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The normal reaction force \(R\) acting on the block is perpendicular to the inclined plane: \(R = m g \cos \theta\). The frictional force opposing the motion is \(F_f = \mu R = \mu m g \cos \theta\). Since the block is moving at a constant speed \(v\) up the slope, the rate of work done against friction (power dissipated) is \(P = F_f v = \mu m g v \cos \theta\).

1 mark for the correct answer (B).

The Young modulus is given by \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{T / A}{x / L} = \frac{TL}{Ax}\). Rearranging for extension gives \(x = \frac{TL}{AE}\). For the second wire: the material is the same, so the Young modulus \(E\) is the same. The length is \(L_2 = 2L\). The radius is half, so the cross-sectional area is \(A_2 = \pi r_2^2 = \pi (r/2)^2 = A/4\). The tension is \(T_2 = 2T\). Substituting these values into the extension equation: \(x_2 = \frac{T_2 L_2}{A_2 E} = \frac{(2T)(2L)}{(A/4)E} = 16 \left(\frac{TL}{AE}\right) = 16x\).

1 mark for the correct answer (D).

The velocity just before impact is \(v_1 = \sqrt{2gh}\) (downwards). The velocity just after impact is \(v_2 = \sqrt{2g(0.64h)} = 0.8\sqrt{2gh}\) (upwards). Taking upwards as the positive direction, the change in velocity is \(\Delta v = v_2 - v_1 = 0.8\sqrt{2gh} - (-\sqrt{2gh}) = 1.8\sqrt{2gh}\). The change in momentum is \(\Delta p = m \Delta v = 1.8 m \sqrt{2gh}\). By Newton's second law, the average force is \(F = \frac{\Delta p}{\Delta t} = \frac{1.8 m \sqrt{2gh}}{\Delta t}\) during collision.

1 mark for the correct answer (C).

At terminal velocity, the upward upthrust force equals the downward viscous drag force (neglecting air weight). Upthrust is \(U = \frac{4}{3}\pi r^3 \rho g\). Viscous drag is \(F_D = 6\pi \eta r v\). Equating the two forces: \(\frac{4}{3}\pi r^3 \rho g = 6\pi \eta r v\). Solving for \(v\) yields \(v = \frac{2 r^2 \rho g}{9 \eta}\).

1 mark for the correct answer (A).

The equation of motion for the falling sphere is \(W - U - F_D = ma\), where \(W\) is the weight, \(U\) is the constant upthrust, and \(F_D = kv\) is the velocity-dependent drag force. Rearranging this gives \(ma = (W - U) - kv\), which simplifies to \(a = \frac{W - U}{m} - \frac{k}{m}v\). This represents a linear relationship with a negative gradient. Thus, the acceleration decreases linearly as the velocity increases.

1 mark for the correct answer (B).

Stiffness is defined by a high Young modulus, which represents a large resistance to elastic deformation. Ductility is the ability of a material to undergo significant plastic deformation before breaking. Therefore, material W is both stiff and ductile.

1 mark for the correct answer (A).

Taking moments about the left end of the rod to eliminate the tension in the left string: the clockwise moments are due to the weight of the block \(2W \times \frac{1}{4}L = \frac{1}{2}WL\) and the weight of the uniform rod acting at its center \(W \times \frac{1}{2}L = \frac{1}{2}WL\). The anticlockwise moment is due to the right-hand string tension \(T_R \times L\). For equilibrium, the sum of clockwise moments equals the anticlockwise moments: \(T_R \times L = \frac{1}{2}WL + \frac{1}{2}WL = WL\). Solving for tension gives \(T_R = W = 1.0W\).

1 mark for the correct answer (B).

Since the car travels up the hill at a constant speed, the engine must supply a driving force \(F\) that balances the component of the gravitational force down the slope and the resistive force: \(F = R + m g \sin \theta\). The useful output power of the engine is the rate of doing work, given by \(P = F v = (R + m g \sin \theta) v\).

1 mark for the correct answer (C).

The horizontal component of the velocity remains constant throughout the motion, so at impact \(v_x = u\).

The ball strikes the ground at an angle of \(45^\circ\) to the horizontal, which means the vertical and horizontal components of the velocity at impact are equal in magnitude:
\(\tan(45^\circ) = \frac{v_y}{v_x} = 1 \implies v_y = v_x = u\)

Using the equations of motion for the vertical direction:
\(v_y^2 = u_y^2 + 2gh\)

Since the ball was projected horizontally, the initial vertical velocity \(u_y = 0\):
\(u^2 = 2gh\)

Rearranging for \(h\):
\(h = \frac{u^2}{2g}\)

Therefore, the correct option is A.

**[1 Mark]**
* Correct answer is A.
* Award 1 mark for selecting option A.
* Reject all other options.

The extension \(\Delta x\) of a wire is given by the formula:
\(\Delta x = \frac{F L}{A E}\)

Where:
* \(F\) is the applied force (same for both)
* \(L\) is the original length
* \(A\) is the cross-sectional area
* \(E\) is the Young modulus of the material (same for both as they are made of the same material)

The cross-sectional area \(A\) is related to the diameter \(d\) by:
\(A = \frac{\pi d^2}{4} \propto d^2\)

Therefore, the extension is proportional to \(\frac{L}{d^2}\):
\(\Delta x \propto \frac{L}{d^2}\)

For wire X:
\(\Delta x_{\text{X}} \propto \frac{L}{d^2}\)

For wire Y:
\(\Delta x_{\text{Y}} \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\)

Taking the ratio of the extension of wire X to wire Y:
\(\frac{\Delta x_{\text{X}}}{\Delta x_{\text{Y}}} = \frac{\frac{L}{d^2}}{\frac{L}{2d^2}} = 2\)

Therefore, the correct option is C.

**[1 Mark]**
* Correct answer is C.
* Award 1 mark for selecting option C.
* Reject all other options.

(a) Since air resistance is neglected, there are no horizontal forces acting on the package after it is released. According to Newton's First Law of Motion, since the net horizontal force is zero, the horizontal acceleration is zero, which means the horizontal component of the velocity remains constant at \( 12\text{ m s}^{-1} \).

(b) For the vertical motion, taking downwards as positive:
Initial vertical velocity \( u_y = 0 \)
Vertical acceleration \( a_y = g = 9.81\text{ m s}^{-2} \)
Vertical displacement \( s_y = 45\text{ m} \)
Using the equation of motion:
\( s_y = u_y t + \frac{1}{2} a_y t^2 \)
\( 45 = 0 + \frac{1}{2} (9.81) t^2 \)
\( t^2 = \frac{90}{9.81} \approx 9.174\text{ s}^2 \)
\( t = \sqrt{9.174} \approx 3.03\text{ s} \)
This is approximately \( 3.0\text{ s} \).

(c) The horizontal velocity component at impact is \( v_x = 12\text{ m s}^{-1} \).
The vertical velocity component at impact is:
\( v_y = u_y + g t = 0 + (9.81 \times 3.029) \approx 29.71\text{ m s}^{-1} \)

The magnitude of the velocity \( v \) is:
\( v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 29.71^2} = \sqrt{144 + 882.7} = \sqrt{1026.7} \approx 32.0\text{ m s}^{-1} \)

The angle \( \theta \) with the horizontal is:
\( \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{29.71}{12}\right) \approx 68.0^{\circ} \)

Part (a) [2 Marks]:
- MP1: States that there are no horizontal forces acting on the package / air resistance is negligible. (1)
- MP2: Links the absence of a horizontal force to Newton's First Law (or \( F = ma \)), stating that there is no acceleration horizontally so velocity remains constant. (1)

Part (b) [3 Marks]:
- MP1: Selects and states the equation of motion \( s = ut + \frac{1}{2}at^2 \) with \( u = 0 \). (1)
- MP2: Substitutes values correctly: \( 45 = 0.5 \times 9.81 \times t^2 \). (1)
- MP3: Calculates \( t = 3.03\text{ s} \) and concludes that \( t \approx 3.0\text{ s} \). (1)

Part (c) [3.75 Marks]:
- MP1: Calculates vertical velocity component \( v_y = 29.7\text{ m s}^{-1} \) (or uses \( v_y^2 = 2as \) to get \( v_y \)). (1)
- MP2: Uses Pythagoras' theorem to find magnitude: \( v = \sqrt{12^2 + 29.7^2} \). (1)
- MP3: Uses trigonometric ratio: \( \theta = \tan^{-1}(v_y / v_x) \). (1)
- MP4: Obtains the final magnitude of \( 32.0\text{ m s}^{-1} \) (accept \( 32.1\text{ m s}^{-1} \)) and direction \( 68.0^{\circ} \) (accept \( 68^{\circ} \)) below the horizontal. (0.75)

(a) A correct free-body diagram must show:
- Weight (\( W \)) acting vertically downwards.
- Upthrust (\( U \)) acting vertically upwards.
- Viscous drag (\( D \)) acting vertically upwards.
- Since the ball-bearing is at terminal velocity, the upward forces must balance the downward forces: \( W = U + D \). The sum of the lengths of the arrows representing \( U \) and \( D \) must equal the length of the arrow representing \( W \).

(b) Calculating the forces:
- Weight \( W = mg = 1.1 \times 10^{-4}\text{ kg} \times 9.81\text{ m s}^{-2} = 1.0791 \times 10^{-3}\text{ N} \)
- Volume of the ball-bearing:
\( V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.5 \times 10^{-3}\text{ m})^3 = 1.4137 \times 10^{-8}\text{ m}^3 \)
- Upthrust \( U = \rho_{\text{fluid}} V g = 1260\text{ kg m}^{-3} \times 1.4137 \times 10^{-8}\text{ m}^3 \times 9.81\text{ m s}^{-2} = 1.7474 \times 10^{-4}\text{ N} \)

Since \( W = U + D \):
\( D = W - U = 1.0791 \times 10^{-3} - 1.7474 \times 10^{-4} = 9.0436 \times 10^{-4}\text{ N} \)

Using Stokes' Law for viscous drag:
\( D = 6\pi \eta r v \)
\( 9.0436 \times 10^{-4} = 6 \pi (0.95) (1.5 \times 10^{-3}) v \)
\( 9.0436 \times 10^{-4} = 0.02686 v \)
\( v = \frac{9.0436 \times 10^{-4}}{0.02686} \approx 0.0337\text{ m s}^{-1} \)

Part (a) [3 Marks]:
- MP1: Draws and labels the Weight force (\( W \), \( mg \), or gravity) pointing vertically downwards. (1)
- MP2: Draws and labels the Upthrust force (\( U \) or buoyancy) and Viscous drag force (\( D \), \( F \), or friction) pointing vertically upwards. (1)
- MP3: The lengths of the arrows are scaled correctly: length of upward arrows (\( U + D \)) is visually equal to the length of the downward arrow (\( W \)). (1)

Part (b) [5.75 Marks]:
- MP1: Correctly calculates Weight: \( W = 1.08 \times 10^{-3}\text{ N} \). (1)
- MP2: Correctly calculates the volume of the sphere: \( V = 1.41 \times 10^{-8}\text{ m}^3 \). (1)
- MP3: Correctly calculates Upthrust using density of liquid: \( U = 1.75 \times 10^{-4}\text{ N} \). (1)
- MP4: Equates forces for constant velocity (\( W = U + D \)) to find \( D = 9.04 \times 10^{-4}\text{ N} \). (1)
- MP5: Equates drag to Stokes' Law formula \( 6\pi\eta r v \) and rearranges for \( v \). (1)
- MP6: Calculates terminal velocity \( v = 0.034\text{ m s}^{-1} \) (accept \( 0.0337\text{ m s}^{-1} \)). (0.75)

(a) According to Newton's Second Law, the net force is proportional to the rate of change of momentum (\( F = \frac{\Delta p}{\Delta t} \)). Because the ball's velocity changes from forward to backward, its momentum changes, which requires a force. This force is exerted on the ball by the wall.
According to Newton's Third Law, when one body exerts a force on another, the second body exerts an equal and opposite force on the first. Therefore, as the wall exerts a force on the ball to change its momentum, the ball exerts a force of equal magnitude and opposite direction on the wall.

(b) Taking the initial direction of motion as positive:
Initial velocity \( u = +35\text{ m s}^{-1} \)
Final velocity \( v = -25\text{ m s}^{-1} \)
Change in momentum \( \Delta p = m(v - u) \)
\( \Delta p = 0.058 \times (-25 - 35) = 0.058 \times (-60) = -3.48\text{ kg m s}^{-1} \)
Therefore, the magnitude of the change in momentum is \( 3.48\text{ kg m s}^{-1} \) (or \( \text{N s} \)).

(c) Using Newton's Second Law:
\( F_{\text{average}} = \frac{\Delta p}{\Delta t} \)
\( F_{\text{average}} = \frac{3.48\text{ kg m s}^{-1}}{8.5 \times 10^{-3}\text{ s}} \approx 409.4\text{ N} \)
To 2 significant figures, the average force is \( 410\text{ N} \).

Part (a) [3.75 Marks]:
- MP1: States that the ball experiences a change in velocity/momentum. (1)
- MP2: References Newton's Second Law to explain that a force is required to produce this rate of change of momentum, which is exerted by the wall. (1)
- MP3: References Newton's Third Law to state that forces always occur in equal and opposite pairs. (1)
- MP4: Connects this to the ball exerting an equal and opposite force on the wall. (0.75)

Part (b) [2 Marks]:
- MP1: Realizes that velocity is a vector and accounts for the direction change (e.g., \( 35 - (-25) = 60 \)). (1)
- MP2: Calculates the magnitude of change in momentum as \( 3.48\text{ kg m s}^{-1} \) (or \( \text{N s} \)). (1)

Part (c) [3 Marks]:
- MP1: Recalls the relationship \( F = \frac{\Delta p}{\Delta t} \). (1)
- MP2: Substitutes values with time correctly converted to seconds (\( 8.5 \times 10^{-3}\text{ s} \)). (1)
- MP3: Calculates average force as \( 410\text{ N} \) (accept \( 409\text{ N} \)). (1)

(a) The cross-sectional area \( A \) is given by:
\( A = \frac{\pi d^2}{4} = \frac{\pi (1.20 \times 10^{-3}\text{ m})^2}{4} \approx 1.131 \times 10^{-6}\text{ m}^2 \)
This is approximately \( 1.1 \times 10^{-6}\text{ m}^2 \).

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

Rearranging to find the extension \( \Delta L \):
\( \Delta L = \frac{F L}{A E} \)
\( \Delta L = \frac{480\text{ N} \times 3.50\text{ m}}{1.131 \times 10^{-6}\text{ m}^2 \times 2.00 \times 10^{11}\text{ Pa}} \)
\( \Delta L = \frac{1680}{2.262 \times 10^5} \approx 7.43 \times 10^{-3}\text{ m} = 7.43\text{ mm} \)

(c) The elastic limit is the maximum force or stress that can be applied to a material such that it will still return to its original length when the load is removed. Once this limit is exceeded, the material undergoes plastic deformation. It will not return to its original length and will have a permanent extension (permanent set) when the load is removed.

Part (a) [2 Marks]:
- MP1: Uses the area formula \( A = \pi r^2 \) or \( A = \frac{\pi d^2}{4} \). (1)
- MP2: Converts the diameter to meters and shows the calculation resulting in \( 1.13 \times 10^{-6}\text{ m}^2 \). (1)

Part (b) [3.75 Marks]:
- MP1: Recalls the Young Modulus equation: \( E = \frac{F L}{A \Delta L} \). (1)
- MP2: Correctly rearranges the equation for extension \( \Delta L \). (1)
- MP3: Substitutes values correctly, using \( 1.13 \times 10^{-6}\text{ m}^2 \) (or \( 1.1 \times 10^{-6}\text{ m}^2 \)). (1)
- MP4: Calculates the extension as \( 7.4\text{ mm} \) (accept \( 7.4 \times 10^{-3}\text{ m} \); if using \( 1.1 \times 10^{-6}\text{ m}^2 \), accept \( 7.6\text{ mm} \)). (0.75)

Part (c) [3 Marks]:
- MP1: Defines the elastic limit as the point beyond which the material will not return to its original shape/length when the force is removed. (1)
- MP2: Explains that exceeding the limit causes plastic deformation. (1)
- MP3: Describes that there will be a permanent extension when the load is removed because atomic layers have slid past each other. (1)

(a) The component of weight acting down the slope is:
\( W_{\text{parallel}} = m g \sin\theta \)
\( W_{\text{parallel}} = 850\text{ kg} \times 9.81\text{ m s}^{-2} \times \sin(8.0^{\circ}) \)
\( W_{\text{parallel}} = 8338.5 \times 0.13917 \approx 1160.5\text{ N} \approx 1.16\text{ kN} \)

(b) Since the vehicle is moving at a constant speed, its acceleration is zero, and the net force along the slope must be zero.
Therefore, the forward force \( F \) exerted by the engine must balance both the component of the weight down the slope and the resistive forces \( R \):
\( F = W_{\text{parallel}} + R \)
\( F = 1160.5\text{ N} + 420\text{ N} = 1580.5\text{ N} \)
This is approximately \( 1.6\text{ kN} \).

(c) The useful power output \( P \) is given by:
\( P = F v \)
Using the calculated forward force:
\( P = 1580.5\text{ N} \times 15\text{ m s}^{-1} = 23707.5\text{ W} \approx 23.7\text{ kW} \)
To 2 significant figures, this is \( 24\text{ kW} \).

Part (a) [2 Marks]:
- MP1: Recalls and uses the component of weight parallel to the slope: \( W = mg \sin\theta \). (1)
- MP2: Calculates the correct weight component as \( 1160\text{ N} \) (or \( 1.16\text{ kN} \)). (1)

Part (b) [3 Marks]:
- MP1: States that because speed is constant, the acceleration is zero and the forces are balanced (net force = 0). (1)
- MP2: Formulates the equilibrium equation: \( F = mg\sin\theta + R \). (1)
- MP3: Substitutes values to obtain \( 1580.5\text{ N} \) and explicitly rounds to show it is \( \approx 1.6\text{ kN} \). (1)

Part (c) [3.75 Marks]:
- MP1: Recalls and uses \( P = F v \). (1)
- MP2: Substitutes the force (either \( 1580\text{ N} \) or \( 1600\text{ N} \)) and speed \( 15\text{ m s}^{-1} \). (1)
- MP3: Obtains a value of \( 2.37 \times 10^4\text{ W} \) (or \( 24\text{ kW} \)). (1)
- MP4: Quotes the final answer to 2 or 3 significant figures with correct units. (0.75)

(a) Pressure in a fluid increases with depth according to the formula \( p = \rho g h \). The bottom face of the submerged cylinder is at a greater depth than its top face. This means the upward pressure on the bottom face is greater than the downward pressure on the top face. Since force is pressure times area, there is a greater upward force on the bottom face than the downward force on the top face, resulting in a net upward force, which is the upthrust.

(b) The volume of the submerged part of the cylinder is:
\( V_{\text{sub}} = A \times h_{\text{sub}} = 4.5 \times 10^{-3}\text{ m}^2 \times 0.18\text{ m} = 8.1 \times 10^{-4}\text{ m}^3 \)
By Archimedes' principle, the upthrust \( U \) is equal to the weight of the displaced water:
\( U = \rho_{\text{water}} V_{\text{sub}} g = 1000\text{ kg m}^{-3} \times 8.1 \times 10^{-4}\text{ m}^3 \times 9.81\text{ m s}^{-2} = 7.946\text{ N} \)
This is approximately \( 7.9\text{ N} \).

(c) Since the cylinder is floating in equilibrium, its weight \( W \) must equal the upthrust \( U \):
\( W = U \implies m g = 7.946\text{ N} \)
\( m = \frac{7.946}{9.81} = 0.81\text{ kg} \)

The total volume of the wooden cylinder is:
\( V_{\text{total}} = A \times h_{\text{total}} = 4.5 \times 10^{-3}\text{ m}^2 \times 0.25\text{ m} = 1.125 \times 10^{-3}\text{ m}^3 \)

The density of the wooden cylinder is:
\( \rho_{\text{wood}} = \frac{m}{V_{\text{total}}} = \frac{0.81\text{ kg}}{1.125 \times 10^{-3}\text{ m}^3} = 720\text{ kg m}^{-3} \)
(Alternatively, \( \rho_{\text{wood}} = \frac{h_{\text{sub}}}{h_{\text{total}}} \times \rho_{\text{water}} = \frac{0.18}{0.25} \times 1000 = 720\text{ kg m}^{-3} \))

Part (a) [3 Marks]:
- MP1: States that pressure in a fluid increases with depth (or quotes \( p = \rho g h \)). (1)
- MP2: Explains that the bottom of the cylinder is at a greater depth than the top, so experiences a greater pressure. (1)
- MP3: Connects pressure difference to a difference in force (since \( F = pA \)), resulting in a net upward force. (1)

Part (b) [2 Marks]:
- MP1: Calculates the submerged volume \( V_{\text{sub}} = 8.1 \times 10^{-4}\text{ m}^3 \). (1)
- MP2: Uses \( U = \rho V g \) to calculate the upthrust and states it is \( 7.946\text{ N} \) which rounds to \( 7.9\text{ N} \). (1)

Part (c) [3.75 Marks]:
- MP1: States that for floating, Weight = Upthrust. (1)
- MP2: Calculates the mass \( m = 0.81\text{ kg} \). (1)
- MP3: Calculates the total volume of the cylinder \( V_{\text{total}} = 1.125 \times 10^{-3}\text{ m}^3 \) (or uses ratio of heights). (1)
- MP4: Calculates the density as \( 720\text{ kg m}^{-3} \). (0.75)

(a) The free-body force diagrams should show:
- For the block on the table (\( m_1 \)):
1. Weight (\( W_1 = m_1 g \)) acting vertically downwards.
2. Normal contact force (\( R \) or \( N \)) acting vertically upwards.
3. Tension (\( T \)) acting horizontally to the right (towards the pulley).
4. Frictional force (\( f = 5.5\text{ N} \)) acting horizontally to the left (opposing motion).
- For the hanging mass (\( m_2 \)):
1. Weight (\( W_2 = m_2 g \)) acting vertically downwards.
2. Tension (\( T \)) acting vertically upwards.

(b) Writing down the equations of motion using Newton's Second Law (\( F_{\text{net}} = m a \)):
- For the block on the table (moving horizontally):
\( T - f = m_1 a \implies T - 5.5 = 2.4 a \quad \text{(Equation 1)} \)
- For the hanging mass (moving downwards):
\( m_2 g - T = m_2 a \implies 1.5 \times 9.81 - T = 1.5 a \implies 14.715 - T = 1.5 a \quad \text{(Equation 2)} \)

Adding Equation 1 and Equation 2 to eliminate \( T \):
\( (T - 5.5) + (14.715 - T) = 2.4 a + 1.5 a \)
\( 9.215 = 3.9 a \)
\( a = \frac{9.215}{3.9} \approx 2.363\text{ m s}^{-2} \approx 2.36\text{ m s}^{-2} \)

(c) To find the tension, substitute \( a \) back into Equation 1:
\( T = 5.5 + 2.4 a \)
\( T = 5.5 + 2.4(2.363) = 5.5 + 5.671 = 11.171\text{ N} \approx 11.2\text{ N} \)

Part (a) [3 Marks]:
- MP1: For block \( m_1 \), draws and labels Weight (down), Normal force (up), Tension (right), and Friction (left). (1.5)
- MP2: For hanging mass \( m_2 \), draws and labels Weight (down) and Tension (up). (1.5)

Part (b) [3.75 Marks]:
- MP1: Formulates equation of motion for \( m_1 \): \( T - f = m_1 a \). (1)
- MP2: Formulates equation of motion for \( m_2 \): \( m_2 g - T = m_2 a \). (1)
- MP3: Eliminates \( T \) algebraically to obtain an expression for \( a \). (1)
- MP4: Calculates acceleration \( a = 2.36\text{ m s}^{-2} \) (accept \( 2.4\text{ m s}^{-2} \)). (0.75)

Part (c) [2 Marks]:
- MP1: Substitutes the calculated acceleration into either equation of motion. (1)
- MP2: Correctly calculates Tension \( T = 11.2\text{ N} \) (accept \( 11.1\text{ N} \) to \( 11.3\text{ N} \)). (1)

(a) Extension \( \Delta x = 8.0\text{ cm} = 0.080\text{ m} \).
According to Hooke's law:
\( F = k \Delta x \)
\( k = \frac{F}{\Delta x} = \frac{120\text{ N}}{0.080\text{ m}} = 1500\text{ N m}^{-1} \)

(b) The work done in stretching the strip is equal to the elastic strain energy stored in the strip. Since the material obeys Hooke's law, we can use:
\( W = \frac{1}{2} F \Delta x \)
\( W = \frac{1}{2} \times 120\text{ N} \times 0.080\text{ m} = 4.8\text{ J} \)
(Or using \( W = \frac{1}{2} k (\Delta x)^2 = 0.5 \times 1500 \times (0.080)^2 = 4.8\text{ J} \)).

(c) The volume \( V \) of the polymer strip is:
\( V = A \times L_0 = 3.2 \times 10^{-5}\text{ m}^2 \times 0.40\text{ m} = 1.28 \times 10^{-5}\text{ m}^3 \)

The elastic strain energy density \( U_{\text{density}} \) is:
\( U_{\text{density}} = \frac{\text{Elastic Strain Energy}}{\text{Volume}} \)
\( U_{\text{density}} = \frac{4.8\text{ J}}{1.28 \times 10^{-5}\text{ m}^3} = 3.75 \times 10^5\text{ J m}^{-3} \)

Part (a) [2 Marks]:
- MP1: Recalls and uses Hooke's Law \( F = k \Delta x \). (1)
- MP2: Converts \( 8.0\text{ cm} \) to \( 0.080\text{ m} \) and shows calculation yielding \( 1500\text{ N m}^{-1} \). (1)

Part (b) [3 Marks]:
- MP1: Recalls work done is area under force-extension graph or uses formula \( W = \frac{1}{2} F \Delta x \) or \( W = \frac{1}{2} k (\Delta x)^2 \). (1)
- MP2: Substitutes the correct values with \( \Delta x \) in meters. (1)
- MP3: Calculates the work done as \( 4.8\text{ J} \). (1)

Part (c) [3.75 Marks]:
- MP1: Recalls the volume formula \( V = A \times L \). (1)
- MP2: Calculates the volume of the strip as \( 1.28 \times 10^{-5}\text{ m}^3 \). (1)
- MP3: Divides energy by volume to find energy density. (1)
- MP4: Calculates the correct energy density as \( 3.75 \times 10^5\text{ J m}^{-3} \). (0.75)

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

1 mark for the correct option A. Correctly links the increase in load resistance to a decrease in current, which leads to a higher terminal potential difference and lower power dissipation in the internal resistance.

From the photoelectric equation, \(E_{k,\max} = hf - \phi\). Doubling the frequency gives \(E'_{k,\max} = 2hf - \phi = 2(E_{k,\max} + \phi) - \phi = 2E_{k,\max} + \phi\). Since the work function \(\phi > 0\), the maximum kinetic energy is more than doubled. Intensity is the energy incident per unit area per second, \(I = \frac{N h f}{A t}\). For constant intensity, doubling \(f\) halves the photon arrival rate \(\frac{N}{t}\), which halves the rate of photoelectron emission.

1 mark for the correct option B. Correctly identifies that maximum kinetic energy is more than doubled due to the work function offset, and that keeping intensity constant while doubling frequency halves the photon arrival rate (and thus emission rate).

Using the equation for drift velocity, \(I = nAve\), we have \(v = \frac{I}{nAe}\). For the second copper wire, the number density \(n\) is the same because the material is the same. The new drift velocity is \(v' = \frac{3I}{n(2A)e} = 1.5 \left(\frac{I}{nAe}\right) = 1.5v\).

1 mark for the correct option B. Correctly uses \(I = nAve\) and applies the ratio of current to area to find the new drift velocity of \(1.5v\).

After passing through the first polarizing filter, the unpolarized light becomes plane-polarized and its intensity is halved to \(I_1 = 0.5 I_0\). When passing through the second filter, Malus's Law applies: \(I_2 = I_1 \cos^2(60^\circ) = 0.5 I_0 \times (0.5)^2 = 0.5 I_0 \times 0.25 = 0.125 I_0\).

1 mark for the correct option A. Correctly applies the factor of 0.5 for the first unpolarized filter, and Malus's law with \(\cos^2(60^\circ)\) for the second filter.

Total internal reflection can only occur when light travels from a medium with a higher refractive index to a medium with a lower refractive index. Therefore, light must travel from X to Y. The critical angle is calculated using \(\sin\theta_c = \frac{n_{\text{less}}}{n_{\text{more}}} = \frac{1.33}{1.62} \approx 0.8210\), which gives \(\theta_c = \arcsin(0.8210) \approx 55.2^\circ\).

1 mark for the correct option B. Correctly identifies the path from higher to lower refractive index and calculates the critical angle as \(55.2^\circ\).

As temperature increases, the lattice ions in the metal vibrate with greater amplitude, increasing the rate of collisions with conduction electrons. This causes the resistance to increase. Since resistance \(R = \frac{V}{I}\) increases, the ratio \(\frac{I}{V}\) (and thus the gradient \(\frac{dI}{dV}\) of the \(I-V\) graph) decreases.

1 mark for the correct option B. Correctly determines that resistance of a metal filament increases with temperature and that the gradient of its \(I-V\) graph decreases.

In the third harmonic, the string forms three half-wavelength loops. Thus, \(L = 3 \left(\frac{\lambda}{2}\right)\), which gives \(\lambda = \frac{2}{3}L\). Statement A is incorrect because there are 4 nodes and 3 antinodes. Statement C is incorrect because adjacent nodes are separated by \(\frac{\lambda}{2} = \frac{1}{3}L\). Statement D is incorrect because adjacent loops are out of phase by \(180^\circ\).

1 mark for the correct option B. Deduces that the length is equal to 1.5 wavelengths for the third harmonic and solves for \(\lambda\).

Using the grating equation \(d \sin\theta = n \lambda\), for the second-order maximum (\(n = 2\)): \(d \sin(30^\circ) = 2\lambda \implies 0.5 d = 2\lambda \implies \lambda = 0.25 d\). For the first-order maximum (\(n = 1\)): \(d \sin\theta_1 = 1\lambda \implies d \sin\theta_1 = 0.25 d \implies \sin\theta_1 = 0.25\). Therefore, \( \theta_1 = \arcsin(0.25) \approx 14.5^\circ \).

1 mark for the correct option B. Correctly applies the diffraction grating equation for both orders and uses the ratio to find the angle of the first-order maximum as \(14.5^\circ\).

The current \(I\) in both wires is the same because they are connected in series.

The relationship between current and drift velocity is given by \(I = nAvq\), where:
- \(n\) is the number density of conduction electrons (which is the same for both wires as they are made of the same material, copper),
- \(A\) is the cross-sectional area of the wire,
- \(v\) is the drift velocity of the conduction electrons,
- \(q\) is the elementary charge.

Since \(I_{\text{P}} = I_{\text{Q}}\):
\(n A_{\text{P}} v_{\text{P}} q = n A_{\text{Q}} v_{\text{Q}} q\)

Therefore:
\(A_{\text{P}} v_{\text{P}} = A_{\text{Q}} v_{\text{Q}}\)

Rearranging for the ratio of the drift velocities:
\(\frac{v_{\text{P}}}{v_{\text{Q}}} = \frac{A_{\text{Q}}}{A_{\text{P}}}\)

Since the cross-sectional area is given by \(A = \pi r^2\):
\(\frac{v_{\text{P}}}{v_{\text{Q}}} = \frac{\pi (2r)^2}{\pi r^2} = \frac{4\pi r^2}{\pi r^2} = 4\)

D is the correct answer.

- Correctly identifies that current \(I\) is constant across series components.
- Correctly applies \(I = nAvq\) and the area of a circle relation to find the ratio is 4.

For a wire fixed at both ends, a stationary wave in the third harmonic consists of three loops (half-wavelengths) along its length \(L\).

Therefore, we can write:
\(L = 3 \left(\frac{\lambda}{2}\right) = 1.5\lambda\)

Rearranging to find the wavelength \(\lambda\):
\(\lambda = \frac{2L}{3} = \frac{2 \times 0.80\text{ m}}{3} = \frac{1.60}{3}\text{ m} \approx 0.533\text{ m}\)

Using the wave equation \(v = f\lambda\):
\(v = 120\text{ Hz} \times \frac{1.60}{3}\text{ m} = 40 \times 1.60 = 64\text{ m s}^{-1}\)

C is the correct answer.

- Identifies that for the third harmonic, \(\lambda = \frac{2L}{3}\).
- Calculates the wave speed using \(v = f\lambda\) to arrive at \(64\text{ m s}^{-1}\).

(a) Electromotive force (emf) is the total energy transferred by a source of electricity per unit charge that passes through it, or the work done per unit charge in driving charge around a complete circuit.

(b) First, calculate the total resistance of the series circuit:
\(R_{\text{total}} = R_{\text{thermistor}} + R_f + r = 450\ \Omega + 150\ \Omega + 2.50\ \Omega = 602.5\ \Omega\)

Then, calculate the current in the circuit:
\(I = \frac{E}{R_{\text{total}}} = \frac{3.00\text{ V}}{602.5\ \Omega} \approx 4.979 \times 10^{-3}\text{ A}\)

Now, calculate the potential difference across the thermistor (\(V_{\text{thermistor}}\)):
\(V_{\text{thermistor}} = I \times R_{\text{thermistor}} = (4.979 \times 10^{-3}\text{ A}) \times 450\ \Omega \approx 2.24\text{ V}\)

(c) As temperature increases, the resistance of the NTC thermistor decreases. This decreases the total resistance of the circuit, causing the current in the circuit to increase. Since the potential difference across the fixed resistor is given by \(V_f = I \times R_f\), and \(R_f\) is constant, the increase in current results in an increase in the potential difference across the fixed resistor.

Part (a):
- MP1: Energy transferred per unit charge (or work done per unit charge) [1 mark]
- MP2: From chemical/other forms to electrical energy (or around a complete circuit) [1 mark]

Part (b):
- MP1: Employs \(R_{\text{total}} = R_{\text{thermistor}} + R_f + r\) to find \(R_{\text{total}} = 602.5\ \Omega\) [1 mark]
- MP2: Calculates the current \(I = 3.00 / 602.5 = 4.98 \times 10^{-3}\text{ A}\) [1 mark]
- MP3: Uses \(V = IR\) with the thermistor resistance [1 mark]
- MP4: Correct final value of \(2.24\text{ V}\) (allow \(2.2\text{ V}\) to \(2.25\text{ V}\)) [1 mark]

Part (c):
- MP1: States that resistance of thermistor decreases as temperature increases [1 mark]
- MP2: Explains that total resistance decreases, so current in circuit increases [1 mark]
- MP3: Concludes that potential difference across the fixed resistor increases [0.75 marks]

(a) The refractive index of glass (\(n_{\text{glass}}\)) relative to a vacuum is:
\(n_{\text{glass}} = \frac{c}{v_{\text{glass}}} = \frac{3.00 \times 10^8\text{ m s}^{-1}}{1.92 \times 10^8\text{ m s}^{-1}} = 1.5625\)

The refractive index of liquid (\(n_{\text{liquid}}\)) relative to a vacuum is:
\(n_{\text{liquid}} = \frac{c}{v_{\text{liquid}}} = \frac{3.00 \times 10^8\text{ m s}^{-1}}{2.25 \times 10^8\text{ m s}^{-1}} = 1.3333\)

The refractive index of glass relative to the liquid (\(_{\text{liquid}}n_{\text{glass}}\)) is:
\(_{\text{liquid}}n_{\text{glass}} = \frac{n_{\text{glass}}}{n_{\text{liquid}}} = \frac{v_{\text{liquid}}}{v_{\text{glass}}} = \frac{2.25 \times 10^8\text{ m s}^{-1}}{1.92 \times 10^8\text{ m s}^{-1}} \approx 1.172\)

(b) Total internal reflection can occur when light travels from a denser medium (glass) to a less dense medium (liquid). The critical angle \(\theta_c\) is given by:
\(\sin \theta_c = \frac{n_{\text{liquid}}}{n_{\text{glass}}} = \frac{v_{\text{glass}}}{v_{\text{liquid}}} = \frac{1}{1.172} \approx 0.8533\)

\(\theta_c = \sin^{-1}(0.8533) \approx 58.6^\circ\)

(c) The frequency of the light remains constant in all media:
\(f = \frac{c}{\lambda_0} = \frac{3.00 \times 10^8\text{ m s}^{-1}}{589 \times 10^{-9}\text{ m}} \approx 5.093 \times 10^{14}\text{ Hz}\)

The wavelength in glass (\(\lambda_{\text{glass}}\)) is given by:
\(\lambda_{\text{glass}} = \frac{v_{\text{glass}}}{f} = \frac{1.92 \times 10^8\text{ m s}^{-1}}{5.093 \times 10^{14}\text{ Hz}} \approx 3.77 \times 10^{-7}\text{ m}\) (or \(377\text{ nm}\)).
Alternatively, using refractive index:
\(\lambda_{\text{glass}} = \frac{\lambda_0}{n_{\text{glass}}} = \frac{589\text{ nm}}{1.5625} \approx 377\text{ nm}\).

Part (a):
- MP1: Recalls relationship between refractive index and wave speed \(n = \frac{c}{v}\) or \(n_{\text{rel}} = \frac{v_1}{v_2}\) [1 mark]
- MP2: Correct substitution of values: \(\frac{2.25 \times 10^8}{1.92 \times 10^8}\) [1 mark]
- MP3: Obtains value 1.17 (at least 3 s.f. shown as 1.172) [1 mark]

Part (b):
- MP1: Recalls \(\sin \theta_c = \frac{1}{n}\) or \(\sin \theta_c = \frac{v_1}{v_2}\) [1 mark]
- MP2: Correct substitution of values \(\sin \theta_c = \frac{1.92 \times 10^8}{2.25 \times 10^8}\) [1 mark]
- MP3: Correct calculation of \(\theta_c = 58.6^\circ\) (allow \(58.5^\circ\) to \(59.0^\circ\)) [1 mark]

Part (c):
- MP1: Recalls that frequency remains constant or uses \(\lambda_{\text{glass}} = \frac{\lambda_0}{n_{\text{glass}}}\) [1 mark]
- MP2: Calculates \(n_{\text{glass}} = 1.56\) or \(f = 5.09 \times 10^{14}\text{ Hz}\) [1 mark]
- MP3: Correct final value of \(\lambda_{\text{glass}} = 3.77 \times 10^{-7}\text{ m}\) (or \(377\text{ nm}\)) [0.75 marks]

(a) When the string is plucked, a progressive wave travels along the string and is reflected at both of the fixed ends. The incident wave and the reflected wave, which have the same frequency, wavelength, and amplitude, travel in opposite directions and superpose (interfere). At points of constructive interference (where waves meet in phase), antinodes with maximum amplitude are formed. At points of destructive interference (where waves meet in antiphase), nodes with zero amplitude are formed, resulting in a standing wave.

(b) For the fundamental frequency, the length of the string \(L\) is equal to half a wavelength:
\(L = \frac{\lambda}{2} \implies \lambda = 2L = 2 \times 0.650\text{ m} = 1.30\text{ m}\)

The wave speed \(v\) is:
\(v = f \lambda = 196\text{ Hz} \times 1.30\text{ m} = 254.8\text{ m s}^{-1}\)

The formula for wave speed on a stretched string is:
\(v = \sqrt{\frac{T}{\mu}}\)

Squaring both sides and solving for tension \(T\):
\(T = v^2 \mu = (254.8\text{ m s}^{-1})^2 \times 4.25 \times 10^{-3}\text{ kg m}^{-1} \approx 276\text{ N}\)

(c) If the temperature rises, the string expands and becomes slightly slack, meaning the tension \(T\) in the string decreases. Since the wave speed \(v = \sqrt{\frac{T}{\mu}}\), a decrease in tension results in a lower wave speed. Because \(f_0 = \frac{v}{2L}\), and the increase in length \(L\) is extremely small while the decrease in \(T\) is more dominant, the fundamental frequency \(f_0\) decreases.

Part (a):
- MP1: Wave is reflected at the fixed ends (creating two waves travelling in opposite directions) [1 mark]
- MP2: Opposing waves superpose / interfere (having same frequency and wavelength) [1 mark]
- MP3: Describes node formation (destructive interference) and antinode formation (constructive interference) [1 mark]

Part (b):
- MP1: Identifies \(\lambda = 2L = 1.30\text{ m}\) [1 mark]
- MP2: Calculates wave speed \(v = f \lambda = 254.8\text{ m s}^{-1}\) [1 mark]
- MP3: Recalls \(v = \sqrt{\frac{T}{\mu}}\) and rearranges to \(T = v^2 \mu\) [1 mark]
- MP4: Correct calculation of \(T = 276\text{ N}\) (allow \(275\text{ N}\) to \(277\text{ N}\)) [1 mark]

Part (c):
- MP1: Identifies that higher temperature causes the string to expand, leading to decreased tension [1 mark]
- MP2: Links decrease in tension to a decrease in wave speed, hence lower fundamental frequency [0.75 marks]

(a) The energy of a photon \(E_p\) is given by:
\(E_p = h f = (6.63 \times 10^{-34}\text{ J s}) \times (7.45 \times 10^{14}\text{ Hz}) = 4.939 \times 10^{-19}\text{ J}\)

To convert this energy into electronvolts:
\(E_p = \frac{4.939 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} \approx 3.09\text{ eV}\)

(b) Using Einstein's photoelectric equation:
\(h f = \Phi + E_{\text{k,max}}\)

Rearranging for maximum kinetic energy \(E_{\text{k,max}}\):
\(E_{\text{k,max}} = h f - \Phi = 3.09\text{ eV} - 2.30\text{ eV} = 0.79\text{ eV}\)

Convert \(E_{\text{k,max}}\) back to joules:
\(E_{\text{k,max}} = 0.79\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 1.264 \times 10^{-19}\text{ J}\)

Since \(E_{\text{k,max}} = \frac{1}{2} m_e v_{\text{max}}^2\):
\(v_{\text{max}} = \sqrt{\frac{2 E_{\text{k,max}}}{m_e}} = \sqrt{\frac{2 \times 1.264 \times 10^{-19}\text{ J}}{9.11 \times 10^{-31}\text{ kg}}} \approx 5.27 \times 10^5\text{ m s}^{-1}\)

(c) Doubling the intensity of the light means there are twice as many photons incident on the plate per second. Since one photon interacts with one electron, the number of photoelectrons emitted per second doubles. However, the maximum kinetic energy of the emitted photoelectrons depends only on the photon energy (frequency) and the work function, so the maximum kinetic energy remains unchanged.

Part (a):
- MP1: Recalls \(E = hf\) and substitutes values correctly [1 mark]
- MP2: Divides energy in joules by \(1.60 \times 10^{-19}\) to convert to eV [1 mark]
- MP3: Obtains \(3.09\text{ eV}\) (allow \(3.08\text{ eV}\) to \(3.10\text{ eV}\)) [1 mark]

Part (b):
- MP1: Recalls \(E_{\text{k,max}} = hf - \Phi\) and calculates \(E_{\text{k,max}} = 0.79\text{ eV}\) [1 mark]
- MP2: Converts \(E_{\text{k,max}}\) to Joules (\(1.26 \times 10^{-19}\text{ J}\)) [1 mark]
- MP3: Uses \(E_{\text{k,max}} = \frac{1}{2} m v^2\) to find velocity [1 mark]
- MP4: Obtains \(v_{\text{max}} = 5.27 \times 10^5\text{ m s}^{-1}\) (allow \(5.2 \times 10^5\) to \(5.3 \times 10^5\text{ m s}^{-1}\)) [1 mark]

Part (c):
- MP1: States that the number of photoelectrons emitted per second doubles because more photons arrive per second [1 mark]
- MP2: States that the maximum kinetic energy remains constant because photon energy (frequency) does not change [0.75 marks]

(a) The radius \(r\) of the wire is half the diameter:
\(r = \frac{0.800\text{ mm}}{2} = 0.400\text{ mm} = 4.00 \times 10^{-4}\text{ m}\)

The cross-sectional area \(A\) is:
\(A = \pi r^2 = \pi (4.00 \times 10^{-4}\text{ m})^2 \approx 5.027 \times 10^{-7}\text{ m}^2\)

(b) The current \(I\) is related to drift velocity \(v\) by:
\(I = n A v q\)

Rearranging for \(v\):
\(v = \frac{I}{n A q}\)

Substitute the known values (where \(q = e = 1.60 \times 10^{-19}\text{ C}\)):
\(v = \frac{3.50\text{ A}}{(8.50 \times 10^{28}\text{ m}^{-3}) \times (5.027 \times 10^{-7}\text{ m}^2) \times (1.60 \times 10^{-19}\text{ C})} \approx 5.12 \times 10^{-4}\text{ m s}^{-1}\)

(c) Since the wires are connected in series, the current \(I\) is the same in both wires. The materials are both copper, so the electron density \(n\) is also the same. The area is proportional to the square of the diameter (\(A \propto d^2\)). If the diameter is doubled, the cross-sectional area increases by a factor of 4. Since \(v = \frac{I}{nAq}\), drift velocity \(v\) is inversely proportional to cross-sectional area \(A\). Therefore, the drift velocity in the second wire is \(\frac{1}{4}\) (or \(0.25\) times) of the drift velocity in the first wire, which is \(1.28 \times 10^{-4}\text{ m s}^{-1}\).

Part (a):
- MP1: Finds radius \(r = 0.400 \times 10^{-3}\text{ m}\) [1 mark]
- MP2: Uses \(A = \pi r^2\) to calculate area [1 mark]
- MP3: Correct calculation showing \(5.03 \times 10^{-7}\text{ m}^2\) (must show at least 3 s.f.) [1 mark]

Part (b):
- MP1: Recalls \(I = nAvq\) [1 mark]
- MP2: Correct substitution with \(q = 1.60 \times 10^{-19}\text{ C}\) [1 mark]
- MP3: Correct calculation of \(v = 5.12 \times 10^{-4}\text{ m s}^{-1}\) (allow \(5.1 \times 10^{-4}\) to \(5.2 \times 10^{-4}\text{ m s}^{-1}\)) [1 mark]

Part (c):
- MP1: States that current \(I\) and electron density \(n\) are identical in both wires because they are in series and of same material [1 mark]
- MP2: Explains that doubling the diameter increases the area by a factor of 4 [1 mark]
- MP3: Concludes that the drift velocity becomes one-quarter (0.25 times) of the initial value [0.75 marks]

(a) The distance from the central fringe to the third-order bright fringe (\(m = 3\)) is \(y_3 = 1.60\text{ cm} = 1.60 \times 10^{-2}\text{ m}\).

Therefore, the fringe spacing (separation between adjacent bright fringes), \(x\), is:
\(x = \frac{y_3}{3} = \frac{1.60 \times 10^{-2}\text{ m}}{3} \approx 5.333 \times 10^{-3}\text{ m}\)

Using the double-slit formula:
\(x = \frac{\lambda D}{d}\)

Rearranging for wavelength \(\lambda\):
\(\lambda = \frac{x d}{D} = \frac{(5.333 \times 10^{-3}\text{ m}) \times (0.180 \times 10^{-3}\text{ m})}{1.50\text{ m}} \approx 6.40 \times 10^{-7}\text{ m}\) (or \(640\text{ nm}\)).

(b) The first dark fringe is formed where destructive interference occurs. For this to happen, the path difference from the two slits to this point on the screen must be exactly half a wavelength (\(\frac{\lambda}{2}\)). This causes the waves arriving from the two slits to meet in antiphase (phase difference of \(\pi\text{ radians}\) or \(180^\circ\)), so they cancel each other out.

(c) According to \(x = \frac{\lambda D}{d}\), the fringe spacing \(x\) is inversely proportional to the slit spacing \(d\). Therefore, if \(d\) is decreased, the fringe spacing \(x\) increases (the fringes become wider and more spread out). The intensity of the fringes remains the same, but the pattern is more spaced out.

Part (a):
- MP1: Calculates fringe spacing \(x = \frac{1.60 \times 10^{-2}}{3} = 5.33 \times 10^{-3}\text{ m}\) [1 mark]
- MP2: Recalls \(x = \frac{\lambda D}{d}\) and rearranges to \(\lambda = \frac{xd}{D}\) [1 mark]
- MP3: Correct calculation of \(\lambda = 6.40 \times 10^{-7}\text{ m}\) (or \(640\text{ nm}\)) [1 mark]

Part (b):
- MP1: Mentions destructive interference [1 mark]
- MP2: Explains that the path difference is half a wavelength (\(\frac{\lambda}{2}\)) [1 mark]
- MP3: Explains that the phase difference is \(\pi\text{ rad}\) or \(180^\circ\) (or waves meet in antiphase) [0.75 marks]

Part (c):
- MP1: States that the fringe spacing \(x\) increases [1 mark]
- MP2: Justifies this using \(x \propto \frac{1}{d}\) [1 mark]

(a) The total resistance of the series combination in bright light is:
\(R_{\text{total}} = R_{\text{LDR}} + R = 350\ \Omega + 2200\ \Omega = 2550\ \Omega\)

Using the potential divider equation:
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R_{\text{total}}} = 9.00\text{ V} \times \frac{350\ \Omega}{2550\ \Omega} \approx 1.24\text{ V}\)

(b) When \(V_{\text{out}} = 6.50\text{ V}\), the potential difference across the fixed resistor \(V_R\) is:
\(V_R = V_{\text{in}} - V_{\text{out}} = 9.00\text{ V} - 6.50\text{ V} = 2.50\text{ V}\)

The current \(I\) in the circuit is:
\(I = \frac{V_R}{R} = \frac{2.50\text{ V}}{2200\ \Omega} \approx 1.136 \times 10^{-3}\text{ A}\)

The resistance of the LDR (\(R_{\text{LDR}}\)) is:
\(R_{\text{LDR}} = \frac{V_{\text{out}}}{I} = \frac{6.50\text{ V}}{1.136 \times 10^{-3}\text{ A}} \approx 5.72 \times 10^3\ \Omega\) (or \(5.72\text{ k}\Omega\)).

Alternatively, using the ratio method:
\(\frac{V_{\text{out}}}{V_{\text{in}}} = \frac{R_{\text{LDR}}}{R_{\text{LDR}} + R} \implies \frac{6.50}{9.00} = \frac{R_{\text{LDR}}}{R_{\text{LDR}} + 2200}\)

\(6.50 R_{\text{LDR}} + 14300 = 9.00 R_{\text{LDR}} \implies 2.50 R_{\text{LDR}} = 14300 \implies R_{\text{LDR}} = 5720\ \Omega\).

(c) When it gets dark, the light intensity decreases, which causes the resistance of the LDR to increase. Because the LDR resistance increases, the output voltage \(V_{\text{out}}\) across the LDR increases. This high voltage signal can be fed into an electronic switch (or electromagnetic relay) that turns the street light on once the potential difference exceeds a preset threshold.

Part (a):
- MP1: Correct conversion of \(2.20\text{ k}\Omega\) to \(2200\ \Omega\) [1 mark]
- MP2: Recalls potential divider equation or calculates current [1 mark]
- MP3: Correct calculation of \(V_{\text{out}} = 1.24\text{ V}\) (allow \(1.2\text{ V}\) to \(1.3\text{ V}\)) [1 mark]

Part (b):
- MP1: Finds voltage across fixed resistor \(V_R = 2.50\text{ V}\) [1 mark]
- MP2: Finds current \(I = 1.14\text{ mA}\) or sets up the ratio equation correctly [1 mark]
- MP3: Correct substitution and calculation of \(R_{\text{LDR}} = 5720\ \Omega\) (or \(5.72\text{ k}\Omega\)) [1.75 marks]

Part (c):
- MP1: Explains that resistance of LDR increases in the dark, so output voltage \(V_{\text{out}}\) increases [1 mark]
- MP2: States that this high voltage can trigger a switch/relay to turn the light on [1 mark]

(a) Transverse waves have oscillations that are perpendicular to the direction of energy transfer (or wave propagation). Since these oscillations can occur in any direction within a plane perpendicular to the wave direction, a polariser can restrict these oscillations to a single plane. Longitudinal waves have oscillations parallel to the direction of energy transfer. Since there is only one direction of oscillation, it cannot be restricted to a single plane perpendicular to the direction of propagation, hence they cannot be polarised.

(b) When unpolarised light passes through the first filter \(P_1\), its intensity is halved:
\(I_1 = \frac{I_0}{2} = \frac{360\text{ W m}^{-2}}{2} = 180\text{ W m}^{-2}\)

According to Malus's Law, the intensity transmitted through the second filter \(P_2\) is:
\(I_2 = I_1 \cos^2 \theta = 180\text{ W m}^{-2} \times \cos^2(30.0^\circ)\)

Since \(\cos(30.0^\circ) = \frac{\sqrt{3}}{2}\), then \(\cos^2(30.0^\circ) = 0.75\).
\(I_2 = 180 \times 0.75 = 135\text{ W m}^{-2}\)

(c) Yes, light is transmitted. When \(P_3\) is introduced at \(45.0^\circ\) between \(P_1\) and \(P_2\):

1. The intensity after \(P_3\) is:
\(I_3 = I_1 \cos^2(45.0^\circ) = 180\text{ W m}^{-2} \times (0.7071)^2 = 180 \times 0.50 = 90.0\text{ W m}^{-2}\)

2. The light incident on \(P_2\) is now polarised at \(45.0^\circ\) to \(P_2\)'s axis. The intensity after \(P_2\) is:
\(I_{\text{final}} = I_3 \cos^2(45.0^\circ) = 90.0\text{ W m}^{-2} \times 0.50 = 45.0\text{ W m}^{-2}\)

Therefore, light is transmitted with an intensity of \(45.0\text{ W m}^{-2}\).

Part (a):
- MP1: States that transverse waves oscillate perpendicular to the direction of wave travel, while longitudinal waves oscillate parallel to it [1 mark]
- MP2: Defines polarisation as restricting oscillations to a single direction/plane perpendicular to energy transfer [1 mark]
- MP3: Explains that because longitudinal oscillations are already restricted to the direction of travel, they cannot be further polarised [1 mark]

Part (b):
- MP1: States that the first polariser halves the intensity (\(I_1 = 180\text{ W m}^{-2}\)) [1 mark]
- MP2: Recalls Malus's Law \(I_2 = I_1 \cos^2 \theta\) [1 mark]
- MP3: Correct calculation of \(I_2 = 135\text{ W m}^{-2}\) [1 mark]

Part (c):
- MP1: Identifies that adding \(P_3\) means light is now transmitted through \(P_2\) [1 mark]
- MP2: Calculates the intensity after the third polariser \(I_3 = 90\text{ W m}^{-2}\) [1 mark]
- MP3: Correct final intensity of \(45.0\text{ W m}^{-2}\) after the second polariser [0.75 marks]

*(a) Experimental Setup:*
The apparatus should be arranged with a steel sphere suspended from an electromagnet. The electromagnet is connected to a low-voltage power supply and a double-pole switch, which simultaneously cuts current to the electromagnet and starts an electronic timer (or digital millisecond counter). A trapdoor is positioned at the bottom of the fall. When the sphere hits the trapdoor, it opens the circuit, stopping the timer.

*(b) Measurements and Instruments:*
1. The distance of fall, \(h\), is measured from the bottom of the sphere (while suspended) to the top of the trapdoor. This distance is measured using a metre rule (precision \(\pm 1\text{ mm}\)).
2. The time of fall, \(t\), is measured from the moment the electromagnet releases the ball until it hits the trapdoor. This is measured directly by the electronic millisecond timer.

*(c) Graphical Analysis:*
Since the sphere falls from rest under gravity, the SUVAT equation applies: \(s = ut + \frac{1}{2}at^2\). With \(u = 0\), \(s = h\), and \(a = g\), we obtain:
\(h = \frac{1}{2}gt^2\)
This is in the linear form \(y = mx + c\), where \(y = h\), \(x = t^2\), \(c = 0\), and the gradient \(m = \frac{1}{2}g\).
Therefore, a graph of \(h\) against \(t^2\) is plotted. A line of best fit is drawn, and its gradient is determined. The acceleration of free fall is calculated as:
\(g = 2 \times \text{gradient}\)

*(d) Systematic Error and Mitigation:*
One potential systematic error is the delay in the release of the steel ball from the electromagnet due to residual magnetism (remanence). This causes the timer to start slightly before the ball actually begins to fall, resulting in an overestimated time \(t\) and an underestimated \(g\).
To minimize this systematic error, the current in the electromagnet should be reduced to the minimum value that is just sufficient to support the weight of the steel sphere. Alternatively, a small piece of non-magnetic paper or tape can be placed over the pole of the electromagnet to act as a spacer.

(a) Diagram [3 Marks]:
- MP1: Electromagnet connected to a power supply/switch and a digital timer. [1]
- MP2: Steel sphere shown suspended beneath the electromagnet. [1]
- MP3: Trapdoor (or light gate at the bottom) connected to the stop input of the timer. [1]

(b) Measurements and Instruments [3 Marks]:
- MP1: Height/distance of fall \(h\) measured using a metre rule. [1]
- MP2: Time of fall \(t\) measured using an electronic timer or millisecond counter. [1]
- MP3: Measure \(h\) from the bottom of the sphere (when suspended) to the top of the trapdoor to ensure accuracy of distance. [1]

(c) Graphical Analysis [4 Marks]:
- MP1: Identifies the equation of motion: \(h = \frac{1}{2}gt^2\). [1]
- MP2: States that a graph of \(h\) (y-axis) against \(t^2\) (x-axis) is plotted. [1]
- MP3: States that the graph should be a straight line through the origin. [1]
- MP4: Explains that the gradient is determined, and \(g = 2 \times \text{gradient}\). [1]

(d) Systematic Error [2 Marks]:
- MP1: Identifies a valid systematic error (e.g., residual magnetism delaying the release of the sphere, or air resistance, or delay in the trapdoor mechanism). [1]
- MP2: Offers a correct corresponding mitigation (e.g., use the minimum current required to hold the ball/use a non-magnetic spacer; or use a dense steel sphere to make air resistance negligible; or calibrate the trapdoor delay). [1]

*(a) Experimental Apparatus:*
The wire is clamped firmly at one end of a long, horizontal bench. It passes over a pulley at the other end, and a mass hanger is attached to its free end. A small marker (such as a piece of tape or paper flag) is taped to the wire near a fixed metre rule on the bench.

*(b) Safety Precautions:*
1. Wear safety goggles to protect the eyes from injury if the wire snaps under high tension.
2. Place a box containing sand or a soft cushion directly underneath the suspended masses to catch them when the wire breaks, preventing injury to feet and damage to the floor.

*(c) Measurements and Accuracy:*
1. **Original length \(L\)**: Measured from the clamp to the marker using a metre rule (precision \(\pm 1\text{ mm}\)). Ensure the wire is straight by applying a small preparatory load before measuring.
2. **Diameter \(d\)**: Measured using a micrometer screw gauge (precision \(\pm 0.01\text{ mm}\)). To ensure accuracy, measure at 3 to 5 different positions along the wire and in two perpendicular directions at each position to account for non-circular cross-section, then find the average.
3. **Applied force \(F\)**: Calculated from the added masses using \(F = mg\).
4. **Extension \(e\)**: Measured by noting the position of the marker relative to the fixed metre rule as masses are added. To ensure accuracy, read the ruler at eye level to avoid parallax error.

*(d) Graphical Analysis:*
The Young modulus is given by:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{e / L} = \frac{FL}{Ae}\)
where \(A = \frac{\pi d^2}{4}\) is the cross-sectional area.
If we plot a graph of applied force \(F\) (y-axis) against extension \(e\) (x-axis), the linear region representing elastic deformation will be a straight line through the origin. The gradient of this line is:
\(\text{gradient} = \frac{F}{e}\)
Thus, the Young modulus can be calculated as:
\(E = \text{gradient} \times \frac{L}{A}\)
Alternatively, plotting stress against strain gives a line of gradient \(E\).

(a) Diagram [3 Marks]:
- MP1: Long horizontal wire clamped at one end of a bench and passing over a pulley at the other end. [1]
- MP2: Masses/mass hanger suspended from the free end of the wire. [1]
- MP3: Marker attached to the wire near a fixed metre rule. [1]

(b) Safety Precautions [2 Marks]:
- MP1: Wearing safety goggles to protect eyes in case the wire snaps. [1]
- MP2: Placing a sand tray/soft cushion under the masses to protect feet/floor if the wire breaks. [1]

(c) Measurements and Accuracy [4 Marks]:
- MP1: Original length \(L\) from clamp to marker measured with a metre rule, ensuring the wire is straight using a small initial load. [1]
- MP2: Diameter \(d\) measured with a micrometer screw gauge in several places and orientations, and averaged. [1]
- MP3: Extension \(e\) determined by reading the marker's position against the metre rule at eye level (to avoid parallax). [1]
- MP4: Force \(F\) calculated using \(F = mg\), where masses are added in equal increments. [1]

(d) Graphical Analysis [4 Marks]:
- MP1: Correctly states the Young modulus formula: \(E = \frac{FL}{Ae}\). [1]
- MP2: Identifies plotting a graph of force \(F\) (y-axis) against extension \(e\) (x-axis) (or stress against strain). [1]
- MP3: Explains that the cross-sectional area \(A\) is calculated using \(A = \frac{\pi d^2}{4}\). [1]
- MP4: Explains that \(E = \text{gradient} \times \frac{L}{A}\) where the gradient is \(\frac{F}{e}\) (or \(E = \text{gradient}\) if stress-strain is plotted). [1]

*(a) Circuit Diagram:*
The circuit must consist of the cell under test in series with an ammeter, a variable resistor (rheostat), and a switch. A voltmeter must be connected in parallel across the terminals of the cell to measure the terminal potential difference.

*(b) Experimental Procedure:*
1. Assemble the circuit and close the switch.
2. Adjust the variable resistor to its maximum resistance to keep the initial current low, and record the current \(I\) from the ammeter and terminal potential difference \(V\) from the voltmeter.
3. Decrease the resistance of the variable resistor in steps to obtain a minimum of 6 different sets of readings of \(I\) and \(V\).
4. Crucially, open the switch between readings to prevent the cell from discharging unnecessarily and to stop current from heating the cell, as temperature changes will alter the internal resistance.

*(c) Graphical Analysis:*
The terminal potential difference \(V\) is related to the EMF \(\varepsilon\) and current \(I\) by the equation:
\(V = \varepsilon - Ir\)
Rearranging this into the form of a straight line (\(y = mx + c\)):
\(V = -rI + \varepsilon\)
By plotting terminal potential difference \(V\) on the y-axis against current \(I\) on the x-axis, we obtain a straight line with a negative gradient.
- The y-intercept of the line is equal to the EMF, \(\varepsilon\).
- The magnitude of the gradient of the line is equal to the internal resistance, \(r\).

*(d) Discussion of Variable Resistor Range:*
The internal resistance of the cell is approximately \(1.5\ \Omega\). If a high-range variable resistor (such as \(0 - 10\text{ k}\Omega\)) is used, the total resistance of the circuit will be dominated by the variable resistor for almost all settings. Consequently, the current in the circuit will be extremely small (on the order of microamperes or milliamperes), and the voltmeter will read almost exactly the EMF \(\varepsilon\) for nearly all settings. The data points on a graph of \(V\) against \(I\) will be bunched very tightly near \(I = 0\), making it impossible to draw an accurate line of best fit.
Using a low-range variable resistor (\(0 - 10\ \Omega\)) allows the load resistance to be of a similar magnitude to the internal resistance, producing a wide and well-distributed range of current values, resulting in an accurate determination of the gradient and intercept.

(a) Circuit Diagram [3 Marks]:
- MP1: Cell, ammeter, variable resistor, and switch connected in a single series loop. [1]
- MP2: Voltmeter connected in parallel across the cell. [1]
- MP3: Correct circuit symbols used for all components (especially cell, variable resistor with diagonal arrow). [1]

(b) Experimental Procedure [3 Marks]:
- MP1: Adjust the variable resistor to vary the resistance and thereby change the current in the circuit. [1]
- MP2: Record at least 6 corresponding readings of current \(I\) and potential difference \(V\). [1]
- MP3: State that the switch must be opened between readings to avoid heating of the cell/internal resistance changing. [1]

(c) Graphical Analysis [3 Marks]:
- MP1: State the equation \(V = -rI + \varepsilon\). [1]
- MP2: Plot \(V\) on the y-axis against \(I\) on the x-axis to obtain a straight line. [1]
- MP3: State that the y-intercept represents the EMF \(\varepsilon\) and the magnitude of the gradient is equal to the internal resistance \(r\). [1]

(d) Variable Resistor Range [3 Marks]:
- MP1: Explain that a high-range variable resistor results in extremely small currents for most settings. [1]
- MP2: Explain that this causes voltmeter readings to be nearly constant (close to \(\varepsilon\)), causing data points to bunch together near the y-axis. [1]
- MP3: Explain that a low-range variable resistor (\(0 - 10\ \Omega\)) ensures that the external resistance is comparable to the internal resistance, leading to a wide, usable range of current and potential difference values. [1]

*(a) Experimental Arrangement:*
The laser is placed on a level surface, aligned to shine a horizontal beam perpendicular to the face of a diffraction grating mounted in a holder. A screen is positioned some distance away, perpendicular to the incident beam. The screen shows a bright central maximum (zeroth order, \(n = 0\)) and symmetric bright spots on both sides representing higher orders (\(n = 1\), \(n = 2\), etc.).

*(b) Safety Precaution:*
Never look directly into the laser beam or its reflection, as intense laser light can cause permanent retinal damage. Ensure there are no reflective surfaces (such as watches or jewelry) in the path of the beam, and set up a warning sign 'Laser in Use'.

*(c) Measurements and Instruments:*
1. **Distance from grating to screen, \(D\)**: Measured using a metre rule (precision \(\pm 1\text{ mm}\)) along the line perpendicular to both the grating and the screen.
2. **Distance from central maximum to first-order maximum, \(x\)**: Measured using a metre rule or digital calipers. To maximize accuracy, measure the total distance between the two first-order maxima on opposite sides of the central spot (\(2x\)) and divide by 2. This reduces the uncertainty and eliminates any potential off-center error.
3. **Calculation of \(\theta\)**: The angle \(\theta\) is calculated using the trigonometry relation:
\(\theta = \tan^{-1}\left(\frac{x}{D}\right)\)

*(d) Slit Separation \(d\):*
The number of lines per millimetre, \(N\), represents the number of slits per millimeter of grating length. Thus, the slit separation \(d\) is the distance between adjacent slits:
\(d = \frac{1\text{ mm}}{N} = \frac{10^{-3}\text{ m}}{N}\)

*(e) Improving Accuracy with Graphical Analysis:*
The grating equation is:
\(d\sin\theta = n\lambda\)
where \(n\) is the order of the diffraction maximum. Rearranging this gives:
\(\sin\theta = n\left(\frac{\lambda}{d}\right)\)
By measuring the diffraction angle \(\theta\) for multiple orders (e.g., \(n = 1\, 2\, 3\)), the student can plot a graph with \(\sin\theta\) on the y-axis against the order \(n\) on the x-axis.
- The graph should be a straight line passing through the origin.
- The gradient of the line is equal to \(\frac{\lambda}{d\)}.
- Therefore, the wavelength is calculated as:
\(\lambda = \text{gradient} \times d\)
Using multiple data points and a graph averages out random experimental errors in measuring individual distances, providing a much more reliable and accurate value for the wavelength than a single-point calculation.

(a) Diagram [3 Marks]:
- MP1: Laser, diffraction grating, and screen aligned in a straight path. [1]
- MP2: Grating aligned perpendicular to the incident laser beam, and screen showing symmetric spots around a central bright spot. [1]
- MP3: Distance \(D\) from grating to screen and distance \(x\) from central maximum to first-order maximum labeled clearly. [1]

(b) Safety Precaution [1 Mark]:
- MP1: Do not stare directly into the laser beam / avoid placing reflective objects in the beam path. [1]

(c) Measurements [4 Marks]:
- MP1: Measure the perpendicular distance \(D\) from grating to screen using a metre rule. [1]
- MP2: Measure the distance \(2x\) between the symmetric first-order maxima on both sides of the central maximum. [1]
- MP3: State that measuring \(2x\) and dividing by 2 improves accuracy / reduces centering and zero-point errors. [1]
- MP4: Calculate the angle \(\theta\) using \(\theta = \tan^{-1}(x/D)\). [1]

(d) Slit Separation [2 Marks]:
- MP1: Explain that slit separation \(d\) is the reciprocal of the number of lines per unit length: \(d = 1/N\). [1]
- MP2: Express \(d\) in metres by converting millimetres to metres, yielding \(d = \frac{10^{-3}}{N}\). [1]

(e) Graphical Analysis [3 Marks]:
- MP1: Rearrange the grating equation to \(\sin\theta = n\frac{\lambda}{d}\). [1]
- MP2: State that a graph of \(\sin\theta\) (y-axis) against order \(n\) (x-axis) is plotted. [1]
- MP3: State that the gradient is \(\frac{\lambda}{d}\), so \(\lambda = \text{gradient} \times d\), which reduces random errors. [1]

At the lowest point of the circular path, the forces acting on the sphere are the tension \(T\) pulling vertically upwards and the weight \(mg\) acting vertically downwards.

The net force acting towards the center of the circle provides the centripetal acceleration:
\(F_{\text{net}} = T - mg = \frac{mv^2}{L}\)

Rearranging for tension \(T\):
\(T = \frac{mv^2}{L} + mg\)

1 mark: Correctly identifies the centripetal force equation at the bottom of the loop and solves for tension.

Let \(v_1\) be the final velocity of the \(3m\) mass and \(v_2\) be the final velocity of the \(m\) mass.

From conservation of linear momentum:
\(3mu + 0 = 3mv_1 + mv_2 \implies 3u = 3v_1 + v_2\)

For an elastic collision, the relative speed of approach equals the relative speed of separation:
\(u = v_2 - v_1 \implies v_2 = u + v_1\)

Substituting this into the momentum equation:
\(3u = 3v_1 + (u + v_1)\)
\(2u = 4v_1 \implies v_1 = 0.50u\)

Now, solving for \(v_2\):
\(v_2 = u + 0.50u = 1.50u\)

Thus, the velocity of the mass \(3m\) is \(0.50u\) and the velocity of the mass \(m\) is \(1.50u\).

1 mark: Correct application of conservation of momentum and relative speed equations to obtain the final velocities.

The electric field between the plates is given by:
\(E = \frac{V}{d}\)

The horizontal electric force acting on the charge is:
\(F_e = qE = \frac{qV}{d}\)

The vertical force acting on the charge is its weight:
\(W = mg\)

In equilibrium, the thread tension \(T\) balances these forces:
\(T \sin\theta = F_e = \frac{qV}{d}\)
\(T \cos\theta = mg\)

Dividing these equations:
\(\tan\theta = \frac{qV}{mgd}\)

1 mark: Correct derivation of the expression for \(\tan\theta\) from the horizontal and vertical forces.

The magnetic force provides the required centripetal force:
\(B e v = \frac{m v^2}{r} \implies r = \frac{m v}{B e}\)

Since kinetic energy is \(E_k = \frac{1}{2} m v^2\), the momentum \(p\) of the electron is:
\(p = m v = \sqrt{2 m E_k}\)

Substituting \(mv\) into the radius equation:
\(r = \frac{\sqrt{2 m E_k}}{B e}\)

1 mark: Correctly relates kinetic energy and momentum, then substitutes into the circular motion equation.

The potential difference \(V\) across the discharging capacitor decays exponentially:
\(V = V_0 e^{-\frac{t}{RC}}\)

At time \(t = 2RC\):
\(V = V_0 e^{-\frac{2RC}{RC}} = V_0 e^{-2}\)

The energy stored in a capacitor is proportional to the square of its potential difference:
\(W = \frac{1}{2} C V^2\)

Substituting the expression for \(V\):
\(W = \frac{1}{2} C (V_0 e^{-2})^2 = \frac{1}{2} C V_0^2 e^{-4} = W_0 e^{-4}\)

Thus, the fraction of initial energy remaining is \(e^{-4}\).

1 mark: Uses the exponential decay equation for voltage and squares it to find the fraction of remaining energy.

Faraday's law states that the induced e.m.f. is equal to the rate of change of magnetic flux linkage:
\(\mathcal{E} = \frac{\Delta (N\Phi)}{\Delta t}\)

Initially, the plane of the coil is perpendicular to the magnetic field, so the initial flux linkage is:
\(N\Phi_i = N B A = N B s^2\)

After a \(90^\circ\) rotation, the plane of the coil is parallel to the magnetic field lines, so the final flux linkage is:
\(N\Phi_f = 0\)

The change in flux linkage is:
\(\Delta (N\Phi) = N B s^2\)

The average induced e.m.f. is:
\(\mathcal{E} = \frac{N B s^2}{\Delta t}\)

1 mark: Correctly identifies the initial and final flux linkage to calculate the rate of change of flux linkage.

By definition, a baryon is a type of hadron (a composite particle made of quarks) that consists of three quarks. Mesons consist of a quark and an antiquark. Leptons are fundamental particles.

1 mark: Identifies that a baryon is a hadron made of three quarks.

Let's check each conservation law:
1. **Baryon Number (\(B\))**:
- Before: \(1\) (neutron)
- After: \(1 + 0 + 0 = 1\) (proton, electron, electron antineutrino)
- Conserved.

2. **Lepton Number (\(L\))**:
- Before: \(0\)
- After: \(0 + 1 + (-1) = 0\)
- Conserved.

3. **Charge (\(Q\))**:
- Before: \(0\)
- After: \(+1 + (-1) + 0 = 0\)
- Conserved.

Therefore, baryon number, lepton number, and charge are all conserved.

1 mark: Correctly identifies that baryon number, lepton number, and charge are all conserved in the given decay equation.

The maximum static frictional force provides the centripetal force and is given by \( F = m \omega^2 r \). Since the mass of the coin and the coefficient of friction between the coin and the turntable are constant, the maximum static frictional force remains constant. Therefore, we can write the relation: \( \omega_1^2 r_1 = \omega_2^2 r_2 \). Given that the new angular velocity is \( \omega_2 = 3\omega \) and the initial distance is \( r_1 = r \), we substitute these into the equation: \( \omega^2 r = (3\omega)^2 r_2 \). This simplifies to \( \omega^2 r = 9\omega^2 r_2 \). Solving for the new distance gives \( r_2 = \frac{r}{9} \).

1 mark for the correct answer A. Reject all other options as they do not correctly apply the inverse-square relationship between distance and angular velocity for a constant centripetal force.

The initial charge on the capacitor is \( Q = C V \). The initial energy stored is given by \( E = \frac{Q^2}{2C} \). When connected in parallel with an uncharged capacitor of capacitance \( 2C \), the total capacitance of the system becomes \( C_{\text{total}} = C + 2C = 3C \). Since the system is isolated from any external power supply, the total charge \( Q \) is conserved and remains the same. The new total energy stored in the system is \( E_{\text{final}} = \frac{Q^2}{2 C_{\text{total}}} = \frac{Q^2}{2(3C)} = \frac{1}{3} \left( \frac{Q^2}{2C} \right) = \frac{E}{3} \).

1 mark for the correct answer B. Reject other options as they either assume potential difference is conserved (which would lead to incorrect energy calculations) or apply incorrect scaling factors.

(a) For a circular orbit, the gravitational force provides the centripetal force:

\(\frac{G M m}{r^2} = \frac{m v^2}{r}\)

Rearranging for orbital speed \(v\):

\(v = \sqrt{\frac{G M}{r}}\)

Substitute the given values into the equation:

\(v = \sqrt{\frac{6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2} \times 4.8 \times 10^{23}\text{ kg}}{1.2 \times 10^7\text{ m}}}\)

\(v = \sqrt{\frac{3.2016 \times 10^{13}}{1.2 \times 10^7}} = \sqrt{2.668 \times 10^6}\)

\(v \approx 1.63 \times 10^3\text{ m s}^{-1}\) (or \(1630\text{ m s}^{-1}\))

(b) The thruster is not necessary to maintain this circular path. Gravity acts perpendicular to the velocity of the probe. It provides the required centripetal acceleration to continuously change the direction of motion without changing the speed. No work is done by the gravitational force, so no fuel consumption or thrust is needed to stay in this stable orbit.

Part (a) [4 Marks]:
- [M1] Use of \(F_g = F_c\) or \(\frac{G M m}{r^2} = \frac{m v^2}{r}\) leading to \(v = \sqrt{\frac{G M}{r}}\).
- [M1] Correct substitution of \(G\), \(M\), and \(r\).
- [A1] Correct calculation of speed value (\(1.63 \times 10^3\text{ m s}^{-1}\) or \(1630\text{ m s}^{-1}\)).
- [A1] Correct units (\ ext{m s}^{-1}\) and appropriate 2 or 3 sig figs.

Part (b) [2.67 Marks]:
- [M1] Explains that gravity is perpendicular to the direction of motion / velocity.
- [A1] Explains that gravity provides the full centripetal acceleration / force.
- [A0.67] Concludes that no work is done on the probe (so no thrust is needed to maintain this constant speed orbit).

By conservation of linear momentum, both horizontal (x-axis) and vertical (y-axis) momentum must be conserved.

Initial momentum of the system:
\(p_{ix} = m_p v_{pi} = (1.67 \times 10^{-27}\text{ kg}) \times (3.50 \times 10^5\text{ m s}^{-1}) = 5.845 \times 10^{-22}\text{ kg m s}^{-1}\)
\(p_{iy} = 0\)

Final momentum of the proton:
\(p_{px} = m_p v_{pf} \cos(60^\circ) = (1.67 \times 10^{-27}\text{ kg}) \times (2.10 \times 10^5\text{ m s}^{-1}) \times 0.5 = 1.7535 \times 10^{-22}\text{ kg m s}^{-1}\)
\(p_{py} = m_p v_{pf} \sin(60^\circ) = (1.67 \times 10^{-27}\text{ kg}) \times (2.10 \times 10^5\text{ m s}^{-1}) \times \sin(60^\circ) = 3.0371 \times 10^{-22}\text{ kg m s}^{-1}\)

By conservation of momentum, for the target nucleus (t):
\(p_{tx} = p_{ix} - p_{px} = (5.845 - 1.7535) \times 10^{-22}\text{ kg m s}^{-1} = 4.0915 \times 10^{-22}\text{ kg m s}^{-1}\)
\(p_{ty} = p_{iy} - p_{py} = -3.0371 \times 10^{-22}\text{ kg m s}^{-1}\)

Calculate the total magnitude of the target nucleus's momentum:
\(p_t = \sqrt{p_{tx}^2 + p_{ty}^2} = \sqrt{(4.0915 \times 10^{-22})^2 + (-3.0371 \times 10^{-22})^2}\)
\(p_t = 10^{-22} \times \sqrt{16.740 + 9.224} = 10^{-22} \times \sqrt{25.964} = 5.0955 \times 10^{-22}\text{ kg m s}^{-1}\)

Now, calculate the final velocity of the target nucleus:
\(v_t = \frac{p_t}{m_t} = \frac{5.0955 \times 10^{-22}\text{ kg m s}^{-1}}{1.13 \times 10^{-26}\text{ kg}} \approx 4.51 \times 10^4\text{ m s}^{-1}\)

- [M1] Statement of conservation of momentum in both x and y directions.
- [M1] Correct calculation of initial x-momentum (\(5.85 \times 10^{-22}\text{ kg m s}^{-1}\)).
- [M1] Correct components of final proton momentum (\(p_{px} = 1.75 \times 10^{-22}\text{ kg m s}^{-1}\) and \(p_{py} = 3.04 \times 10^{-22}\text{ kg m s}^{-1}\)).
- [M1] Calculation of target nucleus momentum components (\(p_{tx} = 4.09 \times 10^{-22}\text{ kg m s}^{-1}\) and \(p_{ty} = -3.04 \times 10^{-22}\text{ kg m s}^{-1}\)).
- [A1] Correct magnitude of target nucleus momentum (\(5.10 \times 10^{-22}\text{ kg m s}^{-1}\)).
- [A1.67] Correct final speed of target nucleus with appropriate unit (\(4.51 \times 10^4\text{ m s}^{-1}\)).

(a) Calculate the electric field strength \(E\):
\(E = \frac{V}{d} = \frac{850\text{ V}}{0.018\text{ m}} \approx 47222\text{ V m}^{-1}\)

For the sphere to be in equilibrium, the upward electric force must balance the downward gravitational force:
\(F_E = F_g \implies q E = m g\)

\(q = \frac{m g}{E} = \frac{4.6 \times 10^{-14}\text{ kg} \times 9.81\text{ m s}^{-2}}{47222\text{ V m}^{-1}} \approx 9.56 \times 10^{-18}\text{ C}\)

Since the upper plate is positive, to produce an upward force (towards the positive plate), the sphere must carry a negative charge. Thus:
\(q = -9.56 \times 10^{-18}\text{ C}\)

(b) If the potential difference \(V\) is increased, the electric field strength \(E\) increases because \(E = V/d\). Since the charge \(q\) remains constant, the upward electric force \(F_E = q E\) will increase and exceed the downward gravitational force. Consequently, there will be a net upward force on the sphere, causing it to accelerate upwards.

Part (a) [4.67 Marks]:
- [M1] Use of \(E = V/d\) to find electric field (\(4.72 \times 10^4\text{ V m}^{-1}\)).
- [M1] Recognises equilibrium condition: \(q E = m g\).
- [M1] Rearranging and substituting values correctly.
- [A1] Correct numerical magnitude of charge (\(9.56 \times 10^{-18}\text{ C}\)).
- [A0.67] Correctly identifies that the sign is negative with justification (attracted to the positive upper plate).

Part (b) [2 Marks]:
- [M1] Explains that increased \(V\) increases \(E\), causing \(F_E\) to exceed gravity.
- [A1] Concludes that there is a net upward force leading to upward acceleration.

(a) The time constant \(\tau\) is given by:
\(\tau = R C = (15 \times 10^3\ \Omega) \times (470 \times 10^{-6}\text{ F}) = 7.05\text{ s}\)

(b) First, calculate the initial charge \(Q_0\):
\(Q_0 = C V = (470 \times 10^{-6}\text{ F}) \times 9.0\text{ V} = 4.23 \times 10^{-3}\text{ C}\)

The discharging equation for charge is:
\(Q = Q_0 e^{-t/\tau}\)

For \(t = 2\tau\):
\(Q = Q_0 e^{-2} = 4.23 \times 10^{-3}\text{ C} \times e^{-2} \approx 4.23 \times 10^{-3} \times 0.1353\)
\(Q \approx 5.72 \times 10^{-4}\text{ C}\) (or \(0.572\text{ mC}\))

(c) When capacitors are connected in parallel, their capacitances add:
\(C_{\text{total}} = C_1 + C_2 = 470\ \mu\text{F} + 470\ \mu\text{F} = 940\ \mu\text{F}\)

Since \(\tau = R C\), and the total capacitance has doubled while the resistance remains constant, the new time constant will double to \(14.1\text{ s}\).

Part (a) [2 Marks]:
- [M1] Use of \(\tau = R C\) with correct unit conversion.
- [A1] Correct answer of \(7.05\text{ s}\).

Part (b) [2.67 Marks]:
- [M1] Calculation of initial charge \(Q_0 = C V = 4.23 \times 10^{-3}\text{ C}\).
- [M1] Use of \(Q = Q_0 e^{-t/\tau}\) with \(t = 2\tau\) (recognising it reduces to \(Q_0 e^{-2}\)).
- [A0.67] Correct final charge (\(5.72 \times 10^{-4}\text{ C}\) or \(0.57\text{ mC}\)).

Part (c) [2 Marks]:
- [M1] Explains that parallel connection doubles the capacitance.
- [A1] States that since \(\tau \propto C\), the time constant will double (to \(14.1\text{ s}\)).

(a) First, calculate the cross-sectional area \(A\) of the circular coil:
\(A = \pi r^2 = \pi \times (0.025\text{ m})^2 \approx 1.963 \times 10^{-3}\text{ m}^2\)

Initial flux linkage through the coil when perpendicular:
\(\Phi_{\text{initial}} = N B A = 150 \times 0.18\text{ T} \times 1.963 \times 10^{-3}\text{ m}^2 \approx 0.0530\text{ Wb-turns}\)

Final flux linkage when parallel to the magnetic field is zero:
\(\Phi_{\text{final}} = 0\)

Change in flux linkage:
\(\Delta \Phi = 0.0530\text{ Wb-turns}\)

According to Faraday's law, the average induced e.m.f. is:
\(\varepsilon = \frac{\Delta \Phi}{\Delta t} = \frac{0.0530\text{ Wb-turns}}{0.12\text{ s}} \approx 0.442\text{ V}\) (or \(0.44\text{ V}\))

(b) Lenz's law states that the direction of an induced e.m.f. is such that it will oppose the change in magnetic flux that produced it.

During rotation, the magnetic flux passing through the coil is decreasing. Therefore, the induced current in the coil flows in a direction that creates its own magnetic field through the coil in the same direction as the external field, attempting to maintain the magnetic flux and oppose the decrease.

Part (a) [4 Marks]:
- [M1] Calculation of area \(A = \pi r^2 = 1.96 \times 10^{-3}\text{ m}^2\).
- [M1] Calculation of initial magnetic flux linkage \(N B A = 0.0530\text{ Wb}\).
- [M1] Use of Faraday's Law: \(\varepsilon = \Delta(NBA)/\Delta t\).
- [A1] Correct value of induced e.m.f. (\(0.44\text{ V}\)).

Part (b) [2.67 Marks]:
- [M1] Accurate statement of Lenz's law (direction of induced e.m.f./current opposes the change in flux).
- [M1] Identification that flux through the coil is decreasing during rotation.
- [A0.67] Explains that the induced current's magnetic field acts in the same direction as the original field to oppose this decrease.

(a) An electron of charge \(e\) accelerated through potential difference \(V\) gains kinetic energy \(E_k\):
\(E_k = e V\)

The kinetic energy is also related to momentum \(p\) and mass \(m_e\) by:
\(E_k = \frac{p^2}{2 m_e}\)

Equating these two expressions:
\(e V = \frac{p^2}{2 m_e} \implies p = \sqrt{2 m_e e V}\)

According to de Broglie's wave-particle duality relation:
\(\lambda = \frac{h}{p}\)

Substituting the expression for momentum \(p\) gives the required relation:
\(\lambda = \frac{h}{\sqrt{2 m_e e V}}\)

(b) Substitute the values into the derived equation:
\(h = 6.63 \times 10^{-34}\text{ J s}\)
\(m_e = 9.11 \times 10^{-31}\text{ kg}\)
\(e = 1.60 \times 10^{-19}\text{ C}\)
\(V = 2.5\text{ kV} = 2500\text{ V}\)

\(\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times (9.11 \times 10^{-31}) \times (1.60 \times 10^{-19}) \times 2500}}\)

\(\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{7.288 \times 10^{-46}}} = \frac{6.63 \times 10^{-34}}{2.70 \times 10^{-23}} \approx 2.46 \times 10^{-11}\text{ m}\) (or \(25\text{ pm}\))

Part (a) [3.67 Marks]:
- [M1] Link work done/electrical potential energy to kinetic energy: \(E_k = eV\).
- [M1] Link kinetic energy to momentum: \(E_k = p^2 / 2m_e\).
- [M1] State de Broglie's relation: \(\lambda = h / p\).
- [A0.67] Algebra showing clear and complete steps to reach the final expression.

Part (b) [3 Marks]:
- [M1] Identifies \(V = 2500\text{ V}\) (converts kV to V correctly).
- [M1] Correct substitution of physics constants (\(h, m_e, e\)) into the equation.
- [A1] Correct final answer with units (\(2.46 \times 10^{-11}\text{ m}\) or \(2.5 \times 10^{-11}\text{ m}\)).

(a) For a charged particle moving perpendicular to a magnetic field, the magnetic force provides the centripetal force:
\(B q v = \frac{m v^2}{r} \implies p = m v = B q r\)

Substitute the values for the electron:
\(p = 0.35\text{ T} \times (1.60 \times 10^{-19}\text{ C}) \times 0.082\text{ m}\)
\(p \approx 4.59 \times 10^{-21}\text{ kg m s}^{-1}\)

(b) The magnetic force equation is \(F = B q v\). The electron (\(e^-\)) and positron (\(e^+\)) have opposite signs of charge. Thus, the magnetic force acts in opposite directions for each particle, causing opposite directions of curvature.
Since the particles are produced with identical initial speeds (due to symmetry in pair production), have identical masses (same momentum magnitude \(p\)), and have the same magnitude of charge \(q = e\), they experience the same magnitude of centripetal force. Thus, the initial radius of curvature \(r = \frac{p}{B q}\) is identical.

(c) As the particles travel through the chamber, they collide with atoms in the bubble chamber medium, causing ionization. During these collisions, the particles lose kinetic energy and momentum. Since \(r = \frac{p}{B q}\), as momentum \(p\) decreases while \(B\) and \(q\) remain constant, the radius of curvature \(r\) must decrease, resulting in spiral-shaped tracks.

Part (a) [2 Marks]:
- [M1] Uses \(p = B q r\).
- [A1] Correct calculation of momentum (\(4.59 \times 10^{-21}\text{ kg m s}^{-1}\)).

Part (b) [2.67 Marks]:
- [M1] Explains that opposite charges experience opposite magnetic forces, causing opposite deflections.
- [M1] States that the radius depends on momentum and charge magnitude (\(r = p/Bq\)).
- [A0.67] Concludes that equal initial momentum and equal charge magnitude lead to equal initial radii.

Part (c) [2 Marks]:
- [M1] Explains that particles lose energy/momentum due to collisions or ionization with atoms.
- [A1] Links this momentum loss directly to a decreasing radius using \(r = p/Bq\).

(a) Since the neutral pion is stationary, its total energy is its rest energy:
\(E_{\text{total}} = 135\text{ MeV}\)

By conservation of momentum, since the initial momentum of the stationary pion is zero, the two photons must travel in opposite directions with equal and opposite momentum. By conservation of energy, the total energy is split equally between the two identical photons:
\(E_{\text{photon}} = \frac{135\text{ MeV}}{2} = 67.5\text{ MeV}\)

Convert this energy to Joules:
\(E_{\text{photon}} = 67.5 \times 10^6\text{ eV} \times (1.60 \times 10^{-19}\text{ J / eV}) = 1.08 \times 10^{-11}\text{ J}\)

Now, use \(E = h f\) to find the frequency:
\(f = \frac{E_{\text{photon}}}{h} = \frac{1.08 \times 10^{-11}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} \approx 1.63 \times 10^{22}\text{ Hz}\)

(b) A photon always carries momentum given by \(p = \frac{E}{c}\). If a single photon were produced, the final state would have non-zero momentum. However, the initial momentum of the stationary pion is zero. Since momentum must be conserved in all interactions, a single-photon decay is impossible as it would violate the conservation of momentum. At least two photons traveling in opposite directions are required so that their vector sum of momentum can be zero.

Part (a) [4 Marks]:
- [M1] Identifies that energy is split equally: each photon has \(67.5\text{ MeV}\) of energy.
- [M1] Converts energy from MeV to Joules correctly (\(1.08 \times 10^{-11}\text{ J}\)).
- [M1] Uses \(E = h f\).
- [A1] Correct calculation of frequency (\(1.63 \times 10^{22}\text{ Hz}\)).

Part (b) [2.67 Marks]:
- [M1] Identifies that a photon must carry linear momentum (\(p = E/c\)).
- [M1] States that the initial momentum of the stationary pion is zero.
- [A0.67] Explains that single-photon decay is impossible because it would violate the conservation of momentum (whereas two photons can cancel momentum).

(a) The two forces acting on the sphere at the top of its path are:
- Weight (gravitational force) acting vertically downwards.
- Tension (of the string) acting downwards (towards the center of the circle).

(b) At the top of the circle, both the weight \(mg\) and the tension \(T\) act downwards, providing the required centripetal force:
\[F_c = T + mg = \frac{mv^2}{r}\]
Rearranging to solve for tension:
\[T = \frac{mv^2}{r} - mg\]
Substitute the values:
\[T = \frac{0.15\text{ kg} \times (3.4\text{ m s}^{-1})^2}{0.85\text{ m}} - 0.15\text{ kg} \times 9.81\text{ m s}^{-2}\]
\[T = 2.04\text{ N} - 1.47\text{ N} = 0.57\text{ N}\]

(c) For minimum speed, the tension in the string becomes zero \((T = 0)\) at the top of the loop.
\[mg = \frac{mv_{\text{min}}^2}{r}\]
\[v_{\text{min}} = \sqrt{gr} = \sqrt{9.81\text{ m s}^{-2} \times 0.85\text{ m}} = 2.89\text{ m s}^{-1}\]

Part (a):
- Weight directed downwards [1 mark]
- Tension directed downwards / towards the center [1 mark]

Part (b):
- Correct equation relating forces to centripetal force: \(T + mg = \frac{mv^2}{r}\) [1 mark]
- Correct substitution of values [1 mark]
- Final tension value \(0.57\text{ N}\) [1 mark] (accept \(0.569\text{ N}\))

Part (c):
- Identification that \(T = 0\) for minimum speed [0.67 marks]
- Correct calculation of minimum speed: \(2.9\text{ m s}^{-1}\) [1 mark] (accept \(2.89\text{ m s}^{-1}\))

(a) The time constant \(\tau\) is given by:
\[\tau = RC\]
\[\tau = (22 \times 10^3\ \Omega) \times (470 \times 10^{-6}\text{ F}) = 10.34\text{ s}\]
This rounds to \(10.3\text{ s}\) (or \(10\text{ s}\) to 2 s.f.).

(b) The initial charge \(Q_0\) on the capacitor is:
\[Q_0 = C V_0 = (470 \times 10^{-6}\text{ F}) \times 12.0\text{ V} = 5.64 \times 10^{-3}\text{ C}\]

After a time \(t = 2\tau = 2RC\), the charge remaining is:
\[Q = Q_0 e^{-t/RC} = Q_0 e^{-2}\]
\[Q = 5.64 \times 10^{-3}\text{ C} \times e^{-2} = 5.64 \times 10^{-3}\text{ C} \times 0.1353 = 7.63 \times 10^{-4}\text{ C}\]

(c) Connecting an identical resistor in parallel halves the total equivalent resistance \(R_{\text{eq}} = R/2\). Since \(\tau = R_{\text{eq}}C\), halving the resistance will halve the time constant, making it \(5.17\text{ s}\).

Part (a):
- Use of \(\tau = RC\) [1 mark]
- Correct calculation to yield \(10.3\text{ s}\) (or \(10\text{ s}\)) with unit [1 mark]

Part (b):
- Correct calculation of initial charge \(Q_0 = 5.64 \times 10^{-3}\text{ C}\) [1 mark]
- Correct use of exponential decay equation with \(t/\tau = 2\) [1 mark]
- Correct value for charge remaining: \(7.6 \times 10^{-4}\text{ C}\) [1 mark] (accept \(7.63 \times 10^{-4}\text{ C}\))

Part (c):
- Explanation that parallel resistors halve the equivalent resistance [1 mark]
- Conclusion that the time constant halves (or decreases to \(5.2\text{ s}\)) [0.67 marks]

(a) The bubble chamber works by ionizing atoms along the paths of charged particles, which act as nucleation centers for bubble formation. The gamma photon is neutral (uncharged) and does not cause direct ionization, leaving no track. The electron and positron are charged, causing ionization and thus leaving visible tracks.

(b) The centripetal force is provided by the magnetic force on the charged particle:
\[\frac{mv^2}{r} = Bqv\]
Rearranging to find the momentum \(p = mv\):
\[p = Bqr\]
Substitute the values:
\[p = 0.12\text{ T} \times (1.60 \times 10^{-19}\text{ C}) \times 0.045\text{ m}\]
\[p = 8.64 \times 10^{-22}\text{ kg m s}^{-1}\]

(c) As the electron travels through the chamber, it collides with atoms, causing ionization and losing kinetic energy. As its velocity and momentum decrease, and since \(r = \frac{p}{Bq}\) where \(B\) and \(q\) are constant, the radius of curvature \(r\) decreases, creating a spiral shape.

Part (a):
- Statement that bubble formation requires charged particles to ionize atoms [1 mark]
- Explanation that the gamma photon is neutral while the electron/positron are charged [1 mark]

Part (b):
- Recalling or deriving \(p = Bqr\) [1 mark]
- Correct conversion of radius \(4.5\text{ cm}\) to \(0.045\text{ m}\) [1 mark]
- Correct calculation of momentum: \(8.6 \times 10^{-22}\text{ kg m s}^{-1}\) (or \(\text{N s}\)) [1 mark] (accept \(8.64 \times 10^{-22}\))

Part (c):
- Identification that the electron loses kinetic energy/momentum due to collisions or ionization [1 mark]
- Linking lower momentum to smaller radius using \(r = \frac{p}{Bq}\) [0.67 marks]

(a) Protons accelerate in the gaps between the drift tubes. Consequently, they travel faster inside each successive tube. To ensure that they arrive at each gap in phase with the alternating voltage (which changes direction at fixed intervals of time), the time spent inside each drift tube must remain constant. Because \(s = vt\), as speed \(v\) increases, the tube length \(s\) must increase to keep the travel time constant.

(b) The proton must spend half a period \(T/2\) of the AC source inside each tube so that the electric field reverses when the proton emerges into the gap:
\[T = \frac{1}{f} = \frac{1}{25 \times 10^6\text{ Hz}} = 4.0 \times 10^{-8}\text{ s}\]
\[\text{Time inside tube } t = \frac{T}{2} = 2.0 \times 10^{-8}\text{ s}\ (\text{or } 20\text{ ns})\]

(c) First, calculate the velocity assuming non-relativistic conditions:
\[E_k = 12\text{ MeV} = 12 \times 10^6 \times 1.60 \times 10^{-19}\text{ J} = 1.92 \times 10^{-12}\text{ J}\]
\[E_k = \frac{1}{2}m_p v^2 \implies v = \sqrt{\frac{2E_k}{m_p}}\]
\[v = \sqrt{\frac{2 \times 1.92 \times 10^{-12}\text{ J}}{1.67 \times 10^{-27}\text{ kg}}} = 4.79 \times 10^7\text{ m s}^{-1}\]

To determine if it is relativistic, we compare this speed to the speed of light \(c = 3.00 \times 10^8\text{ m s}^{-1}\):
\[\frac{v}{c} = \frac{4.79 \times 10^7}{3.00 \times 10^8} \approx 0.16\]
Since \(v\) is only about \(16\%\) of the speed of light, relativistic effects are small. Alternatively, comparing the kinetic energy (\(12\text{ MeV}\)) to the proton's rest energy (\(m_p c^2 \approx 938\text{ MeV}\)) shows \(12\text{ MeV} \ll 938\text{ MeV}\), confirming that treating the proton non-relativistically is a reasonable approximation.

Part (a):
- Statement that protons speed up in the gaps [1 mark]
- Explanation that the AC period is constant, so tubes must be longer to keep the time spent in each tube constant (or to remain in phase) [1 mark]

Part (b):
- Recalling that \(t = \frac{T}{2} = \frac{1}{2f}\) [1 mark]
- Correct calculation yielding \(2.0 \times 10^{-8}\text{ s}\) (or \(20\text{ ns}\)) [1 mark]

Part (c):
- Conversion of kinetic energy to Joules: \(1.92 \times 10^{-12}\text{ J}\) [1 mark]
- Correct calculation of speed: \(4.8 \times 10^7\text{ m s}^{-1}\) [1 mark]
- Valid comparison (comparing speed to \(c\) or kinetic energy to rest energy \(\approx 938\text{ MeV}\)) concluding it is non-relativistic [0.67 marks]

The average kinetic energy of the molecules in an ideal gas is directly proportional to the absolute temperature, \(E_k = \frac{3}{2}kT = \frac{1}{2}m\langle c^2 \rangle\). Therefore, the mean square speed \(\langle c^2 \rangle\) is directly proportional to the absolute temperature \(T\). When the absolute temperature is doubled, the mean square speed must also double. (Note: The root mean square speed, \(\sqrt{\langle c^2 \rangle}\), would increase by a factor of \(\sqrt{2}\)).

1 mark for selecting option B. Reject options referencing the square root factor (A) or square factor (C) as they apply to root mean square speed or are incorrect.

Using the ideal gas equation \(pV = NkT\), since the volume \(V\) of the cylinder is fixed, the pressure is proportional to the product of the number of molecules and the absolute temperature, \(p \propto NT\). The initial pressure is \(p_i = C \cdot N T\) where \(C\) is a constant. The final pressure is \(p_f = C \cdot \left(\frac{1}{2}N\right) \cdot (1.5T) = 0.75 \cdot C \cdot N T = 0.75 p\).

1 mark for selecting option B. Correctly applies the ideal gas law with fixed volume to calculate the proportional change in pressure.

The total energy of the simple harmonic oscillator is given by \(E_T = E_k + E_p = \frac{1}{2} k A^2\). The potential energy is given by \(E_p = \frac{1}{2} k x^2\). Since \(E_k = 3 E_p\), the total energy is \(3 E_p + E_p = 4 E_p = \frac{1}{2} k A^2\). Substituting the expression for \(E_p\) yields \(4 \left(\frac{1}{2} k x^2\right) = \frac{1}{2} k A^2 \implies 4 x^2 = A^2 \implies x = 0.50 A\).

1 mark for selecting option B. Correctly relates total energy, potential energy, and kinetic energy to solve for displacement.

For light damping, the amplitude of the oscillations decreases exponentially over time. The time period and frequency of the oscillations remain approximately constant, while the total mechanical energy decreases as it is dissipated as thermal energy.

1 mark for selecting option A. Reject B, C, and D as they describe incorrect physical behaviors of a lightly damped oscillator.

Let the initial number of X nuclei be \(N_0\). After 3 half-lives, the number of remaining X nuclei is \(N_X = N_0 \cdot \left(\frac{1}{2}\right)^3 = \frac{1}{8} N_0\). Since Y is the stable decay product, the number of Y nuclei formed is \(N_Y = N_0 - N_X = \frac{7}{8} N_0\). The ratio of Y to X is \(\frac{N_Y}{N_X} = \frac{7/8}{1/8} = 7\).

1 mark for selecting option B. Calculates the remaining fraction of parent nuclei and the produced fraction of daughter nuclei to determine the correct ratio.

The binding energy is the energy equivalent of the mass defect. The mass defect \(\Delta m\) is the difference between the total mass of the individual constituent nucleons and the mass of the nucleus: \(\Delta m = (2m_p + 2m_n) - m_{\alpha}\). Therefore, the binding energy is \(E_B = (2m_p + 2m_n - m_{\alpha})c^2\).

1 mark for selecting option A. Identify the correct algebraic expression for the mass defect multiplied by \(c^2\).

Using Wien's displacement law, peak wavelength is inversely proportional to temperature: \(T \propto \frac{1}{\lambda}\). According to the Stefan-Boltzmann law, luminosity is \(L = 4\pi R^2 \sigma T^4\). Since \(L_P = L_Q\), we have \(R_P^2 T_P^4 = R_Q^2 T_Q^4 \implies \left(\frac{R_P}{R_Q}\right)^2 = \left(\frac{T_Q}{T_P}\right)^4 = \left(\frac{\lambda_P}{\lambda_Q}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}\). Taking the square root gives \(\frac{R_P}{R_Q} = 0.25\).

1 mark for selecting option A. Relates radius to temperature and wavelength using the combined laws to find the correct ratio.

First, calculate the recession velocity \(v\) using the redshift relation: \(v = z c = 0.040 \times 3.0 \times 10^5\text{ km s}^{-1} = 12000\text{ km s}^{-1}\). Using Hubble's Law \(v = H_0 d\), the distance is \(d = \frac{v}{H_0} = \frac{12000}{70} \approx 171\text{ Mpc}\), which is closest to \(170\text{ Mpc}\).

1 mark for selecting option C. Calculates velocity from redshift, then applies Hubble's law to solve for distance.

The root-mean-square speed of an ideal gas molecule is given by \(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\). This shows that \(v_{\text{rms}} \propto \sqrt{T}\), where \(T\) is the absolute temperature of the gas. When the gas is heated at constant volume until its absolute temperature is doubled, the root-mean-square speed becomes \(\sqrt{2} v\). When the gas then undergoes an isothermal expansion, the temperature remains constant. Since the temperature does not change during an isothermal process, the root-mean-square speed of the molecules remains unchanged at \(\sqrt{2}v\).

1 mark for correct answer (A).

According to Wien's displacement law, the absolute temperature \(T\) of a star is inversely proportional to its peak wavelength \(\lambda\), so \(T \propto \frac{1}{\lambda}\). Since star Y has a peak wavelength of \(2\lambda\), its temperature is \(T_Y = 0.5 T_X\). According to the Stefan-Boltzmann law, the luminosity of a star is \(L = 4\pi R^2 \sigma T^4\), so \(L \propto R^2 T^4\). For star Y, the radius is \(2R_X\) and the temperature is \(0.5 T_X\), so its luminosity is proportional to \((2R_X)^2 (0.5 T_X)^4 = 4 R_X^2 \times 0.0625 T_X^4 = 0.25 R_X^2 T_X^4\). Therefore, the luminosity of star Y is \(0.25 L\).

1 mark for correct answer (A).

(a) Using the ideal gas equation \(PV = nRT\):
\(n = \frac{P_1 V}{R T_1} = \frac{1.5 \times 10^5\text{ Pa} \times 0.025\text{ m}^3}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 300\text{ K}} = 1.504\text{ mol} \approx 1.50\text{ mol}\)

(b) Since the volume is constant, \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\):
\(T_2 = T_1 \times \frac{P_2}{P_1} = 300\text{ K} \times \frac{2.4 \times 10^5\text{ Pa}}{1.5 \times 10^5\text{ Pa}} = 480\text{ K}\)

The average kinetic energy of an atom is:
\(E_k = \frac{3}{2} k_B T_2 = 1.5 \times (1.38 \times 10^{-23}\text{ J K}^{-1}) \times 480\text{ K} = 9.94 \times 10^{-21}\text{ J}\)

(c) An increase in temperature increases the mean square speed (and average kinetic energy) of the gas molecules. This results in more frequent collisions per second with the container walls and a greater change in momentum per collision. Consequently, the average rate of change of momentum (force) exerted on the walls increases, which increases the pressure.

(a)
- Use of \(PV = nRT\) (1)
- Correct answer to 2 or 3 s.f. [1.5 or 1.50 mol] (1)

(b)
- Use of pressure-temperature relationship to find final temperature of 480 K (1)
- Use of \(E_k = \frac{3}{2} k_B T\) (1)
- Correct calculation of energy [\(9.9 \times 10^{-21}\text{ J}\)] (1)

(c)
- Reference to higher average speed/kinetic energy of molecules (1)
- Reference to more frequent collisions with the walls OR greater change in momentum per collision (1)

(a) According to Hooke's Law, the restoring force is \(F = -kx\). By Newton's second law, \(F = ma\), so \(ma = -kx\), which gives \(a = -\frac{k}{m}x\). Since \(\frac{k}{m}\) is a positive constant, the acceleration \(a\) is directly proportional to displacement \(x\) and acts in the opposite direction (towards the equilibrium position). This defines simple harmonic motion.

(b) The angular frequency is:
\(\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{80\text{ N m}^{-1}}{0.45\text{ kg}}} = 13.33\text{ rad s}^{-1}\)

Maximum speed:
\(v_{\text{max}} = \omega A = 13.33\text{ rad s}^{-1} \times 0.080\text{ m} = 1.07\text{ m s}^{-1}\)

(c) The total energy is \(E_{\text{total}} = \frac{1}{2} k A^2 = 0.5 \times 80 \times 0.080^2 = 0.256\text{ J}\).

The potential energy at \(x = 0.040\text{ m}\) is:
\(E_p = \frac{1}{2} k x^2 = 0.5 \times 80 \times 0.040^2 = 0.064\text{ J}\)

Therefore, the kinetic energy is:
\(E_k = E_{\text{total}} - E_p = 0.256 - 0.064 = 0.192\text{ J}\)

(a)
- Expression of restoring force and Newton's second law showing \(a = -\frac{k}{m}x\) (1)
- Explanation stating acceleration is proportional to displacement and in the opposite direction (1)

(b)
- Calculation of angular frequency \(\omega = 13.3\text{ rad s}^{-1}\) (1)
- Calculation of maximum speed [1.07 m s\(^{-1}\)] (1)
- Correct unit included (1)

(c)
- Use of total energy formula OR kinetic energy expression \(E_k = \frac{1}{2} k (A^2 - x^2)\) (1)
- Correct calculation [0.19 J or 0.192 J] (1)

(a) First convert half-life to seconds:
\(t_{1/2} = 5730\text{ years} \times 365\text{ days/year} \times 24\text{ hours/day} \times 3600\text{ seconds/hour} = 1.807 \times 10^{11}\text{ s}\)

Decay constant:
\(\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.69315}{1.807 \times 10^{11}\text{ s}} = 3.836 \times 10^{-12}\text{ s}^{-1} \approx 3.84 \times 10^{-12}\text{ s}^{-1}\)

(b) Using the decay equation \(A = A_0 e^{-\lambda t}\):
\(\frac{A}{A_0} = e^{-\lambda t} \implies \ln\left(\frac{A}{A_0}\right) = -\lambda t\)

Using decay constant in years to keep the units direct:
\(\lambda_{\text{year}} = \frac{\ln 2}{5730\text{ years}} = 1.2097 \times 10^{-4}\text{ year}^{-1}\)

\(\ln\left(\frac{3.5}{16.0}\right) = -1.5198\)

\(t = \frac{-1.5198}{-1.2097 \times 10^{-4}\text{ year}^{-1}} = 12564\text{ years} \approx 1.26 \times 10^4\text{ years}\)

(a)
- Conversion of half-life in years to seconds (1)
- Use of \(\lambda = \frac{\ln 2}{t_{1/2}}\) (1)
- Correct calculation [\(3.8 \times 10^{-12}\text{ s}^{-1}\) to \(3.84 \times 10^{-12}\text{ s}^{-1}\)] (1)

(b)
- Use of \(A = A_0 e^{-\lambda t}\) (1)
- Mathematical manipulation using natural logs (1)
- Correct substitution of values (1)
- Final value [\(12600\text{ years}\) or \(1.26 \times 10^4\text{ years}\)] (1)

(a) Using Wien's displacement law:
\(\lambda_{\text{max}} T = 2.89 \times 10^{-3}\text{ m K}\)

\(T = \frac{2.89 \times 10^{-3}\text{ m K}}{135 \times 10^{-9}\text{ m}} = 21407\text{ K} \approx 2.14 \times 10^4\text{ K}\) (or \(2.1 \times 10^4\text{ K}\))

(b) Using the Stefan-Boltzmann law:
\(L = 4 \pi R^2 \sigma T^4\)

First find the star's luminosity:
\(L = 2.0 \times 10^4 \times 3.8 \times 10^{26}\text{ W} = 7.6 \times 10^{30}\text{ W}\)

Now rearrange for \(R\):
\(R = \sqrt{\frac{L}{4 \pi \sigma T^4}}\)

\(R = \sqrt{\frac{7.6 \times 10^{30}\text{ W}}{4 \pi \times (5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}) \times (21407\text{ K})^4}}\)

\(T^4 = (21407)^4 = 2.099 \times 10^{17}\text{ K}^4\)

\(R = \sqrt{\frac{7.6 \times 10^{30}}{1.496 \times 10^{11}}} = \sqrt{5.080 \times 10^{19}} = 7.13 \times 10^9\text{ m}\)

(a)
- Use of \(\lambda_{\text{max}} T = 2.89 \times 10^{-3}\text{ m K}\) (1)
- Correct calculated temperature [\(2.14 \times 10^4\text{ K}\) or \(2.1 \times 10^4\text{ K}\)] (1)

(b)
- Correct calculation of the star's luminosity (1)
- Use of Stefan-Boltzmann equation \(L = 4 \pi R^2 \sigma T^4\) (1)
- Rearranging equation for \(R\) (1)
- Correct calculation of \(T^4\) (1)
- Correct final radius [\(7.13 \times 10^9\text{ m}\) or \(7.1 \times 10^9\text{ m}\)] (1)

(a) Thermal energy supplied by the heater:
\(E = P \times t = 120\text{ W} \times (5.0 \times 60\text{ s}) = 36000\text{ J}\)

Change in temperature:
\(\Delta T = 65^\circ\text{C} - 20^\circ\text{C} = 45\text{ K}\)

Using \(E = mc\Delta T\):
\(c = \frac{E}{m\Delta T} = \frac{36000\text{ J}}{0.50\text{ kg} \times 45\text{ K}} = 1600\text{ J kg}^{-1}\text{ K}^{-1}\)

(b) If heat is lost to the surroundings, the actual energy absorbed by the metal block is less than the calculated value of \(36000\text{ J}\). This means the calculated value of \(c\) will be higher than the true value. To minimize this, we can insulate/lag the block, or cool the block below room temperature by a certain amount before heating it to the same amount above room temperature (so heat absorbed from the room during the first half equals heat lost to the room during the second half).

(a)
- Calculate energy supplied by heater using \(E = Pt\) (1)
- Calculate temperature change \(\Delta T = 45\text{ K}\) (1)
- Correct specific heat capacity [\(1600\text{ J kg}^{-1}\text{ K}^{-1}\)] (1)

(b)
- States that heat loss makes calculated value of \(c\) larger than true value (1)
- Explains that actual energy absorbed by block is less than measured electrical energy (1)
- Suggests suitable minimization method (e.g., lagging/insulation or starting below room temperature) (1)
- Gives detail of selected method (e.g., how the cancellation of heat exchange works, or that insulating material has high resistance to heat flow) (1)

(a) Free oscillations occur when a system is initially displaced and then allowed to oscillate without any external driving force; it oscillates at its own natural frequency. Forced oscillations occur when a periodic external driving force acts on the system, forcing it to vibrate at the frequency of the driving force.

(b) Resonance occurs when the driving frequency is equal to the natural frequency of the system, resulting in a maximum transfer of energy and a maximum amplitude of oscillation. As driving frequency increases from below natural frequency, the amplitude starts small, increases to a sharp maximum peak at the natural frequency, and then decreases back to a low value at higher frequencies.

(c) Heavy damping reduces the maximum amplitude at the resonant frequency. It also broadens the resonance curve (making the peak less sharp) and shifts the resonant frequency peak slightly to a lower frequency.

(a)
- Free oscillations: system displaced and oscillates at its natural frequency with no external periodic force (1)
- Forced oscillations: system driven by an external periodic force at the driver's frequency (1)

(b)
- Resonance defined as driving frequency matching natural frequency resulting in maximum amplitude (1)
- Amplitude starts small, peaks at natural frequency, and decreases thereafter (1)

(c)
- Damping reduces maximum peak amplitude (1)
- Damping broadens the resonance curve peak (1)
- Damping shifts the resonant peak to a slightly lower frequency (1)

(a) Calculate change in wavelength:
\(\Delta \lambda = 672.1\text{ nm} - 656.3\text{ nm} = 15.8\text{ nm}\)

Redshift \(z\):
\(z = \frac{\Delta \lambda}{\lambda_0} = \frac{15.8\text{ nm}}{656.3\text{ nm}} = 0.02407 \approx 0.0241\)

Recession velocity \(v\):
\(v = z c = 0.02407 \times 3.00 \times 10^8\text{ m s}^{-1} = 7.221 \times 10^6\text{ m s}^{-1}\) (or \(7220\text{ km s}^{-1}\))

(b) Using Hubble's Law \(v = H_0 d\):
\(d = \frac{v}{H_0} = \frac{7221\text{ km s}^{-1}}{67.8\text{ km s}^{-1}\text{ Mpc}^{-1}} = 106.5\text{ Mpc} \approx 107\text{ Mpc}\)

Converting to meters:
\(d = 106.5 \times 10^6\text{ pc} \times (3.09 \times 10^{16}\text{ m/pc}) = 3.29 \times 10^{24}\text{ m}\)

(a)
- Use of \(z = \frac{\Delta \lambda}{\lambda}\) (1)
- Correct value for redshift \(z = 0.024\) (1)
- Use of \(v = zc\) to get recession velocity [\(7.2 \times 10^6\text{ m s}^{-1}\) or \(7200\text{ km s}^{-1}\)] (1)

(b)
- Use of Hubble's Law \(v = H_0 d\) (1)
- Correct calculation of distance in Mpc [106 to 107 Mpc] (1)
- Use of conversion factor to meters (1)
- Correct distance in meters [\(3.3 \times 10^{24}\text{ m}\)] (1)

(a) Total mass of reactants:
\(m_{\text{reactants}} = 2.014102\text{ u} + 3.016049\text{ u} = 5.030151\text{ u}\)

Total mass of products:
\(m_{\text{products}} = 4.002603\text{ u} + 1.008665\text{ u} = 5.011268\text{ u}\)

Mass defect in atomic mass units:
\(\Delta m = 5.030151\text{ u} - 5.011268\text{ u} = 0.018883\text{ u}\)

Mass defect in kilograms:
\(\Delta m = 0.018883 \times 1.66 \times 10^{-27}\text{ kg} = 3.1346 \times 10^{-29}\text{ kg} \approx 3.13 \times 10^{-29}\text{ kg}\)

(b) Energy released in Joules:
\(E = \Delta m c^2 = (3.1346 \times 10^{-29}\text{ kg}) \times (3.00 \times 10^8\text{ m/s})^2 = 2.8211 \times 10^{-12}\text{ J}\)

Energy in MeV:
\(E = \frac{2.8211 \times 10^{-12}\text{ J}}{1.60 \times 10^{-13}\text{ J/MeV}} = 17.63\text{ MeV} \approx 17.6\text{ MeV}\)

(Alternatively using \(931.5\text{ MeV/u}\):
\(E = 0.018883 \times 931.5\text{ MeV} = 17.59\text{ MeV}\))

(a)
- Calculate reactant and product masses (1)
- Calculate mass defect in u [0.01888 u] (1)
- Convert mass defect to kg [\(3.13 \times 10^{-29}\text{ kg}\)] (1)

(b)
- Use of \(E = \Delta m c^2\) (1)
- Calculation of energy in Joules [\(2.82 \times 10^{-12}\text{ J}\)] (1)
- Division by conversion factor to obtain eV or MeV (1)
- Correct final energy value in range [17.5 to 17.7 MeV] (1)

(a) Using the ideal gas equation:
\(pV = N k_B T\)
\(N = \frac{pV}{k_B T} = \frac{1.20 \times 10^5\text{ Pa} \times 0.0210\text{ m}^3}{1.38 \times 10^{-23}\text{ J K}^{-1} \times 293\text{ K}}\)
\(N = \frac{2520}{4.0434 \times 10^{-21}} = 6.23 \times 10^{23} \approx 6.2 \times 10^{23}\) atoms.

(b) Since volume is constant, the change in temperature \(\Delta T\) is proportional to the change in pressure \(\Delta p\).
\(T_2 = T_1 \times \frac{p_2}{p_1} = 293\text{ K} \times \frac{1.80 \times 10^5}{1.20 \times 10^5} = 439.5\text{ K}\)
\(\Delta T = 439.5\text{ K} - 293\text{ K} = 146.5\text{ K}\)

The change in total kinetic energy of a monoatomic gas is given by:
\(\Delta E_k = \frac{3}{2} N k_B \Delta T\)
\(\Delta E_k = 1.5 \times 6.23 \times 10^{23} \times 1.38 \times 10^{-23}\text{ J K}^{-1} \times 146.5\text{ K} = 1890\text{ J}\)
(Alternatively: \(\Delta E_k = \frac{3}{2} V \Delta p = 1.5 \times 0.0210\text{ m}^3 \times (1.80 \times 10^5 - 1.20 \times 10^5)\text{ Pa} = 1890\text{ J}\))

(c) As temperature increases, the mean kinetic energy of the helium atoms increases, so they move with higher root-mean-square speeds. This results in more frequent collisions with the walls of the canister, and each collision involves a larger change in momentum. Therefore, the average force exerted on the walls increases, leading to an increase in pressure.

(a) [2 Marks]
- Use of \(pV = N k_B T\) with correct substitution of values (1)
- Correct calculation to yield \(6.23 \times 10^{23}\) (1)

(b) [3 Marks]
- Correct identification of \(\Delta T = 146.5\text{ K}\) (or \(T_2 = 439.5\text{ K}\)) OR use of relation \(\Delta E_k = 1.5 V \Delta p\) (1)
- Substitution of values into total kinetic energy formula \(\Delta E_k = 1.5 N k_B \Delta T\) or \(\Delta E_k = 1.5 V \Delta p\) (1)
- Correct final value of \(1890\text{ J}\) (accept range \(1880\text{ J}\) to \(1900\text{ J}\)) (1)

(c) [2 Marks]
- Identifies that higher temperature means higher average speed/kinetic energy of molecules (1)
- Explains that this leads to more frequent collisions with walls AND a greater change in momentum per collision (hence greater force/pressure) (1)

(a) The maximum kinetic energy is equal to the maximum elastic potential energy stored in the spring:
\(E_{\text{max}} = \frac{1}{2} k A^2\)
\(0.450\text{ J} = \frac{1}{2} k (0.0800\text{ m})^2\)
\(0.450 = 0.00320 k\)
\(k = \frac{0.450}{0.00320} = 141\text{ N m}^{-1}\)

(b) The angular frequency is:
\(\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{140.6}{0.350}} = 20.0\text{ rad s}^{-1}\)
Frequency is:
\(f = \frac{\omega}{2\pi} = \frac{20.0}{2\pi} = 3.19\text{ Hz}\)

(c) At \(t = 0\), the displacement is at its maximum positive value, so the velocity is zero. As the block moves towards the equilibrium position, its velocity becomes negative. The velocity-time curve is a negative sine wave:
- Shape: Negative sine curve starting at \(v = 0\).
- Time period: \(T = \frac{1}{f} = \frac{1}{3.19\text{ Hz}} = 0.313\text{ s}\) for one complete cycle.
- Peak velocity: \(v_{\text{max}} = \omega A = 20.04 \times 0.0800 = 1.60\text{ m s}^{-1}\). The velocity oscillates between \(-1.60\text{ m s}^{-1}\) and \(+1.60\text{ m s}^{-1}\).

(a) [2 Marks]
- Equates maximum kinetic energy to maximum potential energy, \(E = 0.5 k A^2\) (1)
- Correct calculation of \(k = 141\text{ N m}^{-1}\) (accept \(140.6\text{ N m}^{-1}\)) (1)

(b) [2 Marks]
- Correct use of \(\omega = \sqrt{\frac{k}{m}}\)
(or \(T = 2\pi \sqrt{\frac{m}{k}}\)) (1)
- Correct calculation of frequency \(f = 3.19\text{ Hz}\) (accept range \(3.18\text{ Hz}\) to \(3.20\text{ Hz}\)) (1)

(c) [3 Marks]
- States that the graph starts at \(v = 0\) and follows a negative sine curve shape (1)
- Calculates and states the period \(T = 0.313\text{ s}\) (1)
- Calculates and states the maximum speed \(v_{\text{max}} = 1.60\text{ m s}^{-1}\) (1)

(a) First, calculate the luminosity of Alpha Centauri A:
\(L = 1.52 \times 3.83 \times 10^{26}\text{ W} = 5.822 \times 10^{26}\text{ W}\)

Using Stefan-Boltzmann's Law:
\(L = 4\pi R^2 \sigma T^4\)

Rearranging for radius \(R\):
\(R = \sqrt{\frac{L}{4\pi \sigma T^4}}\)
\(R = \sqrt{\frac{5.822 \times 10^{26}}{4 \times \pi \times 5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4} \times (5790\text{ K})^4}}\)
\(R = \sqrt{\frac{5.822 \times 10^{26}}{8.006 \times 10^8}} = \sqrt{7.272 \times 10^{17}} = 8.53 \times 10^8\text{ m}\)
Which is approximately \(8.5 \times 10^8\text{ m}\).

(b) Using Wien's displacement law:
\(\lambda_{\text{max}} T = 2.898 \times 10^{-3}\text{ m K}\)
\(\lambda_{\text{max}} = \frac{2.898 \times 10^{-3}\text{ m K}}{5790\text{ K}} = 5.01 \times 10^{-7}\text{ m}\) (or \(501\text{ nm}\))

(c) The absorption lines of distant galaxies are shifted to longer wavelengths (redshifted) because the galaxies are moving away from Earth. This motion causes the wavefronts of the emitted light to be stretched out (cosmological expansion / Doppler effect), increasing the wavelength observed by astronomers on Earth.

(a) [3 Marks]
- Correct calculation of luminosity \(L = 5.82 \times 10^{26}\text{ W}\) (1)
- Rearrangement of Stefan's Law to make \(R\) the subject: \(R = \sqrt{\frac{L}{4\pi\sigma T^4}}\) (1)
- Correct substitution and calculation to show \(R \approx 8.5 \times 10^8\text{ m}\) (1)

(b) [2 Marks]
- Use of Wien's displacement law \(\lambda_{\text{max}} T = 2.898 \times 10^{-3}\text{ m K}\) (1)
- Correct calculation of \(\lambda_{\text{max}} = 5.01 \times 10^{-7}\text{ m}\) (accept range \(5.0 \times 10^{-7}\text{ m}\) to \(5.1 \times 10^{-7}\text{ m}\)) (1)

(c) [2 Marks]
- Identifies that the galaxy is moving away from Earth (or space is expanding) (1)
- Explains that this causes a Doppler shift / stretching of wavelengths towards the red end (redshift) (1)

(a) To measure the resonant length \( L \) accurately:
- Strike the tuning fork on a rubber pad and hold it horizontally just above the open end of the tube.
- Slowly adjust the height of the tube in the water until the sound intensity reaches a clear maximum (resonance).
- Read the level of the water surface and the top of the tube using a metre rule, ensuring the eye is level with the meniscus/top to avoid parallax error. Repeat and take average values.

(b) Plot \( L \) on the vertical y-axis and \( 1/f \) on the horizontal x-axis.
Rearranging the equation yields:
\( L = \frac{v}{4} \left( \frac{1}{f} \right) - e \)
- The gradient \( m \) of the straight line is equal to \( \frac{v}{4} \), so the speed of sound is determined by \( v = 4 \times \text{gradient} \).
- The y-intercept of the line is equal to \( -e \), so the end correction \( e \) is equal to the magnitude of the negative y-intercept.

(c) Convert frequencies and lengths to standard units:
- For \( f_1 = 256 \text{ Hz} \), \( 1/f_1 = \frac{1}{256} = 3.91 \times 10^{-3} \text{ s} \), \( L_1 = 0.305 \text{ m} \)
- For \( f_2 = 512 \text{ Hz} \), \( 1/f_2 = \frac{1}{512} = 1.95 \times 10^{-3} \text{ s} \), \( L_2 = 0.144 \text{ m} \)

\( \text{Gradient} = \frac{L_1 - L_2}{1/f_1 - 1/f_2} = \frac{0.305 - 0.144}{3.91 \times 10^{-3} - 1.95 \times 10^{-3}} = \frac{0.161}{1.96 \times 10^{-3}} = 82.14 \text{ m s} \) (or \( 82.4 \text{ m s} \) using unrounded values)

Using \( v = 4 \times \text{gradient} \):
\( v = 4 \times 82.14 \approx 329 \text{ m s}^{-1} \) (or \( 330 \text{ m s}^{-1} \) to 2 or 3 s.f.).

(d) The major source of uncertainty is locating the exact point of maximum resonance by ear. The sound intensity might appear constant over a small range of lengths.
To minimize this: move the tube up and down through the resonance point several times to narrow down the range, or use a microphone connected to an oscilloscope/sound-level meter to identify the peak amplitude objectively.

(a)
- MP1: Hold vibrating tuning fork close to/just above the top of the tube. (1)
- MP2: Move tube slowly to identify peak loudness/resonance. (1)
- MP3: Measure from water level to top of tube, using a set square to ensure the rule is vertical / avoid parallax. (1)

(b)
- MP1: Plot \( L \) on y-axis and \( 1/f \) on x-axis. (1)
- MP2: State that \( L = \frac{v}{4}\left(\frac{1}{f}\right) - e \) corresponds to \( y = mx + c \). (1)
- MP3: State that gradient \( m = v/4 \implies v = 4m \). (1)
- MP4: State that y-intercept \( c = -e \). (1)

(c)
- MP1: Correctly calculate both values of \( 1/f \) (i.e. \( 3.91 \times 10^{-3} \) and \( 1.95 \times 10^{-3} \)). (1)
- MP2: Calculate gradient to be in the range \( 81.5 - 83.0 \text{ m s} \). (1)
- MP3: Multiply gradient by 4 to obtain speed of sound \( v \approx 326 - 332 \text{ m s}^{-1} \) with correct units. (1)

(d)
- MP1: Identify that judging maximum loudness by ear is subjective / hard to pinpoint. (1)
- MP2: Suggest using a microphone with an oscilloscope/sound-level meter. (1)
- MP3: Detail that the peak on the display indicates the precise resonant length (0.5)

(a) The diagram should show:
- A U-shaped magnet on a digital balance.
- A stiff wire passing between the magnet's poles without touching them, supported by independent retort stands.
- A circuit loop containing the wire, connected in series with a variable DC power supply (or a cell and a rheostat), an ammeter, and a switch.

(b) To ensure the force is purely due to the magnetic field:
- The wire must not touch the magnet or the balance pan.
- Ensure no magnetic materials (like iron stands) are placed close to the setup.
- Zero (tare) the digital balance before turning on the current to ensure only the change in mass \( \Delta m \) due to the magnetic force is measured.
- To vary current safely: adjust the rheostat / power supply to vary current, but keep the current switched off between readings to prevent heating of the wire, which would change its resistance and risk damage.

(c) First, calculate force \( F = \Delta m \cdot g \):
\( F = 2.45 \times 10^{-3} \text{ kg} \times 9.81 \text{ m s}^{-2} = 0.02403 \text{ N} \)

Using \( B = \frac{F}{I L} = \frac{\Delta m \cdot g}{I L} \):
\( B = \frac{2.45 \times 10^{-3} \times 9.81}{4.00 \times 0.050} \)
\( B = \frac{0.02403}{0.200} = 0.12015 \text{ T} \approx 0.120 \text{ T} \)

(d) Calculate the percentage uncertainties:
- For mass: \( \% \Delta(\Delta m) = \frac{0.02}{2.45} \times 100\% = 0.816\% \)
- For current: \( \% \Delta I = \frac{0.05}{4.00} \times 100\% = 1.25\% \)
- For length: \( \% \Delta L = \frac{0.1}{5.0} \times 100\% = 2.00\% \)

Assuming the uncertainty in \( g \) is negligible:
Total percentage uncertainty in \( B = \% \Delta(\Delta m) + \% \Delta I + \% \Delta L \)
\( \text{Total } \% \text{ uncertainty} = 0.816\% + 1.25\% + 2.00\% = 4.066\% \approx 4.1\% \)

(a)
- MP1: Magnet on balance pan clearly separated from the wire. (1)
- MP2: Wire held horizontally between the poles of the magnet. (1)
- MP3: Electrical circuit showing variable power supply/rheostat, ammeter, and switch in series with the wire. (1)

(b)
- MP1: Mention taring/zeroing the balance before turning on the current. (1)
- MP2: Ensure wire does not contact any part of the magnet or balance. (1)
- MP3: Turn current off between readings / keep current small to avoid overheating of the wire. (1)

(c)
- MP1: Convert \( \Delta m \) to kg (i.e., \( 2.45 \times 10^{-3} \text{ kg} \)) and \( L \) to m (i.e., \( 0.050 \text{ m} \)). (1)
- MP2: Correct calculation of force \( F = 0.024 \text{ N} \). (1)
- MP3: Correct calculation of \( B = 0.120 \text{ T} \) (accept answers to 2 or 3 s.f.). (1)
- MP4: Unit of magnetic flux density given correctly as Tesla (T) or \( \text{N A}^{-1} \text{ m}^{-1} \). (0.5)

(d)
- MP1: Correctly calculate the individual percentage uncertainty of at least one quantity. (1)
- MP2: Show the sum of the individual percentage uncertainties (ignoring \( g \)). (1)
- MP3: State correct final percentage uncertainty of \( 4.1\% \) (accept \( 4.07\% \)). (1)

(a) To minimize heat loss:
- Wrap the aluminium block in an insulating jacket (e.g., polystyrene or bubble wrap).
- Place the block on an insulating tile rather than a metal bench.
- Put a drop of thermal paste or oil in the thermometer and heater holes to improve thermal contact and speed up heat transfer to the block.

(b) A graph of \( \theta \) against \( t \) is better because:
- It allows the student to identify and exclude the initial non-linear region (before steady heating is established) and the late region (where heat loss is dominant).
- It averages out random fluctuations in temperature readings.
- The gradient of the linear portion gives a more reliable rate of temperature change than a single calculation.

(c) The thermal energy conservation equation is:
\( P = m c \frac{\Delta \theta}{\Delta t} \)

Since the gradient of the graph of \( \theta \) against \( t \) is \( \frac{\Delta \theta}{\Delta t} \):
\( P = m c \times \text{gradient} \)

Rearranging to find \( c \):
\( c = \frac{P}{m \times \text{gradient}} \)
\( c = \frac{48.0 \text{ W}}{1.00 \text{ kg} \times 0.0533 \text{ }^{\circ}\text{C s}^{-1}} \)
\( c = 900.56 \text{ J kg}^{-1} \text{ }^{\circ}\text{C}^{-1} \approx 901 \text{ J kg}^{-1} \text{ }^{\circ}\text{C}^{-1} \)

(d) The energy delivered by the heater goes into heating both the block and its surroundings (insulator, heater, thermometer):
\( P t = (m c + C_{\text{other}}) \Delta \theta \)
where \( C_{\text{other}} \) is the heat capacity of the other components.

If we assume all energy goes into the block alone (i.e. \( P t = m c_{\text{calculated}} \Delta \theta \)), then:
\( c_{\text{calculated}} = c_{\text{true}} + \frac{C_{\text{other}}}{m} \)

Therefore, the calculated specific heat capacity will be larger than the true value of the aluminium block, as the temperature rise for a given energy input is smaller than expected.

(a)
- MP1: Insulate the outer surface of the block using a poor conductor (e.g. cotton wool, bubble wrap). (1)
- MP2: Place the block on a heat-proof/insulating mat to prevent conduction to the table. (1)
- MP3: Use oil/glycerine in the thermometer/heater holes to improve thermal contact. (1)

(b)
- MP1: Graph allows anomalous data points/non-linear start-up phase to be identified and ignored. (1)
- MP2: Gradient uses multiple data points, reducing random error. (1)
- MP3: Correctly relates the gradient to the rate of temperature rise. (0.5)

(c)
- MP1: Use the correct relationship \( P = m c (\Delta \theta / \Delta t) \). (1)
- MP2: Identify that \( \Delta \theta / \Delta t \) is the gradient. (1)
- MP3: Substitution of values: \( c = 48.0 / (1.00 \times 0.0533) \). (1)
- MP4: Correct evaluation with units: \( 901 \text{ J kg}^{-1} \text{ }^{\circ}\text{C}^{-1} \) (accept \( 900 \text{ J kg}^{-1} \text{ K}^{-1} \)). (0.5)

(d)
- MP1: State that some electrical energy is absorbed by the heater/thermometer/insulation. (1)
- MP2: Explain that the actual temperature rise \( \Delta \theta \) of the block is less than it would be if only the block absorbed the heat. (1)
- MP3: Conclude that the calculated value of \( c \) is larger than the true value. (1)
- MP4: Clear linkage using the rearranged formula or heat equation. (0.5)

(a) Background radiation is always present due to cosmic rays, rocks, and nuclear waste. This must be measured over a long period (e.g. 10 minutes) with no source present to find the average background count rate.
The corrected count rate is calculated by subtracting this background count rate from each measured count rate.

(b) Taking natural logarithms on both sides of \( C = C_0 e^{-\lambda t} \):
\( \ln C = \ln(C_0 e^{-\lambda t}) \)
\( \ln C = \ln C_0 + \ln(e^{-\lambda t}) \)
\( \ln C = -\lambda t + \ln C_0 \)

This is in the linear form \( y = mx + c \), where:
- \( y = \ln C \)
- \( x = t \)
- \( m = -\lambda \) (the gradient)
- \( c = \ln C_0 \) (the y-intercept)

Therefore, the decay constant \( \lambda \) is equal to \( -\text{gradient} \).

(c) Calculate the natural logs:
- At \( t_1 = 0 \text{ s} \), \( \ln C_1 = \ln(120) = 4.787 \)
- At \( t_2 = 100 \text{ s} \), \( \ln C_2 = \ln(35) = 3.555 \)

\( \text{Gradient} = \frac{\ln C_2 - \ln C_1}{t_2 - t_1} = \frac{3.555 - 4.787}{100 - 0} = \frac{-1.232}{100} = -0.01232 \text{ s}^{-1} \)

Since \( \text{Gradient} = -\lambda \):
\( \lambda = 0.0123 \text{ s}^{-1} \)

The half-life \( t_{1/2} \) is:
\( t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.01232 \text{ s}^{-1}} \approx 56.3 \text{ s} \)

(d) Safety precautions:
- Use long tongs to handle the source to maximize distance and reduce intensity of exposure.
- Keep the source in its lead-lined storage container when not actively taking measurements.
- Do not point the source towards anyone.

(a)
- MP1: State that background radiation is present naturally / from cosmic rays etc. (1)
- MP2: Measure background count rate over a long period in the absence of the source. (1)
- MP3: State that corrected count rate = measured count rate minus background count rate. (1)

(b)
- MP1: Correctly apply log rules to obtain \( \ln C = -\lambda t + \ln C_0 \). (1)
- MP2: Match the equation to \( y = mx + c \) to show it's a straight line. (1)
- MP3: State that the decay constant \( \lambda = -\text{gradient} \). (1)

(c)
- MP1: Calculate the values of \( \ln(120) \) and \( \ln(35) \) (or equivalent ratio method). (1)
- MP2: Calculate \( \lambda \approx 0.0123 \text{ s}^{-1} \) (accept \( 0.012 \text{ s}^{-1} \)). (1)
- MP3: Use \( t_{1/2} = \frac{\ln 2}{\lambda} \) correctly. (1)
- MP4: Calculate \( t_{1/2} \approx 56 \text{ s} \) with correct units. (1)

(d)
- MP1: Identify a valid precaution (e.g. use tongs, keep source far, point away). (1)
- MP2: Link the precaution to reducing dose / protecting body tissues. (1)
- MP3: Mention safety wear like safety glasses or keeping distance. (0.5)