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The time of flight \( t \) is determined by vertical motion: \( h = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}} \). The horizontal distance is \( d = v \times t = v \sqrt{\frac{2h}{g}} \). For the second ball, height is \( 2h \) and velocity is \( 2v \), so the new time of flight is \( t' = \sqrt{\frac{2(2h)}{g}} = \sqrt{2} t \). The new horizontal distance is \( d' = (2v) \times t' = 2\sqrt{2} v t = 2\sqrt{2} d \).

Award 1 mark for the correct option C.

The useful work done on the crate (gain in gravitational potential energy) is \( \Delta E_{\text{grav}} = mgh \). The total electrical energy input to the motor is \( E_{\text{in}} = P \times t \). The efficiency is given by the ratio of useful energy output to total energy input: \( \text{Efficiency} = \frac{mgh}{Pt} \).

Award 1 mark for the correct option A.

At terminal velocity, the net force is zero: \( \text{Weight} - \text{Upthrust} = \text{Drag} \). This can be written as \( \frac{4}{3}\pi r^3 (\rho_s - \rho_f)g = 6\pi\eta r v \). Solving for terminal velocity gives \( v = \frac{2r^2(\rho_s - \rho_f)g}{9\eta} \). Since \( \rho_s, \rho_f, g \) and \( \eta \) are constant, \( v \propto r^2 \). If the radius is doubled to \( 2r \), the terminal velocity increases by a factor of \( 2^2 = 4 \), becoming \( 4v \).

Award 1 mark for the correct option B.

The Young modulus 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, \( \Delta L' = \frac{F(2L)}{(A/2)E} = 4 \left( \frac{FL}{AE} \right) = 4\Delta L \).

Award 1 mark for the correct option D.

Since the block is stationary, it is in a state of static equilibrium. The forces acting parallel to the ramp must balance out. The component of the gravitational force acting down the slope is \( mg \sin\theta \). Therefore, the static frictional force acting up the slope must be exactly equal in magnitude to \( mg \sin\theta \). Note that \( \mu R = \mu mg \cos\theta \) represents only the maximum possible limit of static friction, not its active value.

Award 1 mark for the correct option A.

According to Archimedes' principle, the weight of the floating object is equal to the total upthrust exerted by both fluids: \( W = U_{\text{oil}} + U_{\text{water}} \). This can be written as: \( V \rho g = V_{\text{oil}} \rho_{\text{oil}} g + V_{\text{water}} \rho_{\text{water}} g \). Dividing by \( Vg \), we get \( \rho = 0.40 \rho_{\text{oil}} + 0.60 \rho_{\text{water}} = 0.40(800) + 0.60(1000) = 320 + 600 = 920\text{ kg m}^{-3} \).

Award 1 mark for the correct option C.

Initially, the velocity is zero, so the air resistance is zero and the acceleration \( a = g \). As the velocity increases, the drag force increases, which reduces the net downward force and hence the acceleration decreases. At terminal velocity, the net force is zero and \( a = 0 \). Because the velocity approaches terminal velocity asymptotically, the acceleration also decreases exponentially over time according to \( a = g e^{-bt} \). This is a curve where the magnitude of the gradient (steepness) decreases to zero as \( a \) approaches zero.

Award 1 mark for the correct option C.

Under elastic conditions, the force is proportional to extension (\( F = kx \)). The work done \( W \) is represented by the area under the force-extension graph, which is \( W = \frac{1}{2} k x^2 \). For a total extension of \( 2x \), the total work done is \( W' = \frac{1}{2} k (2x)^2 = 4 \left(\frac{1}{2} k x^2\right) = 4W \).

Award 1 mark for the correct option C.

The horizontal component of the velocity remains constant throughout the flight, so \(v_x = v\). The vertical component of the velocity \(v_y\) just before hitting the ground can be found using the equations of motion: \(v_y^2 = u_y^2 + 2as_y\). Since the initial vertical velocity is zero, this gives \(v_y = \sqrt{2gh}\). The angle \(\theta\) to the horizontal is given by \(\tan\theta = \frac{v_y}{v_x} = \frac{\sqrt{2gh}}{v}\).

A is the correct response. [1 mark] for selecting A. Reject all other options as they represent incorrect applications of the kinematics equations or incorrect trigonometric ratios.

The extension \(\Delta x\) is given by \(\Delta x = \frac{FL}{AE}\), where \(F\) is the tension force, \(L\) is the original length, \(A\) is the cross-sectional area, and \(E\) is the Young modulus. The area of a wire is \(A = \frac{\pi d^2}{4}\), which means \(\Delta x \propto \frac{L}{d^2}\) for constant load and material. For wire P, \(\Delta x_P \propto \frac{L}{d^2}\). For wire Q, \(\Delta x_Q \propto \frac{2L}{(2d)^2} = \frac{L}{2d^2}\). Therefore, the ratio of their extensions is \(\frac{\Delta x_P}{\Delta x_Q} = 2\).

C is the correct response. [1 mark] for selecting C. Reject other options due to incorrect scaling of area with diameter or incorrect ratio inversion.

(a) First, calculate the vertical component of the velocity (\(v_y\)) just before impact: Use \(v_y^2 = u_y^2 + 2g s_y\). Since \(u_y = 0\), \(v_y = \sqrt{2 \times 9.81\text{ m s}^{-2} \times 45\text{ m}} = 29.7\text{ m s}^{-1}\). The horizontal component of the velocity (\(v_x\)) remains constant at \(8.0\text{ m s}^{-1}\). The magnitude of the final velocity \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{8.0^2 + 29.7^2} = 30.8\text{ m s}^{-1}\) (rounds to \(31\text{ m s}^{-1}\)). The angle \(\theta\) to the horizontal is given by \(\tan(\theta) = v_y / v_x = 29.7 / 8.0 = 3.71\), so \(\theta = 75^\circ\). (b) Air resistance opposes motion. The vertical drag force acts upwards, reducing the net downward acceleration to less than \(g\). Consequently, the package takes longer to fall through the \(45\text{ m}\) height, increasing the time of flight.

[1 mark] Correct use of equations of motion to find vertical velocity, \(v_y = 29.7\text{ m s}^{-1}\).
[1 mark] Correct use of Pythagoras' theorem to find resultant velocity, \(v = 31\text{ m s}^{-1}\).
[1 mark] Correct calculation of the angle (e.g., \(75^\circ\) to the horizontal or \(15^\circ\) to the vertical).
[1 mark] Statement that air resistance creates a drag force opposing motion.
[1 mark] Identification that the upward drag force reduces the net downward acceleration.
[1 mark] Conclusion that the package takes longer to descend.
[1 mark] Correct spelling, punctuation, and grammar used throughout the explanation.

(a) First, calculate the tension force in the wire: \(F = m g = 15\text{ kg} \times 9.81\text{ m s}^{-2} = 147.2\text{ N}\). Using the Young Modulus equation: \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}\), rearrange for extension \(\Delta L\): \(\Delta L = \frac{F L}{A E} = \frac{147.2 \times 2.5}{1.2 \times 10^{-6} \times 2.0 \times 10^{11}} = \frac{368}{2.4 \times 10^5} = 1.53 \times 10^{-3}\text{ m}\). (b) The elastic strain energy \(E_{\text{el}}\) stored in the wire is given by \(E_{\text{el}} = \frac{1}{2} F \Delta L = 0.5 \times 147.2\text{ N} \times 1.53 \times 10^{-3}\text{ m} = 0.113\text{ J}\), which rounds to \(0.11\text{ J}\).

[1 mark] Correct calculation of tension, \(F = 147\text{ N}\).
[1 mark] Recall of \(E = \frac{F L}{A \Delta L}\) or equivalent.
[1 mark] Substitution of values into the extension equation.
[1 mark] Correct calculation of extension to 2 or 3 sig figs, \(\Delta L = 1.5 \times 10^{-3}\text{ m}\).
[1 mark] Recall of elastic strain energy formula \(E_{\text{el}} = \frac{1}{2} F \Delta L\) or \(E_{\text{el}} = \frac{1}{2} k (\Delta L)^2\).
[1 mark] Substitution of their values into the energy equation.
[1 mark] Correct final answer for energy, \(0.11\text{ J}\) (accept \(0.113\text{ J}\)).

(a) First, calculate the vertical height \(h\) of the ramp: \(h = 1.8\text{ m} \times \sin(30^\circ) = 0.90\text{ m}\). The initial gravitational potential energy \(E_p\) is: \(E_p = m g h = 0.35\text{ kg} \times 9.81\text{ m s}^{-2} \times 0.90\text{ m} = 3.09\text{ J}\). The final kinetic energy \(E_k\) at the bottom is: \(E_k = \frac{1}{2} m v^2 = 0.5 \times 0.35\text{ kg} \times (3.2\text{ m s}^{-1})^2 = 1.792\text{ J}\). The work done against friction \(W_f\) is the loss in mechanical energy: \(W_f = E_p - E_k = 3.09\text{ J} - 1.792\text{ J} = 1.298\text{ J}\), which rounds to \(1.3\text{ J}\). (b) The work done against friction is also equal to the average frictional force \(F_f\) multiplied by the distance along the ramp \(d\): \(W_f = F_f \times d \implies F_f = \frac{W_f}{d} = \frac{1.298\text{ J}}{1.8\text{ m}} = 0.721\text{ N}\), which rounds to \(0.72\text{ N}\).

[1 mark] Calculate vertical height \(h = 0.90\text{ m}\).
[1 mark] Correct calculation of GPE, \(E_p = 3.09\text{ J}\) (accept \(3.1\text{ J}\)).
[1 mark] Correct calculation of KE, \(E_k = 1.79\text{ J}\) (accept \(1.8\text{ J}\)).
[1 mark] Recognition that work done against friction is the difference \(E_p - E_k\).
[1 mark] Correct calculation of work done, \(1.3\text{ J}\).
[1 mark] Recall of \(W = F \times d\) and rearranging for \(F\).
[1 mark] Correct calculation of average frictional force, \(0.72\text{ N}\) (allow ecf from previous parts).

(a) Impulse is defined as the change in momentum: \(\Delta p = m \Delta v = 0.16\text{ kg} \times (24\text{ m s}^{-1} - 0) = 3.84\text{ N s}\) (rounds to \(3.8\text{ N s}\)). (b) Average force is the rate of change of momentum: \(F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{3.84\text{ N s}}{0.012\text{ s}} = 320\text{ N}\). (c) According to Newton's third law, when the stick exerts a force on the ball, the ball exerts an equal and opposite force on the stick. Since both forces act for the exact same contact time, the impulse on the stick is equal in magnitude but opposite in direction to the impulse on the ball. Therefore, the momentum lost by the stick equals the momentum gained by the ball.

[1 mark] Recall of Impulse \(= \Delta p = m \Delta v\).
[1 mark] Correct calculation of impulse with units, \(3.8\text{ N s}\) or \(3.84\text{ kg m s}^{-1}\).
[1 mark] Recall of \(F = \frac{\Delta p}{\Delta t}\).
[1 mark] Correct calculation of average force, \(320\text{ N}\).
[1 mark] State that Newton's third law involves equal and opposite forces acting for the same time interval.
[1 mark] Explain that this results in equal and opposite changes in momentum (impulses) for the stick and ball.
[1 mark] Conclude that the total momentum of the system remains constant.

First, find the volume of the sphere: \(V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (2.0 \times 10^{-3}\text{ m})^3 = 3.351 \times 10^{-8}\text{ m}^3\). Calculate the weight of the sphere: \(W = \rho_{\text{steel}} V g = 7800\text{ kg m}^{-3} \times 3.351 \times 10^{-8}\text{ m}^3 \times 9.81\text{ m s}^{-2} = 2.564 \times 10^{-3}\text{ N}\). Calculate the upthrust acting on the sphere (weight of displaced glycerol): \(U = \rho_{\text{glycerol}} V g = 1260\text{ kg m}^{-3} \times 3.351 \times 10^{-8}\text{ m}^3 \times 9.81\text{ m s}^{-2} = 4.141 \times 10^{-4}\text{ N}\). At terminal velocity, the net force is zero: \(W = U + F_{\text{drag}}\), where \(F_{\text{drag}} = 6 \pi \eta r v\). Therefore, \(F_{\text{drag}} = W - U = 2.564 \times 10^{-3}\text{ N} - 4.141 \times 10^{-4}\text{ N} = 2.150 \times 10^{-3}\text{ N}\). Now, solve for the terminal velocity \(v\): \(v = \frac{F_{\text{drag}}}{6 \pi \eta r} = \frac{2.150 \times 10^{-3}}{6 \pi \times 1.2 \times 2.0 \times 10^{-3}} = \frac{2.150 \times 10^{-3}}{0.04524} = 4.75 \times 10^{-2}\text{ m s}^{-1}\).

[1 mark] Correct calculation of the sphere's volume, \(V = 3.4 \times 10^{-8}\text{ m}^3\).
[1 mark] Correct calculation of the weight of the sphere, \(W = 2.56 \times 10^{-3}\text{ N}\).
[1 mark] Correct calculation of the upthrust, \(U = 4.14 \times 10^{-4}\text{ N}\).
[1 mark] Statement of force balance at terminal velocity: \(W = U + F_{\text{drag}}\).
[1 mark] Correct calculation of the required viscous drag force, \(F_{\text{drag}} = 2.15 \times 10^{-3}\text{ N}\).
[1 mark] Recall of Stokes' law formula \(F = 6 \pi \eta r v\) and rearrangement.
[1 mark] Correct final terminal velocity calculation with units, \(4.7 \times 10^{-2}\text{ m s}^{-1}\) (or \(0.047\text{ m s}^{-1}\); accept \(4.8 \times 10^{-2}\text{ m s}^{-1}\) depending on intermediate rounding).

Let's model the board in equilibrium. The forces acting on the board are: (1) Weight of the board, \(W_b = 35\text{ kg} \times 9.81\text{ m s}^{-2} = 343.35\text{ N}\), acting downwards at the center of gravity (\(2.0\text{ m}\) from A). (2) Weight of the diver, \(W_d = 60\text{ kg} \times 9.81\text{ m s}^{-2} = 588.6\text{ N}\), acting downwards at the right end (\(4.0\text{ m}\) from A). (3) Upward force at support B, \(F_B\), acting at \(1.2\text{ m}\) from A. (4) Force at support A, \(F_A\). Taking moments about point A to find \(F_B\): Clockwise moments = Counter-clockwise moments \(\implies (W_b \times 2.0\text{ m}) + (W_d \times 4.0\text{ m}) = F_B \times 1.2\text{ m}\). \((343.35 \times 2.0) + (588.6 \times 4.0) = 1.2 \times F_B \implies 686.7 + 2354.4 = 1.2 \times F_B \implies 3041.1 = 1.2 \times F_B \implies F_B = 2534.25\text{ N}\) upwards. Now, use translational equilibrium (sum of vertical forces = 0): \(F_A + F_B - W_b - W_d = 0 \implies F_A + 2534.25 - 343.35 - 588.6 = 0 \implies F_A = -1602.3\text{ N}\). The negative sign indicates that the force at support A acts downwards. Thus, the magnitude of the force is \(1.6 \times 10^3\text{ N}\) (to 2 sig figs) acting vertically downwards.

[1 mark] Correct weight of board \(W_b = 343\text{ N}\) acting at \(2.0\text{ m}\) and diver \(W_d = 589\text{ N}\) acting at \(4.0\text{ m}\).
[1 mark] Understanding that the sum of moments about any point is zero in equilibrium.
[1 mark] Correct moment equation about A: \(343.35 \times 2 + 588.6 \times 4 = 1.2 \times F_B\).
[1 mark] Correct calculation of \(F_B = 2530\text{ N}\) (or \(2.5 \times 10^3\text{ N}\)).
[1 mark] Use of vertical force balance equation: \(F_A + F_B = W_b + W_d\).
[1 mark] Calculation of magnitude of \(F_A = 1600\text{ N}\) (or \(1.6 \times 10^3\text{ N}\)).
[1 mark] Correct determination that the direction of the force at A is downwards.

(a) First, calculate the cross-sectional area \(A\): \(A = 4.0 \times 10^{-3}\text{ m} \times 2.0 \times 10^{-3}\text{ m} = 8.0 \times 10^{-6}\text{ m}^2\). Tensile Stress \(\sigma = \frac{F}{A} = \frac{120\text{ N}}{8.0 \times 10^{-6}\text{ m}^2} = 1.5 \times 10^7\text{ Pa}\). Next, calculate the tensile strain \(\epsilon\): Convert units to meters: \(L_0 = 0.080\text{ m}\), \(\Delta L = 0.64 \times 10^{-3}\text{ m}\). Tensile Strain \(\epsilon = \frac{\Delta L}{L_0} = \frac{0.64 \times 10^{-3}\text{ m}}{8.0 \times 10^{-2}\text{ m}} = 8.0 \times 10^{-3}\) (dimensionless). (b) The Young Modulus \(E\) is given by: \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{1.5 \times 10^7\text{ Pa}}{8.0 \times 10^{-3}} = 1.875 \times 10^9\text{ Pa}\), which rounds to \(1.9 \times 10^9\text{ Pa}\).

[1 mark] Calculate cross-sectional area, \(A = 8.0 \times 10^{-6}\text{ m}^2\).
[1 mark] Recall stress formula \(\sigma = F/A\) and substitute values.
[1 mark] Correct calculation of stress with units, \(1.5 \times 10^7\text{ Pa}\).
[1 mark] Recall strain formula \(\epsilon = \Delta L / L\) and convert units correctly.
[1 mark] Correct calculation of strain, \(8.0 \times 10^{-3}\) (accept 0.008).
[1 mark] Recall Young Modulus formula \(E = \text{Stress} / \text{Strain}\).
[1 mark] Correct calculation of Young Modulus with unit, \(1.9 \times 10^9\text{ Pa}\) (accept \(1.88 \times 10^9\text{ Pa}\) or \(\text{N m}^{-2}\)).

(a) Acceleration \(a\) is the rate of change of velocity, so \(a = \frac{dv}{dt}\). Differentiating the velocity equation: \(a = \frac{d}{dt}(3.0 t^2 - 0.20 t^3) = 6.0 t - 0.60 t^2\). At \(t = 4.0\text{ s}\): \(a = 6.0(4.0) - 0.60(4.0)^2 = 24.0 - 0.60(16) = 24.0 - 9.6 = 14.4\text{ m s}^{-2}\). (b) Distance \(s\) is the integral of velocity over time: \(s = \int_{0}^{5.0} v \, dt = \int_{0}^{5.0} (3.0 t^2 - 0.20 t^3) \, dt = \left[ t^3 - 0.050 t^4 \right]_0^{5.0}\). Evaluating the integral at \(t = 5.0\text{ s}\): \(s = (5.0)^3 - 0.050(5.0)^4 = 125 - 0.050(625) = 125 - 31.25 = 93.75\text{ m}\), which rounds to \(94\text{ m}\) (or \(93.8\text{ m}\)).

[1 mark] Recognition that acceleration is the derivative of velocity, \(a = \frac{dv}{dt}\).
[1 mark] Correct differentiation to obtain \(a = 6.0 t - 0.60 t^2\).
[1 mark] Substitution of \(t = 4.0\text{ s}\) to get \(14.4\text{ m s}^{-2}\) (accept \(14\text{ m s}^{-2}\)).
[1 mark] Recognition that distance is the integral of velocity, \(s = \int v \, dt\).
[1 mark] Correct integration to obtain \(t^3 - 0.050 t^4\) (ignore constant of integration due to limits).
[1 mark] Substitution of limits \(0\) and \(5.0\text{ s}\).
[1 mark] Correct final distance calculation with unit, \(94\text{ m}\) (accept \(93.8\text{ m}\)).

(a) The vertical component of the initial velocity is given by:
\( u_y = u \sin(\theta) \)
\( u_y = 14.0\text{ m s}^{-1} \times \sin(25.0^\circ) = 5.917\text{ m s}^{-1} \approx 5.9\text{ m s}^{-1} \).

(b) Using the equation of motion for vertical displacement \( s = ut + \frac{1}{2}at^2 \):
Taking upwards as positive:
\( s = -8.00\text{ m} \)
\( u = 5.917\text{ m s}^{-1} \)
\( a = -9.81\text{ m s}^{-2} \)
\( -8.00 = 5.917t - 4.905t^2 \)
\( 4.905t^2 - 5.917t - 8.00 = 0 \)
Using the quadratic formula:
\( t = \frac{-(-5.917) \pm \sqrt{(-5.917)^2 - 4(4.905)(-8.00)}}{2(4.905)} \)
\( t = \frac{5.917 \pm \sqrt{35.01 + 156.96}}{9.81} = \frac{5.917 \pm 13.855}{9.81} \)
Since time must be positive:
\( t = \frac{19.77}{9.81} = 2.016\text{ s} \approx 2.02\text{ s} \).

(c) The horizontal velocity component is constant:
\( u_x = u \cos(\theta) = 14.0\text{ m s}^{-1} \times \cos(25.0^\circ) = 12.69\text{ m s}^{-1} \).
\( \text{Horizontal distance} = u_x \times t = 12.69\text{ m s}^{-1} \times 2.016\text{ s} = 25.58\text{ m} \approx 25.6\text{ m} \).

(a)
- Use of \( u_y = u \sin(\theta) \) [1M]
- Correct calculation to show \( u_y = 5.92\text{ m s}^{-1} \) [1M]

(b)
- Use of \( s = ut + \frac{1}{2}at^2 \) with appropriate signs for directions [1M]
- Correct substitution of values into the quadratic equation [1M]
- Correct calculation of positive root to give \( 2.02\text{ s} \) (accept range 2.01 to 2.03 s) [1M]

(c)
- Use of \( u_x = 14.0 \cos(25.0^\circ) \) to find horizontal speed (\( 12.7\text{ m s}^{-1} \)) [1M]
- Correct calculation of distance to give \( 25.6\text{ m} \) (accept range 25.4 to 25.8 m) [1M]

(a) First, find the tension (force) in the wire:
\( F = mg = 15.0\text{ kg} \times 9.81\text{ m s}^{-2} = 147.15\text{ N} \)
Now, calculate the tensile stress:
\( \text{Stress} = \frac{F}{A} = \frac{147.15\text{ N}}{1.25 \times 10^{-6}\text{ m}^2} = 1.177 \times 10^8\text{ Pa} \approx 1.18 \times 10^8\text{ Pa} \).

(b) The Young modulus \( E \) is given by:
\( E = \frac{\text{Stress}}{\text{Strain}} \Rightarrow \text{Strain} = \frac{\text{Stress}}{E} \)
\( \text{Strain} = \frac{1.177 \times 10^8\text{ Pa}}{2.00 \times 10^{11}\text{ Pa}} = 5.885 \times 10^{-4} \)
Since \( \text{Strain} = \frac{\Delta L}{L} \):
\( \Delta L = \text{Strain} \times L = 5.885 \times 10^{-4} \times 3.50\text{ m} = 2.060 \times 10^{-3}\text{ m} \approx 2.06 \times 10^{-3}\text{ m} \) (or \( 2.06\text{ mm} \)).

(c) The elastic strain energy \( E_{\text{elastic}} \) is given by:
\( E_{\text{elastic}} = \frac{1}{2} F \Delta L \)
\( E_{\text{elastic}} = 0.5 \times 147.15\text{ N} \times 2.060 \times 10^{-3}\text{ m} = 0.1516\text{ J} \approx 0.152\text{ J} \).

(a)
- Use of \( F = mg \) to find tension [1M]
- Correct calculation of stress to give \( 1.18 \times 10^8\text{ Pa} \) (accept \( 1.2 \times 10^8\text{ Pa} \)) [1M]

(b)
- Use of \( E = \frac{\text{Stress}}{\text{Strain}} \) or \( \Delta L = \frac{F L}{A E} \) [1M]
- Substitution of correct values [1M]
- Correct calculation of extension to give \( 2.06 \times 10^{-3}\text{ m} \) (accept \( 2.1 \times 10^{-3}\text{ m} \)) [1M]

(c)
- Use of \( E_{\text{elastic}} = \frac{1}{2} F \Delta L \) [1M]
- Correct calculation to give \( 0.152\text{ J} \) (accept range 0.150 to 0.153 J) [1M]

The phase difference \(\phi\) in radians is related to the path difference \(d\) by:
\(\phi = \frac{2\pi d}{\lambda}\)

Substituting the given values:
\(\frac{\pi}{3} = \frac{2\pi \times 0.15}{\lambda}\)

Solving for \(\lambda\):
\(\frac{1}{3} = \frac{0.30}{\lambda} \implies \lambda = 0.90\text{ m}\)

Using the wave equation \(v = f\lambda\):
\(v = 50\text{ Hz} \times 0.90\text{ m} = 45\text{ m s}^{-1}\)

1 mark for the correct answer C.

[Method: Calculate wavelength using \(\lambda = \frac{2\pi d}{\phi}\), then calculate wave speed using \(v = f\lambda\).]

The current is given by \(I = nAvq\), where \(A = \pi \frac{d^2}{4}\).
For the first wire:
\(v = \frac{I}{n \left(\pi \frac{d^2}{4}\right) q} = \frac{4I}{n \pi d^2 q}\)

For the second wire, the diameter is \(2d\), so the cross-sectional area \(A_2 = \pi \frac{(2d)^2}{4} = 4 A_1\).
The drift velocity of the second wire \(v_2\) is:
\(v_2 = \frac{3I}{n (4 A_1) q} = \frac{3}{4} \left( \frac{I}{n A_1 q} \right) = 0.75 v\)

1 mark for the correct answer B.

[Method: Recognize that area is proportional to the square of diameter, so area increases by a factor of 4. Use the formula \(I = nAvq\) to deduce that \(v \propto \frac{I}{A}\), yielding a factor of \(0.75\).]

The critical angle \(\theta_c\) at a boundary is given by \(\sin\theta_c = \frac{n_2}{n_1}\), where \(n_1\) is the refractive index of the denser medium (glass) and \(n_2\) is the refractive index of the less dense medium.

Initially, \(n_2 = n_{\text{air}} = 1.00\).
When submerged in water, \(n_2 = n_{\text{water}} > n_{\text{air}}\).
Since \(n_{\text{water}}\) is greater than \(n_{\text{air}}\), the ratio \(\frac{n_2}{n_1}\) increases. This means \(\sin\theta_c\) increases, and therefore the critical angle \(\theta_c\) increases.

The speed of light in a medium is given by \(v = \frac{c}{n}\). Since \(n_{\text{water}} > n_{\text{air}}\), the speed of light in water is less than in air (decreases).

1 mark for the correct answer C.

[Method: Determine that \(n_{\text{water}} > n_{\text{air}}\) leads to an increase in \(\sin\theta_c\) and thus \(\theta_c\), and that higher refractive index means lower speed of light.]

For a negative temperature coefficient (NTC) thermistor, its resistance increases as temperature decreases. Therefore, the resistance of the thermistor increases.

Since the thermistor and the fixed resistor are in series, the total resistance of the circuit increases. With a constant voltage supply \(V\), the current in the circuit decreases.

The output voltage across the thermistor is given by \(V_{\text{out}} = V \times \frac{R_{\text{thermistor}}}{R + R_{\text{thermistor}}}\). As \(R_{\text{thermistor}}\) increases, it takes a larger share of the total resistance, so \(V_{\text{out}}\) increases.

1 mark for the correct answer B.

[Method: Identify NTC relationship (temperature decrease leads to resistance increase), deduce total resistance increases so current decreases, and use potential divider principle to show output voltage increases.]

According to Einstein's photoelectric equation:
\(E_{k,\max} = hf - \phi\)

where \(h\) is the Planck constant, \(f\) is the frequency of the incident light, and \(\phi\) is the work function of the metal.

To increase the maximum kinetic energy \(E_{k,\max}\):
- Increasing the frequency \(f\) directly increases the photon energy \(hf\), which increases \(E_{k,\max}\).
- Increasing intensity increases the rate of emission of photoelectrons but has no effect on their maximum kinetic energy.
- Increasing the exposure time does not change the kinetic energy of individual electrons.
- Using a metal with a larger work function \(\phi\) would decrease \(E_{k,\max}\).

1 mark for the correct answer B.

[Method: Use the equation \(E_{k,\max} = hf - \phi\) to relate frequency directly to maximum kinetic energy.]

The current \(I\) in the circuit is given by the potential difference across the external resistor divided by its resistance:
\(I = \frac{V}{R} = \frac{10.0\text{ V}}{5.0\ \Omega} = 2.0\text{ A}\)

The e.m.f. formula is:
\(E = V + Ir\)

Substituting the values:
\(12.0\text{ V} = 10.0\text{ V} + (2.0\text{ A}) \times r\)
\(2.0\text{ V} = 2.0\text{ A} \times r\)
\(r = 1.0\ \Omega\)

1 mark for the correct answer B.

[Method: Calculate current using Ohm's law on the external resistor, then apply \(E = V + Ir\) to find the internal resistance.]

The diffraction grating equation is:
\(d \sin \theta = n \lambda\)

For the first light source (first-order maximum, \(n = 1\)):
\(d \sin \theta = 1 \times \lambda = \lambda\)

For the second light source (second-order maximum, \(n = 2\)) at the same angle \(\theta\):
\(d \sin \theta = 2 \lambda_2\)

Since \(d \sin \theta\) is constant for both observations:
\(\lambda = 2 \lambda_2 \implies \lambda_2 = 0.5 \lambda\)

1 mark for the correct answer B.

[Method: Set up diffraction grating equations for both cases and equate \(d \sin \theta\).]

The volume of a cylinder is \(V = A L\), where \(A\) is the cross-sectional area and \(L\) is the length.
Since volume is constant:
\(V = A_1 L_1 = A_2 L_2\)

Given \(L_2 = 2L_1\):
\(A_1 L_1 = A_2 (2L_1) \implies A_2 = \frac{A_1}{2}\)

Resistance is given by \(R = \rho \frac{L}{A}\).
The new resistance \(R_2\) is:
\(R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2L_1}{A_1 / 2} = 4 \left( \rho \frac{L_1}{A_1} \right) = 4R\)

1 mark for the correct answer C.

[Method: Deduce that area is halved when length is doubled at constant volume, then use \(R = \rho \frac{L}{A}\) to find the factor of 4 increase.]

Let the total current from the power supply be \(I\). The current passing through the series resistor is \(I\). The power dissipated by this resistor is \(P_{\text{series}} = I^2 R\). The total current \(I\) divides equally between the two identical parallel resistors, so the current passing through each of the parallel resistors is \(\frac{I}{2}\). The power dissipated in one of these parallel resistors is \(P_{\text{parallel}} = \left(\frac{I}{2}\right)^2 R = \frac{I^2 R}{4}\). The ratio of the power in one of the parallel resistors to the series resistor is \(\frac{P_{\text{parallel}}}{P_{\text{series}}} = \frac{\frac{I^2 R}{4}}{I^2 R} = 0.25\).

1 mark for the correct option A.

From the grating equation, \(d \sin\theta = n\lambda\). For the initial second-order maximum, \(d \sin\theta = 2\lambda\), which gives \(\sin\theta = \frac{2\lambda}{d}\). For the new setup, the second-order maximum occurs at \(\theta'\) where \((2d) \sin\theta' = 2(1.5\lambda) = 3\lambda\). Rearranging gives \(\sin\theta' = \frac{3\lambda}{2d}\). Substituting \(\frac{\lambda}{d} = \frac{\sin\theta}{2}\) into this equation yields \(\sin\theta' = \frac{3}{2} \left(\frac{\sin\theta}{2}\right) = \frac{3}{4}\sin\theta = 0.75\sin\theta\).

1 mark for the correct option B.

(a) The equation for current is:
\(I = n A v e\)

The cross-sectional area \(A\) of a wire with diameter \(d\) is:
\(A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}\)

Substituting this into the current equation gives:
\(I = n \left(\frac{\pi d^2}{4}\right) v e\)

Rearranging for \(v\):
\(v = \frac{4I}{n e \pi d^2}\) [As required]

(b) Substitute the given values into the formula:
\(I = 1.5\text{ A}\)
\(n = 8.5 \times 10^{28}\text{ m}^{-3}\)
\(e = 1.6 \times 10^{-19}\text{ C}\)
\(d = 0.80\text{ mm} = 8.0 \times 10^{-4}\text{ m}\)

\(v = \frac{4 \times 1.5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times \pi \times (8.0 \times 10^{-4})^2}\)
\(v = \frac{6}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 3.14159 \times 6.4 \times 10^{-7}}\)
\(v = \frac{6}{2.734 \times 10^4} \approx 2.19 \times 10^{-4}\text{ m s}^{-1}\)

To two significant figures, \(v = 2.2 \times 10^{-4}\text{ m s}^{-1}\).

(c) If temperature increases, the positive ions in the lattice vibrate with greater amplitude, causing more frequent collisions with conduction electrons. Normally this increases resistance and would decrease current, but since the current is maintained constant, all quantities \(I\), \(n\), \(e\), and \(d\) remain unchanged. Therefore, the mean drift velocity remains constant because \(v = \frac{4I}{n e \pi d^2}\) and all quantities on the right side of the equation are constant.

(a) [3 Marks]
- Recall of \(I = nAve\) (1)
- Recall of \(A = \pi \frac{d^2}{4}\) (1)
- Clear algebraic rearrangement to obtain given formula (1)

(b) [3 Marks]
- Substitution of \(e = 1.6 \times 10^{-19}\text{ C}\) and \(d = 8.0 \times 10^{-4}\text{ m}\) in equation (1)
- Correct calculation of denominator as \(2.7 \times 10^4\) or correct intermediate working (1)
- Correct final answer to 2 s.f. with units: \(2.2 \times 10^{-4}\text{ m s}^{-1}\) (1)

(c) [3 Marks]
- State that mean drift velocity remains the same (1)
- Because \(I\), \(n\), \(e\), and \(d\) are all constant (1)
- Explain that increased lattice vibrations/collisions do not affect drift velocity when current is held constant (1)

(a) Light waves from the two slits spread out (diffract) and overlap. Where they overlap, superposition occurs.
For constructive interference (bright fringes), the waves arrive in phase, meaning the path difference is an integer number of wavelengths (\(n\lambda\)). Their displacements add to give maximum amplitude.
For destructive interference (dark fringes), the waves arrive in antiphase, meaning the path difference is an odd number of half wavelengths (\((n + 0.5)\lambda\)). Their displacements cancel each other out.

(b) Use the double-slit formula:
\(w = \frac{\lambda D}{s}\)

Where:
\(\lambda = 633 \times 10^{-9}\text{ m}\)
\(D = 2.4\text{ m}\)
\(s = 0.15 \times 10^{-3}\text{ m}\)

\(w = \frac{633 \times 10^{-9} \times 2.4}{0.15 \times 10^{-3}}\)
\(w = \frac{1.5192 \times 10^{-6}}{1.5 \times 10^{-4}} = 0.010128\text{ m}\)

To 2 s.f., the distance is \(1.0 \times 10^{-2}\text{ m}\) (or \(10\text{ mm}\)).

(c) The green laser has a shorter wavelength (\(532\text{ nm} < 633\text{ nm}\)). Since \(w \propto \lambda\) (for constant \(D\) and \(s\)), a smaller wavelength results in a smaller fringe spacing \(w\). Therefore, the adjacent bright fringes will be closer together.

(a) [4 Marks]
- Light diffracts from slits and overlaps/superposes (1)
- Constructive interference occurs when path difference is \(n\lambda\) / waves are in phase (1)
- Destructive interference occurs when path difference is \((n + 0.5)\lambda\) / waves are in antiphase (1)
- Displacements add for bright fringe and cancel/subtract for dark fringe (1)

(b) [3 Marks]
- Use of \(w = \frac{\lambda D}{s}\) with correct unit conversions (1)
- Correct substitution of values (1)
- Correct calculation to 2 s.f. with unit: \(1.0 \times 10^{-2}\text{ m}\) (accept \(0.010\text{ m}\) or \(10\text{ mm}\)) (1)

(c) [2 Marks]
- Fringe spacing decreases (1)
- Because wavelength is smaller and fringe spacing is directly proportional to wavelength (1)

(a) The circuit diagram must contain: a cell in series with a variable resistor, an ammeter in series to measure the current \(I\), and a voltmeter connected in parallel across the terminals of the cell (or across the variable resistor) to measure the terminal potential difference \(V\).

(b) The terminal potential difference is given by \(V = \mathcal{E} - Ir\), which can be written as:
\(V = -rI + \mathcal{E}\)

Comparing this to the equation of a straight line, \(y = mx + c\):
- A plot of \(V\) against \(I\) yields a straight line.
- The y-intercept represents the electromotive force (\(\text{emf}\), \(\mathcal{E}\)).
- The magnitude of the gradient of the line represents the internal resistance (\(r\)).

(c) Using the equation \(V = \mathcal{E} - Ir\) for both states:
\(V_1 = \mathcal{E} - I_1 r\)
\(V_2 = \mathcal{E} - I_2 r\)

Subtracting the two equations:
\(V_1 - V_2 = r(I_2 - I_1)\)

Rearranging for \(r\):
\(r = \frac{V_1 - V_2}{I_2 - I_1}\)

Substitute the values:
\(r = \frac{1.28 - 1.12}{0.85 - 0.45}\)
\(r = \frac{0.16}{0.40} = 0.40\text{ }\Omega\)

(a) [3 Marks]
- Cell, variable resistor, and ammeter in a single series loop (1)
- Voltmeter connected in parallel across the cell (or across the variable resistor) (1)
- Correct standard symbols used for all components (1)

(b) [3 Marks]
- State or show the relationship \(V = \mathcal{E} - Ir\) (1)
- Identify that \(\text{emf}\) is the y-intercept / intercept on the vertical axis (1)
- Identify that internal resistance \(r\) is equal to \(-\text{gradient}\) / magnitude of gradient (1)

(c) [3 Marks]
- Attempt to use \(\Delta V = r \Delta I\) or set up simultaneous equations (1)
- Correct substitution of values: \(\frac{0.16}{0.40}\) (1)
- Correct final value with unit: \(0.40\text{ }\Omega\) (1)

(a) The work function is the minimum energy required to liberate a conduction electron from the surface of a metal.

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

Where:
\(h = 6.63 \times 10^{-34}\text{ J s}\)
\(f = 1.25 \times 10^{15}\text{ Hz}\)
\(\Phi = 4.31\text{ eV} = 4.31 \times 1.60 \times 10^{-19}\text{ J} = 6.896 \times 10^{-19}\text{ J}\)

Calculate photon energy \(E = hf\):
\(E = 6.63 \times 10^{-34} \times 1.25 \times 10^{15} = 8.2875 \times 10^{-19}\text{ J}\)

Calculate \(E_{\text{k,max}}\):
\(E_{\text{k,max}} = hf - \Phi\)
\(E_{\text{k,max}} = 8.2875 \times 10^{-19} - 6.896 \times 10^{-19} = 1.3915 \times 10^{-19}\text{ J}\)

To 3 s.f., the maximum kinetic energy is \(1.39 \times 10^{-19}\text{ J}\).

(c) If the intensity is doubled, the maximum kinetic energy of the emitted photoelectrons remains unchanged.
This is because there is a one-to-one interaction between a single photon and a single electron. The maximum kinetic energy depends only on the frequency of the incident photons and the work function of the metal, both of which are unchanged. Doubling the intensity only doubles the rate at which photons arrive, which increases the rate of emission of photoelectrons (current) but does not affect the energy of individual photons.

(a) [2 Marks]
- Minimum energy (1)
- Required to release/emit an electron from the surface of a metal (1)

(b) [4 Marks]
- Conversion of work function from eV to Joules: \(4.31 \times 1.6 \times 10^{-19} = 6.9 \times 10^{-19}\text{ J}\) (1)
- Use of \(E = hf\) to calculate photon energy (1)
- Substitution into \(hf = \Phi + E_{\text{k,max}}\) (1)
- Correct evaluation to 3 s.f.: \(1.39 \times 10^{-19}\text{ J}\) (accept \(1.35 \times 10^{-19}\text{ J}\) to \(1.40 \times 10^{-19}\text{ J}\) depending on rounding of \(h\)) (1)

(c) [3 Marks]
- State that there is no change to the maximum kinetic energy (1)
- Explain that there is a one-to-one interaction between photons and electrons (1)
- State that increasing intensity increases the number of photons per second, not their individual energy (1)

(a) A wave travels along the wire and is reflected at the fixed ends. Two progressive waves of the same frequency and amplitude are now travelling in opposite directions. These waves superpose. Where they are in phase, constructive interference occurs, creating points of maximum displacement called antinodes. Where they are in antiphase, destructive interference occurs, creating points of zero displacement called nodes.

(b) (i) For a wire of length \(L\) fixed at both ends vibrating in its fundamental mode:
\(L = \frac{\lambda}{2}\)
\(\lambda = 2 L = 2 \times 1.5\text{ m} = 3.0\text{ m}\)

(ii) The wave speed \(v\) is related to frequency \(f\) and wavelength \(\lambda\) by:
\(v = f \lambda\)

Substitute the values:
\(f = 85\text{ Hz}\)
\(\lambda = 3.0\text{ m}\)
\(v = 85 \times 3.0 = 255\text{ m s}^{-1}\)

To 2 s.f., \(v = 2.6 \times 10^2\text{ m s}^{-1}\) (or \(260\text{ m s}^{-1}\)).

(a) [3 Marks]
- Wave reflects at the ends / waves travel in opposite directions (1)
- Waves have the same frequency/wavelength and superpose (1)
- Constructive interference creates antinodes and destructive interference creates nodes (1)

(b)(i) [2 Marks]
- State or use \(\lambda = 2L\) (1)
- Correct value of \(3.0\text{ m}\) (1)

(b)(ii) [4 Marks]
- Use of \(v = f \lambda\) (1)
- Substitution of \(f = 85\text{ Hz}\) and their calculated \(\lambda\) (1)
- Correct calculation: \(255\text{ m s}^{-1}\) (1)
- Correct final answer to 2 s.f. with unit: \(2.6 \times 10^2\text{ m s}^{-1}\) or \(260\text{ m s}^{-1}\) (1)

(a) The circuit diagram must show: a \(6.0\text{ V}\) DC source (battery/cell) in series with a \(3.3\text{ k}\Omega\) resistor and an LDR. The output terminals or voltmeter representing \(V_{\text{out}}\) must be connected in parallel across the LDR.

(b) When the intensity of light on the LDR decreases, the resistance of the LDR increases.
In a potential divider, the output voltage across a component is proportional to its resistance relative to the total resistance:
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\)
As \(R_{\text{LDR}}\) increases, it takes a larger share of the total resistance, and thus the potential difference across it, \(V_{\text{out}}\), increases.

(c) Use the potential divider formula:
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\)

Where:
\(V_{\text{in}} = 6.0\text{ V}\)
\(R = 3.3\text{ k}\Omega = 3300\text{ }\Omega\)
\(R_{\text{LDR}} = 8.5\text{ k}\Omega = 8500\text{ }\Omega\)

\(V_{\text{out}} = 6.0 \times \frac{8500}{3300 + 8500}\)
\(V_{\text{out}} = 6.0 \times \frac{8500}{11800} = 6.0 \times 0.7203 \approx 4.32\text{ V}\)

To 2 s.f., \(V_{\text{out}} = 4.3\text{ V}\).

(a) [3 Marks]
- Standard symbols for battery, fixed resistor, and LDR in series (1)
- Output voltage \(V_{\text{out}}\) (or voltmeter) connected across the LDR (1)
- Complete and correct circuit with labels (1)

(b) [3 Marks]
- Decreasing light intensity increases the resistance of the LDR (1)
- Total resistance of circuit increases, so current decreases (or LDR has a larger share of total resistance) (1)
- Therefore, \(V_{\text{out}}\) increases (1)

(c) [3 Marks]
- Use of potential divider formula (1)
- Substitution of correct resistances: \(\frac{8.5}{3.3 + 8.5}\) or \(\frac{8500}{11800}\) (1)
- Correct calculation to 2 s.f.: \(4.3\text{ V}\) (1)

(a) The critical angle is the angle of incidence in the more dense medium (glass) for which the angle of refraction in the less dense medium (air) is \(90^\circ\).

(b) Use Snell's law at the critical angle:
\(\sin(\theta_c) = \frac{n_2}{n_1}\)

Where \(n_1 = 1.62\) (glass) and \(n_2 = 1.00\) (air).
\(\sin(\theta_c) = \frac{1.00}{1.62} \approx 0.6173\)
\(\theta_c = \sin^{-1}(0.6173) \approx 38.12^\circ\)

To 3 s.f., the critical angle is \(38.1^\circ\).

(c) The angle of incidence is \(\theta_i = 45.0^\circ\).
Since \(\theta_i > \theta_c\) (\(45.0^\circ > 38.1^\circ\)) and the light is travelling from a more optically dense medium (glass) towards a less optically dense medium (air), total internal reflection (TIR) occurs. No light is refracted into the air; instead, the beam is entirely reflected back into the glass block at an angle of reflection of \(45.0^\circ\) to the normal.

(a) [2 Marks]
- Angle of incidence (in the more dense medium) (1)
- That results in an angle of refraction of \(90^\circ\) / light travelling along the boundary (1)

(b) [3 Marks]
- Use of \(\sin \theta_c = \frac{1}{n}\) (1)
- Substitution of \(n = 1.62\) (1)
- Correct calculation to 3 s.f.: \(38.1^\circ\) (1)

(c) [4 Marks]
- State that total internal reflection (TIR) occurs (1)
- Identify that the angle of incidence is greater than the critical angle (\(45.0^\circ > 38.1^\circ\)) (1)
- Explain that light is travelling from a more dense to a less dense medium (1)
- State that light is reflected back into the glass at \(45.0^\circ\) (1)

(a) Use Ohm's law:
\(R = \frac{V}{I}\)
\(R = \frac{1.50\text{ V}}{0.82\text{ A}} \approx 1.829\text{ }\Omega\)

To 2 s.f., \(R = 1.8\text{ }\Omega\).

(b) Use the resistivity formula:
\(R = \frac{\rho L}{A} \implies \rho = \frac{R A}{L}\)

Where:
\(R = 1.829\text{ }\Omega\)
\(A = 3.5 \times 10^{-7}\text{ m}^2\)
\(L = 2.4\text{ m}\)

\[\rho = \frac{1.829 \times 3.5 \times 10^{-7}}{2.4}\]
\(\rho = \frac{6.4015 \times 10^{-7}}{2.4} \approx 2.667 \times 10^{-7}\text{ }\Omega\text{ m}\)

To 2 s.f., \(\rho = 2.7 \times 10^{-7}\text{ }\Omega\text{ m}\).

(c) A metal consists of a lattice of positive metal ions surrounded by a sea of free (conduction) electrons. As the temperature decreases, the lattice ions vibrate with less thermal energy and lower amplitude. This reduces the frequency of collisions between the drifting conduction electrons and the vibrating lattice ions. As a result, the conduction electrons can drift more easily through the metal, meaning the resistance decreases, and hence the resistivity decreases.

(a) [2 Marks]
- Use of \(R = \frac{V}{I}\) (1)
- Correct value: \(1.8\text{ }\Omega\) (accept \(1.83\text{ }\Omega\)) (1)

(b) [3 Marks]
- Rearranging formula to \(\rho = \frac{RA}{L}\) (1)
- Substitution of values including their unrounded \(R\) (1)
- Correct calculation to 2 s.f. with unit: \(2.7 \times 10^{-7}\text{ }\Omega\text{ m}\) (1)

(c) [4 Marks]
- Lower temperature means metal ions vibrate with smaller amplitude (1)
- Fewer collisions between conduction electrons and lattice ions (1)
- Drift of electrons is easier / less resistance to flow (1)
- Since resistivity is proportional to resistance, resistivity decreases (1)

(a) Mean diameter \(d_{\text{mean}} = \frac{0.38 + 0.37 + 0.39 + 0.38 + 0.38}{5} = 0.38\text{ mm}\). Range = \(0.39 - 0.37 = 0.02\text{ mm}\). Uncertainty in diameter \(= \pm \frac{\text{Range}}{2} = \pm 0.01\text{ mm}\). Percentage uncertainty \(= \frac{0.01}{0.38} \times 100\% \approx 2.63\%\). (b) Cross-sectional area \(A = \frac{\pi d^2}{4} = \frac{\pi (0.38 \times 10^{-3})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2 \approx 1.13 \times 10^{-7}\text{ m}^2\). Percentage uncertainty in area \(A\) is twice that of \(d\) because \(A \propto d^2\). Percentage uncertainty in \(A = 2 \times 2.63\% = 5.26\% \approx 5.3\%\). (c) Resistivity \(\rho = \frac{R A}{l}\). Since gradient of \(R\) vs \(l\) is \(\frac{R}{l} = 4.05\ \Omega\text{ m}^{-1}\), \(\rho = \text{gradient} \times A = 4.05 \times 1.134 \times 10^{-7} = 4.59 \times 10^{-7}\ \Omega\text{ m}\). Percentage uncertainty in \(\rho\) is dominated by area uncertainty (assuming negligible gradient uncertainty), which is \(5.26\%\). Absolute uncertainty in \(\rho = 5.26\% \times 4.59 \times 10^{-7} = 0.24 \times 10^{-7}\ \Omega\text{ m}\). Hence, \(\rho = (4.6 \pm 0.2) \times 10^{-7}\ \Omega\text{ m}\). (d) Precaution: Use a switch in the circuit and only close it briefly when taking readings, or use small currents. Explanation: Current causes heating in the wire, which increases its temperature. Since the resistance/resistivity of metals increases with temperature, continuous current would systematically overestimate the resistance, reducing the accuracy of the determined resistivity.

(a) [3 Marks] 1 mark for correct mean diameter of 0.38 mm. 1 mark for calculating uncertainty using half-range (0.01 mm). 1 mark for correct percentage uncertainty of 2.6% (allow 2.63%). (b) [3 Marks] 1 mark for substituting the mean diameter correctly into \(A = \frac{\pi d^2}{4}\). 1 mark for identifying that percentage uncertainty in \(A\) is twice that of \(d\). 1 mark for correct percentage uncertainty of 5.3% (allow 5.26% or 5%). (c) [4 Marks] 1 mark for recalling \(\rho = \text{gradient} \times A\). 1 mark for calculating resistivity value \(4.6 \times 10^{-7}\ \Omega\text{ m}\) (allow \(4.59 \times 10^{-7}\)). 1 mark for stating percentage uncertainty in resistivity is same as area (5.3%). 1 mark for correct absolute uncertainty \(\pm 0.2 \times 10^{-7}\ \Omega\text{ m}\) with correct units. (d) [2.5 Marks] 1 mark for identifying a valid precaution (e.g., switch open between readings, low current). 1.5 marks for explaining that current causes temperature rise, which increases resistance and results in systematic error/overestimate of resistivity.

(a) Comparing \(h = \frac{1}{2}gt^2\) with the linear equation \(y = mx + c\), where \(y = h\) and \(x = t^2\), we find the y-intercept \(c = 0\) (hence passing through the origin) and the gradient \(m = \frac{1}{2}g\). (b) Calculating \(t^2\) to 3 or 4 significant figures: For \(h = 0.200\text{ m}\), \(t^2 = 0.202^2 = 0.0408\text{ s}^2\). For \(h = 0.400\text{ m}\), \(t^2 = 0.286^2 = 0.0818\text{ s}^2\). For \(h = 0.600\text{ m}\), \(t^2 = 0.350^2 = 0.1225\text{ s}^2\). For \(h = 0.800\text{ m}\), \(t^2 = 0.404^2 = 0.1632\text{ s}^2\). For \(h = 1.000\text{ m}\), \(t^2 = 0.452^2 = 0.2043\text{ s}^2\). (c) Gradient \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1.000 - 0.200}{0.2043 - 0.0408} = \frac{0.800}{0.1635} = 4.893\text{ m s}^{-2}\). Since \(m = \frac{1}{2}g\), \(g = 2m = 2 \times 4.893 = 9.79\text{ m s}^{-2}\) (accept \(9.78\text{ m s}^{-2}\) depending on rounding). (d) Systematic error: Air resistance acts on the ball-bearing during its fall. Explanation: Air resistance opposes motion, slowing the ball-bearing's descent. This increases the measured time \(t\) for each height, resulting in larger \(t^2\) values. A larger horizontal value for the same vertical height reduces the gradient, leading to an underestimate of \(g\).

(a) [2 Marks] 1 mark for matching \(h = \frac{1}{2}gt^2\) with \(y = mx\) where \(c = 0\). 1 mark for stating that \(\text{gradient} = \frac{1}{2}g\). (b) [2 Marks] 2 marks for calculating all five \(t^2\) values correctly to 3 or 4 significant figures (deduct 1 mark if incorrect or inconsistent rounding is used). (c) [5 Marks] 2 marks for correct gradient calculation method with correct substitutions. 1 mark for correct gradient value of \(4.89\text{ m s}^{-2}\) (allow \(4.89 - 4.90\)). 1 mark for multiplying gradient by 2 to get \(g\). 1 mark for final calculated value of \(g = 9.78\text{ m s}^{-2}\) or \(9.79\text{ m s}^{-2}\) with correct units. (d) [3.5 Marks] 1 mark for naming a valid systematic error (e.g., air resistance, delay in release mechanism, or misaligned light gates). 2.5 marks for explaining the mechanism clearly (e.g., air resistance slows the ball, increasing the fall time \(t\), which makes the gradient of the graph too small, causing \(g\) to be underestimated).

(a) \(A = \frac{\pi d^2}{4} = \frac{\pi (0.56 \times 10^{-3})^2}{4} = 2.463 \times 10^{-7}\text{ m}^2 \approx 2.46 \times 10^{-7}\text{ m}^2\). (b) Young modulus \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \left(\frac{F}{\Delta L}\right) \frac{L}{A}\). The gradient of the force-extension graph is \(\frac{F}{\Delta L} = 9.8 \times 10^3\text{ N m}^{-1}\). Thus, \(E = \text{gradient} \times \frac{L}{A} = 9.8 \times 10^3 \times \frac{2.450}{2.463 \times 10^{-7}} = 9.748 \times 10^{10}\text{ Pa} \approx 9.7 \times 10^{10}\text{ Pa}\) (or \(9.8 \times 10^{10}\text{ Pa}\) if using rounded area \(2.46 \times 10^{-7}\text{ m}^2\)). (c) Since \(E = \text{gradient} \times L / A\) and \(A = \frac{\pi d^2}{4}\), the percentage uncertainty in \(E\) is given by: \(\% \text{ uncertainty in } E = \% \text{ uncertainty in gradient} + \% \text{ uncertainty in } L + 2 \times (\% \text{ uncertainty in } d)\). \(\% \text{ uncertainty in } E = 3.5\% + 0.2\% + 2 \times (1.8\%) = 3.5\% + 0.2\% + 3.6\% = 7.3\%\). (d) From \(\Delta L = \frac{FL}{AE}\), for a given force, a longer length \(L\) and smaller cross-sectional area \(A\) (thin wire) result in a significantly larger extension \(\Delta L\). A larger extension can be measured with a much smaller percentage uncertainty, dramatically improving the accuracy and precision of the calculated Young modulus.

(a) [2 Marks] 1 mark for recalling area formula. 1 mark for correct area value of \(2.46 \times 10^{-7}\text{ m}^2\). (b) [4 Marks] 1 mark for recalling stress/strain definition or Young modulus formula \(E = \frac{FL}{A\Delta L}\). 1 mark for recognizing that gradient is \(\frac{F}{\Delta L}\). 1 mark for correct substitution. 1 mark for correct calculated value of \(E = 9.7 \times 10^{10}\text{ Pa}\) or \(9.8 \times 10^{10}\text{ Pa}\) with unit (Pa or \(\text{N m}^{-2}\)). (c) [3.5 Marks] 1 mark for doubling the percentage uncertainty in diameter to find the percentage uncertainty in area (3.6%). 1 mark for correctly setting up the sum of uncertainties. 1.5 marks for correct final answer of 7.3%. (d) [3 Marks] 1 mark for stating that a long, thin wire results in larger extensions. 2 marks for explaining that a larger extension leads to a lower percentage uncertainty in the measurement of extension, improving accuracy of the final value.

(a) Calculating \(\sin i\) and \(\sin r\) values to 3 significant figures: For \(i = 15^\circ, 30^\circ, 45^\circ, 60^\circ, 75^\circ\), the values are: \(\sin i = 0.259, 0.500, 0.707, 0.866, 0.966\). For \(r = 10^\circ, 19.5^\circ, 28^\circ, 35^\circ, 40^\circ\), the values are: \(\sin r = 0.174, 0.334, 0.469, 0.574, 0.643\). (b) Rearranging Snell's law gives \(\sin i = n \sin r\). This matches the equation of a straight line through the origin \(y = mx\), where \(y = \sin i\), \(x = \sin r\), and the gradient is \(n\). Using coordinates for \(30^\circ\) and \(60^\circ\): \((x_1, y_1) = (0.334, 0.500)\) and \((x_2, y_2) = (0.574, 0.866)\). Gradient \(m = \frac{0.866 - 0.500}{0.574 - 0.334} = \frac{0.366}{0.240} = 1.525 \approx 1.53\). Therefore, \(n = 1.53\) (or \(1.52\)). (c) Light directed at the midpoint of the flat face enters the block and travels along a radius towards the curved boundary. Because every radius is perpendicular to the circumference, the light meets the curved boundary along the normal (angle of incidence at the boundary is \(0^\circ\)). Therefore, the light exits without refracting/bending, allowing the angle inside the glass to be measured directly from outside. (d) Using \(\sin \theta_c = \frac{1}{n}\), with \(n = 1.525\), we find \(\sin \theta_c = \frac{1}{1.525} = 0.6557\), giving \(\theta_c = 41.0^\circ\) (allow \(41^\circ\)).

(a) [3 Marks] 1.5 marks for correct \(\sin i\) values. 1.5 marks for correct \(\sin r\) values to 3 significant figures. (b) [4.5 Marks] 1 mark for explaining that \(\sin i = n \sin r\) is linear, so the gradient equals \(n\). 2 marks for substituting \(\sin i\) and \(\sin r\) coordinates correctly into the gradient formula. 1.5 marks for finding gradient \(1.52\) to \(1.53\) with no unit. (c) [3 Marks] 1 mark for stating that rays strike the curved boundary along the normal/radius. 1 mark for stating the angle of incidence at this boundary is \(0^\circ\). 1 mark for concluding that there is no change in direction when exiting, enabling direct measurement. (d) [2 Marks] 1 mark for recalling \(\sin \theta_c = \frac{1}{n}\). 1 mark for calculating \(\theta_c \approx 41^\circ\) (accept range \(40.8^\circ - 41.2^\circ\)).