilibrium, meaning the net force acting on it is zero.

The downward force is the weight:
\(W = mg\)

The upward forces are:
1. Upthrust (\(U\))
2. Viscous drag (\(F = 6\pi\eta r v\))

Balancing these forces:
\(mg = U + 6\pi\eta r v\)

Rearranging for upthrust \(U\):
\(U = mg - 6\pi\eta r v\)

[1 mark] B is the correct response. Formulates the equilibrium equation (weight = upthrust + drag) and rearranges to solve for upthrust.

The useful power output of the motor is the rate at which it does work against gravity:
\(P_{\text{out}} = Fv = Mgv\)

Efficiency \(\eta\) is defined as:
\(\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\)

Rearranging to find the electrical input power \(P_{\text{in}}\):
\(P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{Mgv}{\eta}\)

[1 mark] B is the correct response. Expresses power output as \(Mgv\) and uses efficiency to obtain input power.

To find the force exerted by the right-hand support (\(F_{\text{right}}\)), we take moments about the left-hand support (pivot) to eliminate the force from that support.

The clockwise moments are:
1. Due to the person: \(2W \times \frac{L}{4} = \frac{WL}{2}\)
2. Due to the plank's weight (acting at its midpoint): \(W \times \frac{L}{2} = \frac{WL}{2}\)

The anticlockwise moment is:
- Due to the right support: \(F_{\text{right}} \times L\)

For rotational equilibrium:
\(\sum \text{Clockwise Moments} = \sum \text{Anticlockwise Moments}\)
\(\frac{WL}{2} + \frac{WL}{2} = F_{\text{right}} L\)
\(WL = F_{\text{right}} L\)
\(F_{\text{right}} = W\)

[1 mark] B is the correct response. Correct moment equation about the left support leading to \(F_{\text{right}} = W\).

A ductile material is one that can be permanently stretched or drawn into wires, indicating a large amount of plastic (permanent) deformation before fracture occurs.

- Brittle materials exhibit little or no plastic deformation and break suddenly.
- Hard materials resist scratch, wear, or indentation.
- Stiff materials resist elastic deformation (have a high Young modulus).

[1 mark] B is the correct response. Recalls the definition of a ductile material as having large plastic deformation before fracture.

Resolving the weight of the block parallel and perpendicular to the inclined plane:
1. The component of weight parallel to the slope acts downwards and is given by \(mg \sin\theta\).
2. The component of weight perpendicular to the slope acts into the plane and is given by \(mg \cos\theta\).

Since the block remains at rest (in static equilibrium), the forces must balance in both directions:
- Parallel to the slope: the friction force \(f\) acting up the slope must balance the component of weight down the slope:
\(f = mg \sin\theta\)

- Perpendicular to the slope: the normal contact force \(R\) must balance the component of weight perpendicular to the slope:
\(R = mg \cos\theta\)

[1 mark] B is the correct response. Identifies the parallel component of weight as \(mg \sin\theta\) and equates it to the friction force for static equilibrium.

At the launch point, the kinetic energy of the projectile is given by:
\(E = \frac{1}{2} m u^2\)

At the highest point of its path, the vertical component of the velocity is zero (\(v_y = 0\)). The horizontal component of the velocity remains constant throughout the flight because there is no horizontal acceleration (air resistance is negligible):
\(v_x = u \cos(60^\circ) = 0.5u\)

Therefore, the velocity at the highest point is \(v = v_x = 0.5u\).

The kinetic energy at this point is:
\(E_{\text{top}} = \frac{1}{2} m v^2 = \frac{1}{2} m (0.5u)^2 = 0.25 \left(\frac{1}{2} m u^2\right) = 0.25E\)

Thus, the correct option is B.

1 mark for the correct answer (B).

The Young modulus \(E\) of the material is given by:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}

Rearranging for the extension \)x\):
\(x = \frac{FL}{AE}

Since the cross-sectional area \)A\) is related to the diameter \(d\) by \(A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}\), we can write:
\(x \propto \frac{L}{d^2}\) (since \(F\) and \(E\) are constant for both wires).

For the second wire with length \(L' = 2L\) and diameter \(d' = 2d\), the new extension \(x'\) is:
\(x' \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = 0.5 \left(\frac{L}{d^2}\right)

Therefore:
\)x' = 0.50x\)

Thus, the correct option is B.

1 mark for the correct answer (B).

(a) For the vertical motion: the initial vertical velocity \( u_y = 0\text{ m s}^{-1} \), acceleration \( a = g = 9.81\text{ m s}^{-2} \), and displacement \( s_y = 85\text{ m} \).
Using \( s_y = u_y t + \frac{1}{2} a t^2 \):
\( 85 = 0 + \frac{1}{2} (9.81) t^2 \)
\( t^2 = \frac{170}{9.81} = 17.33 \)
\( t = \sqrt{17.33} = 4.16\text{ s} \), which is approximately \( 4.2\text{ s} \).

(b) For the horizontal motion, velocity remains constant at \( v_x = 42\text{ m s}^{-1} \).
Using \( s_x = v_x \times t \):
\( s_x = 42 \times 4.16 = 175\text{ m} \) (or \( 176\text{ m} \) using \( t = 4.2\text{ s} \)).

(c) The vertical velocity just before impact is:
\( v_y = u_y + a t = 0 + (9.81)(4.16) = 40.8\text{ m s}^{-1} \).
The horizontal velocity is \( v_x = 42\text{ m s}^{-1} \).
Magnitude of velocity \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{42^2 + 40.8^2} = 58.6\text{ m s}^{-1} \).
Direction \( \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{40.8}{42}\right) = 44.2^\circ \) below the horizontal.

(a)
- Use of \( s = ut + \frac{1}{2}at^2 \) vertically with \( u_y = 0 \) (1 mark)
- Correct calculation to show \( t = 4.16\text{ s} \) (1 mark)

(b)
- Use of horizontal distance = speed \(\times\) time (1 mark)
- Correct horizontal distance \( 175\text{ m} \) or \( 176\text{ m} \) (1 mark)

(c)
- Correct calculation of vertical component of velocity \( v_y = 40.8\text{ m s}^{-1} \) (1 mark)
- Correct calculation of magnitude of velocity \( 58.6\text{ m s}^{-1} \) (1 mark)
- Correct direction of velocity \( 44.2^\circ \) below horizontal (1 mark)

(a) Stress \( \sigma = \frac{\text{Force}}{\text{Area}} = \frac{180\text{ N}}{1.5 \times 10^{-6}\text{ m}^2} = 1.20 \times 10^8\text{ Pa} \).

(b) Young Modulus \( E = \frac{\text{Stress}}{\text{Strain}} \), so \( \text{Strain } \epsilon = \frac{\text{Stress}}{E} = \frac{1.2 \times 10^8\text{ Pa}}{2.0 \times 10^{11}\text{ Pa}} = 6.0 \times 10^{-4} \).
Extension \( \Delta L = \text{Strain} \times L_0 = 6.0 \times 10^{-4} \times 2.4\text{ m} = 1.44 \times 10^{-3}\text{ m} \).

(c) If the load is increased beyond the elastic limit, the wire will undergo plastic deformation. When the load is then removed, the wire will not return to its original length and will have a permanent extension.

(a)
- Use of \( \text{stress} = F/A \) (1 mark)
- Correct calculation leading to \( 1.2 \times 10^8\text{ Pa} \) (1 mark)

(b)
- Recall of \( E = \frac{\text{stress}}{\text{strain}} \) or use of \( \Delta L = \frac{F L}{A E} \) (1 mark)
- Substitution of correct values (1 mark)
- Correct answer with units: \( 1.44 \times 10^{-3}\text{ m} \) (1 mark)

(c)
- Wire undergoes plastic/permanent deformation/stretches irreversibly (1 mark)
- Wire retains a permanent extension/does not return to its original length when the load is removed (1 mark)

(a) Taking the initial direction of the first glider as positive:
\( m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \)
\( (0.45 \times 1.8) + (0.25 \times (-1.2)) = (0.45 + 0.25) v \)
\( 0.81 - 0.30 = 0.70 v \)
\( 0.51 = 0.70 v \)
\( v = 0.729\text{ m s}^{-1} \) (in the direction of the first glider).

(b) Initial kinetic energy:
\( E_{ki} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} (0.45)(1.8)^2 + \frac{1}{2} (0.25)(-1.2)^2 = 0.729 + 0.180 = 0.909\text{ J} \).
Final kinetic energy:
\( E_{kf} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} (0.70)(0.729)^2 = 0.186\text{ J} \).
Loss in kinetic energy = \( 0.909\text{ J} - 0.186\text{ J} = 0.723\text{ J} \).
Since kinetic energy is lost, the collision is inelastic.

(c) The total momentum of a system of interacting bodies remains constant, provided no external forces act on the system.

(a)
- Use of conservation of momentum with correct sign for opposing velocity (1 mark)
- Correct calculation of initial momentum as \( 0.51\text{ kg m s}^{-1} \) (1 mark)
- Correct final velocity \( 0.729\text{ m s}^{-1} \) (or \( 0.73\text{ m s}^{-1} \)) (1 mark)

(b)
- Correct calculation of total initial kinetic energy \( 0.909\text{ J} \) (1 mark)
- Correct calculation of final kinetic energy \( 0.186\text{ J} \) (1 mark)
- Calculation of non-zero loss in KE (\( 0.723\text{ J} \)) and stating that this shows the collision is inelastic (1 mark)

(c)
- Correct statement: total momentum is conserved in a closed system / when no external forces act (1 mark)

(a) The component of weight parallel to the slope is:
\( W_{\parallel} = mg \sin\theta = 1200 \times 9.81 \times \sin(6.0^\circ) = 11772 \times 0.1045 = 1230\text{ N} \), which is approximately \( 1200\text{ N} \).

(b) At a constant speed, the net force along the slope is zero.
The engine force \( F_E \) must balance the component of weight down the slope and the resistive force:
\( F_E = W_{\parallel} + F_{\text{resistive}} = 1230\text{ N} + 450\text{ N} = 1680\text{ N} \).
Power \( P = F_E \times v = 1680\text{ N} \times 15\text{ m s}^{-1} = 25200\text{ W} = 25.2\text{ kW} \).
(If using the given \( 1200\text{ N} \) approximation, \( F_E = 1650\text{ N} \) and \( P = 24.8\text{ kW} \). Both are acceptable).

(c) On the flat road, the component of weight acting along the direction of motion becomes zero. The engine's forward force is now greater than the resistive force, producing a net forward force. Therefore, the car will accelerate and its velocity will increase (until a new equilibrium is reached where the resistive force, which increases with speed, balances the engine force).

(a)
- Use of \( W_{\parallel} = mg \sin\theta \) (1 mark)
- Correct calculation leading to \( 1230\text{ N} \) (or \( 1228\text{ N} \) using \( g = 9.8\text{ m s}^{-2} \)) (1 mark)

(b)
- Sum of forces evaluated: \( F_E = W_{\parallel} + 450 \) (1 mark)
- Use of \( P = F v \) (1 mark)
- Correct calculation of power: \( 25.2\text{ kW} \) (or \( 24.8\text{ kW} \) if using the \( 1200\text{ N} \) value) (1 mark)

(c)
- Identifies that there is no longer a component of weight resisting the motion (or net force is now forward) (1 mark)
- Concludes that the car will accelerate / velocity will increase (1 mark)

(a) For springs in parallel, the effective spring constant is the sum of the individual constants:
\( k_p = k + k = 320 + 320 = 640\text{ N m}^{-1} \).

(b) The system consists of the parallel pair in series with the third spring.
The total effective spring constant \( k_{\text{eff}} \) is:
\( \frac{1}{k_{\text{eff}}} = \frac{1}{k_p} + \frac{1}{k} = \frac{1}{640} + \frac{1}{320} = \frac{3}{640} \)
\( k_{\text{eff}} = \frac{640}{3} = 213.3\text{ N m}^{-1} \).

The load force is \( F = mg = 4.5\text{ kg} \times 9.81\text{ m s}^{-2} = 44.15\text{ N} \).
Using \( F = k_{\text{eff}} x \):
\( x = \frac{F}{k_{\text{eff}}} = \frac{44.15}{213.3} = 0.207\text{ m} \).

(Alternatively:
Extension of parallel pair: \( x_p = \frac{F}{k_p} = \frac{44.15}{640} = 0.069\text{ m} \).
Extension of single spring: \( x_s = \frac{F}{k} = \frac{44.15}{320} = 0.138\text{ m} \).
Total extension \( x = x_p + x_s = 0.207\text{ m} \).)

(c) The elastic strain energy \( E_{\text{el}} = \frac{1}{2} F x = \frac{1}{2} (44.15\text{ N})(0.207\text{ m}) = 4.57\text{ J} \).

(a)
- States \( k_p = 320 + 320 = 640\text{ N m}^{-1} \) (1 mark)

(b)
- Use of \( W = mg \) to find load force \( 44.15\text{ N} \) (1 mark)
- Use of series formula for combination to find \( k_{\text{eff}} = 213\text{ N m}^{-1} \) OR calculates individual extensions (1 mark)
- Correct calculation of both contributions or equivalent single step (1 mark)
- Correct total extension \( 0.207\text{ m} \) (or \( 0.21\text{ m} \)) (1 mark)

(c)
- Use of \( E_{\text{el}} = \frac{1}{2} F x \) or \( \frac{1}{2} k x^2 \) (1 mark)
- Correct calculation of energy \( 4.57\text{ J} \) (accept \( 4.6\text{ J} \)) (1 mark)

(a) The diagram should show four forces acting on the crate from a single point:
1. Weight \( W \) acting vertically downwards.
2. Normal contact force \( R \) acting vertically upwards.
3. Tension \( T \) acting upwards and to the right at an angle of \( 25^\circ \) to the horizontal.
4. Friction \( F_{\text{f}} \) acting horizontally to the left (opposing motion).

(b) Using Newton's second law for horizontal motion:
\( T \cos(25^\circ) - F_{\text{f}} = m a \)
\( 150 \cos(25^\circ) - F_{\text{f}} = 35 \times 0.82 \)
\( 135.9 - F_{\text{f}} = 28.7 \)
\( F_{\text{f}} = 135.9 - 28.7 = 107.2\text{ N} \), which is about \( 110\text{ N} \).

(c) The crate is in vertical equilibrium, so the net vertical force is zero:
\( R + T \sin(25^\circ) - W = 0 \)
\( R = W - T \sin(25^\circ) \)
\( W = m g = 35 \times 9.81 = 343.4\text{ N} \)
\( R = 343.4 - 150 \sin(25^\circ) = 343.4 - 63.4 = 280\text{ N} \).

(a)
- Draw and label weight and normal reaction force in opposing vertical directions (1 mark)
- Draw and label tension at an angle and friction horizontally opposing motion (1 mark)

(b)
- Resolves the tension horizontally: \( 150 \cos(25^\circ) \) (1 mark)
- Sets up Newton's second law equation \( T\cos\theta - F_{\text{f}} = ma \) and obtains \( F_{\text{f}} \approx 107\text{ N} \) (1 mark)

(c)
- Resolves the tension vertically: \( 150 \sin(25^\circ) \) (1 mark)
- Formulates the correct vertical equilibrium equation: \( R + T\sin\theta = mg \) (1 mark)
- Correct calculation of normal force: \( 280\text{ N} \) (or \( 279.6\text{ N} \) using \( g = 9.8\text{ m s}^{-2} \)) (1 mark)

(a) The three forces are: Weight (acting downwards), Upthrust (acting upwards), and Viscous drag (acting upwards).

(b) At terminal velocity, the upward forces equal the downward forces:
\( W = U + F_{\text{drag}} \)
Substitute formulas for weight and upthrust using volume \( V = \frac{4}{3} \pi r^3 \), and viscous drag using Stokes' law \( F_{\text{drag}} = 6\pi\eta r v \):
\( \rho_s V g = \rho_f V g + 6\pi\eta r v \)
\( (\rho_s - \rho_f) \left(\frac{4}{3} \pi r^3\right) g = 6\pi\eta r v \)
Rearrange to solve for \( v \):
\( v = \frac{4 \pi r^3 g (\rho_s - \rho_f)}{3 \times 6 \pi \eta r} = \frac{2 r^2 g (\rho_s - \rho_f)}{9 \eta} \).

(c) Substituting the values:
\( v = \frac{2 \times (2.5 \times 10^{-3}\text{ m})^2 \times 9.81\text{ m s}^{-2} \times (7800 - 920)\text{ kg m}^{-3}}{9 \times 0.29\text{ Pa s}} \)
\( v = \frac{2 \times (6.25 \times 10^{-6}) \times 9.81 \times 6880}{2.61} \)
\( v = \frac{0.8439}{2.61} = 0.323\text{ m s}^{-1} \).

(a)
- Identifies Weight / gravity force (downwards) (1 mark for any two, 2 marks for all three)
- Identifies Upthrust and Viscous drag / frictional force (upwards)

(b)
- Equates forces correctly: \( W = U + F_{\text{drag}} \) (1 mark)
- Correct substitution of volume and Stokes' Law expressions, leading to the given formula (1 mark)

(c)
- Correct substitution of all values into the formula (1 mark)
- Correct calculation of numerator or partial evaluation (1 mark)
- Correct final value of terminal velocity: \( 0.32\text{ m s}^{-1} \) to \( 0.323\text{ m s}^{-1} \) (1 mark)

(a) Taking moments about the hinge A:
Clockwise moments are caused by the weight of the beam and the weight of the sign.
Weight of beam \( W_b = 6.5 \times 9.81 = 63.77\text{ N} \) acting at \( 0.75\text{ m} \).
Weight of sign \( W_s = 12 \times 9.81 = 117.72\text{ N} \) acting at \( 1.5\text{ m} \).

Total clockwise moment = \( (63.77 \times 0.75) + (117.72 \times 1.5) = 47.83 + 176.58 = 224.4\text{ N m} \).

Anticlockwise moment is caused by the vertical component of the tension:
\( T_y = T \sin(35^\circ) \) acting at \( 1.5\text{ m} \).
Anticlockwise moment = \( T \sin(35^\circ) \times 1.5 = 0.8604 T \).

Equating moments:
\( 0.8604 T = 224.4 \)
\( T = \frac{224.4}{0.8604} = 260.8\text{ N} \), which is about \( 260\text{ N} \).

(b) The horizontal forces acting on the beam must balance since the system is in equilibrium.
The tension has a horizontal component pointing towards the wall: \( T_x = T \cos(35^\circ) \).
The hinge must exert an equal and opposite horizontal force \( H_x \) pointing away from the wall.
\( H_x = T \cos(35^\circ) \)
Using \( T = 260.8\text{ N} \):
\( H_x = 260.8 \times \cos(35^\circ) = 213.6\text{ N} \approx 214\text{ N} \).
(If using \( T = 260\text{ N} \), then \( H_x = 260 \times \cos(35^\circ) = 213\text{ N} \). Both are acceptable).

(a)
- Identifies the weights of the beam and the sign acting at the correct distances (1 mark)
- Calculates the sum of clockwise moments as \( 224\text{ N m} \) (1 mark)
- Expresses the anticlockwise moment in terms of \( T \): \( T \sin(35^\circ) \times 1.5 \) (1 mark)
- Equates moments to show \( T \approx 261\text{ N} \) (1 mark)

(b)
- Recognizes that the horizontal hinge force must balance the horizontal component of the tension (1 mark)
- Uses \( H_x = T \cos(35^\circ) \) (1 mark)
- Correctly calculates \( 214\text{ N} \) or \( 213\text{ N} \) (1 mark)

(a) Considering the vertical motion of the package with initial vertical velocity \(u_y = 0\):
Using \(s = u_y t + \frac{1}{2}gt^2\):
\(45 \text{ m} = 0 + \frac{1}{2} (9.81 \text{ m s}^{-2}) t^2\)
\(t^2 = \frac{90}{9.81} = 9.174 \text{ s}^2\)
\(t = 3.03 \text{ s}\)
Which is approximately \(3 \text{ s}\).

(b) The horizontal speed of the package relative to the boat is:
\(v_{\text{rel}} = v_{\text{helicopter}} - v_{\text{boat}} = 28 \text{ m s}^{-1} - 12 \text{ m s}^{-1} = 16 \text{ m s}^{-1}\)

Using the time of flight from (a):
\(d = v_{\text{rel}} \times t = 16 \text{ m s}^{-1} \times 3.03 \text{ s} = 48.5 \text{ m}\)
(If using the rounded value of \(3.0 \text{ s}\), \(d = 16 \times 3.0 = 48 \text{ m}\)).

(c) Air resistance opposes the horizontal motion of the package, causing its horizontal velocity to decrease over time. Consequently, the horizontal distance travelled by the package relative to the boat during the fall will be smaller. Therefore, the helicopter must release the package at a smaller horizontal distance behind the boat.

(a)
- Selects and uses vertical equation of motion \(s = \frac{1}{2}gt^2\) (1)
- Obtains value of \(3.03 \text{ s}\) (1)

(b)
- Determines relative horizontal velocity: \(28 - 12 = 16 \text{ m s}^{-1}\) (1)
- Uses distance = speed \(\times\) time (1)
- Obtains a distance of \(48.5 \text{ m}\) or \(48 \text{ m}\) (1)

(c)
- States that air resistance reduces the horizontal velocity/speed of the package (1)
- Concludes that the required horizontal release distance must be smaller (1)

(a) First, calculate the cross-sectional area \(A\) of the wire:
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.56 \times 10^{-3} \text{ m})^2}{4} = 2.463 \times 10^{-7} \text{ m}^2\)

Using the Young modulus formula:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta x / L} = \frac{F L}{A \Delta x}\)

Substitute the values:
\(E = \frac{35 \text{ N} \times 2.40 \text{ m}}{2.463 \times 10^{-7} \text{ m}^2 \times 2.8 \times 10^{-3} \text{ m}}\)
\(E = \frac{84}{6.896 \times 10^{-10}} = 1.22 \times 10^{11} \text{ Pa}\) (accept \(1.2 \times 10^{11} \text{ Pa}\))

(b) For a given load, a longer and thinner wire experiences a greater extension. A larger extension value reduces the percentage uncertainty in the measurement of the extension.

(c) Plastic behaviour means the material undergoes permanent deformation when a deforming force is applied, meaning it does not return to its original length or shape when the load/force is removed.

(a)
- Correct calculation of the cross-sectional area: \(2.46 \times 10^{-7} \text{ m}^2\) (1)
- Uses \(E = \frac{F L}{A \Delta x}\) with appropriate unit conversions (1)
- Calculates Young modulus: \(1.2 \times 10^{11} \text{ Pa}\) (allow range \(1.2 \times 10^{11} \text{ Pa}\) to \(1.22 \times 10^{11} \text{ Pa}\)) (1)

(b)
- States that extension is larger (for the same load) (1)
- Explains that this reduces the percentage uncertainty of the extension measurement (1)

(c)
- States that the wire does not return to its original length/shape when the load is removed (1)
- Mentions that the wire undergoes permanent/inelastic deformation (1)

The relationship between terminal potential difference \(V\), e.m.f. \(\mathcal{E}\), current \(I\), and internal resistance \(r\) is given by the equation: \(V = \mathcal{E} - Ir\). Rearranging this in the form of a straight line equation \(y = mx + c\) gives \(V = -rI + \mathcal{E}\). Comparing the terms, the gradient of a graph of \(V\) against \(I\) is \(-r\). Therefore, the magnitude of the gradient is \(r\).

1 mark for identifying the correct relationship and matching the gradient of the graph to the internal resistance \(r\).

The path difference at point \(P\) is given by \(\Delta x = 5.0\lambda - 3.5\lambda = 1.5\lambda\). Since the path difference is a half-integer multiple of the wavelength (i.e., \(1.5\lambda\)), the waves arrive completely out of phase, leading to destructive interference. The phase difference \(\Delta \phi\) is related to path difference by \(\Delta \phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times 1.5\lambda = 3\pi\text{ radians}\).

1 mark for calculating the path difference and correctly identifying both the phase difference of \(3\pi\text{ radians}\) and destructive interference.

The current in a conductor is given by \(I = nAvq\), where \(n\) is the number density of charge carriers, \(A\) is the cross-sectional area, \(v\) is the drift velocity, and \(q\) is the elementary charge. Rearranging for drift velocity gives \(v = \frac{I}{nAq}\). Since both wires are made of copper, \(n\) and \(q\) are constant, so \(v \propto \frac{I}{A}\). The area of a wire is proportional to the square of its diameter: \(A \propto d^2\). The second wire has twice the diameter, so its cross-sectional area is \(4A\). Thus, the new drift velocity is \(v' \propto \frac{3I}{4A} = 0.75 \left(\frac{I}{A}\right)\). Therefore, \(v' = 0.75v\).

1 mark for calculating the new cross-sectional area and applying the drift velocity equation to find the new drift velocity as \(0.75v\).

According to Einstein's photoelectric equation, the maximum kinetic energy is \(E_k = hf - \phi\), which can be rewritten as \(hf = E_k + \phi\). When the frequency is doubled to \(2f\), the new maximum kinetic energy \(E_k'\) is: \(E_k' = h(2f) - \phi = 2(hf) - \phi\). Substituting \(hf = E_k + \phi\) gives: \(E_k' = 2(E_k + \phi) - \phi = 2E_k + 2\phi - \phi = 2E_k + \phi\).

1 mark for correctly applying the photoelectric equation to the doubled frequency and simplifying to find the correct expression.

When unpolarised light passes through the first polarising filter, its intensity is reduced by half: \(I_1 = \frac{I_0}{2}\). When this polarised light passes through the second filter, we apply Malus's Law: \(I_2 = I_1 \cos^2(\theta)\). Here, \(\theta = 60^\circ\), so \(\cos(60^\circ) = 0.5\), and \(\cos^2(60^\circ) = 0.25\). Therefore, \(I_2 = \frac{I_0}{2} \times 0.25 = 0.125 I_0\).

1 mark for accounting for the initial reduction of intensity to half and correctly applying Malus's Law to find \(0.125 I_0\).

The volume \(V\) of the cylinder is \(V = A \times L\), where \(A\) is the cross-sectional area. Since volume remains constant, if the length increases to \(1.2L\), the cross-sectional area must decrease to \(A' = \frac{A}{1.2}\). The resistance is given by \(R = \rho \frac{L}{A}\). The new resistance \(R'\) is: \(R' = \rho \frac{1.2L}{A/1.2} = 1.44 \left(\rho \frac{L}{A}\right) = 1.44R\).

1 mark for determining the change in area and applying the resistivity formula to calculate the new resistance.

The kinetic energy gained by the electron is \(E_k = eV\). The momentum \(p\) of the electron is related to kinetic energy by \(p = \sqrt{2m E_k} = \sqrt{2m eV}\). The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}\). This shows that \(\lambda \propto \frac{1}{\sqrt{V}}\). If the potential difference is increased to \(4V\), the new wavelength \(\lambda'\) is: \(\lambda' = \frac{\lambda}{\sqrt{4}} = 0.5\lambda\).

1 mark for identifying the inverse square root relationship between wavelength and potential difference, and solving for the new wavelength.

Total internal reflection can occur when light travels from a more optically dense medium to a less optically dense medium (i.e., \(n_1 > n_2\)). The critical angle \(\theta_c\) is calculated using Snell's law with an angle of refraction of \(90^\circ\): \(n_1 \sin\theta_c = n_2 \sin(90^\circ) \implies \sin\theta_c = \frac{n_2}{n_1}\). Substituting the values: \(\sin\theta_c = \frac{1.33}{1.60} \approx 0.83125\). Thus, \(\theta_c = \sin^{-1}(0.83125) \approx 56.2^\circ\).

1 mark for using the critical angle formula correctly and finding the angle of \(56.2^\circ\).

Using Einstein's photoelectric equation: \(E_k = \frac{hc}{\lambda} - \phi\). Rearranging this gives the energy of the initial photon: \(\frac{hc}{\lambda} = E_k + \phi\). When the wavelength is halved to \(\frac{\lambda}{2}\), the energy of the new photon is: \(E_{\text{photon}}' = \frac{hc}{\lambda/2} = 2\left(\frac{hc}{\lambda}\right) = 2(E_k + \phi)\). Using the photoelectric equation again for the new maximum kinetic energy \(E_k'\): \(E_k' = E_{\text{photon}}' - \phi = 2(E_k + \phi) - \phi = 2E_k + \phi\).

A is the correct answer. 1 mark for identifying the correct relationship between photon energy and kinetic energy after halving the wavelength.

The total resistance of the circuit is \(R_{\text{total}} = R + r\). As \(R\) is decreased, the total resistance decreases, so the current \(I = \frac{\varepsilon}{R+r}\) increases. 1. The terminal potential difference is given by \(V = \varepsilon - Ir\). Since the current \(I\) increases, the lost volts \(Ir\) increase, causing \(V\) to decrease. 2. The power dissipated in the internal resistance is given by \(P = I^2 r\). Since the current \(I\) increases, \(P\) must increase. Thus, terminal potential difference decreases and power dissipated in internal resistance increases.

A is the correct answer. 1 mark for analyzing the effect of reducing variable resistance on terminal potential difference and internal power dissipation.

(a) Use Snell's law to find the critical angle: \(\sin(\theta_c) = \frac{n_{\text{liquid}}}{n_{\text{glass}}} = \frac{1.33}{1.52} = 0.875\). Therefore, \(\theta_c = \sin^{-1}(0.875) = 61.04^\circ\), which rounds to \(61.0^\circ\). (b) Since the ray enters normally, it does not refract at the first boundary and continues in a straight line inside the prism. Using simple geometry for a 60-degree prism, the angle of incidence \(\theta_i\) at the second boundary is \(60.0^\circ\). Since \(\theta_i < \theta_c\) (\(60.0^\circ < 61.0^\circ\)), total internal reflection does not occur. The light will refract out into the liquid. (c) A wavefront is a line or surface connecting adjacent points of a wave that are in phase (i.e., oscillating in step with each other).

(a) [1 Mark] Use of \(\sin(\theta_c) = n_2 / n_1\). [1 Mark] Correct substitution of 1.33 and 1.52. [1 Mark] Correct calculation of \(61.0^\circ\) (accept 61 degrees). (b) [1 Mark] Recognition that light enters normal to first face without changing direction. [1 Mark] Determination that the angle of incidence at the second face is \(60.0^\circ\). [1 Mark] Comparison showing \(60^\circ < 61^\circ\). [1 Mark] Conclusion that total internal reflection does not occur. (c) [1 Mark] Reference to points on a wave. [1 Mark] Mentioning that these points are in phase / oscillating in step.

(a) First, find the cross-sectional area of the wire: \(A = \pi r^2 = \pi \left(\frac{0.460 \times 10^{-3}\text{ m}}{2}\right)^2 = 1.662 \times 10^{-7}\text{ m}^2\). Then calculate resistance: \(R = \frac{\rho L}{A} = \frac{1.70 \times 10^{-8}\ \Omega\text{ m} \times 2.40\text{ m}}{1.662 \times 10^{-7}\text{ m}^2} = 0.2455\ \Omega \approx 0.246\ \Omega\). (b) Calculate current using Ohm's law: \(I = \frac{V}{R} = \frac{1.50\text{ V}}{0.2455\ \Omega} = 6.11\text{ A}\) (accept 6.10 A depending on rounding). (c) Use the drift velocity formula \(I = n A v e\): \(v = \frac{I}{n A e} = \frac{6.11\text{ A}}{8.50 \times 10^{28}\text{ m}^{-3} \times 1.662 \times 10^{-7}\text{ m}^2 \times 1.60 \times 10^{-19}\text{ C}} = 2.70 \times 10^{-3}\text{ m s}^{-1}\).

(a) [1 Mark] Correct calculation of cross-sectional area \(A = 1.66 \times 10^{-7}\text{ m}^2\). [1 Mark] Use of \(R = \rho L / A\). [1 Mark] Correct resistance of \(0.246\ \Omega\) (allow \(0.25\ \Omega\) for 2 s.f.). (b) [1 Mark] Use of \(I = V/R\). [1 Mark] Correct calculation of current \(6.10\text{ A}\) or \(6.11\text{ A}\). (c) [1 Mark] Use of \(I = n A v e\). [1 Mark] Rearranging for \(v\). [1 Mark] Substitution of electronic charge \(e = 1.60 \times 10^{-19}\text{ C}\). [1 Mark] Correct calculation of \(2.70 \times 10^{-3}\text{ m s}^{-1}\) (allow ecf from b).

(a) Light passes through the multiple slits of the grating and undergoes diffraction. The diffracted waves spread out and overlap. When waves arrive at a point on the screen in phase (with a path difference of an integer number of wavelengths, \(n\lambda\)), constructive interference occurs, resulting in a bright fringe. (b) Find the slit spacing \(d\): \(d = \frac{1}{500\text{ lines/mm}} = 2.00 \times 10^{-6}\text{ m}\). The distance from the central maximum to the first-order maximum is \(x = 1.62\text{ m} / 2 = 0.81\text{ m}\). The angle \(\theta\) is given by \(\tan(\theta) = \frac{x}{D} = \frac{0.81}{2.50} = 0.324\), so \(\theta = 17.96^\circ\). Alternatively, \(\sin(\theta) = \frac{x}{\sqrt{x^2 + D^2}} = \frac{0.81}{\sqrt{0.81^2 + 2.50^2}} = 0.3082\). Using \(d \sin(\theta) = n \lambda\) with \(n=1\): \[\lambda = d \sin(\theta) = 2.00 \times 10^{-6}\text{ m} \times 0.3082 = 6.164 \times 10^{-7}\text{ m} \approx 6.17 \times 10^{-7}\text{ m}\] (or \(616\text{ nm}\)). (c) A grating with 800 lines/mm has a smaller slit spacing \(d\). Since \(\sin(\theta) = \lambda / d\), decreasing \(d\) increases \(\sin(\theta)\), which increases \(\theta\). The first-order maxima will move further away from the central maximum.

(a) [1 Mark] Mentions diffraction/spreading of light at the slits. [1 Mark] Mentions overlapping/superposition of waves. [1 Mark] Identifies constructive interference occurs when waves are in phase / path difference is \(n\lambda\). (b) [1 Mark] Calculation of \(d = 2.00 \times 10^{-6}\text{ m}\). [1 Mark] Correct division of distance by 2 to get \(x = 0.81\text{ m}\). [1 Mark] Calculation of \(\theta\) or \(\sin(\theta)\). [1 Mark] Correct calculation of wavelength as \(6.16 \times 10^{-7}\text{ m}\) to \(6.17 \times 10^{-7}\text{ m}\). (c) [1 Mark] Identifies that \(d\) decreases. [1 Mark] Explains that \(\theta\) increases, so the fringes are wider spaced.

(a) The diagram must include: a battery in series with an ammeter and a variable resistor (rheostat) to vary the current, and a voltmeter connected in parallel across the terminals of the battery (or across the variable resistor). (b) Use the equation \(V = \mathcal{E} - I r\): For the first data set: \(1.32 = \mathcal{E} - 0.40 r\) (Equation 1). For the second data set: \(1.14 = \mathcal{E} - 0.80 r\) (Equation 2). Subtracting Equation 2 from Equation 1 gives: \(0.18 = 0.40 r \implies r = \frac{0.18}{0.40} = 0.45\ \Omega\). Substitute \(r\) back into Equation 1: \(\mathcal{E} = 1.32 + 0.40 \times 0.45 = 1.32 + 0.18 = 1.50\text{ V}\). (c) When a current flows, charge carriers must pass through the internal resistance of the cell. Work is done against this internal resistance, which converts electrical energy into thermal energy inside the cell. These are referred to as 'lost volts' (\(I r\)), leaving less potential difference available to the external circuit.

(a) [1 Mark] Voltmeter connected in parallel across the battery or variable resistor. [1 Mark] Ammeter and variable resistor in series with the battery. (b) [1 Mark] Use of the formula \(V = \mathcal{E} - I r\). [1 Mark] Sets up two simultaneous equations. [1 Mark] Method to eliminate \(\mathcal{E}\) or \(r\). [1 Mark] Correct calculation of \(r = 0.45\ \Omega\). [1 Mark] Correct calculation of \(\mathcal{E} = 1.50\text{ V}\). (c) [1 Mark] Mentions work done / energy lost due to internal resistance of the cell. [1 Mark] States that this electrical energy is converted to thermal energy / heat inside the cell.

(a) The work function is the minimum energy required to release an electron from the surface of a metal. (b) Energy of an incident photon: \(E = \frac{h c}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{240 \times 10^{-9}\text{ m}} = 8.288 \times 10^{-19}\text{ J}\). Convert the work function to joules: \(\phi = 4.30\text{ eV} \times 1.60 \times 10^{-19}\text{ J/eV} = 6.88 \times 10^{-19}\text{ J}\). Use Einstein's photoelectric equation: \(E_{\text{k, max}} = E - \phi = 8.288 \times 10^{-19}\text{ J} - 6.88 \times 10^{-19}\text{ J} = 1.408 \times 10^{-19}\text{ J}\), which is approximately \(1.4 \times 10^{-19}\text{ J}\). (c) The maximum kinetic energy of the photoelectrons depends only on the photon energy (which depends on frequency/wavelength) and the work function of the metal. Since both remain constant, the maximum kinetic energy is unchanged. Doubling the intensity only increases the number of photons arriving per second, increasing the rate of emission of photoelectrons, but not their individual energies.

(a) [1 Mark] Minimum energy required. [1 Mark] To release/emit an electron from the metal surface. (b) [1 Mark] Use of \(E = h c / \lambda\). [1 Mark] Correct conversion of \(4.30\text{ eV}\) to Joules (\(6.88 \times 10^{-19}\text{ J}\)). [1 Mark] Use of photoelectric equation \(E_{\text{k, max}} = hf - \phi\). [1 Mark] Calculation showing \(1.41 \times 10^{-19}\text{ J}\). (c) [1 Mark] States that maximum kinetic energy remains unchanged. [1 Mark] Explains that intensity only increases the rate of photon arrival / number of photons per second. [1 Mark] States that one photon interacts with one electron (so individual kinetic energy is unaffected).

(a) Calculate total resistance of the potential divider circuit: \(R_{\text{total}} = R_{\text{thermistor}} + R_{\text{fixed}} = 2.8\text{ k}\Omega + 1.5\text{ k}\Omega = 4.3\text{ k}\Omega = 4300\ \Omega\). Use the potential divider formula: \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{fixed}}}{R_{\text{total}}} = 9.0\text{ V} \times \frac{1.5\text{ k}\Omega}{4.3\text{ k}\Omega} = 3.14\text{ V}\). (b) As temperature increases, the resistance of the NTC thermistor decreases. This reduces the total resistance of the series circuit, causing the current in the circuit to increase. Since the potential difference across the fixed resistor is given by \(V = I R_{\text{fixed}}\), and \(R_{\text{fixed}}\) is constant, the increase in current results in an increased potential difference across the fixed resistor. (c) Advantage: Data can be continuously logged automatically and plotted on a computer, or can measure temperature in remote/hazardous environments. Disadvantage: Requires a power supply to operate, or needs calibration before use.

(a) [1 Mark] Summing the resistances to find total resistance \(4.3\text{ k}\Omega\). [1 Mark] Use of potential divider equation or finding current first (\(2.09\text{ mA}\)). [1 Mark] Correct calculation of \(3.14\text{ V}\) (accept 3.1 V). (b) [1 Mark] States that thermistor resistance decreases as temperature increases. [1 Mark] Mentions total resistance decreases OR circuit current increases. [1 Mark] Concludes that voltage across the fixed resistor increases. (c) [1 Mark] Valid advantage (e.g. continuous/automatic logging, remote reading, high resolution). [1 Mark] Valid disadvantage (e.g. needs calibration, requires external electrical power, more complex setup).

(a) Standing waves store energy and do not transfer it through space, whereas progressive waves transfer energy in the direction of wave propagation. Additionally, in a standing wave, all points between adjacent nodes vibrate in phase with differing amplitudes, whereas in a progressive wave, all points have the same amplitude but phase varies continuously. (b) Use the wave speed formula for a stretched string: \(v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{85.0\text{ N}}{1.20 \times 10^{-3}\text{ kg m}^{-1}}} = \sqrt{70833.3} = 266.1\text{ m s}^{-1} \approx 266\text{ m s}^{-1}\). (c) For the fundamental frequency, the length of the string is equal to half a wavelength: \(L = \frac{\lambda}{2} \implies \lambda = 2L = 2 \times 0.650\text{ m} = 1.30\text{ m}\). Calculate fundamental frequency: \(f = \frac{v}{\lambda} = \frac{266.1\text{ m s}^{-1}}{1.30\text{ m}} = 204.7\text{ Hz} \approx 205\text{ Hz}\). (d) Since wave speed \(v \propto \sqrt{T}\) and frequency \(f \propto v\), we have \(f \propto \sqrt{T}\). If tension is doubled, the frequency will increase by a factor of \(\sqrt{2} \approx 1.41\), changing the frequency to \(205 \times 1.41 = 289\text{ Hz}\).

(a) [1 Mark] Energy transfer difference (progressive transfers energy, standing stores it). [1 Mark] Phase or amplitude differences described correctly. (b) [1 Mark] Use of \(v = \sqrt{T/\mu}\). [1 Mark] Correct calculation of wave speed \(266\text{ m s}^{-1}\). (c) [1 Mark] Statement or use of \(\lambda = 2L = 1.30\text{ m}\). [1 Mark] Use of \(f = v / \lambda\). [1 Mark] Correct calculation of fundamental frequency \(205\text{ Hz}\) (allow ecf from b). (d) [1 Mark] Identifies relation \(f \propto \sqrt{T}\). [1 Mark] Explains that frequency increases by a factor of \(\sqrt{2}\) or to \(289\text{ Hz}\).

(a) In polarized light, the electric field oscillations are confined to a single plane containing the direction of wave propagation. (b) Malus's Law is given by \(I = I_0 \cos^2(\theta)\). (c) First, when unpolarized light passes through the first polarizing filter, its intensity is halved: \(I_0 = 0.50 \times I_{\text{inc}}\). Then, as it passes through the second filter at an angle of \(35.0^\circ\), apply Malus's Law: \(I = I_0 \cos^2(35.0^\circ) = (0.50 \times I_{\text{inc}}) \times (0.81915)^2 = 0.50 \times 0.6710 \times I_{\text{inc}} = 0.3355 \times I_{\text{inc}}\). The percentage of the original unpolarized light transmitted is \(0.3355 \times 100\% = 33.6\%\).

(a) [1 Mark] Mentions oscillations are in one plane / single direction. [1 Mark] States this plane includes the direction of wave travel / propagation. (b) [1 Mark] States \(I = I_0 \cos^2(\theta)\) and defines terms clearly (or states the proportional relationship). (c) [1 Mark] States that first polarizer halves the intensity of unpolarized light (\(I_0 = 0.5 I_{\text{inc}}\)). [1 Mark] Uses \(\cos(35^\circ) = 0.819\) or \(\cos^2(35^\circ) = 0.671\). [1 Mark] Multiplies \(0.5\) by \(0.671\). [1 Mark] Obtains final value of \(33.6\%\) (accept range \(33\%\) to \(34\%\)).

*(a) Reason for using a long wire:*
Using a long wire provides a larger, more measurable extension for a given load, which significantly reduces the percentage uncertainty in the extension measurement.

*(b)(i) Determining cross-sectional area:*
1. Check for zero error on the micrometer screw gauge and adjust the readings accordingly.
2. Measure the diameter of the wire at several different positions along its length, and at different orientations (at right angles to each other) at each position to check for a non-circular cross-section, then find the average diameter \(d\).
3. Use the formula \(A = \frac{\pi d^2}{4}\) to calculate the cross-sectional area.

*(b)(ii) Calculation of mean area:*
- Mean diameter \(d = \frac{0.28 + 0.27 + 0.28 + 0.29 + 0.28}{5} = 0.28\text{ mm} = 2.8 \times 10^{-4}\text{ m}\).
- Mean area \(A = \frac{\pi \times (2.8 \times 10^{-4}\text{ m})^2}{4} = 6.16 \times 10^{-8}\text{ m}^2\).
- This is approximately \(6.2 \times 10^{-8}\text{ m}^2\).

*(c) Measuring extension and precaution:*
- Record the initial position of the marker against the metre rule before any masses are added.
- Record the new position of the marker after each weight is added. The extension \(\Delta x\) is calculated as: \(\text{new position} - \text{initial position}\).
- *Precaution*: View the marker and the ruler scale from directly above (at right angles/perpendicularly) to avoid parallax errors, or use a set square to align the marker with the rule.

*(d)(i) Graph analysis derivation:*
- By definition, Young Modulus \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta x/L} = \frac{F L}{A \Delta x}\).
- Rearranging this for force gives: \(F = \left(\frac{E A}{L}\right) \Delta x\).
- This is of the form \(y = mx\), so when \(F\) is plotted against \(\Delta x\), the gradient \(m\) is indeed equal to \(\frac{E A}{L}\).

*(d)(ii) Calculation of Young Modulus:*
- Given: \(\text{gradient} = 2.65 \times 10^3\text{ N m}^{-1}\), \(L = 2.55\text{ m}\), and \(A = 6.16 \times 10^{-8}\text{ m}^2\).
- Since \(\text{gradient} = \frac{E A}{L}\):
\(E = \frac{\text{gradient} \times L}{A} = \frac{2.65 \times 10^3 \times 2.55}{6.16 \times 10^{-8}}\)
\(E = 1.10 \times 10^{11}\text{ Pa}\) (or \(1.1 \times 10^{11}\text{ Pa}\) to 2 significant figures).

**(a) [1 Mark]**
- State that a long wire produces a larger, more measurable extension to reduce the percentage uncertainty in extension. (1)

**(b)(i) [3 Marks]**
- Check and correct for zero error on the micrometer. (1)
- Take measurements at different positions and at right angles (orientations) to find a mean. (1)
- Calculate the area using \(A = \frac{\pi d^2}{4}\) or equivalent. (1)

**(b)(ii) [2 Marks]**
- Calculate mean diameter as \(0.28\text{ mm}\). (1)
- Substitute correct values into area formula to obtain \(6.16 \times 10^{-8}\text{ m}^2\) and state that this is approximately \(6.2 \times 10^{-8}\text{ m}^2\). (1)

**(c) [3 Marks]**
- Note initial position of marker and final position of marker after each load. (1)
- Subtract initial position from final position to find the extension. (1)
- Mention a valid precaution: e.g., view at right angles to avoid parallax error OR use a pointer aligned closely with the rule. (1)

**(d)(i) [1.5 Marks]**
- State \(E = \frac{F L}{A \Delta x}\). (0.5)
- Rearrange to show \(F = \left(\frac{E A}{L}\right) \Delta x\) and state that since \(y = mx\), the gradient is \(\frac{E A}{L}\). (1)

**(d)(ii) [2 Marks]**
- Substitute values correctly: \(E = \frac{2.65 \times 10^3 \times 2.55}{6.16 \times 10^{-8}}\). (1)
- Correct final value \(1.1 \times 10^{11}\text{ Pa}\) (accept \(1.09 \times 10^{11}\text{ Pa}\) up to \(1.12 \times 10^{11}\text{ Pa}\) depending on rounding of area). (1)

*(a) Circuit Diagram:*
The circuit must have the cell connected in a closed series loop with a switch, an ammeter, and a variable resistor. A voltmeter must be connected in parallel across the cell (or across the combination of the cell and its internal resistance).

*(b) Switch purpose:*
The switch is included to break the circuit when readings are not being taken. This prevents the cell from discharging/running down during the experiment and prevents temperature changes inside the cell due to continuous current, which would alter its internal resistance.

*(c) Graph of \(V\) against \(I\):*
- The equation is \(V = -r I + \mathcal{E}\).
- Comparing this to \(y = mx + c\), plotting \(V\) on the y-axis and \(I\) on the x-axis yields a straight-line graph.
- The y-intercept of the line is equal to the e.m.f. \(\mathcal{E}\).
- The gradient of the line is equal to \(-r\) (so the magnitude of the gradient gives the internal resistance \(r\)).

*(d)(i) Internal resistance calculation:*
- Using simultaneous equations:
\(1.38 = \mathcal{E} - 0.15 r\) (Equation 1)
\(1.08 = \mathcal{E} - 0.65 r\) (Equation 2)
- Subtract Equation 2 from Equation 1:
\(1.38 - 1.08 = -0.15 r - (-0.65 r)\)
\(0.30 = 0.50 r \implies r = 0.60\ \Omega\).

*(d)(ii) Electromotive force calculation:*
- Substitute \(r = 0.60\ \Omega\) back into Equation 1:
\(\mathcal{E} = 1.38 + 0.15 \times 0.60 = 1.38 + 0.09 = 1.47\text{ V}\).

*(d)(iii) Percentage uncertainty calculation:*
- The formula for internal resistance is \(r = \frac{\Delta V}{\Delta I}\).
- Change in voltage \(\Delta V = 1.38 - 1.08 = 0.30\text{ V}\).
- Absolute uncertainty in \(\Delta V\) is \(0.01 + 0.01 = 0.02\text{ V}\).
- Percentage uncertainty in \(\Delta V = \frac{0.02}{0.30} \times 100\% \approx 6.67\%\).
- Change in current \(\Delta I = 0.65 - 0.15 = 0.50\text{ A}\).
- Absolute uncertainty in \(\Delta I\) is \(0.02 + 0.02 = 0.04\text{ A}\).
- Percentage uncertainty in \(\Delta I = \frac{0.04}{0.50} \times 100\% = 8.0\%\).
- Total percentage uncertainty in \(r = 6.67\% + 8.0\% = 14.67\% \approx 15\%\).

**(a) [2 Marks]**
- Voltmeter correctly placed in parallel with the cell. (1)
- Cell, switch, ammeter, and variable resistor in a single loop. (1)

**(b) [2 Marks]**
- To prevent the cell from running down/discharging. (1)
- To prevent temperature increase/heating of the cell which would alter its internal resistance. (1)

**(c) [2.5 Marks]**
- Re-write/rearrange equation to the form \(y = mx + c\). (0.5)
- State that the y-intercept of the line represents the e.m.f. \(\mathcal{E}\). (1)
- State that the magnitude of the gradient represents the internal resistance \(r\). (1)

**(d)(i) [2 Marks]**
- Show subtraction of the two equations: \(0.30 = 0.50 r\) or equivalent method. (1)
- Correct value for \(r = 0.60\ \Omega\). (1)

**(d)(ii) [2 Marks]**
- Substitute \(r\) into one of the initial equations to find \(\mathcal{E}\). (1)
- Correct value for \(\mathcal{E} = 1.47\text{ V}\). (1)

**(d)(iii) [2 Marks]**
- Determine percentage uncertainties for both voltage change (\(6.67\%\)) and current change (\(8.0\%\)). (1)
- Sum them to find the total percentage uncertainty of \(14.7\%\) or \(15\%\). (1)

*(a) Why direction does not change:*
The ray enters along a radius of the semi-circle. This means it hits the curved surface normally (at an angle of incidence of \(0^{\circ}\)). Since the angle of incidence is zero at this boundary, no refraction (bending) occurs.

*(b) Diagram:*
- The diagram should depict a semi-circular block.
- An incident ray travels inside the glass, hitting the flat face at point \(O\).
- A normal line (dashed) must be drawn perpendicular to the flat face at \(O\).
- The refracted ray bends away from the normal as it emerges into air.
- Angle \(i\) is labeled between the incident ray and the normal (inside the glass).
- Angle \(r\) is labeled between the refracted ray and the normal (in the air).

*(c) Relationship:*
- Using Snell's law: \(n_1 \sin \theta_1 = n_2 \sin \theta_2\).
- Here, \(n \sin i = 1.00 \times \sin r \implies n = \frac{\sin r}{\sin i}\).

*(d)(i) Graphical analysis:*
- Calculate values:
- For \(i = 15^{\circ}\), \(\sin i = 0.259\); \(r = 23^{\circ}\), \(\sin r = 0.391\)
- For \(i = 25^{\circ}\), \(\sin i = 0.423\); \(r = 39^{\circ}\), \(\sin r = 0.629\)
- For \(i = 35^{\circ}\), \(\sin i = 0.574\); \(r = 58^{\circ}\), \(\sin r = 0.848\)
- For \(i = 40^{\circ}\), \(\sin i = 0.643\); \(r = 71^{\circ}\), \(\sin r = 0.946\)
- Plot a graph of \(\sin r\) on the vertical axis against \(\sin i\) on the horizontal axis.
- Since \(\sin r = n \sin i\), the gradient of the best-fit line represents the refractive index \(n\).

*(d)(ii) Calculation:*
- \(n = \frac{\sin 39^{\circ}}{\sin 25^{\circ}} = \frac{0.6293}{0.4226} = 1.49\).

*(d)(iii) Difficulties:*
1. The light ray from the ray box has a finite thickness, making it difficult to define the exact center of the ray to align with the protractor.
2. It is difficult to locate the exact center point \(O\) on the flat face of the block.
3. Parallax errors may occur when reading angles from a protractor placed under the block.

**(a) [1.5 Marks]**
- State that the ray enters along the normal to the curved surface / angle of incidence is \(0^{\circ}\). (1)
- Conclude that no deviation/refraction occurs because of this. (0.5)

**(b) [3 Marks]**
- Correct semi-circular block drawn with incident ray entering along the radius to \(O\). (1)
- Normal drawn perpendicular to the flat face at \(O\) and refracted ray bending away from normal into air. (1)
- Both angles \(i\) and \(r\) correctly labeled with respect to the normal. (1)

**(c) [1 Mark]**
- State \(n = \frac{\sin r}{\sin i}\). (1)

**(d)(i) [3 Marks]**
- Show calculation of \(\sin i\) and \(\sin r\) for at least two pairs of angles. (1)
- State that a graph of \(\sin r\) on the y-axis against \(\sin i\) on the x-axis is plotted. (1)
- Explain that the gradient of this straight line equals \(n\). (1)

**(d)(ii) [2 Marks]**
- Substitute correct values: \(n = \frac{\sin 39^{\circ}}{\sin 25^{\circ}}\). (1)
- Correct calculation to 2 or 3 significant figures: \(1.49\) (or \(1.5\)). (1)

**(d)(iii) [2 Marks]**
- Any two sensible difficulties listed, such as ray thickness, difficulty locating center \(O\), or parallax errors in protractor reading. (1 per point, max 2)

*(a) Choice of Bob:*
- A small, dense brass bob minimizes the effect of air resistance on the pendulum's motion, ensuring it behaves more like an ideal simple pendulum.
- A smaller bob also makes it easier to measure the distance from the pivot to the center of mass of the bob accurately.

*(b)(i) Timing multiple oscillations:*
- The absolute uncertainty due to human reaction time when starting and stopping the stopwatch is constant (typically \(\pm 0.1\text{ s}\) to \(\pm 0.2\text{ s}\)).
- By timing 20 oscillations, this absolute uncertainty is divided by 20 when calculating the period \(T = t / 20\), which vastly reduces the percentage uncertainty in \(T\).

*(b)(ii) Purpose of a fiducial marker:*
- The pendulum bob travels fastest as it passes through the equilibrium position.
- Because the speed is at its maximum, the bob spends the least amount of time passing the marker, which minimizes timing errors caused by human reaction time.

*(c) Graphical analysis:*
- Squaring both sides of the pendulum equation yields: \(T^2 = \frac{4\pi^2}{g} l\).
- This matches the straight-line equation \(y = mx\) where \(y = T^2\) and \(x = l\).
- Therefore, a graph of \(T^2\) against \(l\) is a straight line through the origin with a gradient \(m = \frac{4\pi^2}{g}\).
- The value of \(g\) is determined by calculating \(g = \frac{4\pi^2}{m}\).

*(d)(i) Value of \(g\):*
- Period \(T = \frac{37.04\text{ s}}{20} = 1.852\text{ s}\).
- Using \(g = \frac{4\pi^2 l}{T^2}\):
\(g = \frac{4 \pi^2 \times 0.850\text{ m}}{(1.852\text{ s})^2} = \frac{33.556}{3.4299} = 9.783\text{ m s}^{-2}\).
- Rounding to 3 significant figures, \(g = 9.78\text{ m s}^{-2}\).

*(d)(ii) Percentage uncertainty in \(g\):*
- Percentage uncertainty in \(l = \frac{0.002}{0.850} \times 100\% = 0.235\%\).
- Percentage uncertainty in time \(t\) (and hence period \(T\)) \(= \frac{0.10}{37.04} \times 100\% = 0.270\%\).
- Since \(g \propto \frac{l}{T^2}\), the percentage uncertainty in \(g\) is:
\(\% \delta g = \% \delta l + 2 \times (\% \delta T)\)
\(\% \delta g = 0.235\% + 2 \times (0.270\%) = 0.235\% + 0.540\% = 0.775\%\).
- This rounds to \(0.78\%\).

**(a) [2 Marks]**
- Use of a small, dense bob minimizes the effect of air resistance. (1)
- Allows more precise determination of length to the center of mass of the bob. (1)

**(b)(i) [2 Marks]**
- Timing error due to human reaction remains constant. (1)
- Dividing total time by 20 reduces the absolute and percentage uncertainty in the period \(T\). (1)

**(b)(ii) [2 Marks]**
- Bob is moving fastest at the equilibrium position. (1)
- This makes the transit time across the marker very short, reducing human reaction time uncertainty in triggering the stopwatch. (1)

**(c) [2.5 Marks]**
- State \(T^2 = \frac{4\pi^2}{g}l\). (0.5)
- Explain that the gradient \(m\) of the graph of \(T^2\) vs \(l\) is \(\frac{4\pi^2}{g}\). (1)
- Show that \(g = \frac{4\pi^2}{m}\). (1)

**(d)(i) [2 Marks]**
- Correctly calculate \(T = 1.852\text{ s}\). (1)
- Calculate \(g = 9.78\text{ m s}^{-2}\) (allow \(9.8\text{ m s}^{-2}\)). (1)

**(d)(ii) [2 Marks]**
- Calculate percentage uncertainty in length (\(0.24\%\)) and period (\(0.27\%\)). (1)
- Double the uncertainty of the period and add to that of the length to get \(0.78\%\) (accept answers in the range \(0.77\% - 0.80\%\)). (1)

Using the principle of conservation of momentum, and defining the direction to the right as positive:

Initial total momentum: \(P_{\text{initial}} = (2m)(3v) + (m)(-v) = 6mv - mv = 5mv\)

Since the two trolleys stick together, the total mass after the collision is \(3m\).

Let \(V\) be the velocity of the combined trolleys:

\(3m \times V = 5mv \implies V = \frac{5}{3}v\)

Since the sign of \(V\) is positive, the direction of motion is to the right.

1 mark for the correct option. (1 mark for correct calculation of final velocity including direction)

The potential difference across a discharging capacitor decreases exponentially according to the formula:
\(V = V_0 e^{-\frac{t}{RC}}\)

Substitute the given time \(t = RC \ln(2)\) into the equation:
\(V = V_0 e^{-\frac{RC \ln(2)}{RC}} = V_0 e^{-\ln(2)}\)

Using exponent rules:
\(e^{-\ln(2)} = e^{\ln(2^{-1})} = 2^{-1} = \frac{1}{2}\)

Therefore, the potential difference is:
\(V = \frac{V_0}{2}\)

1 mark for the correct option. (1 mark for showing substitution into the exponential decay equation and simplifying)

A proton has a quark structure of \(uud\). To conserve charge, the \(\pi^-\) must have a charge of \(-1\). Since pions are mesons, they consist of a quark and an antiquark. Let us analyze the charges of the given options:
- \(u\bar{d}\) has a charge of \(+\frac{2}{3} + \frac{1}{3} = +1\).
- \(d\bar{u}\) has a charge of \(-\frac{1}{3} - \frac{2}{3} = -1\).

Thus, the quark structure of \(\pi^-\) is \(d\bar{u}\).

1 mark for the correct option. (1 mark for identifying the correct meson structure with charge -1)

The centripetal force required to keep the coin moving in a circle is provided by the static frictional force.
\(F_{\text{centripetal}} = m \omega^2 r\)

The maximum frictional force is given by:
\(F_{\text{friction, max}} = \mu m g\)

For no slipping, we require:
\(m \omega^2 r \le \mu m g \implies \omega^2 \le \frac{\mu g}{r}\)

Thus, the maximum angular velocity is:
\(\omega = \sqrt{\frac{\mu g}{r}}\)

1 mark for the correct option. (1 mark for equating centripetal force to maximum friction force and solving for angular velocity)

The magnetic force acts as the centripetal force:
\(q v B = \frac{m v^2}{R} \implies v = \frac{q B R}{m}\)

The kinetic energy \(E_k\) of the particle is:
\(E_k = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{q B R}{m}\right)^2 = \frac{q^2 B^2 R^2}{2m}\)

1 mark for the correct option. (1 mark for substituting the expression for velocity into the kinetic energy equation)

The kinetic energy gained by the proton is \(E_k = q V\).
Since \(E_k = \frac{p^2}{2m}\), the momentum is:
\(p = \sqrt{2 m q V}\)

The de Broglie wavelength is:
\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m q V}}\)

Therefore, \(\lambda \propto \frac{1}{\sqrt{V}}\).

If the potential difference increases by a factor of 4, the wavelength decreases by a factor of \(\sqrt{4} = 2\):
\(\lambda_2 = \frac{\lambda_1}{2}\)

1 mark for the correct option. (1 mark for identifying the inverse square root relationship between wavelength and potential difference)

The point where the electric field is zero must lie where the individual fields are equal in magnitude but opposite in direction. Since the charges are of opposite sign, this point lies outside the region between them, closer to the charge with the smaller magnitude (\(+Q\)). Thus, the position must be at some \(x < 0\).

Let \(x = -s\) where \(s > 0\) is the distance to the left of \(+Q\).

Equating the magnitudes of the electric fields:
\(\frac{k Q}{s^2} = \frac{k (4Q)}{(d+s)^2}\)

Taking the square root on both sides:
\(\frac{1}{s} = \frac{2}{d+s} \implies d+s = 2s \implies s = d\)

So, the position is \(x = -d\).

1 mark for the correct option. (1 mark for setting up the field equation and correctly solving for the position outside the charges)

The maximum e.m.f. induced in a rotating coil is given by:
\(\varepsilon_{\text{max}} = N B A \omega\)
where \(\omega = 2 \pi f\).

Substituting the given values:
\(\varepsilon_{\text{max}} = 150 \times 0.40\text{ T} \times 2.0 \times 10^{-3}\text{ m}^2 \times (2 \pi \times 50\text{ s}^{-1})\)
\(\varepsilon_{\text{max}} = 150 \times 0.40 \times 0.002 \times 100 \pi\)
\(\varepsilon_{\text{max}} = 12 \pi \approx 37.7\text{ V} \approx 38\text{ V}\)

1 mark for the correct option. (1 mark for correctly using the formula for maximum induced e.m.f. including the \(2\pi\) factor for angular frequency)

For an elastic collision between two particles of equal mass (since an alpha particle and a helium-4 nucleus both have mass \(m\)) where one is initially at rest, both momentum and kinetic energy are conserved. By conservation of momentum: \(\vec{p} = \vec{p}_1 + \vec{p}_2\). Squaring this vector equation yields \(p^2 = p_1^2 + p_2^2 + 2 p_1 p_2 \cos(\theta_1 + \theta_2)\). By conservation of kinetic energy: \(\frac{p^2}{2m} = \frac{p_1^2}{2m} + \frac{p_2^2}{2m}\), which simplifies to \(p^2 = p_1^2 + p_2^2\). Substituting this into the momentum equation gives \(2 p_1 p_2 \cos(\theta_1 + \theta_2) = 0\). Since neither particle comes to rest in a non-head-on collision, \(\cos(\theta_1 + \theta_2) = 0\), meaning \(\theta_1 + \theta_2 = 90^\circ\).

1 mark for the correct choice of C.

The charge on the capacitor decays according to \(Q = Q_0 e^{-t/\tau}\), where the time constant is \(\tau = RC\). The current through the resistor also decays as \(I = I_0 e^{-t/RC}\). The power dissipated in the resistor is given by \(P = I^2 R = (I_0 e^{-t/RC})^2 R = I_0^2 R e^{-2t/RC} = P_0 e^{-2t/\tau}\). Comparing this with the standard exponential form for the power decay, \(P = P_0 e^{-t/\tau_P}\), we obtain \(\frac{t}{\tau_P} = \frac{2t}{\tau}\), which simplifies to \(\tau_P = \frac{\tau}{2}\).

1 mark for the correct choice of C.

First, identify the forces acting on the car. The two main forces are the weight of the car \(W = mg\) acting vertically downwards and the normal contact force \(N\) acting perpendicular to the banked track.

Since there is no lateral friction, the horizontal component of the normal contact force provides the required centripetal force:
\(N \sin\theta = \frac{mv^2}{r}\)

The vertical component of the normal contact force balances the weight of the car:
\(N \cos\theta = mg\)

Dividing the first equation by the second gives:
\(\tan\theta = \frac{v^2}{rg}\)

Rearranging for speed \(v\):
\(v = \sqrt{rg \tan\theta}\)

Substitute the given values into the equation:
\(v = \sqrt{85\text{ m} \times 9.81\text{ m s}^{-2} \times \tan(18^\circ)}\)

\(v = \sqrt{833.85 \times 0.3249}\)

\(v = \sqrt{270.93} \approx 16.5\text{ m s}^{-1}\)

- [1 Mark] Identifies the horizontal component of the normal force as the centripetal force: \(N \sin\theta = \frac{mv^2}{r}\) or equivalent.
- [1 Mark] Identifies the vertical balance of forces: \(N \cos\theta = mg\) or equivalent.
- [1 Mark] Derives the expression \(\tan\theta = \frac{v^2}{rg}\).
- [1 Mark] Rearranges the formula to solve for \(v\): \(v = \sqrt{rg \tan\theta}\).
- [1 Mark] Correct substitution of numerical values including \(g = 9.81\text{ m s}^{-2}\).
- [1.66 Marks] Correct final answer of \(16.5\text{ m s}^{-1}\) (allow \(16\text{ m s}^{-1}\) or \(17\text{ m s}^{-1}\) depending on rounding/sig figs). Use of \(g = 9.8\text{ m s}^{-2}\) yields \(16.4\text{ m s}^{-1}\) which is also acceptable.

The absolute electric potential \(V\) at a distance \(r\) from a point charge \(Q\) is given by:
\(V = \frac{Q}{4\pi\varepsilon_0 r}\)

Let the point where the potential is zero be at a distance \(x\) (in meters) from the \(q_1 = +4.0\ \mu\text{C}\) charge. Since the point lies between the two charges, its distance from the \(q_2 = -9.0\ \mu\text{C}\) charge is \(0.150 - x\).

The total potential at this point is the sum of the potentials due to both charges:
\(V = V_1 + V_2 = 0\)

\(\frac{q_1}{4\pi\varepsilon_0 x} + \frac{q_2}{4\pi\varepsilon_0 (0.150 - x)} = 0\)

We can simplify this by dividing out the common constant factors:
\(\frac{q_1}{x} + \frac{q_2}{0.150 - x} = 0\)

Substitute the charge values:
\(\frac{4.0 \times 10^{-6}}{x} + \frac{-9.0 \times 10^{-6}}{0.150 - x} = 0\)

\(\frac{4.0}{x} = \frac{9.0}{0.150 - x}\)

Cross-multiplying gives:
\(4.0(0.150 - x) = 9.0x\)

\(0.600 - 4.0x = 9.0x\)

\(13.0x = 0.600\)

\(x = \frac{0.600}{13.0} \approx 0.0462\text{ m} = 4.6\text{ cm}\)

- [1 Mark] Recalls formula for electric potential \(V = \frac{Q}{4\pi\varepsilon_0 r}\).
- [1 Mark] Expresses the sum of potentials at the point as equal to zero: \(V_1 + V_2 = 0\).
- [1 Mark] Correctly sets up the algebraic relation: \(\frac{4.0}{x} = \frac{9.0}{0.15 - x}\).
- [1 Mark] Performs correct cross-multiplication to obtain \(0.60 - 4x = 9x\).
- [1 Mark] Solves for \(x\) showing clear intermediate steps.
- [1.66 Marks] Correct final answer of \(4.6\text{ cm}\) (or \(0.046\text{ m}\)) to 2 significant figures.

In a cyclotron, the magnetic force provides the necessary centripetal force to keep the protons in a circular path within the dees:
\(Bqv = \frac{mv^2}{r}\)

Rearranging this, we find the angular frequency \(\omega\) or period \(T\) of the motion:
\(v = \frac{Bqr}{m}\)

Since the linear speed \(v\) is related to the period \(T\) of one complete orbit by \(v = \frac{2\pi r}{T}\), we can substitute this in:
\(\frac{2\pi r}{T} = \frac{Bqr}{m} \implies T = \frac{2\pi m}{Bq}\)

The frequency \(f\) of the circular motion (which must match the frequency of the alternating potential difference) is the reciprocal of the period:
\(f = \frac{1}{T} = \frac{Bq}{2\pi m}\)

Substitute the given values into the frequency formula:
\(f = \frac{1.4\text{ T} \times 1.60 \times 10^{-19}\text{ C}}{2\pi \times 1.67 \times 10^{-27}\text{ kg}}\)

\(f = \frac{2.24 \times 10^{-19}}{1.0493 \times 10^{-26}} \approx 2.13 \times 10^7\text{ Hz} = 21.3\text{ MHz}\)

- [1 Mark] Equates magnetic force to centripetal force: \(Bqv = \frac{mv^2}{r}\).
- [1 Mark] Derives or states the equation for the period \(T = \frac{2\pi m}{Bq}\).
- [1 Mark] Relates frequency to period: \(f = \frac{1}{T}\) or directly states \(f = \frac{Bq}{2\pi m}\).
- [1 Mark] Correct substitution of charge \(q = 1.60 \times 10^{-19}\text{ C}\) and mass \(m = 1.67 \times 10^{-27}\text{ kg}\).
- [1 Mark] Shows calculation of numerator and denominator before final division.
- [1.66 Marks] Correct final answer of \(2.13 \times 10^7\text{ Hz}\) or \(21.3\text{ MHz}\) (accept \(2.1 \times 10^7\text{ Hz}\) or \(21\text{ MHz}\)).

First, calculate the cross-sectional area \(A\) of the rectangular coil:
\(A = 0.050\text{ m} \times 0.080\text{ m} = 0.0040\text{ m}^2\)

The initial magnetic flux linkage \(\Phi_{\text{initial}}\) through the coil when it is fully inside the magnetic field is:
\(\Phi_{\text{initial}} = N B A = 250 \times 0.12\text{ T} \times 0.0040\text{ m}^2 = 0.12\text{ Wb-turns}\)

When the coil is completely pulled out of the magnetic field, the final magnetic flux linkage is:
\(\Phi_{\text{final}} = 0\)

According to Faraday's law of electromagnetic induction, the magnitude of the average induced e.m.f. is equal to the rate of change of magnetic flux linkage:
\(\text{e.m.f.} = \frac{\Delta \Phi}{\Delta t} = \frac{\Phi_{\text{initial}} - \Phi_{\text{final}}}{\Delta t}\)

\(\text{e.m.f.} = \frac{0.12\text{ Wb-turns} - 0}{0.15\text{ s}} = 0.80\text{ V}\)

- [1 Mark] Calculates the correct area of the coil: \(A = 0.0040\text{ m}^2\).
- [1 Mark] Recalls the definition of magnetic flux linkage: \(\Phi = N B A\).
- [1 Mark] Calculates the initial flux linkage: \(\Phi_{\text{initial}} = 0.12\text{ Wb-turns}\).
- [1 Mark] Identifies that the change in flux linkage is equal to the initial value since the final value is zero.
- [1 Mark] Recalls Faraday's Law: \(\text{e.m.f.} = \frac{\Delta \Phi}{\Delta t}\).
- [1.66 Marks] Correct final answer with appropriate units: \(0.80\text{ V}\) (or \(0.8\text{ V}\)).

First, use the principle of conservation of linear momentum to find the final velocity \(v_f\) of the combined wagons.

Total initial momentum:
\(p_i = m_1 v_1 + m_2 v_2 = (12000\text{ kg} \times 3.5\text{ m s}^{-1}) + (8000\text{ kg} \times 0) = 42000\text{ kg m s}^{-1}\)

Total mass of the combined system after coupling:
\(M = m_1 + m_2 = 12000\text{ kg} + 8000\text{ kg} = 20000\text{ kg}\)

Using conservation of momentum \(p_i = p_f\):
\(M v_f = 42000\text{ kg m s}^{-1}\)

\(v_f = \frac{42000}{20000} = 2.1\text{ m s}^{-1}\)

Next, calculate the total initial kinetic energy \(E_{k,i}\):
\(E_{k,i} = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \times 12000\text{ kg} \times (3.5\text{ m s}^{-1})^2 = 6000 \times 12.25 = 73500\text{ J}\)

Calculate the total final kinetic energy \(E_{k,f}\):
\(E_{k,f} = \frac{1}{2} M v_f^2 = \frac{1}{2} \times 20000\text{ kg} \times (2.1\text{ m s}^{-1})^2 = 10000 \times 4.41 = 44100\text{ J}\)

Finally, calculate the loss in kinetic energy \(\Delta E_k\):

\(\Delta E_k = E_{k,i} - E_{k,f} = 73500\text{ J} - 44100\text{ J} = 29400\text{ J} = 29.4\text{ kJ}\)

- [1 Mark] Calculates initial momentum: \(42000\text{ kg m s}^{-1}\).
- [1 Mark] Uses conservation of momentum to find the combined speed \(v_f = 2.1\text{ m s}^{-1}\).
- [1 Mark] Calculates the initial kinetic energy: \(73500\text{ J}\).
- [1 Mark] Calculates the final kinetic energy: \(44100\text{ J}\).
- [1 Mark] Identifies that loss in kinetic energy is the difference between initial and final values.
- [1.66 Marks] Correct final answer: \(29400\text{ J}\) or \(2.94 \times 10^4\text{ J}\) (accept \(2.9\text{ kJ}\) if working is correct, though \(29.4\text{ kJ}\) or \(29\text{ kJ}\) is standard).

The potential difference \(V\) across a discharging capacitor at time \(t\) is given by:
\(V = V_0 e^{-\frac{t}{RC}}\)

First, calculate the time constant \(\tau = RC\) of the circuit:
\(\tau = 150 \times 10^3\ \Omega \times 47 \times 10^{-6}\text{ F} = 7.05\text{ s}\)

Substitute the given values into the discharge equation:
\(3.0\text{ V} = 12.0\text{ V} \times e^{-\frac{t}{7.05}}\)

Divide both sides by \(12.0\text{ V}\):
\(0.25 = e^{-\frac{t}{7.05}}\)

Take the natural logarithm of both sides:
\(\ln(0.25) = -\frac{t}{7.05}\)

Since \(\ln(0.25) = -1.3863\):
\(-1.3863 = -\frac{t}{7.05}\)

Solve for \(t\):
\(t = 1.3863 \times 7.05 \approx 9.77\text{ s}\)

- [1 Mark] States or uses formula: \(V = V_0 e^{-t/RC}\).
- [1 Mark] Correctly calculates the time constant \(RC = 7.05\text{ s}\).
- [1 Mark] Correct substitution of \(V = 3.0\text{ V}\) and \(V_0 = 12.0\text{ V}\).
- [1 Mark] Takes natural logs of both sides correctly to obtain \(\ln(3/12) = -t/RC\).
- [1 Mark] Rearranges expression to make \(t\) the subject.
- [1.66 Marks] Correct final answer: \(9.8\text{ s}\) (accept \(9.77\text{ s}\)).

A charged particle moving perpendicular to a uniform magnetic field experiences a magnetic force that acts as the centripetal force.

The magnetic force \(F\) is given by:
\(F = Bqv\)

The centripetal force is given by:
\(F = \frac{mv^2}{r}\)

Equating the two forces:
\(Bqv = \frac{mv^2}{r}\)

We can divide both sides by \(v\):
\(Bq = \frac{mv}{r}\)

Rearranging this equation to solve for the speed \(v\):
\(v = \frac{Bqr}{m}\)

Substitute the given values into the equation:
\(v = \frac{0.80\text{ T} \times 3.20 \times 10^{-19}\text{ C} \times 0.45\text{ m}}{6.64 \times 10^{-27}\text{ kg}}\)

\(v = \frac{1.152 \times 10^{-19}}{6.64 \times 10^{-27}}\)

\(v \approx 1.735 \times 10^7\text{ m s}^{-1}\)

- [1 Mark] Equates magnetic force to centripetal force: \(Bqv = \frac{mv^2}{r}\).
- [1 Mark] Rearranges to make velocity \(v\) the subject: \(v = \frac{Bqr}{m}\).
- [1 Mark] Correct substitution of charge \(3.20 \times 10^{-19}\text{ C}\).
- [1 Mark] Correct substitution of magnetic flux density \(0.80\text{ T}\) and radius \(0.45\text{ m}\).
- [1 Mark] Correct calculation of the numerator \(1.152 \times 10^{-19}\) or intermediate step.
- [1.66 Marks] Correct final answer: \(1.7 \times 10^7\text{ m s}^{-1}\) (or \(1.73 \times 10^7\text{ m s}^{-1}\) or \(1.74 \times 10^7\text{ m s}^{-1}\)).

Let \(T\) be the tension in the string, \(L\) be the length of the string, and \(\theta = 25^\circ\) be the angle to the vertical.

The vertical component of tension balances the weight of the bob:
\(T \cos\theta = mg\)

The horizontal component of tension provides the centripetal force:
\(T \sin\theta = m \omega^2 r\)

where \(r\) is the radius of the circle, given by:
\(r = L \sin\theta = 0.80\text{ m} \times \sin(25^\circ)\)

Substitute \(r\) into the centripetal force equation:
\(T \sin\theta = m \omega^2 L \sin\theta\)

We can divide both sides by \(\sin\theta\):
\(T = m \omega^2 L\)

From the vertical equilibrium equation, we have:
\(T = \frac{mg}{\cos\theta}\)

Equating the two expressions for tension \(T\):
\(\frac{mg}{\cos\theta} = m \omega^2 L\)

Divide both sides by \(m\):
\(\frac{g}{\cos\theta} = \omega^2 L \implies \omega = \sqrt{\frac{g}{L \cos\theta}}\)

The angular velocity \(\omega\) is related to the period \(t_p\) (time for one revolution) by:
\(\omega = \frac{2\pi}{t_p}\)

Therefore:
\(t_p = 2\pi \sqrt{\frac{L \cos\theta}{g}}\)

Substitute the values:
\(t_p = 2\pi \sqrt{\frac{0.80\text{ m} \times \cos(25^\circ)}{9.81\text{ m s}^{-2}}}\)

\(t_p = 2\pi \sqrt{\frac{0.80 \times 0.9063}{9.81}} = 2\pi \sqrt{\frac{0.725}{9.81}} = 2\pi \sqrt{0.0739} \approx 2\pi \times 0.2719 \approx 1.71\text{ s}\)

- [1 Mark] Correctly writes the vertical force equation: \(T \cos\theta = mg\).
- [1 Mark] Correctly writes the centripetal force equation: \(T \sin\theta = m \omega^2 r\) or \(T \sin\theta = \frac{mv^2}{r}\).
- [1 Mark] Identifies the radius of the circular path: \(r = L \sin\theta\).
- [1 Mark] Derives the expression for angular velocity: \(\omega = \sqrt{\frac{g}{L \cos\theta}}\) or period \(t_p = 2\pi \sqrt{\frac{L \cos\theta}{g}}\).
- [1 Mark] Correctly substitutes numerical values including \(g = 9.81\text{ m s}^{-2}\).
- [1.66 Marks] Correct final period of \(1.7\text{ s}\) (accept \(1.71\text{ s}\)).

(a) A block moving in a circle is continuously changing direction, which means its velocity vector is changing. Since acceleration is defined as the rate of change of velocity, the block is accelerating. According to Newton's Second Law, a net force must act in the direction of acceleration (towards the center) to produce this acceleration.

(b) The centripetal force is provided entirely by the tension (restoring force) of the spring.

Spring extension:
\(\Delta x = r - l_0 = 0.25\text{ m} - 0.15\text{ m} = 0.10\text{ m}\)

Using Hooke's Law for the spring force:
\(F = k \Delta x = 32\text{ N m}^{-1} \times 0.10\text{ m} = 3.2\text{ N}\)

Since this spring force provides the centripetal force:
\(F_c = m \omega^2 r\)
\(3.2\text{ N} = 0.15\text{ kg} \times \omega^2 \times 0.25\text{ m}\)
\(3.2 = 0.0375 \omega^2\)
\(\omega^2 = 85.33\text{ rad}^2\text{ s}^{-2}\)
\(\omega = 9.24\text{ rad s}^{-1}\) (or 9.2\text{ rad s}^{-1} to 2 s.f.)

(a)
* State that velocity is a vector / direction is changing, so there is an acceleration [1 mark]
* Link acceleration to a net/unbalanced force using Newton's second law (directed towards the center) [1 mark]

(b)
* Calculate extension \(\Delta x = 0.10\text{ m}\) [1 mark]
* Calculate spring tension \(F = 3.2\text{ N}\) [1 mark]
* Equate spring force to centripetal force formula \(F_c = m\omega^2 r\) [1 mark]
* Obtain \(\omega = 9.2\text{ rad s}^{-1}\) (allow \(9.24\text{ rad s}^{-1}\)) [1 mark]

(a) An electric field line represents the path that a free positive test charge would take when placed in the field.

(b) Let the point where the electric potential is zero be at a distance \(x\) from \(Q_1\) between the charges:
\(V = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q_1}{x} + \frac{Q_2}{0.20 - x} \right) = 0\)

Since \(\frac{1}{4\pi\varepsilon_0} \neq 0\):
\(\frac{3.0 \times 10^{-6}}{x} + \frac{-5.0 \times 10^{-6}}{0.20 - x} = 0\)
\(\frac{3}{x} = \frac{5}{0.20 - x}\)
\(3(0.20 - x) = 5x\)
\(0.60 - 3x = 5x \implies 8x = 0.60 \implies x = 0.075\text{ m}\) (or 7.5 cm).

(c) The electric field strength \(E\) at \(x = 0.075\text{ m}\) is the vector sum of the fields due to both charges. Since \(Q_1\) is positive and \(Q_2\) is negative, both field vectors point in the same direction (towards \(Q_2\)).

\(E_1 = \frac{k|Q_1|}{x^2} = \frac{8.99 \times 10^9 \times 3.0 \times 10^{-6}}{0.075^2} = 4.79 \times 10^6\text{ V m}^{-1}\)
\(E_2 = \frac{k|Q_2|}{(0.20 - x)^2} = \frac{8.99 \times 10^9 \times 5.0 \times 10^{-6}}{0.125^2} = 2.88 \times 10^6\text{ V m}^{-1}\)

\(E_{\text{total}} = E_1 + E_2 = 4.79 \times 10^6 + 2.88 \times 10^6 = 7.67 \times 10^6\text{ V m}^{-1}\) (or \(7.7 \times 10^6\text{ N C}^{-1}\)).

(a)
* State that it is the path of a positive test charge (or direction of force on a positive charge) [1 mark]

(b)
* Use the potential formula \(V = \frac{kQ}{r}\) and state the sum condition \(V_1 + V_2 = 0\) [1 mark]
* Set up equation: \(\frac{3.0}{x} = \frac{5.0}{0.20-x}\) [1 mark]
* Calculate distance \(x = 0.075\text{ m}\) [1 mark]

(c)
* Add the magnitudes of the electric fields because they act in the same direction [1 mark]
* Obtain \(7.7 \times 10^6\text{ V m}^{-1}\) (allow range \(7.6 \times 10^6\) to \(7.7 \times 10^6\)) [1 mark]

(a) The magnetic force acts as the centripetal force:
\(Bqv = \frac{mv^2}{r}\)

Simplifying, the momentum is:
\(p = mv = Bqr\)

Substitute the given values:
\(p = 0.85\text{ T} \times (2 \times 1.60 \times 10^{-19}\text{ C}) \times 0.12\text{ m}\)
\(p = 0.85 \times 3.20 \times 10^{-19} \times 0.12 = 3.264 \times 10^{-20}\text{ kg m s}^{-1}\)

This is approximately \(3.3 \times 10^{-20}\text{ kg m s}^{-1}\).

(b) The kinetic energy \(E_k\) is given by:
\(E_k = \frac{p^2}{2m}\)

Using the calculated momentum:
\(E_k = \frac{(3.264 \times 10^{-20}\text{ kg m s}^{-1})^2}{2 \times 6.64 \times 10^{-27}\text{ kg}}\)
\(E_k = \frac{1.0654 \times 10^{-39}}{1.328 \times 10^{-26}} = 8.02 \times 10^{-14}\text{ J}\)

Convert energy from joules to megaelectronvolts (MeV):
\(E_k = \frac{8.02 \times 10^{-14}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} = 5.01 \times 10^5\text{ eV}\)
\(E_k = 0.501\text{ MeV}\) (or \(0.50\text{ MeV}\) to 2 s.f.)

(a)
* Equate magnetic force to centripetal force to derive \(p = Bqr\) [1 mark]
* Correct substitution showing \(p = 3.26 \times 10^{-20}\text{ kg m s}^{-1}\) [1 mark]

(b)
* Use of \(E_k = \frac{p^2}{2m}\) or \(v = \frac{p}{m}\) and \(E_k = \frac{1}{2}mv^2\) [1 mark]
* Calculate kinetic energy in joules as \(8.0 \times 10^{-14}\text{ J}\) [1 mark]
* Division by \(1.60 \times 10^{-19}\) to convert to eV [1 mark]
* Correct final answer of \(0.50\text{ MeV}\) (allow \(0.50\) to \(0.51\text{ MeV}\)) [1 mark]

(a) Magnetic flux linkage is the product of the magnetic flux through a single loop and the total number of turns in the coil, mathematically expressed as \(N\Phi = BAN\cos\theta\), where \(B\) is the magnetic flux density, \(A\) is the area, and \(N\) is the number of turns.

(b) According to Faraday's law of electromagnetic induction, the induced e.m.f. is equal to the rate of change of magnetic flux linkage.

\(\text{e.m.f.} = - \frac{\Delta (N\Phi)}{\Delta t}\)

Initial magnetic flux linkage:
\(N\Phi_{\text{initial}} = N B A = 500 \times 0.080\text{ T} \times 2.5 \times 10^{-4}\text{ m}^2 = 0.010\text{ Wb-turns}\)

Final magnetic flux linkage:
\(N\Phi_{\text{final}} = 0\)

Change in flux linkage:
\(\Delta (N\Phi) = 0.010\text{ Wb-turns}\)

Induced e.m.f.:
\(\text{e.m.f.} = \frac{0.010\text{ Wb-turns}}{0.040\text{ s}} = 0.25\text{ V}\)

(a)
* State that it is the product of magnetic flux and number of turns [1 mark]
* Define flux as magnetic flux density normal to the area (or mention \(\Phi = BA\cos\theta\)) [1 mark]

(b)
* Use Faraday's Law: \(\text{e.m.f.} = \frac{\Delta (N\Phi)}{\Delta t}\) [1 mark]
* Calculate the initial magnetic flux linkage as \(0.010\text{ Wb-turns}\) [1 mark]
* Substitute the time interval of 0.040 s into the denominator [1 mark]
* State the final induced e.m.f. as \(0.25\text{ V}\) [1 mark]

From the ideal gas equation, \(pV = nRT\), which implies \(\frac{pV}{T} = \text{constant}\). Therefore, \(\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}\). Rearranging for the new pressure \(p_2\) gives: \(p_2 = p_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1}\). Since \(V_2 = 0.5 V_1\) and \(T_2 = 3 T_1\), we have: \(p_2 = p \times 2 \times 3 = 6p\).

1 mark for the correct answer (D).

The velocity of a particle in simple harmonic motion is given by \(v = \frac{dx}{dt} = A \omega \cos(\omega t)\). The maximum velocity is \(v_{\text{max}} = \omega A\). The maximum kinetic energy is given by \(E_{k\text{,max}} = \frac{1}{2} m v_{\text{max}}^2 = \frac{1}{2} m (\omega A)^2 = \frac{1}{2} m \omega^2 A^2\).

1 mark for the correct answer (A).

The stability of a nucleus is determined by its binding energy per nucleon. A more stable nucleus has a higher binding energy per nucleon. Since nucleus X is more stable than nucleus Y, the binding energy per nucleon of X, \(\frac{E_X}{A_X}\), must be greater than that of Y, \(\frac{E_Y}{A_Y}\).

1 mark for the correct answer (B).

Using Wien's displacement law, \(\lambda_{\text{max}} T = \text{constant}\), so temperature \(T \propto \frac{1}{\lambda_{\text{max}}}\). Since \(\lambda_Q = 2\lambda_P\), the temperature of Q is half that of P, i.e., \(T_P = 2 T_Q\). Using Stefan-Boltzmann law, \(L = 4\pi R^2 \sigma T^4\), so \(L \propto R^2 T^4\). Therefore, \(\frac{L_P}{L_Q} = \left(\frac{R_P}{R_Q}\right)^2 \times \left(\frac{T_P}{T_Q}\right)^4 = \left(\frac{R}{2R}\right)^2 \times \left(\frac{2T_Q}{T_Q}\right)^4 = \left(\frac{1}{2}\right)^2 \times (2)^4 = \frac{1}{4} \times 16 = 4\).

1 mark for the correct answer (C).

The total energy supplied by the heater is \(P \Delta t\). This energy is equal to the useful thermal energy absorbed by the liquid to raise its temperature, which is \(m c \Delta \theta\), plus the thermal energy lost to the surroundings, which is \(Q \Delta t\). Thus, \(P \Delta t = m c \Delta \theta + Q \Delta t\).

1 mark for the correct answer (A).

Under light damping, the amplitude of oscillations decreases exponentially with time. The time period (and thus frequency) of the oscillations remains constant.

1 mark for the correct answer (A).

After one half-life, the number of undecayed nuclei remaining is \(\frac{1}{2} N_0\). After two half-lives, the number of undecayed nuclei remaining is \(\frac{1}{4} N_0\). The number of nuclei that have decayed is the initial number minus the remaining number: \(N_0 - \frac{1}{4} N_0 = \frac{3}{4} N_0\).

1 mark for the correct answer (C).

Using the Doppler redshift formula: \(\frac{\Delta \lambda}{\lambda} = \frac{v}{c}\). Here, the galaxy is receding, so the detected wavelength is redshifted (increases). Thus, \(\frac{\lambda_{\text{detected}} - \lambda}{\lambda} = \frac{0.05c}{c} = 0.05\). This gives \(\lambda_{\text{detected}} - \lambda = 0.05 \lambda\), so \(\lambda_{\text{detected}} = 1.05 \lambda\).

1 mark for the correct answer (B).

Using the ideal gas equation: \(pV = NkT\) or \(\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}\).

Rearranging to solve for the final pressure \(p_2\):

\(p_2 = p_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1}\)

Given:
- \(T_2 = 2T_1\)
- \(V_2 = 0.5V_1\)

Substitute these into the equation:

\(p_2 = p \times \frac{1}{0.5} \times \frac{2}{1} = p \times 2 \times 2 = 4p\).

Therefore, the new pressure is \(4p\).

1 mark for the correct answer D.

First, we use Wien's displacement law to find the relationship between the temperatures:

\(\lambda_{\text{max}} T = \text{constant} \implies T \propto \frac{1}{\lambda_{\text{max}}}\)

Therefore:

\(\frac{T_X}{T_Y} = \frac{\lambda_Y}{\lambda_X} = \frac{\lambda_Y}{2\lambda_Y} = \frac{1}{2}\)

Next, we use the Stefan-Boltzmann law for luminosity, \(L = 4\pi R^2 \sigma T^4\):

\(\frac{L_X}{L_Y} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{T_X}{T_Y}\right)^4\)

Substitute the given ratios into this equation:

\(\frac{L_X}{L_Y} = (3)^2 \times \left(\frac{1}{2}\right)^4 = 9 \times \frac{1}{16} = \frac{9}{16}\)

1 mark for the correct answer A.

(a) Convert temperature to Kelvin:
\(T_1 = 27.0 + 273.15 = 300.15\text{ K}\)
Using the ideal gas equation:
\(pV = NkT\)
\(N = \frac{pV}{kT} = \frac{1.20 \times 10^5\text{ Pa} \times 0.045\text{ m}^3}{1.38 \times 10^{-23}\text{ J K}^{-1} \times 300.15\text{ K}} = 1.304 \times 10^{24}\text{ atoms}\)

(b) Since volume and the number of atoms remain constant:
\(\frac{p_1}{T_1} = \frac{p_2}{T_2}\)
\(T_2 = T_1 \times \frac{p_2}{p_1} = 300.15\text{ K} \times \frac{2.00 \times 10^5\text{ Pa}}{1.20 \times 10^5\text{ Pa}} = 500.25\text{ K}\)

The mean kinetic energy of an ideal gas molecule is:
\(\langle E_k \rangle = \frac{3}{2}kT_2\)
\(\langle E_k \rangle = 1.5 \times 1.38 \times 10^{-23}\text{ J K}^{-1} \times 500.25\text{ K} = 1.036 \times 10^{-20}\text{ J}\)

(a) [3 Marks]
- M1: Conversion of temperature to Kelvin (300 K) [1]
- M2: Use of \(pV = NkT\) [1]
- A1: Value of \(N = 1.30 \times 10^{24}\) (accept range 1.30 to 1.31) [1]

(b) [4 Marks]
- M1: Recognition that volume is constant, so \(p \propto T\) to find \(T_2\) [1]
- A1: \(T_2 = 500\text{ K}\) [1]
- M1: Use of \(\langle E_k \rangle = \frac{3}{2}kT\) [1]
- A1: Mean kinetic energy \( = 1.04 \times 10^{-20}\text{ J}\) (accept range 1.03 to 1.04) [1]

(a) Energy required to heat the water:
\(Q = mc\Delta \theta = 0.45\text{ kg} \times 4180\text{ J kg}^{-1}\text{ K}^{-1} \times (100.0 - 20.0)\text{ K}\)
\(Q = 0.45 \times 4180 \times 80 = 150,480\text{ J}\)

Since \(P = \frac{Q}{t}\):
\(t = \frac{150,480\text{ J}}{120\text{ W}} = 1254\text{ s} \approx 1300\text{ s}\)

(b) Energy supplied during boiling:
\(E = P \times t = 120\text{ W} \times 300\text{ s} = 36,000\text{ J}\)

Using latent heat formula:
\(Q = mL\)
\(L = \frac{36000\text{ J}}{1.20 \times 10^{-2}\text{ kg}} = 3.00 \times 10^6\text{ J kg}^{-1}\)

Reason for difference: Not all thermal energy from the heater went into the phase change; some energy was lost to the surroundings or container, which causes the calculated specific latent heat to be higher than the actual value.

(a) [3 Marks]
- M1: Use of \(Q = mc\Delta\theta\) [1]
- M2: Use of \(P = \frac{Q}{t}\) [1]
- A1: Correct calculation showing \(t = 1254\text{ s}\) (which rounds to 1300 s) [1]

(b) [4 Marks]
- M1: Use of \(E = Pt\) to find the electrical energy supplied (36,000 J) [1]
- M1: Use of \(Q = mL\) [1]
- A1: \(L = 3.00 \times 10^6\text{ J kg}^{-1}\) [1]
- A1: Explanation of thermal energy loss to the surroundings / beaker resulting in a larger calculated value [1]

(a) Convert time into seconds:
\(t = 15.0\text{ hours} \times 3600\text{ s/hour} = 54,000\text{ s}\)

Using the exponential decay law for activity:
\(A = A_0 e^{-\lambda t}\)
\(\frac{4.50 \times 10^4}{3.60 \times 10^5} = e^{-\lambda \times 54,000}\)
\(0.125 = e^{-54,000 \lambda}\)

Take the natural logarithm of both sides:
\(\ln(0.125) = -54,000 \lambda\)
\(-2.0794 = -54,000 \lambda\)
\(\lambda = \frac{2.0794}{54,000} = 3.851 \times 10^{-5}\text{ s}^{-1}\)

(b) The relationship between activity and number of nuclei is:
\(A = \lambda N\)
At \(t = 15.0\text{ hours}\), \(A = 4.50 \times 10^4\text{ Bq}\)
\(N = \frac{A}{\lambda} = \frac{4.50 \times 10^4\text{ Bq}}{3.851 \times 10^{-5}\text{ s}^{-1}} = 1.169 \times 10^9\text{ nuclei}\)

(a) [4 Marks]
- M1: Conversion of time to seconds (54,000 s) [1]
- M1: Use of \(A = A_0 e^{-\lambda t}\) [1]
- M1: Logarithmic rearrangement to solve for \(\lambda\) [1]
- A1: \(\lambda = 3.85 \times 10^{-5}\text{ s}^{-1}\) [1]

(b) [3 Marks]
- M1: Use of \(A = \lambda N\) [1]
- M1: Substitution of the remaining activity \(A = 4.50 \times 10^4\text{ Bq}\) [1]
- A1: Correct calculation of \(N = 1.17 \times 10^9\) (accept range 1.16 to 1.18 x 10^9) [1]

(a) Calculate total mass of reactants:
\(m_{\text{reactants}} = 2 \times 2.014102\text{ u} = 4.028204\text{ u}\)

Calculate total mass of products:
\(m_{\text{products}} = 3.016029\text{ u} + 1.008665\text{ u} = 4.024694\text{ u}\)

Calculate mass defect:
\(\Delta m = 4.028204\text{ u} - 4.024694\text{ u} = 0.003510\text{ u}\)

Convert mass defect to energy using \(1\text{ u} = 931.5\text{ MeV}\):
\(E = 0.003510 \times 931.5\text{ MeV} = 3.2696\text{ MeV} \approx 3.27\text{ MeV}\)

(b) Deuterium nuclei contain one proton each, so they have positive charges. Consequently, there is an electrostatic repulsive force between them as they approach. High temperatures mean the nuclei have extremely high kinetic energies, enabling them to overcome this electrostatic repulsion and get close enough (about \(10^{-15}\text{ m}\)) for the attractive strong nuclear force to bind them together.

(a) [4 Marks]
- M1: Calculation of total initial mass and final mass [1]
- M1: Calculation of mass defect \(\Delta m = 0.003510\text{ u}\) [1]
- M1: Use of conversion factor \(1\text{ u} = 931.5\text{ MeV}\) (or equivalent conversion using \(E = \Delta m c^2\)) [1]
- A1: Correct energy \(E = 3.27\text{ MeV}\) (accept range 3.25 to 3.30 MeV) [1]

(b) [3 Marks]
- M1: Identification of electrostatic repulsion between positively charged protons in deuterium nuclei [1]
- M1: High temperature provides high kinetic energy to overcome this electrostatic barrier [1]
- A1: Mention that they must come close enough for the short-range strong nuclear force to pull them together [1]

(a) The angular frequency of a mass-spring system is given by:
\(\omega = \sqrt{\frac{k}{m}}\)
\(\omega = \sqrt{\frac{45\text{ N m}^{-1}}{0.25\text{ kg}}} = \sqrt{180} = 13.42\text{ rad s}^{-1}\)
This is approximately \(13\text{ rad s}^{-1}\).

(b) Maximum velocity \(v_{\text{max}}\) is given by:
\(v_{\text{max}} = \omega A\)
Where amplitude \(A = 6.0\text{ cm} = 0.060\text{ m}\).
Using the calculated value of \(\omega = 13.42\text{ rad s}^{-1}\):
\(v_{\text{max}} = 13.42 \times 0.060 = 0.805\text{ m s}^{-1}\)

Since the mass is released from rest at maximum displacement (\(t = 0\)), it first reaches maximum speed when passing through the equilibrium position.
This corresponds to a time of one-quarter of the time period \(T\):
\(T = \frac{2\pi}{\omega} = \frac{2\pi}{13.42} = 0.468\text{ s}\)
\(t = \frac{T}{4} = \frac{0.468\text{ s}}{4} = 0.117\text{ s}\)

(a) [2 Marks]
- M1: Use of \(\omega = \sqrt{\frac{k}{m}}\) [1]
- A1: Correct substitution showing \(13.4\text{ rad s}^{-1}\) [1]

(b) [5 Marks]
- M1: Use of \(v_{\text{max}} = \omega A\) [1]
- A1: \(v_{\text{max}} = 0.80\text{ m s}^{-1}\) (or \(0.81\text{ m s}^{-1}\)) [1]
- M1: Recognition that maximum velocity occurs at equilibrium, which is at \(t = \frac{T}{4}\) [1]
- M1: Calculation of period \(T = \frac{2\pi}{\omega}\) [1]
- A1: Correct elapsed time \(t = 0.12\text{ s}\) (accept range 0.11 to 0.12 s) [1]

(a) Resonance occurs when the driving frequency of an external periodic force matches the natural frequency of the system, leading to a maximum transfer of energy and a peak in amplitude of oscillation.

As the driving frequency increases from zero:
- At very low frequencies, the amplitude is small and relatively constant.
- As the frequency approaches the natural frequency, the amplitude increases rapidly, reaching a maximum value (the resonance peak) when the driving frequency equals the natural frequency.
- As the driving frequency is increased beyond the natural frequency, the amplitude of oscillation decreases back towards zero.

(b) Increasing damping affects the resonance curve in the following ways:
- The maximum amplitude at the resonant peak is significantly reduced.
- The peak becomes broader (less sharp).
- The frequency at which the maximum amplitude occurs shifts slightly to a lower frequency.

(a) [4 Marks]
- M1: Define resonance as driving frequency matching the natural frequency [1]
- M2: Mention maximum energy transfer / maximum amplitude [1]
- M3: Describe how amplitude increases to a peak at the natural frequency [1]
- A1: Describe how amplitude decreases at frequencies above the natural frequency [1]

(b) [3 Marks]
- M1: State that the peak amplitude decreases [1]
- M2: State that the resonance curve becomes broader/flatter [1]
- A1: State that the peak frequency shifts slightly lower [1]

(a) Using Wien's displacement law:
\(\lambda_{\text{max}} T = 2.898 \times 10^{-3}\text{ m K}\)
\(\lambda_{\text{max}} = \frac{2.898 \times 10^{-3}\text{ m K}}{5790\text{ K}} = 5.005 \times 10^{-7}\text{ m}\) (or \(501\text{ nm}\))

(b) Using the Stefan-Boltzmann law:
\(L = 4 \pi r^2 \sigma T^4\)
Substitute the given values:
\(r = 8.51 \times 10^8\text{ m}\)
\(T = 5790\text{ K}\)
\(\sigma = 5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}\)

\(L = 4 \pi \times (8.51 \times 10^8\text{ m})^2 \times (5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}) \times (5790\text{ K})^4\)
\(L = 4 \pi \times (7.242 \times 10^{17}) \times (5.67 \times 10^{-8}) \times (1.124 \times 10^{15})\)
\(L = 5.794 \times 10^{26}\text{ W}\)

(a) [3 Marks]
- M1: Use of Wien's displacement law \(\lambda_{\text{max}} T = 2.898 \times 10^{-3}\text{ m K}\) [1]
- M1: Correct substitution of \(T = 5790\text{ K}\) [1]
- A1: \(\lambda_{\text{max}} = 5.01 \times 10^{-7}\text{ m}\) (accept range 5.00 to 5.02 x 10^-7 m) [1]

(b) [4 Marks]
- M1: Use of Stefan-Boltzmann equation \(L = 4 \pi r^2 \sigma T^4\) [1]
- M1: Calculation of surface area \(4 \pi r^2 = 9.10 \times 10^{18}\text{ m}^2\) [1]
- M1: Correct power term \(T^4 = 1.12 \times 10^{15}\text{ K}^4\) [1]
- A1: \(L = 5.79 \times 10^{26}\text{ W}\) (accept range 5.75 to 5.82 x 10^26 W) [1]

(a) Calculate change in wavelength:
\(\Delta \lambda = 682.5\text{ nm} - 656.3\text{ nm} = 26.2\text{ nm}\)

Using the Doppler redshift formula:
\(z = \frac{\Delta \lambda}{\lambda} = \frac{v}{c}\)
\(v = c \times \frac{\Delta \lambda}{\lambda}\)
\(v = 3.00 \times 10^8\text{ m s}^{-1} \times \frac{26.2\text{ nm}}{656.3\text{ nm}}\)
\(v = 1.198 \times 10^7\text{ m s}^{-1} \approx 1.20 \times 10^7\text{ m s}^{-1}\)

(b) Convert the velocity to \(\text{km s}^{-1}\):
\(v = 1.20 \times 10^4\text{ km s}^{-1}\)

Using Hubble's law:
\(v = H_0 d\)
\(d = \frac{v}{H_0} = \frac{1.198 \times 10^4\text{ km s}^{-1}}{70.0\text{ km s}^{-1}\text{ Mpc}^{-1}} = 171.1\text{ Mpc}\)

(a) [4 Marks]
- M1: Calculation of wavelength shift \(\Delta \lambda = 26.2\text{ nm}\) [1]
- M1: Use of formula \(z = \frac{\Delta \lambda}{\lambda} = \frac{v}{c}\) [1]
- M1: Substitution of laboratory wavelength in the denominator [1]
- A1: \(v = 1.20 \times 10^7\text{ m s}^{-1}\) (or \(1.2 \times 10^4\text{ km s}^{-1}\)) [1]

(b) [3 Marks]
- M1: Use of Hubble's Law \(v = H_0 d\) [1]
- M1: Correct conversion of unit of velocity to match \(H_0\) (converting to km/s) [1]
- A1: \(d = 171\text{ Mpc}\) (accept range 170 to 173 Mpc) [1]

(a) Convert temperature to Kelvin:
\( T_1 = 22 + 273.15 = 295.15 \text{ K} \)
Using the ideal gas equation \( pV = nRT \):
\( p = \frac{nRT}{V} = \frac{0.90 \text{ mol} \times 8.31 \text{ J mol}^{-1} \text{ K}^{-1} \times 295.15 \text{ K}}{0.020 \text{ m}^3} \)
\( p = 110315 \text{ Pa} \approx 1.1 \times 10^5 \text{ Pa} \)

(b) Since volume and amount of gas are constant:
\( \frac{p_1}{T_1} = \frac{p_2}{T_2} \)
\( T_2 = T_1 \times \frac{p_2}{p_1} = 295.15 \text{ K} \times \frac{1.6 \times 10^5 \text{ Pa}}{110315 \text{ Pa}} = 428.1 \text{ K} \)
Convert back to Celsius:
\( T_2 = 428.1 - 273.15 = 154.95^\circ\text{C} \approx 155^\circ\text{C} \)

(c) Mean kinetic energy of an atom is given by:
\( E_k = \frac{3}{2} k T_2 \)
\( E_k = 1.5 \times 1.38 \times 10^{-23} \text{ J K}^{-1} \times 428.1 \text{ K} \)
\( E_k = 8.86 \times 10^{-21} \text{ J} \approx 8.9 \times 10^{-21} \text{ J} \)

Part (a) [2 Marks]:
- Converts temperature to Kelvin (\( 295 \text{ K} \)) [1]
- Correct substitution into \( pV = nRT \) to show value \( 1.1 \times 10^5 \text{ Pa} \) (at least 3 sig figs, e.g., \( 1.10 \times 10^5 \text{ Pa} \)) [1]

Part (b) [2 Marks]:
- Uses \( \frac{p_1}{T_1} = \frac{p_2}{T_2} \) or recalculates with \( pV = nRT \) [1]
- Correct final temperature in \( ^\circ\text{C} \) (accept range \( 154^\circ\text{C} - 156^\circ\text{C} \)) [1]

Part (c) [3 Marks]:
- Uses \( E_k = \frac{3}{2} k T \) [1]
- Substitutes Boltzmann constant and temperature in Kelvin (ecf from b) [1]
- Correct calculation of energy: \( 8.9 \times 10^{-21} \text{ J} \) (accept \( 8.8 \times 10^{-21} \text{ J} \) to \( 9.0 \times 10^{-21} \text{ J} \)) [1]

(a) Convert half-life to seconds:
\( T_{1/2} = 8.0 \text{ days} = 8.0 \times 24 \times 3600 \text{ s} = 6.912 \times 10^5 \text{ s} \)
Calculate decay constant:
\( \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{6.912 \times 10^5 \text{ s}} = 1.003 \times 10^{-6} \text{ s}^{-1} \approx 1.0 \times 10^{-6} \text{ s}^{-1} \)

(b) Using the exponential decay equation:
\( t = 20 \text{ days} = 20 \times 24 \times 3600 \text{ s} = 1.728 \times 10^6 \text{ s} \)
\( A = A_0 e^{-\lambda t} = 4.5 \times 10^7 \text{ Bq} \times e^{-(1.003 \times 10^{-6} \text{ s}^{-1} \times 1.728 \times 10^6 \text{ s})} \)
\( A = 4.5 \times 10^7 \times e^{-1.733} = 4.5 \times 10^7 \times 0.1767 = 7.95 \times 10^6 \text{ Bq} \approx 8.0 \times 10^6 \text{ Bq} \)
(Alternatively: \( A = A_0 \times (0.5)^{t/T_{1/2}} = 4.5 \times 10^7 \times (0.5)^{20/8} = 7.95 \times 10^6 \text{ Bq} \))

(c) Using the relation between activity and number of nuclei:
\( A = \lambda N \)
\( N = \frac{A}{\lambda} = \frac{7.95 \times 10^6 \text{ Bq}}{1.003 \times 10^{-6} \text{ s}^{-1}} = 7.93 \times 10^{12} \text{ nuclei} \approx 7.9 \times 10^{12} \text{ nuclei} \)

Part (a) [2 Marks]:
- Converts half-life to seconds (\( 6.91 \times 10^5 \text{ s} \)) [1]
- Correct value for decay constant: \( 1.0 \times 10^{-6} \text{ s}^{-1} \) (accept \( 1.00 \times 10^{-6} \text{ s}^{-1} \)) [1]

Part (b) [2 Marks]:
- Uses \( A = A_0 e^{-\lambda t} \) or \( A = A_0 (0.5)^{t / T_{1/2}} \) [1]
- Correct value of activity: \( 8.0 \times 10^6 \text{ Bq} \) (accept range \( 7.9 \times 10^6 \text{ Bq} \) to \( 8.0 \times 10^6 \text{ Bq} \)) [1]

Part (c) [3 Marks]:
- Uses \( A = \lambda N \) [1]
- Substitutes their value of activity \( A \) and decay constant \( \lambda \) [1]
- Correct number of nuclei: \( 7.9 \times 10^{12} \) (accept range \( 7.8 \times 10^{12} \) to \( 8.0 \times 10^{12} \), allowing full ecf from previous parts) [1]

(a) The angular frequency of a spring-mass system is given by:
\( \omega = \sqrt{\frac{k}{m}} \)
\( \omega = \sqrt{\frac{14.0 \text{ N m}^{-1}}{0.350 \text{ kg}}} = \sqrt{40} \approx 6.32 \text{ rad s}^{-1} \approx 6.3 \text{ rad s}^{-1} \)

(b) The maximum velocity is given by:
\( v_{\text{max}} = \omega A \)
Where amplitude \( A = 0.085 \text{ m} \).
\( v_{\text{max}} = 6.325 \text{ rad s}^{-1} \times 0.085 \text{ m} = 0.538 \text{ m s}^{-1} \approx 0.54 \text{ m s}^{-1} \)

(c) The velocity at any displacement \( x \) is given by:
\( v = \pm \omega \sqrt{A^2 - x^2} \)
Given \( v = 0.5 v_{\text{max}} = 0.5 \omega A \):
\( 0.5 \omega A = \omega \sqrt{A^2 - x^2} \)
\( 0.25 A^2 = A^2 - x^2 \)
\( x^2 = 0.75 A^2 \)
\( x = A \sqrt{0.75} = 0.085 \text{ m} \times 0.866 = 0.0736 \text{ m} \approx 0.074 \text{ m} \)

Part (a) [2 Marks]:
- Uses \( \omega = \sqrt{\frac{k}{m}} \) [1]
- Correctly calculates and shows \( \omega = 6.32 \text{ rad s}^{-1} \) [1]

Part (b) [2 Marks]:
- Uses \( v_{\text{max}} = \omega A \) [1]
- Correct value for maximum velocity: \( 0.54 \text{ m s}^{-1} \) (accept \( 0.53 \text{ m s}^{-1} \) to \( 0.54 \text{ m s}^{-1} \)) [1]

Part (c) [3 Marks]:
- Uses \( v^2 = \omega^2(A^2 - x^2) \) or equivalent energy conservation equation (e.g. \( \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant} \)) [1]
- Substitutes \( v = 0.5 v_{\text{max}} \) or value \( 0.27 \text{ m s}^{-1} \) [1]
- Correct calculation of displacement: \( 0.074 \text{ m} \) (accept range \( 0.073 \text{ m} \) to \( 0.075 \text{ m} \)) [1]

(a) Independent variable: Added mass \(M\). Dependent variable: Period of oscillation \(T\) (or time for oscillations \(t\)). (b) To determine \(T\) accurately, the student should: 1. Use a fiducial marker placed at the equilibrium position of the oscillations to provide a clear reference point. 2. Measure the total time \(t\) for a large number of oscillations (e.g., 20 oscillations) using a stopwatch, and then divide by 20 to find \(T\). This reduces the relative uncertainty due to human reaction time. 3. Repeat the measurement for each mass at least three times and calculate an average time. (c) For \(M = 0.100 \text{ kg}\): \(T_1 = 11.00 / 20 = 0.550 \text{ s}\), so \(T_1^2 = 0.3025 \text{ s}^2\). For \(M = 0.300 \text{ kg}\): \(T_2 = 18.20 / 20 = 0.910 \text{ s}\), so \(T_2^2 = 0.8281 \text{ s}^2\). (d) From the equation, the gradient of \(T^2\) against \(M\) is \(m = \frac{4\pi^2}{k}\). \(m = \frac{T_2^2 - T_1^2}{M_2 - M_1} = \frac{0.8281 - 0.3025}{0.300 - 0.100} = \frac{0.5256}{0.200} = 2.628 \text{ s}^2 \text{ kg}^{-1}\). Therefore, \(k = \frac{4\pi^2}{m} = \frac{4\pi^2}{2.628} \approx 15.02 \text{ N m}^{-1}\) (or \(15.0 \text{ N m}^{-1}\)). Using \(T^2 = \frac{4\pi^2}{k} M + \frac{4\pi^2 m_e}{k}\): for \(M = 0.100 \text{ kg}\), \(0.3025 = 2.628 \times 0.100 + 2.628 m_e \implies 0.3025 = 0.2628 + 2.628 m_e \implies 2.628 m_e = 0.0397 \implies m_e \approx 0.0151 \text{ kg}\) or \(15.1 \text{ g}\). (e) Wear safety goggles in case the spring breaks, or place a padded box underneath the mass hanger to catch falling masses.

[Mark Breakdown] (a) Award 1 mark for identifying independent as mass and dependent as period/time. (b) Award 3 marks: 1 mark for timing multiple oscillations (at least 10 or 20); 1 mark for using a fiducial marker at the equilibrium position; 1 mark for repeating and calculating an average. (c) Award 2 marks: 1 mark for correct calculation of both periods \(T_1 = 0.550 \text{ s}\) and \(T_2 = 0.910 \text{ s}\); 1 mark for squaring to get \(T_1^2 = 0.303 \text{ s}^2\) and \(T_2^2 = 0.828 \text{ s}^2\). (d) Award 6 marks: 1 mark for expressing gradient \(m = \frac{4\pi^2}{k}\); 1 mark for calculating gradient as \(2.63 \text{ s}^2 \text{ kg}^{-1}\); 1 mark for calculating \(k \approx 15.0 \text{ N m}^{-1}\) (range 14.9 to 15.1); 1 mark for equating vertical intercept to \(4\pi^2 m_e / k\) or using a data point in the straight-line equation; 1 mark for calculating \(m_e \approx 0.015 \text{ kg}\) (range 15.0 g to 15.2 g). (e) Award 1 mark for a valid safety precaution.

(a) To vary the current, a variable resistor (rheostat) is needed (already mentioned, but DC power supply can also have variable voltage). To measure the current, an ammeter must be connected in series with the wire. (b) According to Fleming's Left-Hand Rule, when current flows through the wire in the magnetic field, a magnetic force acts on the wire. By Newton's Third Law, an equal and opposite force acts on the magnet (and thus on the balance). If the balance reading increases, the force on the magnet is downwards, which means the magnetic force on the wire must be upwards. (c) Comparing \(\Delta m = \frac{B L}{g} I\) with the equation of a straight line \(y = mx + c\): here \(y = \Delta m\) and \(x = I\). Since there is no constant term, the y-intercept \(c = 0\), so the graph is a straight line passing through the origin. The gradient of the graph is given by \(m = \frac{B L}{g}\). (d) Given gradient \(m = 3.42 \times 10^{-3} \text{ kg A}^{-1}\). Since \(m = \frac{B L}{g}\), we have \(B = \frac{m g}{L} = \frac{3.42 \times 10^{-3} \times 9.81}{0.045} = \frac{0.03355}{0.045} = 0.7456 \text{ T}\) (or 0.75 T). Percentage uncertainty in \(m\) is 2%. Percentage uncertainty in \(L\) is \(\frac{0.1}{4.5} \times 100\% \approx 2.22\%\). Assuming \(g\) has negligible uncertainty, the percentage uncertainty in \(B\) is \(\% \Delta B = \% \Delta m + \% \Delta L = 2\% + 2.22\% = 4.22\%\). The absolute uncertainty in \(B\) is \(0.7456 \times 4.22\% \approx 0.031 \text{ T}\). Therefore, \(B = 0.75 \text{ T} \pm 0.03 \text{ T}\).

[Mark Breakdown] (a) Award 1 mark for stating that an ammeter in series is needed to measure the current. (b) Award 3 marks: 1 mark for referencing Fleming's Left-Hand Rule (magnetic force on current-carrying wire); 1 mark for referencing Newton's Third Law (equal and opposite force on the magnet); 1 mark for stating that the magnetic force on the wire is upwards. (c) Award 2 marks: 1 mark for showing that \(\Delta m\) is directly proportional to \(I\) and comparing with \(y = mx\); 1 mark for stating that the gradient is \(B L / g\). (d) Award 6 marks: 1 mark for rearranging the gradient formula to \(B = m g / L\); 1 mark for converting \(L\) to meters; 1 mark for calculating \(B \approx 0.75 \text{ T}\); 1 mark for calculating percentage uncertainty in \(L\) as 2.22%; 1 mark for adding percentage uncertainties to get 4.22%; 1 mark for calculating absolute uncertainty in \(B\) as \(\pm 0.03 \text{ T}\).

(a) Using two different power settings allows the student to eliminate the constant rate of heat loss \(Q\) to the surroundings when subtracting the two energy equations. If a single power setting were used, assuming zero heat loss would lead to a systematic overestimation of \(L_v\). (b) For Trial 1: \(\Delta m_1 = 12.4 \text{ g} = 0.0124 \text{ kg}\), \(\Delta t = 300 \text{ s}\). Rate of mass loss \(R_1 = 0.0124 / 300 \approx 4.133 \times 10^{-5} \text{ kg s}^{-1}\). For Trial 2: \(\Delta m_2 = 22.1 \text{ g} = 0.0221 \text{ kg}\), \(\Delta t = 300 \text{ s}\). Rate of mass loss \(R_2 = 0.0221 / 300 \approx 7.367 \times 10^{-5} \text{ kg s}^{-1}\). (c) The energy balance equations are: \(V_1 I_1 = R_1 L_v + Q\) and \(V_2 I_2 = R_2 L_v + Q\). Subtracting the first equation from the second gives: \(V_2 I_2 - V_1 I_1 = (R_2 - R_1) L_v\). We have: \(P_1 = 11.5 \times 4.2 = 48.3 \text{ W}\). \(P_2 = 14.8 \times 5.5 = 81.4 \text{ W}\). \(P_2 - P_1 = 81.4 - 48.3 = 33.1 \text{ W}\). \(R_2 - R_1 = 7.367 \times 10^{-5} - 4.133 \times 10^{-5} = 3.234 \times 10^{-5} \text{ kg s}^{-1}\). Therefore, \(L_v = 33.1 / (3.234 \times 10^{-5}) \approx 2.023 \times 10^6 \text{ J kg}^{-1}\) (or \(2.02 \times 10^6 \text{ J kg}^{-1}\)). (d) Percentage difference = \(|2.26 \times 10^6 - 2.023 \times 10^6| / (2.26 \times 10^6) \times 100\% \approx 10.5\%\) (accept 10% to 11%). A systematic error that could account for the experimental value being lower than the accepted value is that some water droplets were splashed out of the beaker due to vigorous boiling, which increases the apparent mass loss without absorbing latent heat, thereby reducing the calculated value of \(L_v\).

[Mark Breakdown] (a) Award 2 marks: 1 mark for explaining that using two settings allows the rate of heat loss \(Q\) to be eliminated by subtraction; 1 mark for noting that assuming zero heat loss leads to an inaccurate \(L_v\). (b) Award 2 marks: 1 mark for calculating \(R_1 \approx 4.13 \times 10^{-5} \text{ kg s}^{-1}\); 1 mark for calculating \(R_2 \approx 7.37 \times 10^{-5} \text{ kg s}^{-1}\). (c) Award 5 marks: 1 mark for writing both correct equations; 1 mark for subtracting to obtain \(P_2 - P_1 = (R_2 - R_1) L_v\); 1 mark for calculating \(P_1 = 48.3 \text{ W}\) and \(P_2 = 81.4 \text{ W}\); 1 mark for finding \(\Delta P = 33.1 \text{ W}\) and \(\Delta R \approx 3.23 \times 10^{-5} \text{ kg s}^{-1}\); 1 mark for calculating \(L_v \approx 2.02 \times 10^6 \text{ J kg}^{-1}\). (d) Award 4 marks: 1 mark for calculating percentage difference as approx. 10.5%; 1 mark for identifying a valid systematic error (e.g. water splashing, wet steam escaping); 2 marks for explaining clearly how this error leads to a lower experimental value of \(L_v\).

(a) To measure background radiation, the student should record the count rate with the protactinium source removed far away from the GM tube. The count should be recorded over a long period (e.g., 10 minutes) and divided by the time to obtain the average background count rate. This background count rate is then subtracted from all subsequent measured count rates to obtain the corrected count rate. (b) Protactinium-234m decays primarily by beta minus emission. A thin mica end-window is necessary because beta particles have low penetrating power and would be absorbed by thicker walls (like aluminum or thick glass), preventing them from entering the GM tube. Safety precautions: 1. Do not touch the generator bottle directly; handle it using plastic gloves or tongs. 2. Point the window of the generator away from the body. (c) Taking the natural logarithm of both sides of \(C = C_0 e^{-\lambda t}\) yields \(\ln C = \ln C_0 - \lambda t\). Comparing this with the straight-line equation \(y = mx + c\), a plot of \(\ln C\) on the y-axis against \(t\) on the x-axis gives a straight line. The gradient \(m\) of the graph is equal to \(-\lambda\). Once \(\lambda\) is found, the half-life can be calculated using \(t_{1/2} = \frac{\ln 2}{\lambda}\). (d) Using the values at \(t = 0 \text{ s}\) (\(C_1 = 125 \text{ s}^{-1}\)) and \(t = 120 \text{ s}\) (\(C_2 = 37 \text{ s}^{-1}\)): \(\ln C_1 = \ln 125 \approx 4.828\) and \(\ln C_2 = \ln 37 \approx 3.611\). Gradient \(m = \frac{\ln C_2 - \ln C_1}{120 - 0} = \frac{3.611 - 4.828}{120} = \frac{-1.217}{120} \approx -0.01014 \text{ s}^{-1}\). Since \(m = -\lambda\), the decay constant \(\lambda \approx 0.01014 \text{ s}^{-1}\) (or \(0.0101 \text{ s}^{-1}\)). The half-life is \(t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.69315}{0.01014} \approx 68.3 \text{ s}\) (or \(68 \text{ s}\)).

[Mark Breakdown] (a) Award 2 marks: 1 mark for measuring counts with the source removed/absent over a long period of time to get an average background rate; 1 mark for stating that this background rate is subtracted from the measured count rates. (b) Award 3 marks: 1 mark for identifying that beta particles are emitted and have low penetration, so a thin window is required to let them pass into the tube; 2 marks for stating two appropriate safety precautions (e.g., wear safety glasses/gloves, use tongs, keep a safe distance, point away from eyes, wash hands). (c) Award 3 marks: 1 mark for taking logs to show \(\ln C = \ln C_0 - \lambda t\); 1 mark for stating that the gradient of the graph of \(\ln C\) against \(t\) is equal to \(-\lambda\); 1 mark for relating half-life to decay constant using \(t_{1/2} = \ln 2 / \lambda\). (d) Award 4 marks: 1 mark for calculating \(\ln(125) = 4.83\) and \(\ln(37) = 3.61\); 1 mark for calculating gradient as \(-0.0101 \text{ s}^{-1}\); 1 mark for determining \(\lambda \approx 0.0101 \text{ s}^{-1}\); 1 mark for calculating half-life \(t_{1/2} \approx 68 \text{ s}\).