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The horizontal component of the velocity remains constant throughout the motion, so \(v_x = v\). The vertical component of the velocity immediately before impact is \(v_y\). Since the velocity vector makes an angle \(\theta\) with the horizontal, we have \(\tan\theta = \frac{v_y}{v_x} = \frac{v_y}{v}\), which gives \(v_y = v \tan\theta\). Using the equation of motion for the vertical direction, \(v_y^2 = u_y^2 + 2gh\). Since the ball is projected horizontally, its initial vertical velocity \(u_y = 0\). Substituting the expressions gives \((v \tan\theta)^2 = 2gh\). Rearranging for \(h\) yields \(h = \frac{v^2 \tan^2\theta}{2g}\).

1 mark for the correct option (A).

The impulse on an object is equal to the area under the force-time graph, and it also equals the change in momentum. The force-time graph consists of: 1) A rectangular section from \(t = 0\text{ s}\) to \(t = 2.0\text{ s}\) with area \(15\text{ N} \times 2.0\text{ s} = 30\text{ N s}\). 2) A triangular section from \(t = 2.0\text{ s}\) to \(t = 6.0\text{ s}\) with area \(\frac{1}{2} \times 4.0\text{ s} \times 15\text{ N} = 30\text{ N s}\). The total impulse is \(30\text{ N s} + 30\text{ N s} = 60\text{ N s}\). Since impulse equals change in momentum, \(\Delta p = m \Delta v\), we have \(60\text{ N s} = 3.0\text{ kg} \times (v_f - 0)\), which gives \(v_f = 20\text{ m s}^{-1}\).

1 mark for the correct option (C).

The extension \(\Delta L\) of a wire is given by \(\Delta L = \frac{F L}{A E}\), where \(A = \frac{\pi d^2}{4}\). Thus, \(\Delta L \propto \frac{L}{d^2 E}\). We can set up the ratio: \(\frac{\Delta L_{\text{P}}}{\Delta L_{\text{Q}}} = \left(\frac{L_{\text{P}}}{L_{\text{Q}}}\right) \times \left(\frac{d_{\text{Q}}}{d_{\text{P}}}\right)^2 \times \left(\frac{E_{\text{Q}}}{E_{\text{P}}}\right)\). Substituting the given relations where \(L_{\text{P}} = 2 L_{\text{Q}}\), \(d_{\text{P}} = 0.5 d_{\text{Q}}\), and \(E_{\text{P}} = 2 E_{\text{Q}}\), we get: \(\frac{\Delta L_{\text{P}}}{\Delta L_{\text{Q}}} = 2 \times \left(\frac{1}{0.5}\right)^2 \times \frac{1}{2} = 2 \times 4 \times 0.5 = 4.0\).

1 mark for the correct option (D).

The useful power output needed to lift the load at constant speed is \(P_{\text{out}} = F v = m g v = 80\text{ kg} \times 9.81\text{ m s}^{-2} \times 1.5\text{ m s}^{-1} = 1177.2\text{ W}\). The efficiency \(\eta\) is defined as \(\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\). Therefore, the electrical input power is \(P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{1177.2\text{ W}}{0.60} = 1962\text{ W} \approx 2.0\text{ kW}\).

1 mark for the correct option (C).

At terminal velocity, the downward force (weight) equals the upward viscous drag force: \(W = F_{\text{drag}}\). The weight of the sphere is \(W = m g = \rho V g = \rho \left(\frac{4}{3}\pi r^3\right) g \propto r^3\). According to Stokes' Law, the drag force is \(F_{\text{drag}} = 6 \pi \eta r v \propto r v\). Setting these proportional to each other: \(r^3 \propto r v \implies v \propto r^2\). Since the radius of the second sphere is doubled to \(2r\), its terminal velocity becomes \((2)^2 v = 4v\).

1 mark for the correct option (C).

Let the tension at the left end be \(T_{\text{L}} = 120\text{ N}\) and at the right end be \(T_{\text{R}}\). For vertical equilibrium, the sum of the upward forces must equal the downward weight: \(T_{\text{L}} + T_{\text{R}} = W \implies 120\text{ N} + T_{\text{R}} = 300\text{ N} \implies T_{\text{R}} = 180\text{ N}\). Taking moments about the left end: \(\sum M_{\text{left}} = 0 \implies (W \times x) - (T_{\text{R}} \times 4.0\text{ m}) = 0\), where \(x\) is the distance of the center of gravity from the left end. This gives \(300\text{ N} \times x = 180\text{ N} \times 4.0\text{ m} \implies 300x = 720 \implies x = 2.4\text{ m}\).

1 mark for the correct option (C).

The ultimate tensile strength (UTS) is defined as the maximum value of stress on the stress-strain curve, representing the maximum nominal tensile stress the material can withstand. Option A is incorrect because the Young modulus is the gradient of the linear region, not the area. Option C is incorrect because beyond the elastic limit, the material undergoes permanent plastic deformation. Option D is incorrect because the Young modulus (the gradient of the linear region) is an intensive material property and is independent of the wire's cross-sectional area.

1 mark for the correct option (B).

Since the block slides at a constant velocity, the forces acting on it are in equilibrium. Resolving parallel to the incline: \(f = m g \sin\theta\), where \(f\) is the frictional force. Resolving perpendicular to the incline: \(R = m g \cos\theta\), where \(R\) is the normal reaction force. The coefficient of dynamic friction is \(\mu = \frac{f}{R}\). Substituting the components: \(\mu = \frac{m g \sin\theta}{m g \cos\theta} = \tan\theta\).

1 mark for the correct option (C).

Let the tension in the rope at end Y be \(T\). The tension in the rope at end X is therefore \(3T\). For vertical equilibrium, the sum of upward forces must equal the downward force (the weight \(W\)): \(3T + T = W \implies 4T = W \implies T = \frac{W}{4}\). Taking moments about end X: Clockwise moment due to weight is \(W \times x\), where \(x\) is the distance of the centre of gravity from end X. Anticlockwise moment due to the tension at Y is \(T \times L = \frac{W}{4} \times L\). For rotational equilibrium: \(W x = \frac{W}{4} L \implies x = 0.25 L\). Thus, the distance of the centre of gravity from end X is \(0.25 L\).

A is the correct answer (1 mark).

The Young modulus \(E\) is given by \(E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{\Delta x / L} = \frac{F L}{A \Delta x}\). Since the cross-sectional area of a wire of diameter \(d\) is \(A = \frac{\pi d^2}{4}\), we have \(\Delta x = \frac{4 F L}{\pi d^2 E}\). Because both wires are made of the same metal, their Young modulus \(E\) is the same. Under the same tensile force \(F\), the extension \(\Delta x\) is proportional to \(\frac{L}{d^2}\). For wire P, \(\Delta x_{\text{P}} \propto \frac{L}{d^2}\). For wire Q, \(\Delta x_{\text{Q}} \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\). Taking the ratio gives \(\frac{\Delta x_{\text{P}}}{\Delta x_{\text{Q}}} = \frac{L/d^2}{L/(2d^2)} = 2\).

C is the correct answer (1 mark).

(a) Consider vertical motion:
\(u_y = 0\text{ m s}^{-1}\)
\(s_y = 80\text{ m}\)
\(a_y = 9.81\text{ m s}^{-2}\)
Using \(s = ut + \frac{1}{2}at^2\):
\(80 = 0 + 0.5 \times 9.81 \times t^2\)
\(t^2 = \frac{160}{9.81} = 16.31\)
\(t = 4.04\text{ s}\) (which is approximately \(4\text{ s}\)).

(b) Consider horizontal motion:
\(s_x = v_x \times t\)
\(s_x = 25\text{ m s}^{-1} \times 4.04\text{ s} = 101\text{ m}\) (or \(100\text{ m}\) using \(t = 4\text{ s}\)).

(c) Vertical component of velocity just before impact:
\(v_y = u_y + a_y t = 0 + (9.81 \times 4.04) = 39.6\text{ m s}^{-1}\)
Horizontal component of velocity remains constant:
\(v_x = 25\text{ m s}^{-1}\)
Magnitude of resultant velocity:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{25^2 + 39.6^2} = \sqrt{625 + 1568.16} = \sqrt{2193.16} = 46.8\text{ m s}^{-1}\) (or \(47\text{ m s}^{-1}\)).

(a)
- Use of \(s = ut + \frac{1}{2}at^2\) with vertical components (1)
- Correct calculation leading to \(t = 4.04\text{ s}\) (1)

(b)
- Use of \(s = vt\) with horizontal component (1)
- Correct horizontal distance \(101\text{ m}\) [accept \(100\text{ m}\) if using \(4\text{ s}\)] (1)

(c)
- Correct calculation of vertical velocity component \(v_y = 39.6\text{ m s}^{-1}\) (1)
- Use of Pythagoras' theorem to find resultant velocity (1)
- Correct magnitude of \(47\text{ m s}^{-1}\) [accept range \(46.8 - 47\text{ m s}^{-1}\)] (1)

(a) The free-body force diagram should show:
1. Weight \(W\) (or \(mg\)) pointing vertically downwards.
2. Normal contact force \(R\) (or \(N\)) pointing perpendicular to the ramp surface.
3. Pulling force \(F\) pointing parallel to and up the ramp.
4. Friction \(f\) pointing parallel to and down the ramp.

(b) Since the crate moves at constant velocity, the forces are in equilibrium.
Resolving forces parallel to the ramp:
\(F - W\sin(30^\circ) - f = 0\)
\(F = W\sin(30^\circ) + f\)
Where \(W = mg = 12 \times 9.81 = 117.72\text{ N}\)
\(F = (117.72 \times \sin(30^\circ)) + 25\)
\(F = 58.86 + 25 = 83.86\text{ N}\)
\(F = 84\text{ N}\) (to 2 significant figures).

(a)
- Weight arrow pointing vertically downwards (1)
- Normal contact force perpendicular to the slope, pointing away from surface (1)
- Friction pointing down the slope AND pulling force parallel to and up the slope, both clearly labeled (1)

(b)
- State or imply that the resultant force parallel to the slope is zero (1)
- Correct component of weight down the slope represented by \(mg\sin(30^\circ)\) (1)
- Substitution of correct values: \(12 \times 9.81 \times \sin(30^\circ) + 25\) (1)
- Correct final answer of \(84\text{ N}\) (or \(83.9\text{ N}\)) (1)

(a) Tensile strain is defined as the change in length per unit original length (or \(\Delta L / L\)).

(b) Using Young Modulus formula:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}\)
Rearranging for extension \(\Delta L\):
\(\Delta L = \frac{FL}{AE}\)
\(\Delta L = \frac{90 \times 2.5}{4.5 \times 10^{-7} \times 2.0 \times 10^{11}}\)
\(\Delta L = \frac{225}{9.0 \times 10^4} = 2.5 \times 10^{-3}\text{ m}\) (or \(2.5\text{ mm}\)).

(c) Since Hooke's Law is obeyed:
Elastic strain energy \(E_{\text{el}} = \frac{1}{2} F \Delta L\)
\(E_{\text{el}} = 0.5 \times 90 \times 2.5 \times 10^{-3} = 0.1125\text{ J}\)
\(E_{\text{el}} = 0.11\text{ J}\) (to 2 significant figures).

(a)
- Change in length divided by original length (accept ratio of extension to original length) (1)

(b)
- Recall of \(E = \frac{FL}{A\Delta L}\) or \(\text{Stress} = \frac{F}{A}\) and \(\text{Strain} = \frac{\Delta L}{L}\) (1)
- Correct algebraic rearrangement for \(\Delta L\) (1)
- Correct calculation to yield \(2.5 \times 10^{-3}\text{ m}\) (1)

(c)
- Recall of \(E_{\text{el}} = \frac{1}{2} F \Delta L\) (1)
- Substitution of their value of \(\Delta L\) (1)
- Correct final answer of \(0.11\text{ J}\) (or \(0.113\text{ J}\)) (1)

(a) Useful power output is the rate of increase of gravitational potential energy:
\(P_{\text{out}} = \frac{\Delta E_p}{\Delta t} = \frac{m g h}{\Delta t} = \left(\frac{m}{\Delta t}\right) g h\)
Where \(\frac{m}{\Delta t} = 8.5\text{ kg s}^{-1}\), \(g = 9.81\text{ m s}^{-2}\), and \(h = 15\text{ m}\).
\(P_{\text{out}} = 8.5 \times 9.81 \times 15 = 1250.775\text{ W}\)
\(P_{\text{out}} = 1.25\text{ kW}\) (or \(1300\text{ W}\)).

(b) Efficiency \(= \frac{\text{Useful power output}}{\text{Power input}} \times 100\%\)
\(\text{Efficiency} = \frac{1251}{1800} \times 100\% = 69.5\%\)
\(\text{Efficiency} = 70\%\) (or \(69.5\%\)).

(c) The energy that is not transferred to gravitational potential energy is dissipated. It is converted to thermal energy (heat) and sound energy in the pump's motor and moving parts due to friction and electrical resistance.

(a)
- Recall of \(\Delta E_p = mgh\) (1)
- Realisation that useful power is \(\left(\frac{m}{t}\right)gh\) (1)
- Correct calculation of power output \(= 1250\text{ W}\) (or \(1.25\text{ kW}\)) (1)

(b)
- Recall of \(\text{Efficiency} = \frac{\text{Power out}}{\text{Power in}}\) (1)
- Correct calculation of efficiency \(70\%\) (accept range \(69\% - 70\%\)) (1)

(c)
- Dissipated to the surroundings as thermal energy / sound (1)
- Mention of friction in moving parts / resistance in electrical components as the cause (1)

(a) The total momentum of a system remains constant, provided no external resultant force acts on the system.

(b) Total initial momentum = total final momentum
\(m_A u_A + m_B u_B = (m_A + m_B) v\)
\((0.80 \times 3.0) + 0 = (0.80 + 1.2) \times v\)
\(2.4 = 2.0 v\)
\(v = 1.2\text{ m s}^{-1}\).

(c) Calculate initial kinetic energy:
\(E_{k\text{, initial}} = \frac{1}{2} m_A u_A^2 = 0.5 \times 0.80 \times (3.0)^2 = 3.6\text{ J}\)
Calculate final kinetic energy:
\(E_{k\text{, final}} = \frac{1}{2} (m_A + m_B) v^2 = 0.5 \times 2.0 \times (1.2)^2 = 1.44\text{ J}\)
Since the final kinetic energy is less than the initial kinetic energy, kinetic energy is not conserved. Therefore, the collision is inelastic.

(a)
- Total momentum of a closed system is constant (1)
- In the absence of external forces / net external force is zero (1)

(b)
- Substitution of correct values into conservation of momentum equation (1)
- Correct calculation of velocity \(v = 1.2\text{ m s}^{-1}\) (1)

(c)
- Correct calculation of initial Kinetic Energy \(= 3.6\text{ J}\) (1)
- Correct calculation of final Kinetic Energy \(= 1.44\text{ J}\) (1)
- Comparison of values leading to the correct conclusion that the collision is inelastic (1)

(a) The diagram should show:
1. Weight (gravity) acting vertically downwards.
2. Upthrust acting vertically upwards.
3. Drag (viscous drag/friction) acting vertically upwards.

(b) At terminal velocity, the net force is zero:
\(W = U + D\) (or \(\text{Weight} = \text{Upthrust} + \text{Drag}\)).

(c) Upthrust is equal to the weight of the oil displaced by the ball bearing.
\(U = \rho_{\text{fluid}} \times V \times g\)
Calculate the volume of the sphere:
\(V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.5 \times 10^{-3})^3 = 1.4137 \times 10^{-8}\text{ m}^3\)
Now calculate upthrust:
\(U = 900 \times 1.4137 \times 10^{-8} \times 9.81 = 1.248 \times 10^{-4}\text{ N}\)
\(U = 1.3 \times 10^{-4}\text{ N}\) (to 2 significant figures).

(a)
- Weight arrow pointing vertically down (1)
- Upthrust arrow pointing vertically up (1)
- Drag arrow pointing vertically up (1)

(b)
- Correct equation: \(W = U + D\) (or equivalent symbols clearly identified) (1)

(c)
- Correct calculation of the volume of the sphere \(V = 1.41 \times 10^{-8}\text{ m}^3\) (1)
- Use of \(U = \rho V g\) (1)
- Correct final answer of \(1.3 \times 10^{-4}\text{ N}\) (accept \(1.2 \times 10^{-4}\text{ N}\) if using \(g = 9.8\text{ m s}^{-2}\)) (1)

(a) A uniform shelf has its mass/weight distributed evenly along its length, so its center of gravity acts at its geometric center (i.e., at \(0.6\text{ m}\) from the hinge).

(b) Let the hinge be the pivot.
Clockwise moment due to the shelf's weight:
\(\text{Moment}_{\text{cw}} = W \times \frac{L}{2} = 45\text{ N} \times 0.6\text{ m} = 27\text{ N m}\)
Anticlockwise moment due to the tension \(T\):
\(\text{Moment}_{\text{acw}} = T \sin(35^\circ) \times L = T \sin(35^\circ) \times 1.2\text{ m}\)
For equilibrium, \(\text{Moment}_{\text{cw}} = \text{Moment}_{\text{acw}}\):
\(27 = T \sin(35^\circ) \times 1.2\)
\(T \sin(35^\circ) = 22.5\)
\(T = \frac{22.5}{\sin(35^\circ)} = \frac{22.5}{0.5736} = 39.2\text{ N}\)
\(T = 39\text{ N}\) (to 2 significant figures).

(c) The vertical forces must balance:
\(V_{\text{hinge}} + T\sin(35^\circ) = W\)
Since \(T\sin(35^\circ) = 22.5\text{ N}\) and \(W = 45\text{ N}\):
\(V_{\text{hinge}} = 45 - 22.5 = 22.5\text{ N}\).
Thus, the vertical component of the force exerted by the hinge is exactly half of the weight of the shelf (or is less than the weight).

(a)
- Center of gravity/mass is at the midpoint of the shelf (1)

(b)
- Use of \(\text{moment} = \text{force} \times \text{perpendicular distance}\) (1)
- Correct calculation of clockwise moment \(= 27\text{ N m}\) (1)
- Identification of the perpendicular component of tension \(T\sin(35^\circ)\) (1)
- Correct calculation of tension \(T = 39\text{ N}\) (accept \(39.2\text{ N}\)) (1)

(c)
- States that the vertical force from the hinge is \(22.5\text{ N}\) or half the weight (1)
- Explains this by stating that the vertical component of tension already supports half the weight (1)

(a) Under increasing stress, copper behaves as a ductile material. It undergoes elastic deformation initially, but then undergoes significant plastic (permanent) deformation before fracturing. Glass behaves as a brittle material; it undergoes only elastic deformation and fractures suddenly at its elastic limit without undergoing any significant plastic deformation.

(b) Copper is ductile and can undergo large plastic deformation under tension, allowing its shape to be permanently changed (drawn into wires) without breaking. Glass is brittle and cannot undergo plastic deformation; any attempt to deform it causes stress concentrations to form, leading to immediate fracture.

(c) The Young Modulus is defined as stress divided by strain. In the linear region of a stress-strain graph, stress is proportional to strain. Therefore, the Young Modulus is equal to the gradient of the straight-line portion of the stress-strain graph.

(a)
- Identifies copper as ductile and glass as brittle (1)
- States that copper undergoes significant plastic/permanent deformation before breaking (1)
- States that glass has little or no plastic deformation and fractures suddenly (at its elastic limit) (1)

(b)
- Relates ductility of copper to its ability to be drawn into wires due to plastic flow/deformation (1)
- Relates brittleness of glass to crack propagation and lack of plastic deformation leading to fracture (1)

(c)
- Identifies that \(E = \frac{\text{stress}}{\text{strain}}\) (1)
- States that Young Modulus is the gradient of the straight line / linear / elastic section of the graph (1)

(a)
To find the acceleration down the slope, we resolve forces parallel to the slope.

The component of the skier's weight parallel to the slope is:
\(W_{\parallel} = mg \sin(\theta)\)
\(W_{\parallel} = 75\text{ kg} \times 9.81\text{ m s}^{-2} \times \sin(25^\circ) = 311.0\text{ N}\)

Applying Newton's second law parallel to the slope:
\(F_{\text{net}} = W_{\parallel} - F_{\text{resistive}} = ma\)
\(311.0\text{ N} - 120\text{ N} = 75\text{ kg} \times a\)
\(191.0\text{ N} = 75\text{ kg} \times a\)
\(a = \frac{191.0}{75} = 2.55\text{ m s}^{-2}\)
This is approximately \(2.6\text{ m s}^{-2}\).

(b)
First, find the speed \(v\) of the skier at the bottom of the slope using \(v^2 = u^2 + 2as\):
\(v^2 = 0 + 2 \times 2.55\text{ m s}^{-2} \times 40\text{ m} = 204\text{ m}^2\text{ s}^{-2}\)
\(v = 14.3\text{ m s}^{-1}\) (using \(a = 2.6\text{ m s}^{-2}\) gives \(v = 14.4\text{ m s}^{-1}\))

Next, find the time of flight \(t\) in the air using vertical motion:
\(s_y = u_y t + \frac{1}{2}g t^2\)
Since the skier leaves horizontally, \(u_y = 0\).
\(1.2\text{ m} = 0.5 \times 9.81\text{ m s}^{-2} \times t^2\)
\(t^2 = \frac{1.2}{4.905} = 0.2446\text{ s}^2\)
\(t = 0.495\text{ s}\)

Now, calculate the horizontal distance \(s_x\):
\(s_x = v_x t = 14.3\text{ m s}^{-1} \times 0.495\text{ s} = 7.08\text{ m}\) (or \(7.1\text{ m}\) to 2 s.f.)
(Using \(a = 2.6\text{ m s}^{-2}\) gives \(14.4\text{ m s}^{-1} \times 0.495\text{ s} = 7.13\text{ m}\))

(a)
- M1: Calculation of the component of weight parallel to the slope \(mg \sin(25^\circ) = 311\text{ N}\)
- M1: Use of \(F_{\text{net}} = ma\) with resistive force subtracted (e.g., \(311 - 120 = 75 \times a\))
- A1: Correct calculation showing \(a = 2.55\text{ m s}^{-2}\) (must show 3 significant figures to justify "approximately \(2.6\text{ m s}^{-2}\)")

(b)
- M1: Use of \(v^2 = u^2 + 2as\) to find the speed at the bottom of the slope
- A1: Correct value of speed \(v = 14.3\text{ m s}^{-1}\) (or \(14.4\text{ m s}^{-1}\) using show-that value)
- M1: Use of \(s = \frac{1}{2}g t^2\) to find time of flight (\(t = 0.495\text{ s}\) or \(0.50\text{ s}\))
- A1: Correct horizontal distance in range \(7.1\text{ m}\) to \(7.2\text{ m}\) with appropriate units

(a)
First, calculate the cross-sectional area \(A\) of the wire:
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.80 \times 10^{-3}\text{ m})^2}{4}\)
\(A = 5.03 \times 10^{-7}\text{ m}^2\)

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

Rearranging to make the extension \(\Delta x\) the subject:
\(\Delta x = \frac{F L}{A E}\)
\(\Delta x = \frac{45.0\text{ N} \times 2.50\text{ m}}{(5.03 \times 10^{-7}\text{ m}^2) \times (1.0 \times 10^{11}\text{ Pa})}\)
\(\Delta x = \frac{112.5}{5.03 \times 10^4} = 2.24 \times 10^{-3}\text{ m} = 2.2\text{ mm}\)

(b)
Definition of elastic limit: The maximum force/stress that can be applied to a material such that it returns to its original length when the force/stress is removed (i.e., no permanent plastic deformation occurs).

Experimental method:
- Measure the initial length of the wire with no load.
- Apply the load, then remove it completely.
- Re-measure the length of the wire. If the wire is longer than its original length (permanent extension has occurred), then the elastic limit has been exceeded.

(a)
- M1: Use of \(A = \frac{\pi d^2}{4}\) (or \(A = \pi r^2\))
- A1: Correct cross-sectional area \(A = 5.0 \times 10^{-7}\text{ m}^2\) (accept \(5.03 \times 10^{-7}\text{ m}^2\))
- M1: Rearrangement of \(E = \frac{FL}{A\Delta x}\) to make \(\Delta x\) the subject
- A1: Correct extension of \(2.2 \times 10^{-3}\text{ m}\) (or \(2.2\text{ mm}\))

(b)
- B1: Definition of elastic limit: the point/force beyond which a material will not return to its original length when the load is removed
- B1: Description of measurement: Measure original length, load the wire, and then measure the length again after removing the load
- B1: Conclusion: If the length is greater than the original length (or permanent extension is observed), the elastic limit has been exceeded

The resistance is given by \(R = \rho \frac{L}{A}\), where the cross-sectional area is \(A = \frac{\pi d^2}{4}\). Thus, \(R = \frac{4\rho L}{\pi d^2}\). For the second wire, the resistance \(R' = \rho \frac{3L}{\pi (2d)^2/4} = \frac{12\rho L}{4\pi d^2} = \frac{3}{4} \left(\frac{4\rho L}{\pi d^2}\right) = 0.75 R\).

1 mark: Correctly identifies that resistance is proportional to length and inversely proportional to the square of diameter, leading to the factor of 0.75.

Using the diffraction grating equation \(d \sin \theta = n \lambda\), where the grating spacing is \(d = \frac{1}{N}\). Substituting the given values: \(\frac{1}{N} \sin(30^\circ) = 3\lambda\). Since \(\sin(30^\circ) = 0.5\), we get \(\frac{1}{2N} = 3\lambda\). Rearranging for \(N\) gives \(N = \frac{1}{6\lambda}\).

1 mark: Correct application of the grating equation with substituted values to solve for N.

For an NTC thermistor, an increase in temperature leads to a decrease in its resistance. Since the potential divider splits voltage in proportion to resistance, a lower thermistor resistance means a smaller fraction of the total voltage is dropped across it. Therefore, \(V_{\text{out}}\) across the thermistor decreases.

1 mark: Identifies that thermistor resistance decreases and the potential difference across it decreases.

Only transverse waves can undergo polarisation because they oscillate perpendicular to the direction of wave travel. Ultrasound waves are longitudinal sound waves, which oscillate parallel to the direction of travel, and therefore cannot be polarised.

1 mark: Correctly identifies ultrasound waves as longitudinal waves which cannot be polarised.

Since the wires are in series, they carry the same current \(I\). The current is given by \(I = nAvq\). Since they are made of the same material, the charge carrier density \(n\) is the same. Thus, \(A_X v_X = A_Y v_Y\). The cross-sectional area is proportional to the square of the diameter, so \(A_Y = 4A_X\). This gives \(A_X v = 4A_X v_Y\), which simplifies to \(v_Y = 0.25v\).

1 mark: Uses the drift velocity formula and series current relation to find the inverse-square dependency on diameter.

According to Einstein's photoelectric equation, \(E_k = hf - \phi\), where \(\phi\) is the work function of the metal. If the frequency is doubled, the new maximum kinetic energy is \(E_k' = 2hf - \phi = 2(E_k + \phi) - \phi = 2E_k + \phi\). Since the work function \(\phi\) is a positive constant, \(2E_k + \phi\) is strictly greater than \(2E_k\).

1 mark: Correctly applies Einstein's photoelectric equation to show that the new kinetic energy is greater than twice the original kinetic energy.

A fundamental distinction between progressive and stationary waves is that progressive waves transmit energy away from the source through space, whereas stationary waves store vibrational energy within the boundaries of the wave system without net energy transfer.

1 mark: Identifies the correct energy transfer difference between progressive and stationary waves.

When the external resistance \(R\) is decreased, the total resistance of the circuit \(R + r\) decreases. According to Ohm's law, the current \(I = \frac{\varepsilon}{R+r}\) must increase. The terminal potential difference is given by \(V = \varepsilon - Ir\). Since the current \(I\) increases, the lost volts \(Ir\) increase, causing the terminal potential difference \(V\) to decrease.

1 mark: Correctly deduces that current increases and terminal potential difference decreases.

When the resistance of the variable resistor \(R\) is decreased, the total resistance of the circuit \(R_{\text{total}} = R + r\) decreases. Since the electromotive force \(E\) is constant, the current \(I\) in the circuit, given by \(I = \frac{E}{R+r}\), must increase. Let us examine each statement:
- A is incorrect: The terminal potential difference is given by \(V = E - Ir\). Since \(I\) increases, the 'lost volts' \(Ir\) increases, which means \(V\) must decrease.
- B is correct: The electrical power dissipated in the internal resistance is \(P = I^2 r\). Since \(I\) increases and \(r\) remains constant, the power dissipated in the internal resistance increases.
- C is incorrect: The quantity \(E - V\) represents the lost volts, which is equal to \(Ir\). Since \(I\) increases, \(E - V\) increases.
- D is incorrect: The total power supplied by the battery is \(P_{\text{total}} = EI\). Since \(I\) increases, \(P_{\text{total}}\) increases.

1 mark: B - The power dissipated in the internal resistance increases.

[Incorrect options breakdown: A is incorrect because terminal potential difference decreases; C is incorrect because lost volts increase; D is incorrect because total power supplied increases.]

For a string of length \(L\) fixed at both ends, a stationary wave can only be established if there is a node at each end. This boundary condition means that the length of the string must be an integer multiple of half-wavelengths:
\(L = n \frac{\lambda}{2}\) where \(n = 1, 2, 3, \dots\)
Rearranging for wavelength \(\lambda\) gives:
\(\lambda = \frac{2L}{n}\)
Let us test the given options to find which one yields an integer value for \(n\):
- Option A: \(\frac{3L}{2} = \frac{2L}{n} \implies n = \frac{4}{3}\) (not an integer)
- Option B: \(\frac{2L}{3} = \frac{2L}{n} \implies n = 3\) (integer, representing the 3rd harmonic)
- Option C: \(\frac{4L}{3} = \frac{2L}{n} \implies n = \frac{6}{4} = 1.5\) (not an integer)
- Option D: \(\frac{3L}{4} = \frac{2L}{n} \implies n = \frac{8}{3}\) (not an integer)
Therefore, \(\frac{2L}{3}\) is the only possible wavelength.

1 mark: B - \(\frac{2L}{3}\)

[Incorrect options breakdown: A, C, and D do not satisfy the boundary conditions because they do not correspond to an integer number of half-wavelengths fitting on the string.]

(a) A correct circuit diagram must show: a cell in series with a variable resistor, an ammeter in series, and a voltmeter connected in parallel across either the cell or the variable resistor.

(b) Using the equation \(V = \varepsilon - Ir\), a graph of \(V\) on the y-axis against \(I\) on the x-axis will yield a straight line. The y-intercept of this line represents the EMF (\(\varepsilon\)) of the cell. The magnitude of the gradient of the line represents the internal resistance (\(r\)) of the cell.

(c) Using the potential difference equation twice:
\(1.42 = \varepsilon - 0.35r\)
\(1.22 = \varepsilon - 0.85r\)
Subtracting the second equation from the first:
\(0.20 = 0.50r\)
\(r = 0.40\ \Omega\)
Substitute \(r\) back into the first equation:
\(\varepsilon = 1.42 + 0.35(0.40) = 1.56\text{ V}\).

(a) [1 mark] Voltmeter connected in parallel across the cell or variable resistor.
[1 mark] Ammeter connected in series with the cell and variable resistor.

(b) [1 mark] State that a graph of \(V\) against \(I\) is plotted.
[1 mark] State that the y-intercept represents the EMF (\(\varepsilon\)).
[1 mark] State that the gradient represents \(-r\) (or the magnitude of the gradient is the internal resistance \(r\)).

(c) [1 mark] Use of \(V = \varepsilon - Ir\) to set up two simultaneous equations.
[1 mark] Correct subtraction/elimination step to find \(r\).
[1 mark] Correct calculation of \(r = 0.40\ \Omega\).
[0.75 mark] Correct calculation of \(\varepsilon = 1.56\text{ V}\).

(a) Using \(v = c / n\):
\(v = 3.00 \times 10^8\text{ m s}^{-1} / 1.52 = 1.97 \times 10^8\text{ m s}^{-1}\).

(b) Total internal reflection can only occur when light is traveling from a medium of higher refractive index to one of lower refractive index. Since \(n_{\text{water}} = 1.33\) and \(n_{\text{glass}} = 1.52\), water has a lower refractive index than glass. Therefore, light traveling from water into glass is entering a more optically dense medium, and total internal reflection is impossible.

(c) Using \(\sin \theta_c = n_2 / n_1\):
\(\sin \theta_c = 1.33 / 1.52 = 0.875\)
\(\theta_c = \sin^{-1}(0.875) = 61.0^\circ\).

(a) [1 mark] Use of \(v = c / n\).
[1 mark] Correct final speed: \(1.97 \times 10^8\text{ m s}^{-1}\).

(b) [1 mark] State that TIR requires light to travel from a high refractive index to a lower refractive index.
[1 mark] Identify that \(n_{\text{water}}\) (1.33) is less than \(n_{\text{glass}}\) (1.52).
[0.75 mark] Conclude that TIR cannot occur because light is moving into a more dense medium.

(c) [1 mark] Use of \(\sin \theta_c = n_2 / n_1\).
[1 mark] Correct substitution of 1.33 and 1.52.
[1 mark] Calculation of \(\sin \theta_c = 0.875\).
[1 mark] Correct final angle: \(61.0^\circ\) (accept \(61^\circ\)).

(a) Using \(R = \rho L / A\):
\(\rho = R A / L = 4.50 \times 2.40 \times 10^{-7} / 1.50 = 7.20 \times 10^{-7}\ \Omega\text{ m}\).

(b) As temperature increases, the positive lattice ions in the metal vibrate with greater amplitude. This increases the frequency of collisions between the free/conduction electrons and the vibrating lattice ions. Consequently, the drift velocity of the electrons decreases, which leads to an increase in the electrical resistance of the wire.

(c) In an NTC thermistor, as temperature increases, the thermal energy is sufficient to release a large number of charge carriers (electrons) from the valence band into the conduction band. This massive increase in the number density (n) of charge carriers outweighs the effect of lattice vibrations, causing the overall resistance of the thermistor to decrease.

(a) [1 mark] Recall of \(R = \rho L / A\).
[1 mark] Rearrangement to \(\rho = R A / L\) and substitution.
[1 mark] Correct final answer: \(7.20 \times 10^{-7}\ \Omega\text{ m}\) with unit.

(b) [1 mark] Resistance of the wire increases.
[1 mark] Metal lattice ions vibrate with larger amplitude.
[1 mark] Frequency of electron-ion collisions increases.
[0.75 mark] Reduces the drift velocity of charge carriers.

(c) [1 mark] Resistance of NTC thermistor decreases with temperature.
[1 mark] Temperature rise releases significantly more charge carriers (increases n).

(a) Line spacing \(d = 1 / (500 \times 10^3) = 2.00 \times 10^{-6}\text{ m}\).

(b)(i) Using trigonometry: \(\tan \theta = x / D\)
\(\tan \theta = 1.45 / 2.50 = 0.58\)
\(\theta = \tan^{-1}(0.58) = 30.11^\circ \approx 30.1^\circ\).

(b)(ii) Using \(d \sin \theta = n \lambda\) for \(n = 2\):
\(2 \lambda = 2.00 \times 10^{-6} \times \sin(30.11^\circ)\)
\(2 \lambda = 2.00 \times 10^{-6} \times 0.5017 = 1.003 \times 10^{-6}\text{ m}\)
\(\lambda = 5.02 \times 10^{-7}\text{ m}\) (or 502 nm).

(a) [1 mark] Use of \(d = 1 / N\) with correct conversion to lines per meter.
[1 mark] Correct calculation of \(d = 2.00 \times 10^{-6}\text{ m}\).

(b)(i) [1 mark] Recognize trig relationship \(\tan \theta = x / D\).
[1 mark] Substitution of 1.45 and 2.50.
[0.75 mark] Correct angle of \(30.1^\circ\).

(b)(ii) [1 mark] Recall of \(d \sin \theta = n \lambda\).
[1 mark] Substitution of \(n = 2\).
[1 mark] Substitution of calculated \(d\) and \(\theta\).
[1 mark] Correct wavelength: \(5.02 \times 10^{-7}\text{ m}\) (accept \(5.0 \times 10^{-7}\text{ m}\) to \(5.1 \times 10^{-7}\text{ m}\)).

(a) The circuit diagram should display: a 9.00 V DC source, with a series loop containing a fixed resistor of \(1.20\text{ k}\Omega\) and an LDR. Output connections for \(V_{\text{out}}\) should be marked in parallel across the LDR symbol.

(b) Using the potential divider formula:
\(V_{\text{out}} = V_{\text{in}} \times [ R_{\text{LDR}} / (R_{\text{LDR}} + R_{\text{fixed}}) ]\)
\(V_{\text{out}} = 9.00 \times [ 350 / (350 + 1200) ] = 9.00 \times [ 350 / 1550 ] = 2.03\text{ V}\).

(c) Under night conditions:
\(7.50 = 9.00 \times [ R / (R + 1200) ]\)
\(7.50 / 9.00 = R / (R + 1200)\)
\(5 / 6 = R / (R + 1200)\)
\(5(R + 1200) = 6R\)
\(5R + 6000 = 6R\)
\(R = 6000\ \Omega = 6.00\text{ k}\Omega\).

(a) [1 mark] Fixed resistor and LDR in series across the supply.
[1 mark] \(V_{\text{out}}\) correctly shown in parallel across the LDR.

(b) [1 mark] Recall potential divider equation.
[1 mark] Correct substitution of 350 and 1200 (including \(k\Omega\) to \(\Omega\) conversion).
[1 mark] Correct output voltage: \(2.03\text{ V}\) (accept \(2.0\text{ V}\)).

(c) [1 mark] Set up equation with 7.50 V as \(V_{\text{out}}\).
[1 mark] Correct algebraic rearrangement.
[1 mark] Calculation of \(R = 6000\ \Omega\).
[0.75 mark] Correct unit and final format (\(6.0\text{ k}\Omega\) or \(6000\ \Omega\)).

(a) Progressive waves transfer energy through the medium from one point to another without transferring matter. Standing waves do not transfer energy; instead, they store energy within the wave pattern.

(b) First, find mass per unit length \(\mu\):
\(\mu = 4.80 \times 10^{-3}\text{ kg} / 1.20\text{ m} = 4.00 \times 10^{-3}\text{ kg m}^{-1}\)
Next, use \(v = \sqrt{T / \mu}\):
\(v = \sqrt{36.0 / (4.00 \times 10^{-3})} = \sqrt{9000} = 94.9\text{ m s}^{-1}\).

(c)(i) The first harmonic has a single loop with nodes at both fixed ends and a single antinode in the middle.

(c)(ii) For the first harmonic, \(\lambda = 2L = 2 \times 1.20\text{ m} = 2.40\text{ m}\).
Using \(f = v / \lambda\):
\(f = 94.87\text{ m s}^{-1} / 2.40\text{ m} = 39.5\text{ Hz}\) (accept 39.5 Hz to 40 Hz depending on rounding of v).

(a) [1 mark] Progressive waves transfer energy.
[1 mark] Standing waves store energy/do not transfer energy.

(b) [1 mark] Calculate \(\mu = 4.00 \times 10^{-3}\text{ kg m}^{-1}\) (including g to kg conversion).
[1 mark] Use of \(v = \sqrt{T / \mu}\).
[1 mark] Correct speed: \(94.9\text{ m s}^{-1}\).

(c)(i) [1 mark] Wave profile drawn with one loop, nodes at ends, and central antinode labeled.

(c)(ii) [1 mark] Identify that \(\lambda = 2L = 2.40\text{ m}\).
[1 mark] Use of \(f = v / \lambda\).
[0.75 mark] Correct final frequency: \(39.5\text{ Hz}\) (accept 39.5 - 40 Hz).

(a) Energy of photon \(E = hc / \lambda\):
\(E = (6.63 \times 10^{-34} \times 3.00 \times 10^8) / (365 \times 10^{-9}) = 5.449 \times 10^{-19}\text{ J}\)
Convert to eV:
\(E = 5.449 \times 10^{-19}\text{ J} / (1.60 \times 10^{-19}\text{ J eV}^{-1}) = 3.41\text{ eV}\), which is approximately 3.40 eV.

(b) Using Einstein's photoelectric equation:
\(hf = \phi + E_{\text{k,max}}\)
\(E_{\text{k,max}} = 3.406\text{ eV} - 2.36\text{ eV} = 1.046\text{ eV}\)
Convert to Joules:
\(E_{\text{k,max}} = 1.046 \times 1.60 \times 10^{-19}\text{ J} = 1.67 \times 10^{-19}\text{ J}\) (or \(1.68 \times 10^{-19}\text{ J}\) if using 3.41 eV).

(c) Using \(\phi = h f_0\):
\(\phi = 2.36 \times 1.60 \times 10^{-19}\text{ J} = 3.776 \times 10^{-19}\text{ J}\)
\(f_0 = 3.776 \times 10^{-19} / (6.63 \times 10^{-34}) = 5.70 \times 10^{14}\text{ Hz}\).

(a) [1 mark] Use of \(E = hc / \lambda\).
[1 mark] Conversion of wavelength to meters (\(365 \times 10^{-9}\text{ m}\)).
[1 mark] Correct conversion of Joules to eV showing 3.41 eV.

(b) [1 mark] Use of Einstein's photoelectric equation.
[1 mark] Calculation of kinetic energy in eV (1.05 eV).
[1 mark] Correct conversion to Joules: \(1.67 \times 10^{-19}\text{ J}\) to \(1.68 \times 10^{-19}\text{ J}\).

(c) [1 mark] Use of \(\phi = h f_0\).
[1 mark] Convert \(\phi\) to Joules (\(3.78 \times 10^{-19}\text{ J}\)).
[0.75 mark] Correct frequency: \(5.70 \times 10^{14}\text{ Hz}\).

(a) Area \(A = \pi r^2 = \pi \times (D / 2)^2\)
\(A = \pi \times (1.20 \times 10^{-3}\text{ m} / 2)^2 = \pi \times (6.00 \times 10^{-4})^2 = 1.131 \times 10^{-6}\text{ m}^2 \approx 1.13 \times 10^{-6}\text{ m}^2\).

(b) Using \(I = n A v e\):
\(v = I / (n A e)\)
\(v = 4.50 / (8.50 \times 10^{28} \times 1.131 \times 10^{-6} \times 1.60 \times 10^{-19})\)
\(v = 4.50 / (1.538 \times 10^4) = 2.93 \times 10^{-4}\text{ m s}^{-1}\).

(c) The rate of energy conversion is electrical power \(P\):
\(P = V I\)
\(P = 0.150\text{ V} \times 4.50\text{ A} = 0.675\text{ W}\) (or \(\text{J s}^{-1}\)).

(a) [1 mark] Correct radius \(r = 0.60 \times 10^{-3}\text{ m}\).
[1 mark] Correct substitution into \(A = \pi r^2\) to get \(1.13 \times 10^{-6}\text{ m}^2\).

(b) [1 mark] Use of \(I = n A v e\).
[1 mark] Substitution of n, A, e.
[1 mark] Correct rearrangement for v.
[0.75 mark] Correct final velocity: \(2.93 \times 10^{-4}\text{ m s}^{-1}\).

(c) [1 mark] Use of \(P = V I\) (or \(P = I^2 R\)).
[1 mark] Substitution of V = 0.150 V and I = 4.50 A.
[1 mark] Correct power value of 0.675 W (with unit).

(a) To measure the diameter accurately, the student should use a digital micrometer screw gauge. They must measure the diameter at at least three different positions along the wire and at right angles (perpendicular orientations) at each position, then calculate the average diameter. The micrometer should also be checked for zero error before taking measurements.

(b) A low current prevents the wire from heating up. If the wire heats up, its temperature increases, which increases its resistance and therefore changes its resistivity, introducing systematic error.

(c) Calculate cross-sectional area \(A\):
\(A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}\)
\(A = \frac{\pi (0.35 \times 10^{-3} \text{ m})^2}{4} = 9.621 \times 10^{-8} \text{ m}^2\)

Use the resistivity formula:
\(R = \frac{\rho L}{A} \implies \rho = \frac{R A}{L}\)
\(\rho = \frac{3.65 \times 9.621 \times 10^{-8}}{0.850} = 4.131 \times 10^{-7} \ \Omega \text{ m}\)
Rounding to 2 significant figures (due to diameter \(d\) being 2 s.f.):
\(\rho = 4.1 \times 10^{-7} \ \Omega \text{ m}\).

(d) Hazard: The wire can become very hot if the current is too high or if left connected, which can cause skin burns or melt insulation.
Precaution: Switch off the power supply between readings and do not touch the wire directly while the current is flowing.

(e) The circuit diagram should show: a power supply/battery, a switch, an ammeter in series with the test wire, and a voltmeter in parallel across the length of wire being measured (usually using a jockey/sliding contact to vary the length).

Part (a) [3 Marks]:
- MP1: Use a micrometer screw gauge / digital micrometer [1 mark]
- MP2: Measure at multiple (at least 3) different positions and perpendicular orientations [1 mark]
- MP3: Calculate average / check for and subtract zero error [1 mark]

Part (b) [2 Marks]:
- MP1: Current causes heating / temperature rise [1 mark]
- MP2: Resistance / resistivity depends on temperature (which would invalidate the constant temperature assumption) [1 mark]

Part (c) [4 Marks]:
- MP1: Conversion of diameter to metres \(3.5 \times 10^{-4} \text{ m}\) and calculation of Area \(A = 9.62 \times 10^{-8} \text{ m}^2\) [1 mark]
- MP2: Rearrangement of resistivity formula to \(\rho = \frac{R A}{L}\) [1 mark]
- MP3: Correct substitution of values into the formula [1 mark]
- MP4: Final answer \(4.1 \times 10^{-7} \ \Omega \text{ m}\) (accept range \(4.1 \times 10^{-7}\) to \(4.14 \times 10^{-7}\)) with correct unit [1 mark]

Part (d) [1.5 Marks]:
- MP1: Identifies hazard (hot wire / burn hazard) [1 mark]
- MP2: Appropriate precaution (switch off between readings / use a resistor to limit current / do not touch the wire) [0.5 marks]

Part (e) [2 Marks]:
- MP1: Ammeter in series and voltmeter in parallel with the test wire [1 mark]
- MP2: Battery/power source, switch, and variable length indicator (like a jockey or sliding contact) shown correctly [1 mark]

(a) A double-interrupt card has two solid sections of known length separated by a gap. As it passes through a single light gate, it provides two time intervals and the time between them. This allows the data logger to calculate two velocities (initial and final) and the time interval between them, thus directly calculating acceleration without needing to measure \(h\).

(b) Using \(v^2 = u^2 + 2as\):
- Assume the card is dropped from rest, so initial velocity \(u = 0\).
- The acceleration \(a = g\) (acceleration of free fall).
- The displacement \(s = h\).
Substituting these gives \(v^2 = 2gh\).

(c) The light gate contains an infrared beam. When the card passes through, it blocks the beam.
- The data logger measures the duration of time \(\Delta t\) that the beam is interrupted.
- It calculates velocity using \(v = \frac{w}{\Delta t}\), where \(w\) is the width of the card's segment.
- Since \(\Delta t\) is very small, this average velocity is extremely close to the instantaneous velocity as it passes.

(d)(i) The equation is \(v^2 = 2gh\). Comparing this to \(y = mx + c\), a plot of \(v^2\) against \(h\) yields a straight line with gradient \(m = 2g\). Therefore, \(g = \frac{\text{gradient}}{2}\).
(ii) A non-zero y-intercept indicates a systematic error, such as the card being released with an initial velocity (not from rest), or a systematic error in measuring the height \(h\).

(e) From \(g = \frac{v^2}{2h}\):
Percentage uncertainty in \(g\) is:
\(\% \Delta g = 2 \times \% \Delta v + \% \Delta h\)
\(\% \Delta v = \frac{0.05}{3.10} \times 100\% = 1.613\%\)
\(\% \Delta h = \frac{0.002}{0.500} \times 100\% = 0.400\%\)
\(\% \Delta g = 2 \times 1.613\% + 0.400\% = 3.226\% + 0.400\% = 3.626\%\)
Rounding to 3 s.f. gives \(3.63\%\).

Part (a) [2 Marks]:
- MP1: Allows the measurement of two velocities (or initial and final velocity) using a single gate [1 mark]
- MP2: To find acceleration directly / eliminates the need to measure falling distance \(h\) manually [1 mark]

Part (b) [2 Marks]:
- MP1: Quotes \(v^2 = u^2 + 2as\) and sets \(u = 0\) [1 mark]
- MP2: Sets \(a = g\) and \(s = h\) to get \(v^2 = 2gh\) [1 mark]

Part (c) [3 Marks]:
- MP1: Infrared beam is blocked by the card, timing the duration of interruption \(\Delta t\) [1 mark]
- MP2: Card width \(w\) is entered into the software, and velocity is calculated as \(v = w / \Delta t\) [1 mark]
- MP3: Short interruption time means average velocity is a very close approximation of instantaneous velocity [1 mark]

Part (d) [2.5 Marks]:
- (i) MP1: Compares \(v^2 = 2gh\) to \(y = mx\) and states gradient \(m = 2g\) [1 mark]
- (i) MP2: \(g = \text{gradient} / 2\) [0.5 marks]
- (ii) MP3: Identifies y-intercept represents initial velocity \(u \neq 0\) / systematic error in height measurement [1 mark]

Part (e) [3 Marks]:
- MP1: Formula for combining uncertainties: \(\%\Delta g = 2\%\Delta v + \%\Delta h\) [1 mark]
- MP2: Calculates \(\%\Delta v = 1.61\%\) AND \(\%\Delta h = 0.40\%\) [1 mark]
- MP3: Correct final value of \(3.63\%\) (accept range \(3.6\%\) to \(3.63\%\)) [1 mark]

(a) A long, thin wire produces a larger, more measurable extension for a given force because extension \(\Delta x = \frac{F L}{A E}\). Increasing \(L\) and decreasing \(A\) (thin wire) increases \(\Delta x\), which reduces the percentage uncertainty in the extension measurement.

(b) Measurement of extension:
- Attach a marker (e.g., a small piece of tape or paper flag) to the wire near a reference scale (e.g., a metre rule clamped alongside).
- Read the position of the marker against the scale before and after adding weights.
- To avoid parallax error, look directly perpendicular (at eye level) to the scale and marker, or use a mirror behind the scale to align the marker with its reflection.
- Alternatively, use a travelling microscope focused on a reference mark on the wire, which offers high precision and eliminates parallax.

(c) Area \(A\):
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.32 \times 10^{-3} \text{ m})^2}{4} = 8.042 \times 10^{-8} \text{ m}^2\)

Young Modulus formula:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta x / L} = \frac{F L}{A \Delta x}\)
\(E = \frac{22.5 \times 2.850}{8.042 \times 10^{-8} \times 6.6 \times 10^{-3}}\)
\(E = \frac{64.125}{5.308 \times 10^{-10}} = 1.208 \times 10^{11} \text{ Pa}\)
Rounding to 2 s.f. gives \(1.2 \times 10^{11} \text{ Pa}\).

(d)(i) Elastic limit (or limit of proportionality).
(ii) Permanent deformation means the wire has undergone plastic deformation. The atoms have slid past one another, changing the micro-structure. Hooke's law no longer applies, and the cross-sectional area has decreased, making subsequent stress calculations and elasticity measurements inaccurate.

Part (a) [2 Marks]:
- MP1: Larger extension produced for a given force [1 mark]
- MP2: Reduces percentage uncertainty in the extension measurement [1 mark]

Part (b) [4 Marks]:
- MP1: Mark/tape attached to the wire near a scale OR use of travelling microscope [1 mark]
- MP2: Read position before and after adding loads [1 mark]
- MP3: Minimize parallax: eye level perpendicular to scale OR mirror behind scale OR microscope focus [1 mark]
- MP4: Measurement of original length from clamp to marker [1 mark]

Part (c) [4 Marks]:
- MP1: Correct calculation of area: \(8.04 \times 10^{-8} \text{ m}^2\) [1 mark]
- MP2: Correct substitution into \(E = \frac{F L}{A \Delta x}\) [1 mark]
- MP3: Correct calculation: \(1.2 \times 10^{11} \text{ Pa}\) (accept range \(1.2 \times 10^{11}\) to \(1.21 \times 10^{11}\)) [1 mark]
- MP4: Correct unit (\(\text{Pa}\) or \(\text{N m}^{-2}\)) [1 mark]

Part (d) [2.5 Marks]:
- (i) MP1: Elastic limit (or limit of proportionality) [1 mark]
- (ii) MP2: Wire has deformed plastically / Hooke's law no longer applies [1 mark]
- (ii) MP3: Cross-sectional area has permanently decreased (changing the stress calculation) [0.5 marks]

(a) Procedure:
1. Place the semi-circular glass block on a sheet of paper and trace its outline, marking the flat boundary and the center of the flat face (point of incidence).
2. Draw a normal line perpendicular to the flat face at this center point.
3. Direct a narrow ray of light from the ray box through the curved face so it strikes the center of the flat face at an angle of incidence \(i\).
4. Mark the path of the incident ray and the refracted ray using pencil dots.
5. Remove the block, draw the rays, and measure angles \(i\) and \(r\) relative to the normal using a protractor.
6. Repeat for at least 5 different angles of incidence (e.g., from \(10^\circ\) to \(60^\circ\)).

(b) When light enters the curved surface of the semi-circular block along a radial line, it strikes the curved surface normal to the boundary (at \(90^\circ\)). Therefore, it does not undergo refraction at the curved face. Refraction only occurs at the flat boundary, which simplifies the experiment.

(c) According to Snell's Law, \(n_1 \sin i_1 = n_2 \sin i_2\). Since light travels from glass (refractive index \(n\)) to air (refractive index \(n_{\text{air}} \approx 1.0\)), if \(i\) is the angle in air and \(r\) is the angle in glass: \(\sin i = n \sin r\). Comparing this with the linear equation \(y = mx + c\), a plot of \(\sin i\) (y-axis) against \(\sin r\) (x-axis) gives a straight line through the origin with gradient \(m = n\).

(d) Calculate \(\sin i\) for \(i = 50.0^\circ\):
\(\sin(50.0^\circ) = 0.7660\)
Calculate maximum and minimum values of \(\sin i\):
\(\sin(51.0^\circ) = 0.7771\)
\(\sin(49.0^\circ) = 0.7547\)
Uncertainty in \(\sin i\):
\(\Delta(\sin i) = \frac{0.7771 - 0.7547}{2} = 0.0112\)
Percentage uncertainty:
\(\%\Delta(\sin i) = \frac{0.0112}{0.7660} \times 100\% = 1.46\%\) (which rounds to \(1.5\%\)).

Part (a) [4 Marks]:
- MP1: Draw outline of the block and mark normal line [1 mark]
- MP2: Use ray box to direct a narrow beam of light at the flat face [1 mark]
- MP3: Mark rays with dots and use protractor to measure angles \(i\) and \(r\) [1 mark]
- MP4: Repeat for a range of at least 5 different angles [1 mark]

Part (b) [2.5 Marks]:
- MP1: Ray entering the curved boundary along the normal/radius does not refract [1 mark]
- MP2: Refraction only occurs at the flat exit boundary [1 mark]
- MP3: Eliminates need to measure or correct for refraction at a second boundary [0.5 marks]

Part (c) [3 Marks]:
- MP1: Snell's law: \(n = \frac{\sin i}{\sin r}\) where \(i\) is in air and \(r\) is in glass [1 mark]
- MP2: Rearranges to \(\sin i = n \sin r\) [1 mark]
- MP3: Gradient of \(\sin i\) vs \(\sin r\) is equal to \(n\) [1 mark]

Part (d) [3 Marks]:
- MP1: Calculates \(\sin(50.0^\circ) = 0.766\) [1 mark]
- MP2: Determines absolute uncertainty in \(\sin i\) (e.g., \(\sin(51^\circ) - \sin(50^\circ) = 0.011\)) [1 mark]
- MP3: Correct percentage uncertainty \(1.5\%\) (accept range \(1.4\%\) to \(1.5\%\)) [1 mark]