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The horizontal velocity is constant: \(v_x = u \cos\theta\). The vertical velocity at any time \(t\) is \(v_y = u \sin\theta - gt\). Since \(x = (u \cos\theta)t\), we have \(t = \frac{x}{u \cos\theta}\), which gives \(v_y = u \sin\theta - \frac{gx}{u \cos\theta}\). The kinetic energy is \(E_k = \frac{1}{2}m(v_x^2 + v_y^2) = \frac{1}{2}m\left(u^2\cos^2\theta + \left(u\sin\theta - \frac{gx}{u \cos\theta}\right)^2\right)\). Expanding this yields a quadratic equation in \(x\) with a positive coefficient for the \(x^2\) term, which describes a parabola that opens upwards. At the peak of the trajectory, \(v_y = 0\), so the kinetic energy is at its minimum value: \(E_{k,\text{min}} = \frac{1}{2}m u^2 \cos^2\theta\). Since \(\theta < 90^\circ\), this minimum value is non-zero. Therefore, the correct option is A.

[1 mark] A - A parabola that opens upwards, with a non-zero minimum value. Award 1 mark for the correct choice.

At terminal velocity, the upward forces balance the downward force: \(U + F_d = W\) where weight \(W = \frac{4}{3}\pi r^3 \rho_s g\), upthrust \(U = \frac{4}{3}\pi r^3 \rho_l g\), and viscous drag \(F_d = 6\pi\eta r v\) (by Stokes' Law). Substituting these into the balance equation: \(\frac{4}{3}\pi r^3 \rho_l g + 6\pi\eta r v = \frac{4}{3}\pi r^3 \rho_s g\), which simplifies to \(6\pi\eta r v = \frac{4}{3}\pi r^3 (\rho_s - \rho_l)g\). Solving for \(v\) gives: \(v = \frac{2r^2(\rho_s - \rho_l)g}{9\eta}\). Therefore, the correct option is A.

[1 mark] A - Correctly derived terminal velocity formula. Award 1 mark for the correct choice.

1. Calculate the useful work output: \(W_{\text{out}} = mgh = 120\text{ kg} \times 9.81\text{ m s}^{-2} \times 8.0\text{ m} = 9417.6\text{ J}\). 2. Calculate the useful power output: \(P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{9417.6\text{ J}}{5.0\text{ s}} = 1883.5\text{ W}\). 3. Calculate the required electrical input power: \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{P_{\text{out}}}{0.65} = \frac{1883.5\text{ W}}{0.65} = 2897.7\text{ W} \approx 2.9\text{ kW}\). Therefore, the correct option is C.

[1 mark] C - 2.9 kW. Award 1 mark for the correct choice.

The Young's Modulus \(E\) of the material is given by \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}\). Rearranging for extension gives \(\Delta L = \frac{FL}{AE}\). For the second wire of the same material, the Young's Modulus is the same. The extension under the same force is \(\Delta L_2 = \frac{F(2L)}{(2A)E} = \frac{FL}{AE} = \Delta L\). Therefore, the correct option is B.

[1 mark] B - \(\Delta L\). Award 1 mark for the correct choice.

Using Newton's Second Law, the average force \(F\) on the ball is given by the rate of change of momentum: \(F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}\). Letting the direction away from the wall be positive: \(u = -12\text{ m s}^{-1}\) and \(v = +8.0\text{ m s}^{-1}\). Thus, \(\Delta p = m(v - u) = 0.15\text{ kg} \times (8.0\text{ m s}^{-1} - (-12\text{ m s}^{-1})) = 3.0\text{ N s}\). The average force is \(F = \frac{3.0\text{ N s}}{0.040\text{ s}} = 75\text{ N}\). By Newton's Third Law, the force exerted on the wall is also \(75\text{ N}\). Therefore, the correct option is D.

[1 mark] D - 75 N. Award 1 mark for the correct choice.

1. Work done during loading: Work in elastic region is \(W_1 = \frac{1}{2} \times 100\text{ N} \times 2.0 \times 10^{-3}\text{ m} = 0.10\text{ J}\). Work in plastic loading region is \(W_2 = \frac{100\text{ N} + 120\text{ N}}{2} \times (5.0 - 2.0) \times 10^{-3}\text{ m} = 0.33\text{ J}\). Total loading work is \(W_{\text{load}} = 0.10 + 0.33 = 0.43\text{ J}\). 2. Work recovered during unloading: \(W_{\text{unload}} = \frac{1}{2} \times 120\text{ N} \times (5.0 - 1.5) \times 10^{-3}\text{ m} = 0.21\text{ J}\). 3. Net work done: \(\Delta W = 0.43\text{ J} - 0.21\text{ J} = 0.22\text{ J}\). Therefore, the correct option is C.

[1 mark] C - 0.22 J. Award 1 mark for the correct choice.

Let \(T_L\) and \(T_R\) be the tensions in the left and right ropes respectively. The weight of the uniform beam acts at its midpoint (\(2.0\text{ m}\) from either end). Taking moments about the left end: \(\sum M_{\text{left}} = 0 \implies (300\text{ N} \times 1.0\text{ m}) + (150\text{ N} \times 2.0\text{ m}) - (T_R \times 4.0\text{ m}) = 0\), which simplifies to \(300 + 300 = 4 T_R \implies T_R = 150\text{ N}\). Therefore, the correct option is B.

[1 mark] B - 150 N. Award 1 mark for the correct choice.

Using the equation of motion \(s = ut + \frac{1}{2}gt^2\) with \(u = 0\): For the first interval \(t = T\), \(s_1 = h = \frac{1}{2}gT^2\). For the total interval up to \(t = 2T\), \(s_2 = \frac{1}{2}g(2T)^2 = 4 \left(\frac{1}{2}gT^2\right) = 4h\). The distance covered between \(t = T\) and \(t = 2T\) is \(\Delta s = s_2 - s_1 = 4h - h = 3h\). Therefore, the correct option is C.

[1 mark] C - \(3h\). Award 1 mark for the correct choice.

Using the equation of motion with constant acceleration:
\[v^2 = u^2 + 2as\]
Since the vehicle starts from rest, the initial velocity \(u = 0\), so:
\[v = \sqrt{2as}\]

The net force accelerating the vehicle is given by Newton's second law:
\[F = ma\]

Instantaneous power \(P\) is the product of the force and the instantaneous velocity:
\[P = Fv = ma \sqrt{2as}\]

This can be simplified by bringing \(a\) inside the square root:
\[P = m \sqrt{a^2 \times 2as} = m \sqrt{2a^3s}\]

Therefore, option A is the correct expression.

**[1 Mark]**
* **A** - Correctly identifies the expression for instantaneous power by combining \(v = \sqrt{2as}\) and \(F = ma\) to obtain \(P = m\sqrt{2a^3s}\).

The elastic strain energy \(E\) stored in a wire under tension is given by:
\[E = \frac{1}{2} F \Delta L\]

The extension \(\Delta L\) is related to the Young modulus \(E_y\) of the material by:
\[\Delta L = \frac{FL}{A E_y}\]

Substituting \(\Delta L\) into the energy expression gives:
\[E = \frac{F^2 L}{2 A E_y}\]

Since both wires are made of the same material, they have the same Young modulus \(E_y\). Since they are also subjected to the same force \(F\), the energy stored is proportional to:
\[E \propto \frac{L}{A}\]

Since cross-sectional area \(A = \frac{\pi d^2}{4} \propto d^2\), we can write:
\[E \propto \frac{L}{d^2}\]

For wire X:
\[E_{\text{X}} \propto \frac{L}{d^2}\]

For wire Y:
\[E_{\text{Y}} \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\]

Dividing the two values:
\[\frac{E_{\text{X}}}{E_{\text{Y}}} = \frac{L / d^2}{L / (2d^2)} = 2\]

Therefore, option C is correct.

**[1 Mark]**
* **C** - Correctly calculates the ratio of the stored elastic strain energies as 2.

Let upwards be the positive vertical direction.

(a) The initial vertical component of velocity is given by:
\( v_{iy} = v \sin\theta = 25.0\text{ m s}^{-1} \times \sin(30.0^\circ) = 12.5\text{ m s}^{-1} \)

(b) To find the total flight time, we use the vertical displacement formula:
\( s = ut + \frac{1}{2}at^2 \)
Here, vertical displacement \( s = -12.0\text{ m} \), vertical initial velocity \( u = 12.5\text{ m s}^{-1} \), and vertical acceleration \( a = -9.81\text{ m s}^{-2} \).
\( -12.0 = 12.5t - \frac{1}{2}(9.81)t^2 \)
\( 4.905t^2 - 12.5t - 12.0 = 0 \)
Using the quadratic formula:
\( t = \frac{-(-12.5) \pm \sqrt{(-12.5)^2 - 4(4.905)(-12.0)}}{2 \times 4.905} \)
\( t = \frac{12.5 \pm \sqrt{156.25 + 235.44}}{9.81} \)
\( t = \frac{12.5 \pm 19.79}{9.81} \)
Taking the positive root:
\( t = \frac{32.29}{9.81} = 3.29\text{ s} \)

(c) The horizontal velocity is constant because air resistance is negligible:
\( v_x = v \cos\theta = 25.0\text{ m s}^{-1} \times \cos(30.0^\circ) = 21.65\text{ m s}^{-1} \)

The horizontal distance travelled is:
\( s_x = v_x \times t = 21.65\text{ m s}^{-1} \times 3.291\text{ s} = 71.3\text{ m} \) (or \(71.2\text{ m}\) if using rounded value of \(t = 3.29\text{ s}\)).

**Part (a)** [1 Mark]
* Correct calculation showing \(v_{iy} = 25.0 \sin(30.0^\circ) = 12.5\text{ m s}^{-1}\) (1)

**Part (b)** [4 Marks]
* Uses vertical SUVAT with correct signs, e.g., \(-12 = 12.5t - 4.905t^2\) (1)
* Rearranges to form a standard quadratic equation (1)
* Correct substitution of values into the quadratic formula (1)
* Obtains a final time of \(3.29\text{ s}\) (1) *(accept answers rounding to 3.3 s)*

**Part (c)** [2 Marks]
* Calculates the horizontal component of velocity: \(v_x = 21.65\text{ m s}^{-1}\) (1)
* Calculates correct horizontal distance: \(71.2\text{ m}\) or \(71.3\text{ m}\) (1)

(a) For two springs connected in series, the reciprocal of the effective spring constant \(k_{\text{eff}}\) is the sum of the reciprocals of individual spring constants:
\( \frac{1}{k_{\text{eff}}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{120\text{ N m}^{-1}} \)
\( k_{\text{eff}} = \frac{120}{2} = 60\text{ N m}^{-1} \)

(b) The tension force in the combination is equal to the gravitational force acting on the suspended load:
\( F = mg = 0.850\text{ kg} \times 9.81\text{ m s}^{-2} = 8.3385\text{ N} \)
Using Hooke's Law:
\( F = k_{\text{eff}} \Delta x \implies \Delta x = \frac{F}{k_{\text{eff}}} = \frac{8.3385\text{ N}}{60\text{ N m}^{-1}} = 0.139\text{ m} \)

(c) The elastic strain energy \(E_{\text{el}}\) stored in the system is given by:
\( E_{\text{el}} = \frac{1}{2} F \Delta x = \frac{1}{2} \times 8.3385\text{ N} \times 0.139\text{ m} = 0.579\text{ J} \)
Alternatively:
\( E_{\text{el}} = \frac{1}{2} k_{\text{eff}} (\Delta x)^2 = \frac{1}{2} \times 60\text{ N m}^{-1} \times (0.139\text{ m})^2 = 0.579\text{ J} \)

**Part (a)** [2 Marks]
* Uses series combination equation: \(\frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2}\) (1)
* Obtains value of \(60\text{ N m}^{-1}\) (1)

**Part (b)** [2 Marks]
* Calculates suspended load force \(F = 8.34\text{ N}\) (1)
* Calculates extension \(\Delta x = 0.139\text{ m}\) (or \(0.14\text{ m}\)) (1)

**Part (c)** [3 Marks]
* Recalls elastic strain energy equation \(E_{\text{el}} = \frac{1}{2}F\Delta x\) or \(E_{\text{el}} = \frac{1}{2}k(\Delta x)^2\) (1)
* Substitutes correct values into selected equation (1)
* Correct calculation of stored energy \(E_{\text{el}} = 0.579\text{ J}\) (1) *(accept range 0.578 J - 0.583 J depending on rounding)*

(a) A free-body force diagram must include three distinct forces with arrows originating on the block:
1. Weight (or force of gravity, \(W\)) acting vertically downwards.
2. Normal contact force (\(R\) or \(N\)) acting perpendicular to, and outwards from, the ramp surface.
3. Friction (\(F_{\text{friction}}\)) acting parallel to the ramp pointing up the ramp (opposing the direction of motion).

(b) First, resolve the weight component acting parallel to the ramp (downwards):
\( W_{\parallel} = mg \sin\theta = 3.20\text{ kg} \times 9.81\text{ m s}^{-2} \times \sin(25.0^\circ) = 31.392\text{ N} \times 0.4226 = 13.27\text{ N} \)

Next, apply Newton's second law along the direction parallel to the ramp:
\( F_{\text{net}} = W_{\parallel} - F_{\text{friction}} = ma \)
\( 13.27\text{ N} - 5.50\text{ N} = 3.20\text{ kg} \times a \)
\( 7.77\text{ N} = 3.20\text{ kg} \times a \)
\( a = \frac{7.77\text{ N}}{3.20\text{ kg}} = 2.43\text{ m s}^{-2} \)

**Part (a)** [3 Marks]
* Draw and label Weight vertically downwards (1)
* Draw and label Normal contact force perpendicular to the ramp (1)
* Draw and label Friction parallel to and up the ramp (1)
*(Note: Arrows must point in the correct directions and start on the block)*

**Part (b)** [4 Marks]
* Identifies gravity component down the slope as \(mg \sin\theta\) (1)
* Calculates value of gravity component down slope \(= 13.27\text{ N}\) (1)
* Applies \(F_{\text{net}} = ma\) with both forces subtracted (1)
* Obtains final acceleration \(a = 2.43\text{ m s}^{-2}\) (1) *(accept 2.4 m s^-2)*

(a) The tension force in the wire is equal to the weight of the suspended mass:
\( F = mg = 4.50\text{ kg} \times 9.81\text{ m s}^{-2} = 44.145\text{ N} \)
Tensile stress is defined as:
\( \sigma = \frac{F}{A} = \frac{44.145\text{ N}}{3.50 \times 10^{-7}\text{ m}^2} = 1.261 \times 10^8\text{ Pa} \)

(b) The definition of the Young modulus is:
\( E = \frac{\text{stress}}{\text{strain}} = \frac{\sigma}{\varepsilon} \implies \varepsilon = \frac{\sigma}{E} \)
\( \varepsilon = \frac{1.261 \times 10^8\text{ Pa}}{1.20 \times 10^{11}\text{ Pa}} = 1.051 \times 10^{-3} \)
Since strain is defined as \( \varepsilon = \frac{\Delta L}{L_0} \):
\( \Delta L = \varepsilon \times L_0 = 1.051 \times 10^{-3} \times 2.40\text{ m} = 2.52 \times 10^{-3}\text{ m} = 2.52\text{ mm} \)

(c) The elastic limit of a material refers to the maximum stress or force that can be applied to it such that, when the load is removed, the material returns to its original shape and length. If the stress exceeds this limit, the material undergoes permanent (plastic) deformation and will be permanently stretched.

**Part (a)** [2 Marks]
* Uses \(F = mg\) and recalls \(\sigma = \frac{F}{A}\) (1)
* Calculates stress \(\sigma = 1.26 \times 10^8\text{ Pa}\) (1) *(accept 1.3 × 10^8 Pa)*

**Part (b)** [3 Marks]
* Uses \(E = \frac{\text{stress}}{\text{strain}}\) to find strain \(= 1.05 \times 10^{-3}\) (1)
* Relates strain to extension: \(\text{strain} = \frac{\Delta L}{L_0}\) (1)
* Calculates extension \(\Delta L = 2.52 \times 10^{-3}\text{ m}\) (or \(2.52\text{ mm}\)) (1)

**Part (c)** [2 Marks]
* Defines elastic limit as the point/stress value beyond which permanent/plastic deformation occurs (1)
* Explains that the material will not return to its original length when the force is removed (1)

(a) The component of weight parallel to the slope is given by:
\( W_{\parallel} = mg \sin\theta \)
\( W_{\parallel} = 1400\text{ kg} \times 9.81\text{ m s}^{-2} \times \sin(6.0^\circ) = 1435.6\text{ N} \)
Which is approximately \(1400\text{ N}\) to two significant figures.

(b) Since the car travels at a constant velocity, the forward force \(F\) provided by the motor must balance the sum of the resistive forces and the weight component down the slope:
\( F = W_{\parallel} + F_{\text{resistive}} = 1435.6\text{ N} + 450\text{ N} = 1885.6\text{ N} \)
(If using the rounded value of \(1400\text{ N}\), then \(F = 1400\text{ N} + 450\text{ N} = 1850\text{ N}\))

The useful power output is:
\( P_{\text{out}} = F \times v = 1885.6\text{ N} \times 15.0\text{ m s}^{-1} = 28284\text{ W} = 28.3\text{ kW} \)
(Using the rounded force, \( P_{\text{out}} = 1850\text{ N} \times 15.0\text{ m s}^{-1} = 27.8\text{ kW} \))

(c) Efficiency is defined as:
\( \text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} \)
Using the unrounded power output:
\( P_{\text{in}} = \frac{28284\text{ W}}{0.85} = 33275\text{ W} = 33.3\text{ kW} \)
(Using \(27.8\text{ kW}\), \( P_{\text{in}} = \frac{27750\text{ W}}{0.85} = 32.6\text{ kW} \))

**Part (a)** [2 Marks]
* Recalls component of weight is \(mg \sin\theta\) (1)
* Calculates \(1435.6\text{ N}\) and shows it is approximately \(1400\text{ N}\) (1)

**Part (b)** [3 Marks]
* Sums weight component and resistive force to find total forward force \(F = 1886\text{ N}\) (or \(1850\text{ N}\)) (1)
* Recalls and uses \(P = Fv\) (1)
* Obtains useful power of \(28.3\text{ kW}\) (or \(27.8\text{ kW}\)) (1)

**Part (c)** [2 Marks]
* Recalls and rearranges \(\text{efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\) (1)
* Calculates power input: \(33.3\text{ kW}\) (or \(32.6\text{ kW}\)) (1)

(a) Let the direction to the right be positive.
Initial velocities:
\( u_A = +1.2\text{ m s}^{-1} \)
\( u_B = -0.40\text{ m s}^{-1} \)

Final velocity of A:
\( v_A = +0.20\text{ m s}^{-1} \)

According to the conservation of linear momentum:
\( m_A u_A + m_B u_B = m_A v_A + m_B v_B \)
\( 0.45(1.2) + 0.30(-0.40) = 0.45(0.20) + 0.30(v_B) \)
\( 0.54 - 0.12 = 0.09 + 0.30 v_B \)
\( 0.42 = 0.09 + 0.30 v_B \)
\( 0.33 = 0.30 v_B \)
\( v_B = \frac{0.33}{0.30} = 1.1\text{ m s}^{-1} \)
Since the result is positive, the velocity is directed to the right.

(b) To determine if the collision is elastic, compare total kinetic energy before and after:
\( E_{k,\text{before}} = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 = \frac{1}{2}(0.45)(1.2)^2 + \frac{1}{2}(0.30)(-0.40)^2 \)
\( E_{k,\text{before}} = 0.324\text{ J} + 0.024\text{ J} = 0.348\text{ J} \)

\( E_{k,\text{after}} = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.45)(0.20)^2 + \frac{1}{2}(0.30)(1.1)^2 \)
\( E_{k,\text{after}} = 0.009\text{ J} + 0.1815\text{ J} = 0.1905\text{ J} \)

Since \( E_{k,\text{after}} < E_{k,\text{before}} \), kinetic energy is not conserved, meaning the collision is inelastic.

**Part (a)** [4 Marks]
* States conservation of linear momentum equation (1)
* Uses correct sign for velocity directed to the left (e.g., \(-0.40\text{ m s}^{-1}\)) (1)
* Rearranges equation correctly to make \(v_B\) the subject (1)
* Obtains final answer \(1.1\text{ m s}^{-1}\) and states direction is to the right (1)

**Part (b)** [3 Marks]
* Calculates initial total kinetic energy \(= 0.348\text{ J}\) (1)
* Calculates final total kinetic energy \(= 0.191\text{ J}\) (1)
* Concludes collision is inelastic because kinetic energy is not conserved (1)

(a) When first released, the downward force (weight) is greater than the upward force (upthrust), resulting in a net downward force that causes the ball to accelerate.
As velocity increases, the viscous drag force increases (since drag is proportional to velocity in laminar flow).
Eventually, the sum of the upward forces (upthrust + drag) equals the downward weight. The net force becomes zero, so acceleration becomes zero, and terminal velocity is reached.

(b) At terminal velocity, the forces are in equilibrium:
\( W = U + F_D \implies W - U = F_D \)
Weight \( W = \rho_{\text{ball}} V g \)
Upthrust \( U = \rho_{\text{fluid}} V g \)
Therefore:
\( (\rho_{\text{ball}} - \rho_{\text{fluid}}) V g = 6\pi \eta r v \)
Volume of the sphere \( V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.50 \times 10^{-3}\text{ m})^3 = 1.4137 \times 10^{-8}\text{ m}^3 \)
Substituting values:
\( (7800 - 1260)\text{ kg m}^{-3} \times (1.4137 \times 10^{-8}\text{ m}^3) \times 9.81\text{ m s}^{-2} = 6 \pi \eta (1.50 \times 10^{-3}\text{ m}) (0.18\text{ m s}^{-1}) \)
\( 6540 \times 1.4137 \times 10^{-8} \times 9.81 = 6\pi \eta (2.70 \times 10^{-4}) \)
\( 9.070 \times 10^{-4} = 5.089 \times 10^{-3} \eta \)
\( \eta = \frac{9.070 \times 10^{-4}}{5.089 \times 10^{-3}} = 0.178\text{ Pa s} \) (or \(0.18\text{ Pa s}\))

**Part (a)** [3 Marks]
* Mentions weight is initially greater than upthrust, creating a downward net force / acceleration (1)
* States drag force increases as velocity increases (1)
* Explains that terminal velocity occurs when net force is zero (weight = upthrust + drag) (1)

**Part (b)** [4 Marks]
* States equilibrium condition: \(W = U + F_D\) (1)
* Calculates sphere volume \(V = 1.41 \times 10^{-8}\text{ m}^3\) (1)
* Substitutes densities, volume, and terminal velocity correctly into force balance equation (1)
* Calculates viscosity \(\eta = 0.178\text{ Pa s}\) or \(0.18\text{ Pa s}\) (1) *(accept units of N s m^-2 or kg m^-1 s^-1)*

(a) During Stage 1 (first \(3.0\text{ s}\)), the athlete accelerates from rest (\(u = 0\)) to \(v\).
Since acceleration is constant, the average velocity is:
\( v_{\text{avg}} = \frac{u+v}{2} = \frac{0+v}{2} = 0.5v \)
Distance \( s_1 \) travelled during Stage 1 is:
\( s_1 = v_{\text{avg}} \times t_1 = 0.5v \times 3.0\text{ s} = 1.5v \)

(b) During Stage 2, the athlete travels at a constant velocity \(v\).
Time spent in Stage 2 is:
\( t_2 = 11.5\text{ s} - 3.0\text{ s} = 8.5\text{ s} \)
Distance \( s_2 \) travelled during Stage 2 is:
\( s_2 = v \times t_2 = 8.5v \)

The total distance of the race is \( 100\text{ m} \):
\( s_1 + s_2 = 100\text{ m} \)
\( 1.5v + 8.5v = 100 \)
\( 10.0v = 100 \)
\( v = 10.0\text{ m s}^{-1} \)

(c) Any valid real-world physical limitation, such as:
* Real runners do not maintain a perfectly constant acceleration from the start.
* Real runners cannot instantly transition from accelerating to a constant velocity at \(3.0\text{ s}\).
* Air resistance would make maintaining constant maximum speed require varying force.

**Part (a)** [2 Marks]
* Uses average velocity of \(0.5v\) or applies SUVAT with \(a = \frac{v}{3}\) (1)
* Obtains expression \(1.5v\) (1)

**Part (b)** [4 Marks]
* Deduces time for Stage 2 is \(8.5\text{ s}\) (1)
* Formulates expression for second stage distance: \(8.5v\) (1)
* Equates total distance to \(100\text{ m}\): \(1.5v + 8.5v = 100\) (1)
* Calculates maximum velocity \(v = 10.0\text{ m s}^{-1}\) (1)

**Part (c)** [1 Mark]
* Mentions a sensible physical limitation (e.g. non-constant acceleration, non-instantaneous change in acceleration, air resistance effects) (1)

For part (a), the principle of conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. For part (b), the initial height of the car is h = 0.80 m * sin(30°) = 0.40 m. The initial gravitational potential energy lost is E_p = mgh = 0.15 kg * 9.81 m s\(^{-2}\) * 0.40 m = 0.589 J. The work done against friction is W = F * d = 0.25 N * 0.80 m = 0.20 J. The kinetic energy of the car at the bottom of the ramp is E_k = E_p - W = 0.589 J - 0.20 J = 0.389 J. Using E_k = 0.5 * m * v\(^2\), we find v = \(\sqrt{(2 * 0.389 J) / 0.15 kg}\) = 2.28 m s\(^{-1}\), which is approximately 2.3 m s\(^{-1}\). For part (c), we first calculate the time t taken to fall vertically to the floor: s = 0.5 * g * t\(^2\) implies 0.85 m = 0.5 * 9.81 m s\(^{-2}\) * t\(^2\), so t = \(\sqrt{0.1733}\) = 0.416 s. The horizontal distance is x = v * t. Using v = 2.28 m s\(^{-1}\), x = 2.28 * 0.416 = 0.95 m. Using the rounded value of 2.3 m s\(^{-1}\), x = 2.3 * 0.416 = 0.96 m.

Part (a): [1 Mark] State that energy cannot be created or destroyed, only converted/transferred from one form to another. Part (b): [3 Marks] MP1: Calculate vertical height h = 0.40 m and potential energy lost = 0.59 J (or use Ep = mgh directly in energy equation). MP2: Calculate work done against friction = 0.20 J. MP3: Set Ek = Ep - W and solve to show v = 2.28 m s\(^{-1}\). Part (c): [3 Marks] MP1: Use of vertical motion equation s = 0.5 * g * t\(^2\) to find time. MP2: Calculate t = 0.42 s. MP3: Calculate horizontal range x = 0.95 m (accept 0.96 m).

For part (a), the student must measure: 1) Original length of the wire using a metre rule or tape measure; 2) Diameter of the wire using a micrometer screw gauge. To ensure accuracy, the diameter should be measured at several different points and directions along the wire to calculate a mean diameter. For part (b), the cross-sectional area of the wire is A = \(\pi\) * r\(^2\) = \(\pi\) * (0.28 * 10\(^{-3}\) m)\(^2\) = 2.46 * 10\(^{-7}\) m\(^2\). Since Young modulus E = (F * L) / (A * \(\Delta\)L), the extension is \(\Delta\)L = (F * L) / (A * E) = (35 N * 2.40 m) / (2.46 * 10\(^{-7}\) m\(^2\) * 1.2 * 10\(^{11}\) Pa) = 84 / 29520 = 2.84 * 10\(^{-3}\) m, which is 2.8 mm.

Part (a): [3 Marks] MP1: Measure original length using a metre rule / tape measure. MP2: Measure diameter using a micrometer screw gauge. MP3: Detail of measuring the diameter at multiple points/orientations and finding the average. Part (b): [4 Marks] MP1: Use of Area = \(\pi\) * d\(^2\) / 4 or \(\pi\) * r\(^2\) to get A = 2.46 * 10\(^{-7}\) m\(^2\). MP2: Recall of Young modulus formula E = stress / strain. MP3: Rearrangement of formula to make extension the subject: \(\Delta\)L = (F * L) / (A * E). MP4: Correct calculation to give 2.8 * 10\(^{-3}\) m or 2.8 mm.

When light travels from one medium to another, its frequency \(f\) remains constant because it depends only on the source of the light. The speed of light in the medium is given by \(v = \frac{c}{n}\). Since \(v = f \lambda'\), the new wavelength \(\lambda'\) is given by \(\lambda' = \frac{v}{f} = \frac{c}{n f} = \frac{\lambda}{n}\).

1 mark for the correct option A. Correctly identifying that frequency is unchanged and wavelength is scaled by \(1/n\).

The volume of the wire \(V = A \times L\) is constant, so if length increases to \(1.10 L\), the cross-sectional area must decrease to \(A / 1.10\). Resistance is given by \(R = \frac{\rho L}{A}\). The new resistance \(R' = \frac{\rho (1.10 L)}{A / 1.10} = 1.21 \frac{\rho L}{A} = 1.21 R\).

1 mark for the correct option B. Correctly applying volume conservation to relate area and length, then calculating the new resistance.

Using Einstein's photoelectric equation, \(hf = \Phi + E_k\), where \(\Phi\) is the work function of the metal. For the doubled frequency \(2f\), the new maximum kinetic energy \(E_k'\) is given by \(E_k' = h(2f) - \Phi = 2(hf) - \Phi = 2(E_k + \Phi) - \Phi = 2E_k + \Phi\). Since the work function \(\Phi\) is a positive quantity, \(E_k'\) must be greater than \(2E_k\).

1 mark for the correct option B. Applying Einstein's photoelectric equation to show that the new kinetic energy is \(2E_k + \Phi\), which is greater than \(2E_k\).

An NTC thermistor has a resistance that decreases as its temperature increases. In a series potential divider circuit, the ratio of the potential differences across the components is equal to the ratio of their resistances. Since the thermistor's resistance decreases relative to the fixed resistor \(R\), the voltage drop across the thermistor decreases. Therefore, the voltmeter reading decreases.

1 mark for the correct option A. Recalling that NTC thermistor resistance decreases with temperature and applying the potential divider rule to deduce that the voltmeter reading decreases.

For a string fixed at both ends vibrating in its third harmonic, there are three loops, so \(L = 3 \times \frac{\lambda}{2}\), which gives \(\lambda = \frac{2L}{3}\). The distance between a node and an adjacent antinode is a quarter of a wavelength, \(\frac{\lambda}{4}\). Thus, the distance is \(\frac{1}{4} \times \frac{2L}{3} = \frac{L}{6}\).

1 mark for the correct option A. Determining the wavelength in terms of \(L\) for the third harmonic and calculating the node-to-antinode distance as \(\lambda / 4\).

Since the wires are connected in series, the same current \(I\) flows through both wires. Using the equation for current \(I = nAvq\), we have \(n A_X v_X q = n A_Y v_Y q\). Since they are both copper, \(n\) is the same. Thus, \(A_X v_X = A_Y v_Y\), which gives \(\frac{v_X}{v_Y} = \frac{A_Y}{A_X}\). The area \(A\) is proportional to the square of the diameter \(d^2\). Since \(d_X = 2d_Y\), then \(A_X = 4A_Y\). Therefore, \(\frac{v_X}{v_Y} = \frac{A_Y}{4A_Y} = 0.25\).

1 mark for the correct option D. Recognizing that current is the same in series, linking area to diameter squared, and solving for the ratio of drift velocities.

The fringe spacing is given by \(x = \frac{\lambda D}{d}\). If the new screen distance is \(D' = 2D\) and the new slit separation is \(d' = \frac{d}{2}\), the new fringe spacing is \(x' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{d/2} = 4 \frac{\lambda D}{d} = 4x\).

1 mark for the correct option D. Using the double-slit formula and substituting the changes to find the new fringe spacing.

The terminal potential difference \(V\) is related to the e.m.f. \(\mathcal{E}\) and the current \(I\) by the equation \(V = \mathcal{E} - Ir\). Comparing this to the equation of a straight line, \(y = mx + c\), we see that a graph of \(V\) against \(I\) is a straight line with a negative gradient of \(-r\) and a vertical intercept of \(\mathcal{E}\).

1 mark for the correct option A. Using the equation \(V = \mathcal{E} - Ir\) to identify that the graph of \(V\) against \(I\) is a straight line with a negative gradient and an intercept of \(\mathcal{E}\).

The current \(I\) in both wires must be the same because they are connected in series:
\[I_X = I_Y\]

The relationship between current and drift velocity is given by:
\[I = nAvq\]

Since both wires are made of copper, they have the same number density of conduction electrons \(n\) and the same charge carrier charge \(q\). This means:
\[A_X v_X = A_Y v_Y\]

Since the cross-sectional area of a wire of diameter \(d\) is \(A = \frac{\pi d^2}{4}\), we can substitute this into the equation:
\[\frac{\pi d_X^2}{4} v_X = \frac{\pi d_Y^2}{4} v_Y\]

\[d_X^2 v_X = d_Y^2 v_Y\]

Given that the diameter of wire \(X\) is twice that of wire \(Y\) (\(d_X = 2d_Y\)):
\[(2d_Y)^2 v_X = d_Y^2 v_Y\]

\[4 d_Y^2 v_X = d_Y^2 v_Y\]

\[4 v_X = v_Y\]

Rearranging to find the ratio \(\frac{v_X}{v_Y}\):
\[\frac{v_X}{v_Y} = \frac{1}{4}\]

Therefore, option D is the correct answer.

D is the correct answer (1 mark).

Award 1 mark for selecting option D, identifying that \(I\) is constant in series and that drift velocity is inversely proportional to cross-sectional area (the square of the diameter).

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

For the first-order maximum (\(n = 1\)):
\[d \sin\theta = \lambda \implies \sin\theta = \frac{\lambda}{d}\]

In the second setup, we have a new wavelength \(\lambda' = 1.5\lambda\) and a new slit spacing \(d' = 2d\). The equation for the new first-order maximum is:
\[d' \sin\theta' = \lambda'\]

Substituting the new values into the equation:
\[(2d) \sin\theta' = 1.5\lambda\]

Rearranging for \(\sin\theta'\):
\[\sin\theta' = \frac{1.5\lambda}{2d} = 0.75 \left(\frac{\lambda}{d}\right)\]

Since \(\sin\theta = \frac{\lambda}{d}\) from the first setup:
\[\sin\theta' = 0.75\sin\theta = \frac{3}{4}\sin\theta\]

A is the correct answer (1 mark).

Award 1 mark for selecting option A, by successfully setting up the ratio of the diffraction grating equations for both cases.

(a) First, calculate the photon energy using \(E = \frac{hc}{\lambda}\):\
\(E = \frac{(6.63 \times 10^{-34}\text{ J s}) \times (3.00 \times 10^8\text{ m s}^{-1})}{220 \times 10^{-9}\text{ m}} = 9.04 \times 10^{-19}\text{ J}\).\
\
Convert the work function from eV to Joules:\
\(\Phi = 4.3 \times 1.60 \times 10^{-19}\text{ J} = 6.88 \times 10^{-19}\text{ J}\).\
\
Using Einstein's photoelectric equation:\
\(E_k = E - \Phi = 9.04 \times 10^{-19}\text{ J} - 6.88 \times 10^{-19}\text{ J} = 2.16 \times 10^{-19}\text{ J}\).\
This is approximately \(2.2 \times 10^{-19}\text{ J}\).\
\
(b) Calculate the de Broglie wavelength of the emitted electrons:\
First, find the momentum \(p = \sqrt{2 m_e E_k}\) of the electrons:\
\(p = \sqrt{2 \times (9.11 \times 10^{-31}\text{ kg}) \times (2.16 \times 10^{-19}\text{ J})} = 6.273 \times 10^{-25}\text{ kg m s}^{-1}\).\
\
Now, find the de Broglie wavelength using \(\lambda = \frac{h}{p}\):\
\(\lambda = \frac{6.63 \times 10^{-34}\text{ J s}}{6.273 \times 10^{-25}\text{ kg m s}^{-1}} = 1.06 \times 10^{-9}\text{ m}\) (or \(1.1 \times 10^{-9}\text{ m}\)).

Part (a):\
- [1 Mark] Use of \(E = hc/\lambda\) to calculate photon energy.\
- [1 Mark] Conversion of work function into Joules (\(6.88 \times 10^{-19}\text{ J}\)).\
- [1 Mark] Subtraction of work function from photon energy to get correct KE of \(2.16 \times 10^{-19}\text{ J}\).\
\
Part (b):\
- [1 Mark] Correct formula linking de Broglie wavelength to momentum, \(\lambda = h/p\).\
- [1 Mark] Use of \(p = \sqrt{2mE_k}\) or equivalent to calculate momentum.\
- [1 Mark] Correct calculation of momentum as \(6.27 \times 10^{-25}\text{ kg m s}^{-1}\).\
- [1 Mark] Correct final answer of \(1.1 \times 10^{-9}\text{ m}\) (accept \(1.06\text{ nm}\) to \(1.1\text{ nm}\)).\
- [1.75 Marks] Quality of written communication, appropriate unit usage, and rounding consistency.

(a) The equation is: \(V = E - Ir\). \
As the current \(I\) increases, more charge carriers flow through the internal resistance \(r\) of the battery per second. Consequently, the potential difference dropped across the internal resistance (the lost volts, \(Ir\)) increases. Since the e.m.f. \(E\) remains constant, the remaining potential difference available to the external circuit (the terminal potential difference \(V\)) must decrease.\
\
(b) We are given two sets of data:\
For \(I_1 = 1.20\text{ A}\), \(V_1 = 5.40\text{ V}\):\
\(5.40 = E - 1.20 r\) (Equation 1)\
\
For \(I_2 = 2.00\text{ A}\), \(V_2 = 3.00\text{ V}\):\
\(3.00 = E - 2.00 r\) (Equation 2)\
\
Subtract Equation 2 from Equation 1:\
\(2.40 = 0.80 r\)\
\(r = \frac{2.40}{0.80} = 3.0 \\ \Omega\).\
\
Substitute \(r = 3.0 \\ \Omega\) back into Equation 2:\
\(3.00 = E - 2.00(3.0)\) \
\(3.00 = E - 6.00 \implies E = 9.0\text{ V}\).\
\
Therefore, the e.m.f. of the battery is \(9.0\text{ V}\) and the internal resistance is \(3.0 \\ \Omega\).

Part (a):\
- [1 Mark] Stating correct equation: \(V = E - Ir\).\
- [1 Mark] Explaining that increasing current increases 'lost volts' or voltage across internal resistance (\(Ir\)).\
- [1 Mark] Explaining that because e.m.f. remains constant, the terminal potential difference must decrease as a result.\
\
Part (b):\
- [1 Mark] Setting up two simultaneous equations based on experimental data.\
- [1 Mark] Eliminating e.m.f. \(E\) to find a relation for internal resistance \(r\).\
- [1 Mark] Correct calculation of \(r = 3.0 \\ \Omega\).\
- [1 Mark] Substituting \(r\) back to find e.m.f. \(E = 9.0\text{ V}\).\
- [1.75 Marks] Units shown correctly (\(\text{V}\) and \(\Omega\)) with consistent significant figures throughout calculations.

(a) Use the grating equation: \(d \sin\theta = n\lambda\).\
For the first-order maximum (\(n = 1\)):\
\(d \sin(18.5^\circ) = 1 \times 632.8 \times 10^{-9}\text{ m}\).\
\
Calculate \(d\):\
\(d = \frac{632.8 \times 10^{-9}}{\sin(18.5^\circ)} = \frac{632.8 \times 10^{-9}}{0.3173} \approx 1.994 \times 10^{-6}\text{ m}\).\
\
Calculate the number of lines per millimetre, \(N\):\
\(N = \frac{1 \times 10^{-3}\text{ m}}{d} = \frac{10^{-3}}{1.994 \times 10^{-6}} \approx 501.5\text{ lines per millimetre}\) (approximately \(500\text{ lines per millimetre}\)).\
\
(b) The maximum angle of diffraction is \(\theta = 90^\circ\) (since \(\sin\theta \le 1\)).\
Using \(d \sin\theta = n\lambda\) with \(\sin\theta = 1\):\
\(n_{\max} = \frac{d}{\lambda} = \frac{1.994 \times 10^{-6}\text{ m}}{632.8 \times 10^{-9}\text{ m}} \approx 3.15\).\
\
Since the order must be an integer, the maximum possible order is \(n = 3\).\
\
The total number of visible bright fringes on both sides of the central maximum is:\
\(N_{\text{fringes}} = 2n_{\max} + 1 = 2(3) + 1 = 7\).

Part (a):\
- [1 Mark] State and use the formula \(d \sin\theta = n\lambda\).\
- [1 Mark] Correct conversion of wavelength to metres (\(6.328 \times 10^{-7}\text{ m}\) or \(632.8 \times 10^{-9}\text{ m}\)).\
- [1 Mark] Correct calculation of the grating spacing \(d = 1.99 \times 10^{-6}\text{ m}\).\
- [1 Mark] Correct calculation of lines per mm (\(500 \pm 3\text{ lines/mm}\) depending on rounding).\
\
Part (b):\
- [1 Mark] Stating that maximum angle occurs when \(\sin\theta = 1\) or \(\theta = 90^\circ\).\
- [1 Mark] Calculation of non-integer maximum order as \(3.15\) or substitution to show \(n=3\) is allowed but \(n=4\) is not.\
- [1 Mark] Identifying maximum order \(n = 3\).\
- [1 Mark] Finding total fringes by formula \(2n + 1 = 7\).\
- [0.75 Marks] Clear explanation of physical limits of diffraction (why light cannot diffract beyond \(90^\circ\)).

(a) An LDR is made from a semiconductor. As light intensity increases, the LDR absorbs more photons. These photons transfer energy to electrons in the valence band, allowing them to overcome the bandgap and move into the conduction band. This process significantly increases the number density \(n\) of free charge carriers, which increases conductivity and thus decreases the resistance of the LDR.\
\
(b) In dark conditions, \(R_{\text{LDR}} = 4.5\text{ k}\Omega = 4500 \\ \Omega\).\
Using the potential divider equation:\
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\) \
\(V_{\text{out}} = 9.0\text{ V} \times \frac{4.5\text{ k}\Omega}{1.5\text{ k}\Omega + 4.5\text{ k}\Omega} = 9.0\text{ V} \times \frac{4.5}{6.0} = 6.75\text{ V}\).\
\
(c) In bright light, \(V_{\text{out}} = 1.5\text{ V}\).\
\(1.5 = 9.0 \times \frac{R_{\text{LDR}}}{1.5\text{ k}\Omega + R_{\text{LDR}}}\) \
\
Divide both sides by \(1.5\):\
\(1 = 6 \times \frac{R_{\text{LDR}}}{1.5 + R_{\text{LDR}}}\) (working in \(\text{k}\Omega\))\
\(1.5 + R_{\text{LDR}} = 6 R_{\text{LDR}}\) \
\(1.5 = 5 R_{\text{LDR}}\) \
\(R_{\text{LDR}} = 0.3\text{ k}\Omega = 300 \\ \Omega\).

Part (a):\
- [1 Mark] Stating that LDR resistance decreases as light intensity increases.\
- [1 Mark] Explaining that absorbed photons release electrons / free carriers into conduction band.\
- [1 Mark] Referring to increase in carrier concentration (\(n\)).\
\
Part (b):\
- [1 Mark] Correct formula for potential divider circuit.\
- [1 Mark] Calculation showing total resistance is \(6.0\text{ k}\Omega\).\
- [1 Mark] Correct output voltage \(V_{\text{out}} = 6.75\text{ V}\) (or \(6.8\text{ V}\)).\
\
Part (c):\
- [1 Mark] Setting up correct equation with \(V_{\text{out}} = 1.5\text{ V}\).\
- [1 Mark] Correct algebraic steps to solve for \(R_{\text{LDR}}\).\
- [0.75 Marks] Correct final answer of \(300 \\ \Omega\) (or \(0.3\text{ k}\Omega\)) with correct units.

(a) The critical angle \(\theta_c\) is calculated using Snell's law when the refracted angle is \(90^\circ\):\
\(\sin\theta_c = \frac{n_2}{n_1} = \frac{1.41}{1.52} \approx 0.9276\)\
\(\theta_c = \sin^{-1}(0.9276) = 68.1^\circ\).\
\
(b) Using Snell's law at the air-core boundary:\
\(n_{\text{air}} \sin\theta_i = n_1 \sin\theta_r\)\
\(1.00 \times \sin(30.0^\circ) = 1.52 \times \sin\theta_r\)\
\(\sin\theta_r = \frac{0.500}{1.52} \approx 0.3289\)\
\(\theta_r = \sin^{-1}(0.3289) = 19.2^\circ\).\
\
(c) The ray meets the side wall (core-cladding interface). From geometry, the normal to the flat end-face is parallel to the axis of the fiber, and the normal to the core-cladding boundary is perpendicular to the axis. Thus, the angle of incidence \(\theta\) at the core-cladding boundary is:\
\(\theta = 90.0^\circ - \theta_r = 90.0^\circ - 19.2^\circ = 70.8^\circ\).\
\
Compare this angle of incidence to the critical angle:\
Since \(\theta = 70.8^\circ\) is greater than the critical angle \(\theta_c = 68.1^\circ\), the ray will undergo total internal reflection (TIR) and will not escape into the cladding.

Part (a):\
- [1 Mark] Stating or using \(\sin\theta_c = n_2 / n_1\).\
- [1 Mark] Correct calculation of critical angle \(\theta_c = 68.1^\circ\).\
\
Part (b):\
- [1 Mark] Using Snell's law: \(n_1 \sin\theta_1 = n_2 \sin\theta_2\) at the entrance boundary.\
- [1 Mark] Substituting correct numbers: \(1.00 \sin(30^\circ) = 1.52 \sin\theta_r\).\
- [1 Mark] Finding refraction angle \(\theta_r = 19.2^\circ\).\
\
Part (c):\
- [1 Mark] Correct geometric relationship: \\theta = 90.0^\\circ - \\theta_r\\.\
- [1 Mark] Correct angle at boundary: \(70.8^\circ\) (accept \(70.8^\circ\) to \(71.0^\circ\) depending on rounding).\
- [1 Mark] Explicit comparison showing \(70.8^\circ > 68.1^\circ\) leading to total internal reflection.\
- [0.75 Marks] Statement of final conclusion (TIR occurs).

(a) Calculate cross-sectional area \(A\) using the resistivity formula \(R = \frac{\rho L}{A}\):\
\(A = \frac{\rho L}{R} = \frac{1.72 \times 10^{-8} \\ \Omega\text{ m} \times 2.40\text{ m}}{3.20 \\ \Omega} = 1.29 \times 10^{-8}\text{ m}^2\).\
\
Since the wire has a circular cross-section, \(A = \frac{\pi d^2}{4}\):\
\(1.29 \times 10^{-8}\text{ m}^2 = \frac{\pi d^2}{4}\) \
\(d^2 = \frac{4 \times 1.29 \times 10^{-8}}{\pi} = 1.6425 \times 10^{-8}\text{ m}^2\) \
\(d = \sqrt{1.6425 \times 10^{-8}\text{ m}^2} = 1.28 \times 10^{-4}\text{ m}\) (or \(0.128\text{ mm}\)).\
\
(b) When the temperature of the metal increases, the positive lattice ions gain kinetic energy and vibrate with a larger amplitude. This increased vibration increases the frequency of collisions between the conducting free electrons (charge carriers) and the lattice ions. Consequently, the drift velocity of the electrons decreases, which results in an increase in the resistance of the wire.

Part (a):\
- [1 Mark] Rearranging resistivity formula to make area \(A\) the subject.\
- [1 Mark] Substitution of values into formula for \(A\).\
- [1 Mark] Correct calculation of area \(A = 1.29 \times 10^{-8}\text{ m}^2\).\
- [1 Mark] Using circle area formula to solve for diameter \(d\).\
- [1 Mark] Correct calculation of diameter \(d = 1.28 \times 10^{-4}\text{ m}\) (or equivalent \(0.128\text{ mm}\)).\
\
Part (b):\
- [1 Mark] Identifying that lattice ions vibrate with larger amplitude at higher temperature.\
- [1 Mark] Describing more frequent collisions between conduction electrons and lattice ions.\
- [1 Mark] Explaining that this limits the drift velocity of electrons, thereby increasing resistance.\
- [0.75 Marks] Use of clear physical terminology and consistent reasoning throughout.

(a) Calculate the wave speed using \(v = \sqrt{\frac{T}{\mu}}\):\
\(v = \sqrt{\frac{45\text{ N}}{1.2 \times 10^{-3}\text{ kg m}^{-1}}} = \sqrt{37500} \approx 193.65\text{ m s}^{-1}\) (or \(194\text{ m s}^{-1}\)).\
\
(b) For the fundamental mode of vibration (first harmonic), the length of the wire \(L\) corresponds to half a wavelength:\
\(L = \frac{\lambda}{2} \implies \lambda = 2L = 2 \times 0.80\text{ m} = 1.60\text{ m}\).\
\
Calculate the frequency \(f\) using the wave equation \(v = f \lambda\):\
\(f = \frac{v}{\lambda} = \frac{193.65\text{ m s}^{-1}}{1.60\text{ m}} = 121\text{ Hz}\) (to 3 s.f.).\
\
(c) When the wire is plucked, progressive waves are generated and travel to both fixed ends of the wire. At the fixed ends, the waves undergo reflection (with a phase change of \(180^\circ\)). These reflected waves travel back along the wire in opposite directions. The original and reflected waves, which have the same frequency and amplitude, superpose (or interfere) with each other. This superposition creates stationary wave patterns, characterized by nodes (where destructive interference occurs and amplitude is zero) and antinodes (where constructive interference occurs and amplitude is maximum).

Part (a):\
- [1 Mark] Recall and use formula \(v = \sqrt{T/\mu}\).\
- [1 Mark] Correct substitution and calculation of wave speed as \(194\text{ m s}^{-1}\) (or \(193.6\text{ m s}^{-1}\) to \(193.7\text{ m s}^{-1}\)).\
\
Part (b):\
- [1 Mark] Identifying that \(\lambda = 2L\) for the first harmonic.\
- [1 Mark] Calculating wavelength as \(1.60\text{ m}\).\
- [1 Mark] Using \(f = v/\lambda\) with calculated velocity and wavelength.\
- [1 Mark] Calculating frequency as \(121\text{ Hz}\) (accept \(120.6\text{ Hz}\) to \(121.2\text{ Hz}\) depending on rounding).\
\
Part (c):\
- [1 Mark] Mentioning reflection of waves at the fixed boundaries/ends.\
- [1 Mark] Explaining that two waves travelling in opposite directions superpose/interfere.\
- [0.75 Marks] Explicitly mentioning nodes (zero displacement/destructive interference) and antinodes (maximum displacement/constructive interference).

(a) Use the equation for current: \(I = nAve\).\
First, calculate the cross-sectional area \(A\) of the wire:\
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.60 \times 10^{-3}\text{ m})^2}{4} = 2.827 \times 10^{-7}\text{ m}^2\).\
\
Rearrange the current equation for drift velocity \(v\):\
\(v = \frac{I}{nAe}\) \
\
Substitute the values (where \(e = 1.60 \times 10^{-19}\text{ C}\)):\
\(v = \frac{0.45\text{ A}}{8.5 \times 10^{28}\text{ m}^{-3} \times 2.827 \times 10^{-7}\text{ m}^2 \times 1.60 \times 10^{-19}\text{ C}}\) \
\(v = \frac{0.45}{3844.72} \approx 1.17 \times 10^{-4}\text{ m s}^{-1}\).\
\
(b) When the switch is closed, an electric field is established throughout the entire circuit almost instantaneously (travelling close to the speed of light). This electric field exerts a force on all free conduction electrons throughout the entire length of the wire simultaneously. Consequently, all the electrons begin to drift at the same time, producing a current throughout the circuit almost instantly, rather than waiting for individual electrons to travel from the power supply to the lamp.

Part (a):\
- [1 Mark] Stating or using \(I = nAve\).\
- [1 Mark] Calculating area \(A = 2.83 \times 10^{-7}\text{ m}^2\).\
- [1 Mark] Substituting value of elementary charge \(e = 1.60 \times 10^{-19}\text{ C}\).\
- [1 Mark] Correct calculation of drift velocity \(1.17 \times 10^{-4}\text{ m s}^{-1}\) (accept \(1.2 \times 10^{-4}\text{ m s}^{-1}\) or \(1.15 \times 10^{-4}\text{ m s}^{-1}\) depending on rounding).\
\
Part (b):\
- [1 Mark] Mentioning that an electric field is established throughout the circuit almost instantly (at the speed of light).\
- [1 Mark] Explaining that all free conduction electrons experience a force at the same time.\
- [1 Mark] Explaining that charge starts moving simultaneously everywhere in the circuit, hence instantaneous current flow.\
- [1.75 Marks] Clear logical argument without contradiction, using appropriate scientific terminology (e.g., electric field, drift, simultaneous).

(a) The circuit diagram should show:
- A power supply connected in series with a switch, a variable resistor, and the test wire.
- An ammeter connected in series with the test wire.
- A voltmeter connected in parallel across the variable length of the wire being measured.
- (The variable resistor is included to keep the current low to prevent heating of the wire, which would change its resistance.)

(b)
- Use a micrometer screw gauge.
- First, check for a zero error and subtract/add it to all subsequent readings.
- Measure the diameter at several different positions along the wire.
- At each position, take measurements at different orientations (i.e., at right angles to each other) to account for any non-circularity.
- Calculate the mean diameter \(d\) and use \(A = \frac{\pi d^2}{4}\) to find the cross-sectional area. This averaging minimizes random uncertainty in the area.

(c)
- The relationship between resistance \(R\), resistivity \(\rho\), length \(L\), and cross-sectional area \(A\) is:
\[R = \frac{\rho L}{A}\]
- Comparing this to \(y = mx + c\), a plot of \(R\) against \(L\) gives a straight line through the origin with a gradient \(m = \frac{\rho}{A}\).
- Therefore, the resistivity can be calculated using:
\[\rho = m \times A\]
where \(m\) is the gradient of the graph.

(d)
First, calculate the cross-sectional area:
\[d = 0.38 \times 10^{-3}\text{ m}\]
\[A = \frac{\pi (0.38 \times 10^{-3})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\]

The resistivity is:
\[\rho = \text{gradient} \times A = 4.15 \times 1.134 \times 10^{-7} = 4.71 \times 10^{-7}\text{ }\Omega\text{ m}\]

To find the uncertainty:
Percentage uncertainty in gradient:
\[\% \Delta m = \frac{0.08}{4.15} \times 100\% = 1.93\%\]

Percentage uncertainty in diameter:
\[\% \Delta d = \frac{0.01}{0.38} \times 100\% = 2.63\%\]

Since \(A = \frac{\pi d^2}{4}\), the percentage uncertainty in area is:
\[\% \Delta A = 2 \times \% \Delta d = 2 \times 2.63\% = 5.26\%\]

The percentage uncertainty in resistivity is:
\[\% \Delta \rho = \% \Delta m + \% \Delta A = 1.93\% + 5.26\% = 7.19\%\]

The absolute uncertainty in resistivity is:
\[\Delta \rho = 7.19\% \times 4.71 \times 10^{-7}\text{ }\Omega\text{ m} = 0.34 \times 10^{-7}\text{ }\Omega\text{ m}\]

So, \(\rho = (4.7 \pm 0.3) \times 10^{-7}\text{ }\Omega\text{ m}\) (or \(4.71 \pm 0.34 \times 10^{-7}\text{ }\Omega\text{ m}\)).

- (a) [2 Marks]
- MP1: Voltmeter connected in parallel across the variable length of wire, and ammeter in series with the cell and wire. (1)
- MP2: Correct symbols used for voltmeter, ammeter, cell/power supply, and a method to vary length shown (e.g., sliding contact / arrow on the wire). (1)
- (b) [4 Marks]
- MP1: Use a micrometer screw gauge. (1)
- MP2: Check for and correct for zero error. (1)
- MP3: Measure at multiple positions along the wire and at perpendicular orientations. (1)
- MP4: Calculate mean diameter, then use \(A = \frac{\pi d^2}{4}\). (1)
- (c) [3 Marks]
- MP1: Recall \(R = \frac{\rho L}{A}\). (1)
- MP2: State that the gradient is \(m = \frac{\rho}{A}\). (1)
- MP3: Rearrange to give \(\rho = m \times A\) where \(m\) is the gradient. (1)
- (d) [3.5 Marks]
- MP1: Calculate \(A = 1.13 \times 10^{-7}\text{ m}^2\) and \(\rho = 4.71 \times 10^{-7}\text{ }\Omega\text{ m}\) (accept \(4.7 \times 10^{-7}\)). (1)
- MP2: Calculate percentage uncertainty in \(d\) as \(2.63\%\) and percentage uncertainty in \(A\) as \(5.26\%\). (1)
- MP3: Add percentage uncertainties: \(\% \Delta \rho = 1.93\% + 5.26\% = 7.19\%\). (0.5)
- MP4: Calculate absolute uncertainty as \(\pm 0.3 \times 10^{-7}\text{ }\Omega\text{ m}\) (or \(\pm 0.34 \times 10^{-7}\)), and state final answer with consistent units and decimal places. (1)

(a)
- The light gate measures the time \(t\) for which the beam is interrupted by the card of length \(x\) as the card passes through.
- The data logger calculates the velocity using:
\[v = \frac{x}{t}\]

(b)
- The acceleration along the ramp is given by \(a = g \sin\theta\).
- Using the equation of motion \(v^2 = u^2 + 2as\) where \(u = 0\) and \(s = d\):
\[v^2 = 2(g \sin\theta)d\]
\[v^2 = 2gd \sin\theta\]

(c)
- Plot a graph of \(v^2\) on the y-axis against \(d\) on the x-axis.
- The graph should be a straight line passing through the origin.
- Since \(v^2 = (2g \sin\theta) d\), the gradient \(m\) of this line is \(2g \sin\theta\).
- To find \(g\), measure the angle \(\theta\) using a protractor (or via trigonometry using ruler measurements of the height and length of the inclined plane).
- Calculate \(g\) using:
\[g = \frac{m}{2 \sin\theta}\]
where \(m\) is the gradient.

(d)
From \(g = \frac{v^2}{2 d \sin\theta}\):
- Percentage uncertainty in \(v^2\) is:
\[2 \times \% \Delta v = 2 \times 2.0\% = 4.0\%\]
- Percentage uncertainty in \(d\) is:
\[\% \Delta d = \frac{0.005}{0.800} \times 100\% = 0.625\%\]
- To find the percentage uncertainty in \(\sin\theta\), we look at the absolute variation:
\[\sin(10.0^\circ) = 0.1736\]
\[\sin(10.5^\circ) = 0.1822\]
\[\text{Difference } \Delta(\sin\theta) = 0.1822 - 0.1736 = 0.0086\]
\[\% \Delta (\sin\theta) = \frac{0.0086}{0.1736} \times 100\% = 4.95\%\]
- The total percentage uncertainty in \(g\) is the sum of these percentage uncertainties:
\[\% \Delta g = \% \Delta (v^2) + \% \Delta d + \% \Delta (\sin\theta)\]
\[\% \Delta g = 4.0\% + 0.625\% + 4.95\% = 9.575\% \approx 9.6\%\]

- (a) [2 Marks]
- MP1: Light gate measures the time interval \(t\) for which the card breaks the light beam. (1)
- MP2: Velocity calculated using \(v = \frac{x}{t}\) where \(x\) is the length of the card. (1)
- (b) [2 Marks]
- MP1: Recognize \(a = g \sin\theta\). (1)
- MP2: Correct substitution into \(v^2 = 2ad\) to get \(v^2 = 2gd \sin\theta\). (1)
- (c) [5 Marks]
- MP1: Plot a graph of \(v^2\) on the y-axis against \(d\) on the x-axis (accept \(v\) against \(\sqrt{d}\)). (1)
- MP2: State that the graph is a straight line through the origin. (1)
- MP3: State that the gradient \(m = 2g \sin\theta\) (or equivalent for other plots). (1)
- MP4: Explain how to measure \(\theta\) using a protractor (or trigonometry on the ramp dimensions). (1)
- MP5: Rearrange to find \(g = \frac{m}{2\sin\theta}\). (1)
- (d) [3.5 Marks]
- MP1: Determine percentage uncertainty in \(v^2\) as \(4.0\%\). (1)
- MP2: Determine percentage uncertainty in \(d\) as \(0.625\%\) (or \(0.6\%\)). (0.5)
- MP3: Determine percentage uncertainty in \(\sin\theta\) as \(\approx 5.0\%\) (accept range \(4.9\% - 5.0\%\)). (1)
- MP4: Sum percentage uncertainties to get \(9.6\%\) (accept range \(9.5\% - 9.7\%\)). (1)

(a) A well-labelled diagram should include:
- A long test wire clamped firmly between two wooden blocks at one end of the bench.
- The wire passing over a low-friction pulley at the other end of the bench with a weight hanger attached.
- A paper/tape marker attached to the wire near a meter ruler clamped to the bench to measure extension.
- Clearly labeled length \(L_0\) from the clamped wooden blocks to the marker.

(b)
- Goggles/Eye protection: The wire is under high tension and can snap suddenly, which could cause serious eye injuries.
- Sand tray / padded box on the floor: Placed directly below the suspended masses to catch them safely if the wire snaps, preventing injury to feet and damage to the floor.

(c) (i)
- First, calculate the cross-sectional area \(A\):
\[d = 0.28 \times 10^{-3}\text{ m}\]
\[A = \frac{\pi d^2}{4} = \frac{\pi (0.28 \times 10^{-3})^2}{4} = 6.158 \times 10^{-8}\text{ m}^2\]
- Young Modulus \(E\) is given by:
\[E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L_0} = \frac{F L_0}{A \Delta L}\]
\[E = \frac{15.0 \times 2.450}{6.158 \times 10^{-8} \times 5.0 \times 10^{-3}}\]
\[E = \frac{36.75}{3.079 \times 10^{-10}} = 1.19 \times 10^{11}\text{ Pa}\]
So, \(E \approx 1.2 \times 10^{11}\text{ Pa}\) (or \(\text{N m}^{-2}\)).

(c) (ii)
- Percentage uncertainty in the extension \(\Delta L\):
\[\% \Delta (\Delta L) = \frac{0.2}{5.0} \times 100\% = 4.0\%\]
- Percentage uncertainty in the diameter \(d\):
\[\% \Delta d = \frac{0.01}{0.28} \times 100\% = 3.57\%\]
- Since \(A \propto d^2\), the percentage uncertainty in the cross-sectional area \(A\) is:
\[\% \Delta A = 2 \times \% \Delta d = 2 \times 3.57\% = 7.14\%\]
- The total percentage uncertainty in the Young modulus \(E\) is the sum of the percentage uncertainties in \(A\) and \(\Delta L\) (since other uncertainties are negligible):
\[\% \Delta E = \% \Delta A + \% \Delta (\Delta L) = 7.14\% + 4.0\% = 11.14\% \approx 11\%\]

- (a) [3 Marks]
- MP1: Wire clamped firmly at one end and passing over a pulley with hanging masses. (1)
- MP2: Marker attached to wire, with a parallel scale (meter ruler) positioned nearby. (1)
- MP3: Scale and clamp labelled, including the reference length \(L_0\) clearly shown from the clamped end to the marker. (1)
- (b) [3 Marks]
- MP1: Wear safety goggles / eye protection. (1)
- MP2: Place a protective box/sand tray underneath the weights. (1)
- MP3: Explanation for safety: goggles protect eyes if the wire snaps; box/tray prevents injury to feet/damage to floor if masses fall. (1)
- (c)(i) [3 Marks]
- MP1: Correct calculation of cross-sectional area \(A = 6.16 \times 10^{-8}\text{ m}^2\). (1)
- MP2: Use of \(E = \frac{F L_0}{A \Delta L}\) with substitution. (1)
- MP3: Correct value of \(E \approx 1.2 \times 10^{11}\text{ Pa}\) (accept \(1.19 \times 10^{11}\text{ Pa}\) or \(\text{N m}^{-2}\)). (1)
- (c)(ii) [3.5 Marks]
- MP1: Calculate percentage uncertainty in \(\Delta L\) as \(4.0\%\). (1)
- MP2: Calculate percentage uncertainty in \(d\) as \(3.57\%\) (accept \(3.6\%\)). (0.5)
- MP3: Double percentage uncertainty in \(d\) to get percentage uncertainty in area \(A\) as \(7.14\%\) (accept \(7.1\%\) or \(7.2\%\)). (1)
- MP4: Sum percentage uncertainties to get \(11.1\%\) (accept range \(11\% - 11.2\%\)). (1)

(a)
- For the fundamental frequency, the length of the column plus the end correction is equal to one quarter of a wavelength:
\[L + c = \frac{\lambda}{4}\]
\[\lambda = 4(L + c)\]

(b)
- Select a tuning fork of known frequency \(f\) and strike it against a rubber block to cause it to vibrate.
- Hold the vibrating tuning fork horizontally just above the open end of the tube.
- Slowly adjust the vertical position of the tube in the water until a distinct maximum in loudness (resonance) is heard.
- Measure the length \(L\) from the top of the tube to the water level using a meter ruler.
- Repeat this measurement to obtain a mean value of \(L\) for that frequency.
- Repeat the entire procedure for at least 5 different tuning forks with a range of frequencies (e.g., between \(256\text{ Hz}\) and \(512\text{ Hz}\)).

(c)
- Substituting \(\lambda = \frac{v}{f}\) into the equation \(L + c = \frac{\lambda}{4}\):
\[L + c = \frac{v}{4f}\]
\[L = \frac{v}{4}\left(\frac{1}{f}\right) - c\]
- Comparing this to the equation of a straight line, \(y = mx + c'\), where \(y = L\) and \(x = 1/f\):
- The gradient \(m = \frac{v}{4}\), so the speed of sound is given by:
\[v = 4 \times \text{gradient}\]
- The y-intercept is \(-c\), so the end correction is:
\[c = -\text{y-intercept}\]

(d)
- (i) The speed of sound in air is:
\[v = 4 \times 84.5\text{ m s}^{-1} = 338\text{ m s}^{-1}\]
- (ii) The end correction is:
\[c = -(-1.2\text{ cm}) = 1.2\text{ cm}\text{ (or } 0.012\text{ m)}\]

- (a) [2 Marks]
- MP1: State that \(L + c = \frac{\lambda}{4}\). (1)
- MP2: Correctly express relationship as \(\lambda = 4(L + c)\). (1)
- (b) [5 Marks]
- MP1: Strike a tuning fork of known frequency and hold it over the open tube. (1)
- MP2: Move the tube up/down in the water until maximum sound intensity/loudness is heard. (1)
- MP3: Measure \(L\) from top of tube to the water meniscus using a ruler. (1)
- MP4: Repeat the measurement of \(L\) and find a mean for that frequency. (1)
- MP5: Repeat for a range of different frequencies (at least 5 different tuning forks). (1)
- (c) [3.5 Marks]
- MP1: Substitute \(v = f\lambda\) to obtain \(L = \frac{v}{4f} - c\). (1)
- MP2: State that this matches \(y = mx + c\) where \(y = L\), \(x = 1/f\), and the graph is a straight line. (1)
- MP3: State that \(v = 4 \times \text{gradient}\). (1)
- MP4: State that the end correction \(c = -\text{y-intercept}\) (or y-intercept is \(-c\)). (0.5)
- (d) [2 Marks]
- MP1: Speed of sound \(v = 4 \times 84.5 = 338\text{ m s}^{-1}\). (1)
- MP2: End correction \(c = 1.2\text{ cm}\) (or \(0.012\text{ m}\)). (1)

The momentum perpendicular to the wall changes from \(mv \cos \theta\) (towards the wall) to \(-mv \cos \theta\) (away from the wall). The change in the perpendicular component of momentum is \(\Delta p = mv \cos \theta - (-mv \cos \theta) = 2mv \cos \theta\). The momentum parallel to the wall is \(mv \sin \theta\) both before and after the collision, so it remains unchanged. Therefore, the magnitude of the total change in momentum is \(2mv \cos \theta\).

1 mark for correct option A.

The charge on a discharging capacitor is given by \(Q = Q_0 e^{-\frac{t}{RC}}\). Substituting \(t = 2RC\) gives \(Q = Q_0 e^{-\frac{2RC}{RC}} = Q_0 e^{-2}\). The fraction of the initial charge remaining is \(\frac{Q}{Q_0} = e^{-2}\).

1 mark for correct option A.

A strangeness of \(-2\) means the baryon contains two strange quarks (\(s\)), each with a strangeness of \(-1\). A baryon consists of three quarks, so its composition is \(qss\), where \(q\) is a third quark. The charge of a strange quark is \(-\frac{1}{3}e\), so two strange quarks have a total charge of \(-\frac{2}{3}e\). For the baryon to have a net charge of \(0\), the third quark must have a charge of \(+\frac{2}{3}e\), which is the charge of an up quark (\(u\)). Therefore, the quark composition is \(uss\).

1 mark for correct option A.

The radius of the path of a charged particle in a magnetic field is given by \(r = \frac{p}{qB} = \frac{\sqrt{2m E_k}}{qB}\). For the second particle: \(m' = 2m\), \(q' = 2q\), and \(E_k' = 2E_k\). The new radius \(r'\) is: \(r' = \frac{\sqrt{2 m' E_k'}}{q' B} = \frac{\sqrt{2(2m)(2E_k)}}{2qB} = \frac{\sqrt{8 m E_k}}{2qB} = \frac{2\sqrt{2 m E_k}}{2qB} = \frac{\sqrt{2 m E_k}}{qB} = r\). Thus, the radius of the path remains \(r\).

1 mark for correct option A.

The frequency of the alternating voltage source is constant, meaning the polarity of the tubes changes at regular time intervals. For the protons to always experience an accelerating electric field in the gaps between tubes, the transit time through each tube must be constant (equal to half the period of the alternating voltage). As the protons accelerate and their speed increases, they travel a greater distance in this same time interval. Therefore, the tubes must become progressively longer.

1 mark for correct option A.

At the highest point of the bridge, the centripetal acceleration is directed vertically downwards towards the center of the circle. The forces acting on the car are the weight \(mg\) downwards and the normal contact force \(R\) upwards. The resultant downward force is the centripetal force: \(F_c = mg - R = \frac{mv^2}{r}\). Rearranging for \(R\) gives \(R = mg - \frac{mv^2}{r}\).

1 mark for correct option A.

Initially, the magnetic flux linkage is maximum because the field is perpendicular to the plane of the coil (the normal to the plane is parallel to the field, so \(\theta = 0^\circ\)): \(\Phi_1 = BAN \cos(0^\circ) = BAN\). When the coil is rotated through \(60^\circ\), the angle between the normal to the coil and the magnetic field becomes \(60^\circ\). The new flux linkage is: \(\Phi_2 = BAN \cos(60^\circ) = 0.50 BAN\). The magnitude of the change in flux linkage is: \(\Delta \Phi = \Phi_1 - \Phi_2 = BAN - 0.50 BAN = 0.50 BAN\).

1 mark for correct option A.

Let the position of the charge \(+Q\) be at \(x = 0\) and the charge \(-4Q\) be at \(x = d\). For a point between the charges at distance \(x\) from \(+Q\), where \(0 < x < d\), the electric potential \(V\) is the sum of the potentials due to both charges: \(V = \frac{kQ}{x} + \frac{k(-4Q)}{d - x} = 0\). This simplifies to \(\frac{1}{x} = \frac{4}{d - x}\). Solving for \(x\): \(d - x = 4x \implies 5x = d \implies x = 0.20 d\).

1 mark for correct option A.

The charge \(Q\) on a discharging capacitor at time \(t\) is given by:
\(Q = Q_0 e^{-\frac{t}{RC}}\)

The energy \(E\) stored in the capacitor is given by:
\(E = \frac{Q^2}{2C}\)

Substituting the expression for \(Q\) into the energy formula gives:
\(E = \frac{(Q_0 e^{-\frac{t}{RC}})^2}{2C} = \frac{Q_0^2}{2C} e^{-\frac{2t}{RC}} = E_0 e^{-\frac{2t}{RC}}\)

We want to find the time \(t\) at which the energy has decreased to \(\frac{1}{e}\) of its initial value \(E_0\):
\(E = \frac{E_0}{e} = E_0 e^{-1}\)

Equating the two expressions for \(E\):
\(E_0 e^{-\frac{2t}{RC}} = E_0 e^{-1}\)
\(e^{-\frac{2t}{RC}} = e^{-1}\)

Taking the natural logarithm of both sides:
\(-\frac{2t}{RC} = -1\)
\(t = \frac{RC}{2}\)

Award 1 mark for the correct answer A.
Method: Deduce energy decay formula \(E = E_0 e^{-2t/RC}\) using \(E \propto Q^2\), substitute \(E = E_0/e\), and solve for \(t = RC/2\).

At the highest point of the vertical circular path, the forces acting on the object are its weight \(mg\) and the tension \(T\) in the string, both pointing downwards towards the centre of the circle. The centripetal force equation at the top is:
\(T + mg = \frac{m v_{\text{top}}^2}{L}\)

Since the tension \(T\) at the highest point is zero:
\(mg = \frac{m v_{\text{top}}^2}{L} \implies v_{\text{top}}^2 = gL\)

Using conservation of mechanical energy between the highest point and the lowest point, and taking the lowest point as the reference level for gravitational potential energy:
\(E_{\text{top}} = E_{\text{bottom}}\)
\(m g (2L) + \frac{1}{2} m v_{\text{top}}^2 = \frac{1}{2} m v_{\text{bottom}}^2\)

Substitute \(v_{\text{top}}^2 = gL\):
\(2mgL + \frac{1}{2}mgL = \frac{1}{2} m v_{\text{bottom}}^2\)
\(2.5mgL = \frac{1}{2} m v_{\text{bottom}}^2\)
\(v_{\text{bottom}}^2 = 5gL\)
\(v_{\text{bottom}} = \sqrt{5gL}\)

Award 1 mark for the correct answer C.
Method: Apply circular motion criteria at the top to find \(v^2_{\text{top}} = gL\), apply conservation of mechanical energy between top and bottom, and solve for speed at the bottom.

**(a)** At the top of the loop, the forces acting downwards are the weight of the car \( mg \) and the normal reaction force \( N \).
For circular motion: \( mg + N = \frac{m v^2}{R} \).
To maintain contact, the minimum speed occurs when \( N = 0 \).
Therefore, \( mg = \frac{m v^2}{R} \implies v = \sqrt{g R} \).
\( v = \sqrt{9.81 \text{ m s}^{-2} \times 0.40 \text{ m}} = 1.98 \text{ m s}^{-1} \approx 2.0 \text{ m s}^{-1} \).

**(b)** By conservation of energy, the elastic potential energy stored in the spring must equal the sum of the gravitational potential energy and kinetic energy at the top of the loop:
\( E_{\text{elastic}} = E_{k} + E_{p} \)
\( \frac{1}{2} k x^2 = \frac{1}{2} m v^2 + m g (2R) \)
Substituting minimum speed \( v^2 = g R \):
\( \frac{1}{2} k x^2 = \frac{1}{2} m (g R) + 2 m g R = 2.5 m g R \)
\( \frac{1}{2} (180) x^2 = 2.5 \times 0.25 \times 9.81 \times 0.40 \)
\( 90 x^2 = 2.4525 \text{ J} \)
\( x^2 = 0.02725 \implies x \approx 0.165 \text{ m} \approx 0.17 \text{ m} \).

**(c)** The minimum speed required at the top is determined solely by gravity and track geometry (i.e., \( v = \sqrt{gR} \)), so the required speed *at the top* remains unchanged. However, because energy is dissipated as heat and sound by resistive forces, a greater initial launch speed (and thus greater spring compression) is needed. If the car is launched with a speed slightly less than this required minimum, it will not have enough kinetic energy to reach the top at \( 1.98 \text{ m s}^{-1} \). The normal force will drop to zero before the top, and the car will lose contact and fall in a parabolic projectile trajectory.

**(a)**
- Equates centripetal force to gravitational force for minimum speed (1)
- Rearranges to show \( v = \sqrt{g R} \) (1)
- Calculates \( 1.98 \text{ m s}^{-1} \) and concludes \( \approx 2.0 \text{ m s}^{-1} \) (1)

**(b)**
- Identifies conservation of energy equation: \( \frac{1}{2} k x^2 = \frac{1}{2} m v^2 + m g h \) (1)
- Recognizes height at the top is \( 2R = 0.80 \text{ m} \) (1)
- Substitutes values correctly: \( 90 x^2 = 2.45 \) (1)
- Calculates \( x = 0.17 \text{ m} \) (or \( 0.165 \text{ m} \)) (1)

**(c)**
- States that the minimum speed required at the top is unchanged as it depends only on \( g \) and \( R \) (1)
- Explains that the launch speed/energy must increase to compensate for work done against friction/resistive forces (1)
- Explains that if launched below this speed, the normal force becomes zero before the top and the car falls/loses contact (1)

**(a)**
Let us resolve momentum along the x-axis (parallel to original motion):
\( P_{i,x} = m_A u_A = 0.40 \times 3.0 = 1.20 \text{ kg m s}^{-1} \)
\( P_{f,x} = m_A v_A \cos(53.13^\circ) + m_B v_B \cos(\theta) \)
\( 1.20 = 0.40 \times 1.8 \times 0.60 + 0.60 v_{B,x} \)
\( 1.20 = 0.432 + 0.60 v_{B,x} \implies 0.60 v_{B,x} = 0.768 \implies v_{B,x} = 1.28 \text{ m s}^{-1} \)

Now resolve momentum along the y-axis (perpendicular to original motion):
\( P_{i,y} = 0 \)
\( P_{f,y} = m_A v_A \sin(53.13^\circ) - m_B v_B \sin(\theta) \)
\( 0 = 0.40 \times 1.8 \times 0.80 - 0.60 v_{B,y} \)
\( 0 = 0.576 - 0.60 v_{B,y} \implies 0.60 v_{B,y} = 0.576 \implies v_{B,y} = 0.96 \text{ m s}^{-1} \)

Calculate speed \( v_B \):
\( v_B = \sqrt{v_{B,x}^2 + v_{B,y}^2} = \sqrt{1.28^2 + 0.96^2} = \sqrt{1.6384 + 0.9216} = \sqrt{2.56} = 1.60 \text{ m s}^{-1} \).

Calculate angle \( \theta \):
\( \tan(\theta) = \frac{v_{B,y}}{v_{B,x}} = \frac{0.96}{1.28} = 0.75 \implies \theta = 36.87^\circ \approx 37^\circ \).

**(b)**
An elastic collision conserves kinetic energy.
Initial kinetic energy:
\( E_{k,i} = \frac{1}{2} m_A u_A^2 = \frac{1}{2} (0.40) (3.0)^2 = 1.80 \text{ J} \).
Final kinetic energy:
\( E_{k,f} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = \frac{1}{2} (0.40) (1.8)^2 + \frac{1}{2} (0.60) (1.6)^2 \)
\( E_{k,f} = 0.20 \times 3.24 + 0.30 \times 2.56 = 0.648 + 0.768 = 1.416 \text{ J} \).
Since \( E_{k,f} < E_{k,i} \) (\( 1.416 \text{ J} < 1.80 \text{ J} \)), kinetic energy is not conserved. Hence, the collision is inelastic.

**(a)**
- Calculates initial momentum along the x-axis as \( 1.20 \text{ kg m s}^{-1} \) (1)
- Uses conservation of momentum in the x-direction to find \( v_{B,x} = 1.28 \text{ m s}^{-1} \) (1)
- Uses conservation of momentum in the y-direction to find \( v_{B,y} = 0.96 \text{ m s}^{-1} \) (1)
- Uses Pythagoras' theorem to find total speed \( v_B \) (1)
- Obtains \( v_B = 1.60 \text{ m s}^{-1} \) (1)
- Calculates the angle \( \theta \approx 37^\circ \) (or \( 36.9^\circ \)) (1)

**(b)**
- Evaluates initial kinetic energy of the system as \( 1.80 \text{ J} \) (1)
- Evaluates final kinetic energy of the system as \( 1.42 \text{ J} \) (1)
- Compares the two values and states they are unequal (1)
- Concludes that the collision is inelastic because kinetic energy is not conserved (1)

**(a)** The equation for capacitor discharge is:
\( V = V_0 e^{-t/RC} \)
Given \( V_0 = 12 \text{ V} \), \( V = 4.4 \text{ V} \), \( t = 4.5 \text{ s} \), and \( R = 15000 \\ \Omega \):
\( 4.4 = 12 e^{-4.5 / (15000 C)} \)
\( \frac{4.4}{12} = e^{-4.5 / (15000 C)} \)
\( \ln(0.3667) = -\frac{4.5}{15000 C} \)
\( -1.0033 = -\frac{4.5}{15000 C} \)
\( C = \frac{4.5}{15000 \times 1.0033} \approx 2.99 \times 10^{-4} \text{ F} = 299 \\ \mu\text{F} \approx 300 \\ \mu\text{F} \).

**(b)**
Energy stored when fully charged (using \( C = 2.99 \times 10^{-4} \text{ F} \)):
\( E_0 = \frac{1}{2} C V_0^2 = \frac{1}{2} \times (2.99 \times 10^{-4} \text{ F}) \times (12 \text{ V})^2 = 0.0215 \text{ J} = 21.5 \text{ mJ} \).
(If using \( C = 300 \\ \mu\text{F} \), \( E_0 = 21.6 \text{ mJ} \)).

Energy stored at \( t = 4.5 \text{ s} \):
\( E = \frac{1}{2} C V^2 = \frac{1}{2} \times (2.99 \times 10^{-4} \text{ F}) \times (4.4 \text{ V})^2 = 2.90 \times 10^{-3} \text{ J} = 2.90 \text{ mJ} \).
(If using \( C = 300 \\ \mu\text{F} \), \( E = 2.90 \text{ mJ} \)).

**(c)**
We want the energy to drop to \( 15 \text{ mJ} = 0.015 \text{ J} \).
Since \( E = \frac{1}{2} C V^2 \) and \( V = V_0 e^{-t/RC} \), the energy as a function of time is:
\( E = E_0 e^{-2t/RC} \)
\( 15 \text{ mJ} = 21.5 \text{ mJ} \times e^{-2t / RC} \)
\( \frac{15}{21.5} = e^{-2t / (15000 \times 2.99 \times 10^{-4})} \)
\( 0.6977 = e^{-2t / 4.485} \)
\( \ln(0.6977) = -\frac{2t}{4.485} \)
\( -0.3599 = -\frac{2t}{4.485} \)
\( 2t = 1.614 \implies t \approx 0.81 \text{ s} \).
(If using \( C = 300 \\ \mu\text{F} \), \( E_0 = 21.6 \text{ mJ} \), \( 15/21.6 = e^{-2t/4.5} \implies \ln(0.694) = -2t/4.5 \implies 2t = 1.64 \implies t = 0.82 \text{ s} \)).

**(a)**
- Selects and uses \( V = V_0 e^{-t/RC} \) (1)
- Substitutes values correctly into equation: \( 4.4 = 12 e^{-4.5 / (15000 C)} \) (1)
- Applies natural logarithm correctly to isolate \( C \) (1)
- Arrives at \( C = 2.99 \times 10^{-4} \text{ F} \) and states that this is \( \approx 300 \\ \mu\text{F} \) (1)

**(b)**
- Uses \( E = \frac{1}{2} C V^2 \) (1)
- Calculates initial energy as \( 21.5 \text{ mJ} \) (or \( 21.6 \text{ mJ} \)) (1)
- Calculates remaining energy at \( t = 4.5 \text{ s} \) as \( 2.90 \text{ mJ} \) (1)

**(c)**
- Identifies that \( E \propto V^2 \) leads to \( E = E_0 e^{-2t/RC} \) (or calculates the required voltage \( V = 10.0 \text{ V} \) corresponding to \( 15 \text{ mJ} \)) (1)
- Substitutes values into decay formula (1)
- Obtains a time in the range \( 0.81 \text{ s} \) to \( 0.82 \text{ s} \) (1)

**(a)** An ion entering the selector experiences an electric force \( F_E = q E \) and a magnetic force \( F_B = q v B_1 \). These fields are oriented so that the forces act in opposite directions. Only ions with a velocity where these forces are equal and opposite will pass straight through undeflected.
Setting forces equal:
\( q E = q v B_1 \implies v = \frac{E}{B_1} \).

**(b)** In the region with magnetic field \( B_2 \), the magnetic force provides the centripetal force:
\( q v B_2 = \frac{m v^2}{r} \implies r = \frac{m v}{q B_2} \).

Let \( m_1 \) be the mass of \( {}^{6}\text{Li}^{+} \):
\( m_1 = 6.015 \times 1.66 \times 10^{-27} \text{ kg} = 9.985 \times 10^{-27} \text{ kg} \).
\( r_1 = \frac{(9.985 \times 10^{-27} \text{ kg}) \times (2.4 \times 10^5 \text{ m s}^{-1})}{(1.60 \times 10^{-19} \text{ C}) \times 0.35 \text{ T}} = 0.0428 \text{ m} \).

Let \( m_2 \) be the mass of \( {}^{7}\text{Li}^{+} \):
\( m_2 = 7.016 \times 1.66 \times 10^{-27} \text{ kg} = 1.165 \times 10^{-26} \text{ kg} \).
\( r_2 = \frac{(1.165 \times 10^{-26} \text{ kg}) \times (2.4 \times 10^5 \text{ m s}^{-1})}{(1.60 \times 10^{-19} \text{ C}) \times 0.35 \text{ T}} = 0.0499 \text{ m} \).

For a semi-circular path, the displacement is the diameter \( d = 2r \). The separation \( \Delta d \) at the detector is:
\( \Delta d = 2(r_2 - r_1) = 2 \times (0.0499 - 0.0428) = 2 \times 0.0071 \text{ m} = 0.0142 \text{ m} \approx 1.4 \text{ cm} \).

**(c)** Since \( r = \frac{m v}{q B} \), doubling the charge \( q \) to \( 2e \) will halve the radius of the circular path. The ion will still curve in the same direction but will trace out a path with a smaller radius (exactly half, i.e., \( 2.14 \text{ cm} \)).

**(a)**
- Mentions electric force is \( qE \) and magnetic force is \( qvB \) acting in opposite directions (1)
- Explains that only undeflected ions pass through, meaning net force is zero (1)
- Equates forces: \( qE = qvB_1 \) (1)
- Correctly derives \( v = E/B_1 \) (1)

**(b)**
- Equates magnetic force to centripetal force to obtain expression for radius: \( r = \frac{m v}{q B_2} \) (1)
- Calculates mass values in kg or calculates mass difference \( \Delta m = 1.001 \text{ u} = 1.66 \times 10^{-27} \text{ kg} \) (1)
- Identifies that separation is \( 2 \Delta r \) (difference in diameters) (1)
- Calculates separation to be \( 1.4 \text{ cm} \) (or \( 1.42 \text{ cm} \) or \( 0.014 \text{ m} \)) (1)

**(c)**
- States that the radius of the path will be smaller / halved (1)
- Explains that the direction of deflection is unchanged (deflects to the same side) (1)

**(a)** Magnetic flux linkage is the product of the magnetic flux passing normally through a loop and the number of turns in the coil. Mathematically, it is given by \( N \Phi = N B A \cos(\theta) \) where \( B \) is the magnetic flux density, \( A \) is the area, and \( N \) is the number of turns.

**(b)**
Maximum flux linkage occurs when \( B = B_0 \) and \( \cos(\theta) = 1 \):
\( (N \Phi)_{\text{max}} = N B_0 A \)
Convert area to \( \text{m}^2 \):
\( A = 4.0 \text{ cm}^2 = 4.0 \times 10^{-4} \text{ m}^2 \).
\( (N \Phi)_{\text{max}} = 250 \times (15 \times 10^{-3} \text{ T}) \times (4.0 \times 10^{-4} \text{ m}^2) \)
\( (N \Phi)_{\text{max}} = 1.5 \times 10^{-3} \text{ Wb-turns} \).

**(c)**
According to Faraday's law of electromagnetic induction, the induced e.m.f. \( \varepsilon \) is equal to the rate of change of magnetic flux linkage:
\( \varepsilon = -\frac{d(N\Phi)}{dt} \).

Since \( B \) varies as a sine wave \( B = B_0 \sin(2\pi f t) \), the flux linkage also varies as a sine wave.
Its rate of change (and therefore the induced e.m.f.) varies as a cosine wave (shifted by \( 90^\circ \) or \( \pi/2 \text{ rad} \) relative to the flux density).

The maximum rate of change of the magnetic flux density is:
\( \left(\frac{dB}{dt}\right)_{\text{max}} = 2\pi f B_0 \).

Thus, the maximum induced e.m.f. is:
\( \varepsilon_{\text{max}} = N A \left(\frac{dB}{dt}\right)_{\text{max}} = N A (2\pi f B_0) \)
\( \varepsilon_{\text{max}} = 250 \times (4.0 \times 10^{-4} \text{ m}^2) \times 2\pi \times 50 \text{ s}^{-1} \times 0.015 \text{ T} \)
\( \varepsilon_{\text{max}} = 0.10 \times 314.16 \times 0.015 = 0.471 \text{ V} \approx 0.47 \text{ V} \).

**(a)**
- Defines flux linkage as product of flux and number of turns (1)
- Specifies that the magnetic field must be perpendicular to the loop area (or includes \( \cos\theta \)) (1)

**(b)**
- Converts area correctly to \( 4.0 \times 10^{-4} \text{ m}^2 \) (1)
- Uses formula \( N B A \) (1)
- Calculates maximum flux linkage to be \( 1.5 \times 10^{-3} \text{ Wb-turns} \) (1)

**(c)**
- States Faraday's Law: induced e.m.f. is proportional to the rate of change of flux linkage (1)
- Identifies that the induced e.m.f. will vary sinusoidally / as a cosine wave (shifted by \( 90^\circ \)) (1)
- Recognizes that maximum rate of change occurs when flux is passing through zero (1)
- Employs derivative or rate-of-change formula: \( \varepsilon_{\text{max}} = N A \omega B_0 \) (1)
- Calculates maximum e.m.f. correctly as \( 0.47 \text{ V} \) (1)

**(a)**
- The magnetic field is uniform and perpendicular to the plane of the dees. It exerts a magnetic force \( F = qvB \) perpendicular to the motion of the protons, which acts as a centripetal force and keeps the protons in a circular path.
- The electric field is located in the gap between the two dees. It accelerates the protons each time they cross the gap, increasing their kinetic energy and speed.
- An alternating potential difference is applied so that the electric field reverses direction in phase with the half-period of the proton's circular path, ensuring it always accelerates the protons across the gap.

**(b)**
Convert kinetic energy to Joules:
\( E_k = 15 \text{ MeV} = 15 \times 10^6 \times 1.60 \times 10^{-19} \text{ J} = 2.40 \times 10^{-12} \text{ J} \).

Using the non-relativistic formula for kinetic energy:
\( E_k = \frac{1}{2} m v^2 \)
\( 2.40 \times 10^{-12} = \frac{1}{2} \times (1.67 \times 10^{-27} \text{ kg}) \times v^2 \)
\( v^2 = \frac{4.80 \times 10^{-12}}{1.67 \times 10^{-27}} = 2.874 \times 10^{15} \)
\( v = 5.36 \times 10^7 \text{ m s}^{-1} \approx 5.4 \times 10^7 \text{ m s}^{-1} \).

**(c)**
The de Broglie wavelength \( \lambda \) is:
\( \lambda = \frac{h}{p} = \frac{h}{m v} \)
\( \lambda = \frac{6.63 \times 10^{-34} \text{ J s}}{1.67 \times 10^{-27} \text{ kg} \times 5.36 \times 10^7 \text{ m s}^{-1}} \)
\( \lambda = \frac{6.63 \times 10^{-34}}{8.95 \times 10^{-20}} = 7.41 \times 10^{-15} \text{ m} \).

For a probing particle to resolve the structure of a target, its de Broglie wavelength must be comparable to or smaller than the size of the structure being investigated (\( \lambda \le 10^{-15} \text{ m} \)). Since the calculated wavelength \( 7.4 \times 10^{-15} \text{ m} \) is larger than the diameter of the proton (\( 10^{-15} \text{ m} \)), these protons are not suitable for resolving details of its internal structure.

**(a)**
- States that the magnetic field causes the circular motion / deflects the path of protons (1)
- States that the electric field in the gap accelerates the protons / increases their speed (1)
- Mentions that the electric field must alternate in direction (1)
- Explains that the frequency of the alternating voltage is constant because the time spent in each dee is independent of speed (1)

**(b)**
- Converts MeV to Joules: \( 15 \times 10^6 \times 1.60 \times 10^{-19} = 2.4 \times 10^{-12} \text{ J} \) (1)
- Uses \( E_k = \frac{1}{2} m v^2 \) (1)
- Obtains speed \( v = 5.36 \times 10^7 \text{ m s}^{-1} \) and concludes \( \approx 5.4 \times 10^7 \text{ m s}^{-1} \) (1)

**(c)**
- Uses de Broglie equation \( \lambda = h / (m v) \) (1)
- Calculates \( \lambda = 7.4 \times 10^{-15} \text{ m} \) (1)
- Explains that to resolve structures, \( \lambda \le \text{structure size} \), and since \( 7.4 \times 10^{-15} \text{ m} > 10^{-15} \text{ m} \), these protons are not suitable (1)

**(a)**
- Pion-minus \( \pi^{-} \) is a meson: \( d\bar{u} \) (charge = \( -1/3 - 2/3 = -1 \)).
- Proton \( p \) is a baryon: \( uud \) (charge = \( 2/3 + 2/3 - 1/3 = +1 \)).
- Neutral Kaon \( K^{0} \) is a meson with strangeness +1: \( d\bar{s} \) (charge = \( -1/3 + 1/3 = 0 \)).
- Lambda baryon \( \Lambda^{0} \) is a baryon with strangeness -1: \( uds \) (charge = \( 2/3 - 1/3 - 1/3 = 0 \)).

**(b)**
- **Charge (Q):**
LHS: \( (-1) + (+1) = 0 \)
RHS: \( 0 + 0 = 0 \)
Since \( 0 = 0 \), charge is conserved.
- **Baryon number (B):**
LHS: \( 0 + 1 = 1 \)
RHS: \( 0 + 1 = 1 \)
Since \( 1 = 1 \), baryon number is conserved.
- **Strangeness (S):**
LHS: \( 0 + 0 = 0 \)
RHS: \( (+1) + (-1) = 0 \)
Since \( 0 = 0 \), strangeness is conserved.

**(c)**
Let us look at the strangeness of the particles in the decay \( \Lambda^{0} \rightarrow p + \pi^{-} \):
- Before decay: \( \Lambda^{0} \) has strangeness \( S = -1 \).
- After decay: Proton \( p \) has strangeness \( S = 0 \) and \( \pi^{-} \) has strangeness \( S = 0 \). Total final strangeness is \( 0 \).

Since \( S \) changes from \( -1 \) to \( 0 \), strangeness is not conserved (\( \Delta S = 1 \)). Only the weak interaction can violate the conservation of strangeness (by changing the flavor of the strange quark \( s \) to an up quark \( u \)). Therefore, this decay must proceed via the weak interaction.

**(a)**
- Identifies \( \pi^{-} \) as \( d\bar{u} \) (1)
- Identifies \( p \) as \( uud \) (1)
- Identifies \( K^{0} \) as \( d\bar{s} \) (1)
- Identifies \( \Lambda^{0} \) as \( uds \) (1)

**(b)**
- Shows conservation of charge (with explicit numbers: \( -1 + 1 = 0 + 0 \)) (1)
- Shows conservation of baryon number (with explicit numbers: \( 0 + 1 = 0 + 1 \)) (1)
- Shows conservation of strangeness (with explicit numbers: \( 0 + 0 = 1 - 1 \)) (1)

**(c)**
- Identifies that strangeness of \( \Lambda^0 \) is \( -1 \) and products have strangeness \( 0 \) (1)
- Concludes that strangeness is not conserved (change of \( +1 \)) (1)
- States that only the weak interaction can change quark flavor and violate strangeness conservation (1)

**(a)** Momentum must be conserved in the collision. In the center-of-mass frame where the electron and positron collide head-on with equal and opposite velocities, the total initial momentum is zero. A single photon always carries non-zero momentum (\( p = E/c \)), so a single photon cannot be produced without violating conservation of momentum. Two photons can travel in opposite directions with equal and opposite momenta, summing to zero.

**(b)** The total energy of the system is conserved.
For each particle:
Total energy \( E = E_k + m_0 c^2 \).
Rest mass of electron/positron \( m_0 = 9.11 \times 10^{-31} \text{ kg} \).
Rest mass energy in Joules: \( m_0 c^2 = 9.11 \times 10^{-31} \times (3.00 \times 10^8)^2 = 8.199 \times 10^{-14} \text{ J} \).
In MeV: \( \frac{8.199 \times 10^{-14} \text{ J}}{1.60 \times 10^{-13} \text{ J/MeV}} \approx 0.512 \text{ MeV} \).

Total energy of one particle: \( E = 2.0 \text{ MeV} + 0.512 \text{ MeV} = 2.512 \text{ MeV} \).
Since two identical photons are produced from two particles of equal energy, each photon carries the energy of one of the particles:
\( E_{\text{photon}} = 2.512 \text{ MeV} \approx 2.51 \text{ MeV} \).

In Joules:
\( E_{\text{photon}} = 2.512 \times 10^6 \times 1.60 \times 10^{-19} \text{ J} = 4.02 \times 10^{-13} \text{ J} \).

**(c)**
Frequency of each photon:
\( E = h f \implies f = \frac{E}{h} = \frac{4.02 \times 10^{-13} \text{ J}}{6.63 \times 10^{-34} \text{ J s}} = 6.06 \times 10^{20} \text{ Hz} \).

Wavelength of each photon:
\( \lambda = \frac{c}{f} = \frac{3.00 \times 10^8 \text{ m s}^{-1}}{6.06 \times 10^{20} \text{ Hz}} = 4.95 \times 10^{-13} \text{ m} \).
(Or using \( \lambda = \frac{hc}{E} \))

**(a)**
- Mentions that momentum must be conserved (1)
- Explains that the initial momentum of a head-on collision is zero, whereas a single photon must have momentum, so at least two photons moving in opposite directions are required (1)

**(b)**
- Identifies that total energy includes kinetic energy and rest energy (1)
- Calculates rest mass energy of an electron/positron as \( 0.51 \text{ MeV} \) (or \( 8.2 \times 10^{-14} \text{ J} \)) (1)
- Calculates energy of one photon as \( 2.51 \text{ MeV} \) (1)
- Converts energy to Joules: \( 4.02 \times 10^{-13} \text{ J} \) (allow \( 4.0 \times 10^{-13} \text{ J} \)) (1)

**(c)**
- Uses \( E = h f \) to calculate frequency (1)
- Obtains \( f = 6.1 \times 10^{20} \text{ Hz} \) (or \( 6.06 \times 10^{20} \text{ Hz} \)) (1)
- Uses \( c = f \lambda \) or \( \lambda = h c / E \) to find wavelength (1)
- Obtains \( \lambda = 4.95 \times 10^{-13} \text{ m} \) (allow \( 5.0 \times 10^{-13} \text{ m} \)) (1)

The mean kinetic energy of a gas molecule is given by \(\frac{3}{2} k T\). Since the helium and argon gases are in thermal equilibrium in the same container, they are at the same temperature \(T\). Therefore, their mean kinetic energies are equal: \(\frac{1}{2} m_{\text{He}} \langle c_{\text{He}}^2 \rangle = \frac{1}{2} m_{\text{Ar}} \langle c_{\text{Ar}}^2 \rangle\). This can be rearranged to find the ratio of their mean square speeds: \(\frac{\langle c_{\text{He}}^2 \rangle}{\langle c_{\text{Ar}}^2 \rangle} = \frac{m_{\text{Ar}}}{m_{\text{He}}\). Substituting the molar masses (which are proportional to the molecular masses): \(\frac{\langle c_{\text{He}}^2 \rangle}{\langle c_{\text{Ar}}^2 \rangle} = \frac{40\text{ g mol}^{-1}}{4\text{ g mol}^{-1}} = 10\). Hence, the correct option is D.

D (1 mark)

The thermal energy transferred to the liquid by the heater in a time interval \(\Delta t\) is given by: \(\Delta E = P \Delta t\). Assuming there are no heat losses to the surroundings, this thermal energy increases the temperature of the liquid according to: \(\Delta E = m c \Delta \theta\). Equating these two expressions gives: \(P \Delta t = m c \Delta \theta\). Rearranging to express the rate of temperature increase (the gradient of the temperature-time graph, \(G\)): \(G = \frac{\Delta \theta}{\Delta t} = \frac{P}{m c}\). Solving for the specific heat capacity \(c\): \(c = \frac{P}{mG}\). Hence, the correct option is A.

A (1 mark)

For isotope X, the number of half-lives that have elapsed in time \(4T\) is: \(n_{\text{X}} = \frac{4T}{T} = 4\). The number of remaining nuclei of X is: \(N_{\text{X}} = N_0 \left(\frac{1}{2}\right)^4 = \frac{N_0}{16}\). For isotope Y, the number of half-lives that have elapsed in time \(4T\) is: \(n_{\text{Y}} = \frac{4T}{2T} = 2\). The number of remaining nuclei of Y is: \(N_{\text{Y}} = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4}\). The ratio of the number of remaining nuclei of X to Y is: \(\frac{N_{\text{X}}}{N_{\text{Y}}} = \frac{N_0 / 16}{N_0 / 4} = \frac{4}{16} = 0.25\). Hence, the correct option is A.

A (1 mark)

The displacement of an object starting from the equilibrium position at \(t = 0\) is given by the equation: \(x = A \sin(\omega t)\), where the angular frequency is \(\omega = \frac{2\pi}{T}\). To find the time taken to reach a displacement of \(\frac{A}{2}\), we substitute this into the displacement equation: \(\frac{A}{2} = A \sin(\omega t) \implies \sin(\omega t) = \frac{1}{2}\). The smallest positive solution for this equation is: \(\omega t = \frac{\pi}{6}\). Substitute \(\omega = \frac{2\pi}{T}\) into this relation: \(\frac{2\pi}{T} t = \frac{\pi}{6} \implies t = \frac{T}{12}\). Hence, the correct option is A.

A (1 mark)

Damping dissipates energy from the oscillating system. As the degree of damping is increased: 1. The maximum amplitude of the oscillations decreases, resulting in a broader and flatter resonance peak. 2. The frequency at which this maximum amplitude occurs (the resonant frequency) shifts slightly to a lower frequency. Therefore, the resonance peak becomes broader and shifts to a lower frequency. Hence, the correct option is B.

B (1 mark)

According to Stefan-Boltzmann's law, the luminosity of a star is given by: \(L = 4\pi R^2 \sigma T^4\). This means that the luminosity \(L\) is proportional to the square of its radius \(R\) and the fourth power of its surface temperature \(T\): \(L \propto R^2 T^4\). For Star P: \(L_{\text{P}} \propto R^2 T^4\). For Star Q: \(L_{\text{Q}} \propto (3R)^2 (2T)^4 = 9R^2 \times 16T^4 = 144 R^2 T^4\). Thus, the ratio of the luminosities is: \(\frac{L_{\text{P}}}{L_{\text{Q}}} = \frac{R^2 T^4}{144 R^2 T^4} = \frac{1}{144}\). Hence, the correct option is A.

A (1 mark)

The distance \(d\) to a star in parsecs (pc) is inversely proportional to its parallax \(p\) in arcseconds: \(d = \frac{1}{p}\). Using the given value: \(d = \frac{1}{0.025} = 40\text{ pc}\). To find the distance in light-years, we multiply the distance in parsecs by the conversion factor \(3.26\text{ light-years per parsec}\): \(d = 40\text{ pc} \times 3.26\text{ light-years pc}^{-1} = 130.4\text{ light-years}\). This is approximately \(130\text{ light-years}\), which corresponds to option D.

D (1 mark)

Redshift \(z\) for a galaxy moving away at non-relativistic speeds is given by the formula: \(z = \frac{v}{c}\), where \(v\) is the recession velocity and \(c\) is the speed of light. Rearranging the equation to solve for the recession velocity \(v\): \(v = z c\). Substitute the given values into the equation: \(v = 0.040 \times 3.0 \times 10^8\text{ m s}^{-1} = 1.2 \times 10^7\text{ m s}^{-1}\). Hence, the correct option is C.

C (1 mark)

The total energy of a simple harmonic oscillator is given by \(E_{\text{total}} = \frac{1}{2} k A^2\), where \(k\) is the force constant of the spring and \(A\) is the amplitude.

The potential energy of the oscillator at a displacement \(x\) is given by \(E_{\text{p}} = \frac{1}{2} k x^2\).

Substituting \(x = \frac{1}{3}A\) into the potential energy equation yields:
\(E_{\text{p}} = \frac{1}{2} k \left(\frac{1}{3}A\right)^2 = \frac{1}{9} \left(\frac{1}{2} k A^2\right) = \frac{1}{9} E_{\text{total}}\)

The kinetic energy \(E_{\text{k}}\) is the difference between the total energy and the potential energy:
\(E_{\text{k}} = E_{\text{total}} - E_{\text{p}} = E_{\text{total}} - \frac{1}{9} E_{\text{total}} = \frac{8}{9} E_{\text{total}}\)

Therefore, the ratio of the kinetic energy to the total energy is \(\frac{E_{\text{k}}}{E_{\text{total}}} = \frac{8}{9}\).

C is the correct answer.
- 1 mark: Correctly determines that potential energy is \(\frac{1}{9}\) of the total energy, and subtracts this from 1 to obtain the kinetic energy fraction of \(\frac{8}{9}\).

According to the ideal gas equation, \(pV = NkT\), which means the absolute temperature \(T\) is proportional to the product of pressure and volume:
\(T \propto pV\)

If the pressure is halved (\(p_{\text{final}} = 0.5 p_{\text{initial}}\)) and the volume is tripled (\(V_{\text{final}} = 3 V_{\text{initial}}\)), the new temperature \(T_{\text{final}}\) is:
\(T_{\text{final}} \propto (0.5 p_{\text{initial}}) \times (3 V_{\text{initial}}) = 1.5 p_{\text{initial}} V_{\text{initial}}\)

Thus, the absolute temperature increases by a factor of \(1.5\), so \(T_{\text{final}} = 1.5 T_{\text{initial}} = \frac{3}{2} T_{\text{initial}}\).

The root-mean-square (r.m.s.) speed of the gas molecules is related to absolute temperature by:
\(v_{\text{rms}} = \sqrt{\frac{3kT}{m}} \propto \sqrt{T}\)

Therefore, the ratio of the final r.m.s. speed to the initial r.m.s. speed is:
\(\frac{v_{\text{rms, final}}}{v_{\text{rms, initial}}} = \sqrt{\frac{T_{\text{final}}}{T_{\text{initial}}}} = \sqrt{\frac{3}{2}}\)

C is the correct answer.
- 1 mark: Correctly links temperature change to the product of pressure and volume, finding that temperature increases by a factor of \(1.5\), and applies the proportional relationship \(v_{\text{rms}} \propto \sqrt{T}\) to find the ratio is \(\sqrt{\frac{3}{2}}\).

To find the total energy required to melt the ice, we must calculate two separate stages:
1. Heating the ice from \(-10^\circ\text{C}\) to \(0^\circ\text{C}\):
\(Q_1 = m c_{\text{ice}} \Delta \theta = 0.15\text{ kg} \times 2100\text{ J kg}^{-1}\text{ K}^{-1} \times (0 - (-10))\text{ K} = 3150\text{ J}\).
2. Melting the ice at \(0^\circ\text{C}\):
\(Q_2 = m L_f = 0.15\text{ kg} \times 3.3 \times 10^5\text{ J kg}^{-1} = 49500\text{ J}\).

Total thermal energy required:
\(Q_{\text{total}} = Q_1 + Q_2 = 3150\text{ J} + 49500\text{ J} = 52650\text{ J}\).

Using the relationship between power, energy, and time:
\(t = \frac{Q_{\text{total}}}{P} = \frac{52650\text{ J}}{80\text{ W}} = 658.125\text{ s}\).
Rounding to 3 significant figures gives \(658\text{ s}\).

(a) Specific latent heat of fusion is defined as the energy per unit mass required to change a substance from solid to liquid phase at a constant temperature. (2 marks)
(b)
- Use of \(Q = m c \Delta \theta\) to calculate heating energy: \(Q_1 = 3150\text{ J}\). (1 mark)
- Use of \(Q = m L\) to calculate phase change energy: \(Q_2 = 49500\text{ J}\). (1 mark)
- Addition of both energies to find total energy: \(Q_{\text{total}} = 52650\text{ J}\). (1 mark)
- Use of \(P = \frac{E}{t}\) to calculate time: \(t = \frac{52650}{80}\). (1 mark)
- Correct final answer with units: \(658\text{ s}\) (accept \(660\text{ s}\) for 2 s.f.). (1 mark)

(a) Convert temperature to Kelvin:
\(T = 27 + 273.15 = 300.15\text{ K}\) (or \(300\text{ K}\)).
Using the ideal gas equation:
\(pV = N k T \Rightarrow N = \frac{pV}{kT}\)
\(N = \frac{1.20 \times 10^5\text{ Pa} \times 0.045\text{ m}^3}{1.38 \times 10^{-23}\text{ J K}^{-1} \times 300\text{ K}} = \frac{5400}{4.14 \times 10^{-21}} \approx 1.30 \times 10^{24}\text{ atoms}\).

(b) The average kinetic energy of gas molecules is:
\(\frac{1}{2} m \langle c^2 \rangle = \frac{3}{2} k T \Rightarrow v_{\text{rms}} = \sqrt{\frac{3 k T}{m}}\).
Mass of one helium atom, \(m = \frac{\text{molar mass}}{N_A} = \frac{4.00 \times 10^{-3}\text{ kg mol}^{-1}}{6.02 \times 10^{23}\text{ mol}^{-1}} = 6.64 \times 10^{-27}\text{ kg}\).
Therefore:
\(v_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23}\text{ J K}^{-1} \times 300\text{ K}}{6.64 \times 10^{-27}\text{ kg}}} = \sqrt{1.871 \times 10^6} \approx 1368\text{ m s}^{-1}\).
Using 3 significant figures, \(v_{\text{rms}} = 1370\text{ m s}^{-1}\).

(a)
- Convert temperature to Kelvin: \(300\text{ K}\). (1 mark)
- Correct rearrangement of \(pV = NkT\) to make \(N\) the subject. (1 mark)
- Correct calculation leading to \(1.3 \times 10^{24}\). (1 mark)

(b)
- Calculate the mass of a single helium atom: \(6.64 \times 10^{-27}\text{ kg}\). (1 mark)
- Recall and substitute into the equation \(\frac{1}{2} m \langle c^2 \rangle = \frac{3}{2} k T\) or \(v_{\text{rms}} = \sqrt{\frac{3 k T}{m}}\). (2 marks)
- Correct final calculation of \(1370\text{ m s}^{-1}\) (allow range \(1360 - 1380\text{ m s}^{-1}\)). (1 mark)

(a) Lead-206 is stable, and standard geological models assume no lead was trapped in the molten magma at the time the rock solidified, meaning all current lead must have originated from the parent uranium isotope.

(b) To find the age of the rock, we calculate the initial amount of uranium-238.
Number of moles of U-238 remaining:
\(n_{\text{U}} = \frac{1.50 \times 10^{-3}\text{ g}}{238\text{ g mol}^{-1}} = 6.303 \times 10^{-6}\text{ mol}\).

Number of moles of Pb-206 formed:
\(n_{\text{Pb}} = \frac{0.45 \times 10^{-3}\text{ g}}{206\text{ g mol}^{-1}} = 2.184 \times 10^{-6}\text{ mol}\).

Since one U-238 atom decays to form one Pb-206 atom, the initial moles of U-238 was:
\(n_{\text{initial}} = n_{\text{U}} + n_{\text{Pb}} = 6.303 \times 10^{-6} + 2.184 \times 10^{-6} = 8.487 \times 10^{-6}\text{ mol}\).

Using the decay equation:
\(N = N_0 e^{-\lambda t} \Rightarrow \frac{N}{N_0} = \frac{n_{\text{U}}}{n_{\text{initial}}} = \frac{6.303 \times 10^{-6}}{8.487 \times 10^{-6}} = 0.7427\).

The decay constant \(\lambda\) is:
\(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{4.47 \times 10^9\text{ years}} = 1.551 \times 10^{-10}\text{ year}^{-1}\).

Solving for \(t\):
\(\ln(0.7427) = -\lambda t \Rightarrow -0.2975 = - (1.551 \times 10^{-10}) t\)
\(t = \frac{0.2975}{1.551 \times 10^{-10}} \approx 1.918 \times 10^9\text{ years}\).
This is approximately \(1.92 \times 10^9\text{ years}\).

(a)
- Lead has a lower melting point / is excluded during solidification, or rock is assumed to have no initial Pb-206. (1 mark)

(b)
- Convert mass of both U-238 and Pb-206 to moles: \(n_{\text{U}} = 6.30 \times 10^{-6}\text{ mol}\) and \(n_{\text{Pb}} = 2.18 \times 10^{-6}\text{ mol}\). (1 mark)
- Calculate the initial number of moles of U-238: \(n_{\text{initial}} = 8.49 \times 10^{-6}\text{ mol}\). (1 mark)
- Use \(\frac{N}{N_0}\) ratio: \(0.743\). (1 mark)
- Use \(\lambda = \frac{\ln 2}{T_{1/2}}\) to find decay constant: \(1.55 \times 10^{-10}\text{ yr}^{-1}\). (1 mark)
- Correctly apply decay equation \(N = N_0 e^{-\lambda t}\). (1 mark)
- Final answer: \(1.92 \times 10^9\text{ years}\) (accept range \(1.90 \times 10^9\text{ yr} - 1.95 \times 10^9\text{ yr}\)). (1 mark)

(a) Binding energy is the minimum work/energy required to completely separate a nucleus into its constituent individual protons and neutrons (nucleons).

(b) Total energy released is calculated from the difference in the total binding energy of the products and the reactants.

Total binding energy of reactants (only \(^{235}\text{U}\) as neutron has zero binding energy):
\(E_{\text{reactants}} = 235 \times 7.59\text{ MeV} = 1783.65\text{ MeV}\).

Total binding energy of products:
\(E_{\text{products}} = (141 \times 8.33\text{ MeV}) + (92 \times 8.51\text{ MeV}) = 1174.53\text{ MeV} + 782.92\text{ MeV} = 1957.45\text{ MeV}\).

Energy released:
\(\Delta E = E_{\text{products}} - E_{\text{reactants}} = 1957.45\text{ MeV} - 1783.65\text{ MeV} = 173.8\text{ MeV}\).
Rounding to 3 significant figures gives \(174\text{ MeV}\).

(a)
- Energy required to separate a nucleus into its constituent nucleons (or energy released when nucleons form a nucleus). (1 mark)
- Must mention 'constituent nucleons' or 'individual protons and neutrons' to secure second mark. (1 mark)

(b)
- Calculate total binding energy of U-235: \(1783.65\text{ MeV}\). (1 mark)
- Calculate total binding energy of Ba-141: \(1174.53\text{ MeV}\). (1 mark)
- Calculate total binding energy of Kr-92: \(782.92\text{ MeV}\). (1 mark)
- Sum the binding energies of the products: \(1957.45\text{ MeV}\). (1 mark)
- Calculate the difference to find energy released: \(174\text{ MeV}\) (accept \(173.8\text{ MeV}\)). (1 mark)

(a) The angular frequency \(\omega\) is given by:

\(\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{25\text{ N m}^{-1}}{0.35\text{ kg}}} \approx 8.452\text{ rad s}^{-1}\).

The maximum acceleration \(a_{\text{max}}\) is:
\(a_{\text{max}} = \omega^2 A\) where \(A = 0.080\text{ m}\).
\(a_{\text{max}} = 71.43\text{ s}^{-2} \times 0.080\text{ m} \approx 5.71\text{ m s}^{-2}\), which is approximately \(5.7\text{ m s}^{-2}\).

(b) The velocity \(v\) of an oscillator at displacement \(x\) is given by:
\(v = \pm \omega \sqrt{A^2 - x^2}\)
\(v = 8.452 \times \sqrt{0.080^2 - 0.040^2} = 8.452 \times \sqrt{0.0064 - 0.0016} = 8.452 \times \sqrt{0.0048}\)
\(v = 8.452 \times 0.06928 \approx 0.586\text{ m s}^{-1}\).
Rounding to 2 significant figures gives \(0.59\text{ m s}^{-1}\).

(a)
- Use of \(\omega^2 = \frac{k}{m}\) or \(\omega = \sqrt{\frac{k}{m}}\). (1 mark)
- Recall of \(a = -\omega^2 x\) to identify \(a_{\text{max}} = \omega^2 A\). (1 mark)
- Calculate \(a_{\text{max}} = 5.71\text{ m s}^{-2}\) clearly showing substitution. (1 mark)

(b)
- Recall and substitute into \(v = \pm \omega \sqrt{A^2 - x^2}\). (2 marks)
- Correct substitution of values: \(v = 8.45 \times \sqrt{0.080^2 - 0.040^2}\). (1 mark)
- Final speed: \(0.59\text{ m s}^{-1}\) (or \(0.586\text{ m s}^{-1}\)). (1 mark)

Free oscillations occur when a system is displaced from its equilibrium position and allowed to oscillate at its natural frequency without any periodic external force acting on it.
Forced oscillations occur when a periodic external driving force acts on a system, forcing it to oscillate at the frequency of the driving force rather than its natural frequency.
Resonance occurs when the driving frequency is equal to (or very close to) the natural frequency of the oscillating system, resulting in a maximum transfer of energy from the driver to the oscillator, causing a large peak in amplitude.
When a heavy damping mechanism is introduced:
1. The peak amplitude of the resonance curve is greatly reduced.
2. The resonance curve becomes broader and flatter, meaning the system responds significantly over a wider range of frequencies.
3. The resonant frequency (the frequency at which peak amplitude occurs) shifts slightly to a lower frequency.

- Free oscillations: oscillation at natural frequency / without external periodic force. (1 mark)
- Forced oscillations: system is driven by an external periodic force at the driver's frequency. (1 mark)
- Resonance: occurs when driving frequency matches natural frequency, leading to maximum energy transfer / maximum amplitude. (1 mark)
- Damping definition: resistive forces that dissipate mechanical energy of the system as thermal energy. (1 mark)
- Effect on peak amplitude: Peak amplitude is lower/reduced. (1 mark)
- Effect on width: Peak of the curve is broader/flatter. (1 mark)
- Effect on peak frequency: Resonant frequency shifts slightly to a lower frequency. (1 mark)

(a) According to Wien's displacement law:
\(\lambda_{\text{max}} T = 2.898 \times 10^{-3}\text{ m K}\).
\(\lambda_{\text{max}} = \frac{2.898 \times 10^{-3}\text{ m K}}{12100\text{ K}} = 2.395 \times 10^{-7}\text{ m}\) (or \(240\text{ nm}\)).

(b) Luminosity of Rigel is:
\(L = 1.20 \times 10^5 \times 3.85 \times 10^{26}\text{ W} = 4.62 \times 10^{31}\text{ W}\).

According to Stefan-Boltzmann law:
\(L = 4 \pi r^2 \sigma T^4\) where \(\sigma = 5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}\).

Rearranging for radius \(r\):
\(r = \sqrt{\frac{L}{4 \pi \sigma T^4}}\)
\(r = \sqrt{\frac{4.62 \times 10^{31}}{4 \pi \times (5.67 \times 10^{-8}) \times 12100^4}}\)
\(r = \sqrt{\frac{4.62 \times 10^{31}}{7.125 \times 10^{-7} \times 2.1436 \times 10^{16}}} = \sqrt{\frac{4.62 \times 10^{31}}{1.5273 \times 10^{10}}} = \sqrt{3.025 \times 10^{21}}\)
\(r \approx 5.50 \times 10^{10}\text{ m}\).

(a)
- Use of Wien's displacement law \(\lambda_{\text{max}} T = 2.898 \times 10^{-3}\text{ m K}\). (1 mark)
- Correct calculation: \(2.40 \times 10^{-7}\text{ m}\) (accept \(2.39 - 2.40 \times 10^{-7}\text{ m}\)). (1 mark)

(b)
- Calculate luminosity of Rigel: \(4.62 \times 10^{31}\text{ W}\). (1 mark)
- Recall Stefan-Boltzmann law \(L = 4 \pi r^2 \sigma T^4\). (1 mark)
- Correct rearrangement of equation for \(r\). (1 mark)
- Substitution of correct numerical values including \(T^4\). (1 mark)
- Correct final answer with unit: \(5.50 \times 10^{10}\text{ m}\) (accept range \(5.45 \times 10^{10}\text{ m} - 5.55 \times 10^{10}\text{ m}\)). (1 mark)

(a) The wavelength has increased (redshifted), indicating that the galaxy is moving away from the Earth (receding).

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

Using the Doppler equation:
\(z = \frac{\Delta \lambda}{\lambda} = \frac{v}{c}\)
\(v = c \times \frac{\Delta \lambda}{\lambda} = 3.00 \times 10^8\text{ m s}^{-1} \times \frac{15.8\text{ nm}}{656.3\text{ nm}} = 7.222 \times 10^6\text{ m s}^{-1}\) (or \(7222\text{ km s}^{-1}\)).

Using Hubble's Law:
\(v = H_0 d \Rightarrow d = \frac{v}{H_0}\).

Using \(H_0 = 70\text{ km s}^{-1}\text{ Mpc}^{-1}\):
\(d = \frac{7222\text{ km s}^{-1}}{70\text{ km s}^{-1}\text{ Mpc}^{-1}} \approx 103\text{ Mpc}\).

Alternatively, using SI units:
\(d = \frac{7.222 \times 10^6\text{ m s}^{-1}}{2.27 \times 10^{-18}\text{ s}^{-1}} \approx 3.18 \times 10^{24}\text{ m}\).

(a)
- Galaxy is moving away / receding from Earth because the light is redshifted. (1 mark)

(b)
- Calculate the change in wavelength: \(\Delta \lambda = 15.8\text{ nm}\). (1 mark)
- Use \(z = \frac{\Delta \lambda}{\lambda}\) to find redshift \(z = 0.02407\). (1 mark)
- Use \(v = z c\) to calculate velocity: \(7.22 \times 10^6\text{ m s}^{-1}\) (or \(7220\text{ km s}^{-1}\)). (1 mark)
- Use Hubble's law \(v = H_0 d\). (1 mark)
- Substitution of correct values to find distance. (1 mark)
- Correct final answer with unit: \(103\text{ Mpc}\) (allow \(100 - 105\text{ Mpc}\)) or \(3.18 \times 10^{24}\text{ m}\) (allow \(3.15 - 3.25 \times 10^{24}\text{ m}\)). (1 mark)

(a) Convert half-life to seconds: \(T_{1/2} = 5730 \times 3.16 \times 10^7\text{ s} = 1.81 \times 10^{11}\text{ s}\). Use the decay constant equation: \(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{1.81 \times 10^{11}\text{ s}} = 3.83 \times 10^{-12}\text{ s}^{-1}\). (b) Use the radioactive decay law: \(A = A_0 e^{-\lambda t}\). Rearranging gives \(\ln\left(\frac{A}{A_0}\right) = -\lambda_{year} t\). Find the decay constant in year^{-1}: \(\lambda_{year} = \frac{\ln 2}{5730\text{ years}} = 1.21 \times 10^{-4}\text{ year}^{-1}\). Substitute values: \(\ln\left(\frac{6.2}{15.3}\right) = - (1.21 \times 10^{-4}\text{ year}^{-1}) t\), so \(-0.903 = - (1.21 \times 10^{-4}) t\), which gives \(t \approx 7460\text{ years}\). Allow rounding to \(7500\text{ years}\).

Part (a): MP1: Correct conversion of half-life to seconds (1.81 * 10^11 s). (1 mark) MP2: Uses lambda = ln(2) / T_1/2. (1 mark) MP3: Correct calculation of decay constant in range 3.81 * 10^-12 to 3.84 * 10^-12 s^-1. (1 mark) Part (b): MP4: Uses A = A_0 e^(-lambda * t) or equivalent logarithmic form. (1 mark) MP5: Calculates lambda in year^-1 (1.21 * 10^-4 year^-1) OR calculates decay time in seconds (2.36 * 10^11 s). (1 mark) MP6: Shows appropriate algebraic rearrangement to make t the subject. (1 mark) MP7: Correct age of the artifact in the range 7400 to 7500 years. (1 mark)

(a) Use the ideal gas equation: \(pV = nRT\). Convert temperature to Kelvin: \(T = 20 + 273 = 293\text{ K}\). Solve for volume: \(V = \frac{nRT}{p} = \frac{0.045 \times 8.31 \times 293}{1.01 \times 10^5} = 1.08 \times 10^{-3}\text{ m}^3\). (b) Since the volume is constant, \(\frac{p_1}{T_1} = \frac{p_2}{T_2}\). Solve for final temperature: \(T_2 = T_1 \times \frac{p_2}{p_1} = 293 \times \frac{2.50 \times 10^5}{1.01 \times 10^5} = 725.2\text{ K}\). The temperature change is \(\Delta T = 725.2 - 293 = 432.2\text{ K}\). Find the mass of the gas: \(m = n \times M = 0.045 \times 4.00 \times 10^{-3} = 1.80 \times 10^{-4}\text{ kg}\). Calculate thermal energy: \(\Delta E = m c \Delta T = (1.80 \times 10^{-4}) \times (3.12 \times 10^3) \times 432.2 = 243\text{ J}\). This rounds to \(240\text{ J}\) to two significant figures.

Part (a): MP1: Converts Celsius temperature to Kelvin (293 K). (1 mark) MP2: Rearranges pV = nRT to find V. (1 mark) MP3: Correct calculation of volume in range 1.08 * 10^-3 to 1.1 * 10^-3 m^3. (1 mark) Part (b): MP4: Calculates the final temperature (725 K) or the temperature change (432 K). (1 mark) MP5: Calculates mass of helium gas (1.80 * 10^-4 kg). (1 mark) MP6: Uses thermal energy equation Delta E = m * c * Delta T. (1 mark) MP7: Correct value of thermal energy in range 240 J to 243 J. (1 mark)

(a) Apply Wien's displacement law: \(\lambda_{\text{max}} T = 2.898 \times 10^{-3}\text{ m K}\). Therefore, \(T = \frac{2.898 \times 10^{-3}}{380 \times 10^{-9}} = 7626\text{ K}\). (b) Apply Stefan-Boltzmann law: \(L = 4\pi r^2 \sigma T^4\). Rearranging for radius: \(r = \sqrt{\frac{L}{4\pi \sigma T^4}}\). Substitute values: \(r = \sqrt{\frac{4.5 \times 10^{28}}{4 \pi \times (5.67 \times 10^{-8}) \times (7626)^4}} = \sqrt{\frac{4.5 \times 10^{28}}{2.41 \times 10^9}} = 4.3 \times 10^9\text{ m}\). (c) The distance can be determined using trigonometric parallax. By measuring the apparent change in position of the star relative to more distant background stars at six-month intervals (opposite ends of the Earth's orbit), the parallax angle can be determined. Using trigonometry or the relationship \(d = 1/p\), the distance can then be calculated.

Part (a): MP1: Uses lambda_max * T = 2.898 * 10^-3 m K. (1 mark) MP2: Correct temperature in the range 7600 K to 7630 K. (1 mark) Part (b): MP3: Uses Stefan-Boltzmann law L = 4 * pi * r^2 * sigma * T^4. (1 mark) MP4: Shows rearrangement to solve for r. (1 mark) MP5: Calculates radius in the range 4.3 * 10^9 m to 4.4 * 10^9 m. (1 mark) Part (c): MP6: Describes measuring the position of the star against distant background stars at opposite sides of Earth's orbit (6 months apart). (1 mark) MP7: Explains how the parallax angle is used to calculate the distance d using d = 1/p or trigonometry. (1 mark)

(a) The circuit should show an alternating current (a.c.) power supply in series with the solenoid, an a.c. ammeter, and a variable resistor (or rheostat) to keep the current amplitude constant. The search coil must be connected directly to the input of an oscilloscope. (b) An oscilloscope displays the voltage waveform, allowing the student to directly and accurately measure the peak voltage (amplitude) V_0. A standard digital multimeter is typically calibrated for low frequencies (e.g. 50-60 Hz) and measures root-mean-square (r.m.s.) voltage, which would not give the direct peak value at varying frequencies. (c) Since the number of turns N, area A, and peak magnetic flux density B_0 are held constant, the equation V_0 = (2\pi N A B_0) f is of the form y = mx, which represents a straight line through the origin with a gradient of 2\pi N A B_0. (d) Percentage uncertainty = (absolute uncertainty / measured value) \times 100 = (2 / 24) \times 100 = 8.3%. (e) Since gradient m = 2\pi N A B_0, we have B_0 = m / (2\pi N A). Substituting the values: B_0 = 2.45 \times 10^{-4} / (2 \times \pi \times 500 \times 1.2 \times 10^{-4}) = 2.45 \times 10^{-4} / 0.377 = 6.50 \times 10^{-4} T. (f) The solenoid can become very hot due to high currents over time. The student should switch off the current between taking measurements to prevent overheating and potential burns.

(a) [3 marks]: 1 mark for correct symbol of a.c. power supply, solenoid, and series ammeter. 1 mark for showing a variable resistor in series to control current. 1 mark for search coil connected to an oscilloscope. (b) [2 marks]: 1 mark for stating that the oscilloscope allows direct measurement of peak voltage from the displayed waveform. 1 mark for stating that standard multimeters measure r.m.s. or are inaccurate at high frequencies. (c) [2 marks]: 1 mark for identifying that 2, \pi, N, A, and B_0 are constants. 1 mark for relating the equation to y = mx, which has a y-intercept of zero. (d) [1 mark]: 1 mark for 8.3% (accept 8% or 8.33%). (e) [3 marks]: 1 mark for identifying that gradient = 2\pi N A B_0. 1 mark for rearranging to make B_0 the subject. 1 mark for correct calculation showing 6.5 \times 10^{-4} T (accept 6.50 \times 10^{-4} T). (f) [1.5 marks]: 1 mark for stating safety precaution (e.g., switch off current between readings) and 0.5 mark for the reason (to prevent overheating / burns).

(a) The student should stir the water bath thoroughly before taking any temperature and pressure readings, and wait for a short period after heating has stopped to ensure thermal equilibrium between the water bath and the flask of air. (b) Absolute zero can be found by extrapolating the straight-line graph to the horizontal axis where the pressure p is zero. The temperature coordinate at this intercept is the value of absolute zero in degrees Celsius. (c) Pressure is only directly proportional to absolute temperature if both volume and mass are kept constant, according to the pressure law. When using a U-tube manometer, heating the gas causes it to expand and push down the liquid on the flask side. To keep the volume constant, the open side of the U-tube must be adjusted vertically so that the liquid level on the flask side returns to a fixed reference mark before each reading. (d) The equation of the line is p = m \theta + c. Setting p = 0 gives 0 = m \theta_0 + c, which rearranges to \theta_0 = -c / m. Substituting the values: \theta_0 = - (9.89 \times 10^4 Pa) / (362 Pa ^\circ C^{-1}) = -273.2 ^\circ C. (e) The percentage uncertainty in c is (0.15 \times 10^4 / 9.89 \times 10^4) \times 100% = 1.52%. The percentage uncertainty in m is (12 / 362) \times 100% = 3.31%. Since \theta_0 = c / m, the percentage uncertainty in \theta_0 is the sum of these percentage uncertainties: 1.52% + 3.31% = 4.83%.

(a) [2 marks]: 1 mark for stating that the water bath must be stirred. 1 mark for explaining that this ensures thermal equilibrium/uniform temperature throughout. (b) [1 mark]: 1 mark for stating that absolute zero is the x-intercept (or temperature where pressure is zero) obtained by extrapolating the line of best fit. (c) [3 marks]: 1 mark for referencing the Pressure Law / Gas Law where p is proportional to T only if volume is constant. 1 mark for explaining that gas expansion pushes liquid level down. 1 mark for explaining that the adjustable tube must be raised/lowered to return the liquid to the index mark. (d) [2.5 marks]: 1 mark for setting p = 0 and writing the expression \theta_0 = -c / m. 1.5 marks for correct calculation resulting in -273 ^\circ C (or -273.2 ^\circ C). (e) [4 marks]: 1 mark for percentage uncertainty in c = 1.52%. 1 mark for percentage uncertainty in m = 3.31%. 1 mark for stating that percentage uncertainties should be added. 1 mark for final percentage uncertainty of 4.8% (or 4.83%).

(a) Reaction time error (typically around 0.1 to 0.2 seconds) is a constant random error. By timing 20 oscillations, this absolute timing error is distributed over 20 cycles, which reduces the percentage uncertainty in the determined period T by a factor of 20. (b) A fiducial marker (such as a pin or card) should be placed at the equilibrium position of the cantilever's oscillation. The student should start and stop the stopwatch at the instant the cantilever passes this marker, because the cantilever is moving at its maximum speed at the equilibrium position, which minimizes the timing judgment uncertainty. (c) The mean time for 20 oscillations is t_mean = (14.82 + 14.76 + 14.88) / 3 = 14.82 s. The range of the measurements is 14.88 - 14.76 = 0.12 s. The absolute uncertainty in the total time is range / 2 = 0.12 / 2 = 0.06 s. The mean time period T for a single oscillation is T = 14.82 / 20 = 0.741 s. The absolute uncertainty in T is 0.06 / 20 = 0.003 s. Thus, T = (0.741 \pm 0.003) s. (d) From the equation, the gradient g = 4\pi^2 / k. Therefore, k = 4\pi^2 / g = 4\pi^2 / 1.12 = 35.25 N m^{-1}$. The vertical intercept c = 4\pi^2 m_{eff} / k. Since c = g \times m_{eff}, we have m_{eff} = c / g = 0.045 / 1.12 = 0.0402 kg (or 40.2 g).

(a) [2 marks]: 1 mark for stating that reaction time error is constant. 1 mark for explaining that timing more oscillations reduces the percentage uncertainty in T. (b) [2 marks]: 1 mark for placing the marker at the equilibrium (center) position. 1 mark for explaining that the cantilever moves fastest here, reducing judgment error. (c) [4 marks]: 1 mark for calculating mean time t = 14.82 s. 1 mark for finding the absolute uncertainty in total time as range/2 = 0.06 s. 1 mark for dividing mean time by 20 to get T = 0.741 s. 1 mark for dividing the uncertainty by 20 to get 0.003 s. (d) [4.5 marks]: 1 mark for identifying gradient = 4\pi^2/k. 1 mark for calculating stiffness k = 35.3 N m^{-1} (accept 35 to 35.3). 1 mark for identifying vertical intercept = 4\pi^2 m_{eff}/k. 1 mark for showing m_{eff} = c / gradient. 0.5 mark for calculating m_{eff} = 0.040 kg (or 40 g, accept 0.040 to 0.0402 kg).

(a) To measure the background count rate, the radioactive gamma source is placed inside its lead-lined storage container far away from the Geiger-Muller (GM) tube. The count rate is recorded for a long time period (e.g., 10 minutes) and then divided by the time. It must be subtracted from all subsequent readings because background radiation is always present from cosmic rays and rocks, and subtracting it ensures we only measure the radiation coming from the source itself. (b) Taking the natural logarithm of both sides of C = C_0 e^{-\mu x} gives \ln C = \ln C_0 - \mu x. This matches the linear equation y = mx + c, where y = \ln C, x = x, and gradient m = -\mu. Therefore, the graph of \ln C against x is a straight line, and the linear attenuation coefficient \mu is equal to the negative of the gradient. (c) Table of values: For x = 0.0 mm, \ln C = \ln(124) = 4.82. For x = 5.0 mm, \ln C = \ln(76) = 4.33. For x = 10.0 mm, \ln C = \ln(45) = 3.81. For x = 15.0 mm, \ln C = \ln(28) = 3.33. For x = 20.0 mm, \ln C = \ln(17) = 2.83. (d) When the count rate is halved, C = 0.5 C_0. Substituting this into the equation: 0.5 C_0 = C_0 e^{-\mu x_{1/2}} \implies 0.5 = e^{-\mu x_{1/2}}. Taking natural logs: \ln(0.5) = -\mu x_{1/2} \implies x_{1/2} = \ln 2 / \mu. Given gradient = -\mu = -0.100 mm^{-1}$, we have \mu = 0.100 mm^{-1}. Therefore, x_{1/2} = 0.693 / 0.100 mm^{-1} = 6.93 mm. (e) Safety measures include: 1. Keep the source in a lead-lined container when not in use. 2. Handle the source only using long-handled tongs to maximize distance. 3. Point the source away from the body and never point it towards anyone else. 4. Minimize the duration of exposure to the source.

(a) [2.5 marks]: 1 mark for stating that the source must be removed/stored away during background measurement. 0.5 mark for measuring counts over a long time and dividing by time to get rate. 1 mark for stating that background radiation is always present and subtraction gives the true/corrected count rate of the source. (b) [2 marks]: 1 mark for correctly showing the mathematical step \ln C = \ln C_0 - \mu x. 1 mark for stating that comparing this to y = mx + c shows gradient = -\mu (or \mu = -gradient). (c) [2 marks]: 2 marks for calculating all 5 values of \ln C correctly (4.82, 4.33, 3.81, 3.33, 2.83) to 2 or 3 decimal places (deduct 1 mark if table structure is missing or values are incorrectly rounded). (d) [3 marks]: 1 mark for using the relationship C / C_0 = 0.5 to show x_{1/2} = \ln 2 / \mu. 1 mark for identifying \mu = 0.100 mm^{-1} from the gradient. 1 mark for calculating x_{1/2} = 6.93 mm (accept 6.9 mm). (e) [3 marks]: 1 mark each for any three valid safety measures: using long-handled tongs, keeping the source in a lead container when not in use, pointing the source away from people, keeping a safe distance, or minimizing exposure time (maximum 3 marks).