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By conservation of energy, the total energy of the object is constant: \(E_{\text{total}} = E_{\text{k}} + E_{\text{p}} = E_{\text{k0}}\). The gravitational potential energy at height \(h\) is \(E_{\text{p}} = mgh\). Therefore, the kinetic energy at height \(h\) is \(E_{\text{k}} = E_{\text{k0}} - mgh\). This is an equation of a straight line where the gradient is \(-mg\) (constant and negative) and the intercept on the vertical axis is \(E_{\text{k0}}\). Thus, the correct graph is a straight line with a negative gradient.

1 mark for selecting option A.

For the frame to be in vertical equilibrium, the sum of the vertical components of the tension in both strings must equal the weight of the frame. Each string has a vertical component of tension equal to \(T \sin\theta\). Therefore: \(2T \sin\theta = mg\). Solving for \(T\) gives \(T = \frac{mg}{2\sin\theta}\).

1 mark for selecting option D.

The Young modulus \(E\) is given by \(E = \frac{\sigma}{\varepsilon} = \frac{F/A}{\Delta x/L} = \frac{FL}{A\Delta x}\), where \(A = \pi d^2 / 4\). Rearranging for extension gives: \(\Delta x = \frac{4FL}{\pi d^2 E}\). Since both wires are of the same material, \(E\) is constant. The load \(F\) is also the same. Therefore, \(\Delta x \propto \frac{L}{d^2}\). For wire X: \(L_{\text{X}} = 2L_{\text{Y}}\) and \(d_{\text{X}} = 2d_{\text{Y}}\). Thus, \(\Delta x_{\text{X}} \propto \frac{2L_{\text{Y}}}{(2d_{\text{Y}})^2} = \frac{2L_{\text{Y}}}{4d_{\text{Y}}^2} = \frac{1}{2}\left(\frac{L_{\text{Y}}}{d_{\text{Y}}^2}\right)\). This means \(\Delta x_{\text{X}} = 0.5 \Delta x_{\text{Y}}\), so the ratio is 0.5.

1 mark for selecting option C.

Because the block is moving at a constant velocity, its acceleration is zero, meaning the net force acting on it along the slope must be zero. The component of the weight acting down the slope is \(mg \sin\theta\). The frictional force acts up the slope to oppose motion. Therefore, the magnitude of the frictional force must balance the component of the weight down the slope: \(F_{\text{friction}} = mg \sin\theta\).

1 mark for selecting option A.

For a sphere falling at terminal velocity in a fluid, the weight \(W\) is balanced by the upthrust \(U\) and the viscous drag \(F_{\text{D}}\). Thus, \(W - U = F_{\text{D}}\). This gives \(\frac{4}{3}\pi r^3 (\rho_{\text{s}} - \rho_{\text{f}})g = 6\pi \eta r v\). Rearranging for terminal velocity \(v\): \(v = \frac{2r^2 g(\rho_{\text{s}} - \rho_{\text{f}})}{9\eta}\). Since the material (density \(\rho_{\text{s}}\)) and fluid (density \(\rho_{\text{f}}\), viscosity \(\eta\)) are the same, \(v \propto r^2\). If the radius of the second sphere is \(2r\), its terminal velocity will be proportional to \((2r)^2 = 4r^2\). Therefore, the ratio of the terminal velocities is 4.

1 mark for selecting option C.

Efficiency is defined as the ratio of useful power output to total power input. The useful work done in lifting the load is the increase in gravitational potential energy, \(\Delta E_{\text{p}} = Mgh\). The useful power output is the rate of doing this work: \(P_{\text{out}} = \frac{Mgh}{t}\). The power input is \(P\). Therefore, efficiency \(= \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{Mgh/t}{P} = \frac{Mgh}{Pt}\).

1 mark for selecting option A.

The work done in stretching the wire elastically, which is stored as elastic strain energy \(U\), is given by \(U = \frac{1}{2} F \Delta x\). From the definition of Young modulus, \(E = \frac{FL}{A\Delta x}\), we can express the force as \(F = \frac{EA\Delta x}{L}\). Substituting this expression for \(F\) into the energy formula gives: \(U = \frac{1}{2} \left(\frac{EA\Delta x}{L}\right) \Delta x = \frac{EA(\Delta x)^2}{2L}\).

1 mark for selecting option B.

The initial kinetic energy is \(E_{\text{k}} = \frac{1}{2} m u^2\), where \(u\) is the initial launch speed. Since air resistance is negligible, the horizontal component of velocity remains constant throughout the flight: \(v_{\text{x}} = u \cos\theta\). At the highest point of the trajectory, the vertical component of velocity is zero, so the speed of the projectile is equal to its horizontal component, \(u \cos\theta\). The kinetic energy at this point is therefore \(E_{\text{highest}} = \frac{1}{2} m (u \cos\theta)^2 = \left(\frac{1}{2} m u^2\right) \cos^2\theta = E_{\text{k}} \cos^2\theta\).

1 mark for selecting option C.

At the highest point of its trajectory, the vertical component of the projectile's velocity is zero, while the horizontal component remains constant at \(u \cos \theta\), where \(u\) is the initial launch speed. The initial kinetic energy is \(E_k = \frac{1}{2} m u^2\). The kinetic energy at the highest point is \(E_{\text{top}} = \frac{1}{2} m (u \cos \theta)^2 = \frac{1}{2} m u^2 \cos^2 \theta = E_k \cos^2 \theta\). Therefore, the correct option is B.

1 mark for the correct option B. Incorrect options: A is incorrect because the projectile still has horizontal motion; C is incorrect because kinetic energy is proportional to velocity squared; D is incorrect because it corresponds to the vertical component of the initial kinetic energy.

The elastic strain energy \(E\) stored in a wire is given by \(E = \frac{1}{2} F \Delta x\), where \(\Delta x\) is the extension. The extension is given by \(\Delta x = \frac{F L}{A E_y}\), where \(E_y\) is the Young modulus and \(A\) is the cross-sectional area. Substituting this gives \(E = \frac{F^2 L}{2 A E_y}\). Since both wires have the same material (same \(E_y\)) and are subjected to the same force \(F\), the energy is proportional to \(\frac{L}{A}\). Since area \(A = \frac{\pi d^2}{4}\), we have \(A \propto d^2\), so \(E \propto \frac{L}{d^2}\). For wire X, \(E_X \propto \frac{L}{d^2}\). For wire Y, \(E_Y \propto \frac{2L}{(2d)^2} = \frac{L}{2d^2}\). Taking the ratio gives \(\frac{E_X}{E_Y} = 2\). Therefore, the correct option is C.

1 mark for the correct option C. Incorrect options: A is incorrect due to inverting the relationship; B is incorrect due to neglecting that area is proportional to the square of diameter; D is incorrect due to neglecting the factor of 2 in length.

(a) The vertical component of the initial velocity is given by:
\(u_y = u \sin\theta\)
\(u_y = 12.4 \times \sin(38.0^\circ) = 7.634\text{ m s}^{-1}\)
This is approximately \(7.6\text{ m s}^{-1}\).

(b) Using the equation of motion for vertical displacement \(s = u_y t + \frac{1}{2} a t^2\), taking upwards as positive:
\(s = -1.85\text{ m}\)
\(u_y = 7.634\text{ m s}^{-1}\)
\(a = -9.81\text{ m s}^{-2}\)

\(-1.85 = 7.634 t - 4.905 t^2\)
\(4.905 t^2 - 7.634 t - 1.85 = 0\)

Using the quadratic formula:
\(t = \frac{7.634 \pm \sqrt{(-7.634)^2 - 4(4.905)(-1.85)}}{2(4.905)}
\)t = \frac{7.634 \pm \sqrt{58.28 + 36.30}}{9.81}
\(t = \frac{7.634 \pm \sqrt{94.58}}{9.81}
\)t = \frac{7.634 + 9.725}{9.81} = 1.770\text{ s}\)

(c) The horizontal component of the velocity is:
\(u_x = u \cos\theta = 12.4 \times \cos(38.0^\circ) = 9.771\text{ m s}^{-1}\)

The horizontal distance travelled is:
\(x = u_x \times t = 9.771 \times 1.770 = 17.29\text{ m}\)
Rounding to 3 significant figures gives \(17.3\text{ m}\).

(a) [2 Marks]
- Uses \(u_y = u \sin\theta\) [1M]
- Correct calculation leading to \(7.63\text{ m s}^{-1}\) (must show at least 3 s.f.) [1M]

(b) [3 Marks]
- Selects \(s = u t + \frac{1}{2} a t^2\) with appropriate signs (e.g., \(s = -1.85\text{ m}\)) [1M]
- Formulates quadratic equation or uses alternative multi-step SUVAT method [1M]
- Obtains \(t = 1.77\text{ s}\) (accept \(1.8\text{ s}\)) [1M]

(c) [2 Marks]
- Calculates horizontal velocity component \(u_x = 9.77\text{ m s}^{-1}\) [1M]
- Multiplies horizontal velocity by their time from (b) to find \(17.3\text{ m}\) (accept range \(17.1 - 17.5\text{ m}\) depending on rounding) [1M]

(a) First, calculate the force (tension) exerted by the suspended mass:
\(F = m g = 6.50 \times 9.81 = 63.765\text{ N}\)

Stress (\(\sigma\)) is defined as force per unit cross-sectional area:
\(\sigma = \frac{F}{A} = \frac{63.765}{3.50 \times 10^{-7}} = 1.8219 \times 10^8\text{ Pa}\)
To 3 significant figures, \(\sigma = 1.82 \times 10^8\text{ Pa}\) (or \(\text{N m}^{-2}\)).

(b) The relationship between stress, strain, and Young modulus is:
\(E = \frac{\text{Stress}}{\text{Strain}}

Therefore, \)\text{Strain} = \frac{\text{Stress}}{E} = \frac{1.8219 \times 10^8}{1.20 \times 10^{11}} = 1.518 \times 10^{-3}\)

Since \(\text{Strain} = \frac{\Delta x}{L}\):
\(\Delta x = \text{Strain} \times L = 1.518 \times 10^{-3} \times 2.40 = 3.643 \times 10^{-3}\text{ m}\) (or \(3.64\text{ mm}\)).

(c) The elastic limit is the maximum stress or force that can be applied to a material such that it returns to its original length when the stress or force is removed. Beyond this point, plastic (permanent) deformation occurs.

(a) [3 Marks]
- Uses \(W = m g\) to find force \(F = 63.8\text{ N}\) [1M]
- Uses \(\text{Stress} = \frac{F}{A}\) [1M]
- Correct calculation to give \(1.82 \times 10^8\text{ Pa}\) [1M]

(b) [3 Marks]
- Uses \(\text{Strain} = \frac{\text{Stress}}{E}\) or \(E = \frac{F L}{A \Delta x}\) [1M]
- Rearranges for extension \(\Delta x\) [1M]
- Correct calculation to give \(3.64 \times 10^{-3}\text{ m}\) (or \(3.64\text{ mm}\)) [1M]

(c) [2 Marks]
- Identifies that the material will return to its original length when the load is removed (if below limit) [1M]
- Identifies that beyond this limit, permanent/plastic deformation occurs [1M]

(a) The gain in gravitational potential energy (\(\Delta E_p\)) is:
\(\Delta E_p = m g h\)
\(\Delta E_p = 85.0\text{ kg} \times 9.81\text{ m s}^{-2} \times 14.5\text{ m} = 12090.8\text{ J}\)
Rounding to 3 significant figures, \(\Delta E_p = 1.21 \times 10^4\text{ J}\) (or \(12.1\text{ kJ}\)).

(b) The useful power output (\(P_{\text{out}}\)) of the motor is the rate of gain of gravitational potential energy:
\(P_{\text{out}} = \frac{\Delta E_p}{t} = \frac{12090.8\text{ J}}{15.0\text{ s}} = 806.1\text{ W}\)

The power input (\(P_{\text{in}}\)) is \(1.10\text{ kW} = 1100\text{ W}\).

The efficiency (\(\eta\)) is:
\(\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% = \frac{806.1}{1100} \times 100\% = 73.3\%\)

(c) Reasons why efficiency is less than \(100\%\):
1. Mechanical friction in the moving parts (e.g., bearings or pulleys) of the hoist/motor system, which dissipates energy as thermal energy.
2. Electrical resistance in the coils of the motor (Joule heating), causing energy to be lost as thermal energy.

(a) [2 Marks]
- Uses \(\Delta E_p = m g h\) [1M]
- Correct calculation to give \(1.21 \times 10^4\text{ J}\) [1M]

(b) [3 Marks]
- Uses \(P = \frac{E}{t}\) to find useful power output (\(806\text{ W}\)) [1M]
- Uses \(\text{efficiency} = \frac{\text{useful power output}}{\text{total power input}}\) [1M]
- Correct calculation to give \(73.3\%\) [1M]

(c) [2 Marks]
- Identifies friction in moving parts / pulley / bearings (resulting in thermal energy loss) [1M]
- Identifies electrical resistance in motor coils / Joule heating (resulting in thermal energy loss) [1M]

(a) Upthrust is equal to the weight of the displaced fluid (glycerol):
\(V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.50 \times 10^{-3})^3 = 1.4137 \times 10^{-8}\text{ m}^3\)

\(U = \rho_{\text{glycerol}} \times V \times g = 1260 \times 1.4137 \times 10^{-8} \times 9.81 = 1.747 \times 10^{-4}\text{ N}\)
This is approximately \(1.7 \times 10^{-4}\text{ N}\).

(b) When travelling at terminal velocity, the forces are in equilibrium:
\(W = U + F_{\text{drag}}\)

Calculate the weight of the steel sphere:
\(W = \rho_{\text{steel}} \times V \times g = 7800 \times 1.4137 \times 10^{-8} \times 9.81 = 1.0817 \times 10^{-3}\text{ N}\)

Therefore, the drag force \(F_{\text{drag}}\) is:
\(F_{\text{drag}} = W - U = 1.0817 \times 10^{-3} - 1.747 \times 10^{-4} = 9.070 \times 10^{-4}\text{ N}\)

According to Stokes' law:
\(F_{\text{drag}} = 6 \pi \eta r v\)

Solve for terminal velocity \(v\):
\(v = \frac{F_{\text{drag}}}{6 \pi \eta r} = \frac{9.070 \times 10^{-4}}{6 \pi \times 1.41 \times 1.50 \times 10^{-3}}
\)v = \frac{9.070 \times 10^{-4}}{0.039867} = 0.02275\text{ m s}^{-1}\) (or \(2.28 \times 10^{-2}\text{ m s}^{-1}\)).

(c) Conditions for Stokes' law to apply (any one):
- The flow around the sphere must be laminar (non-turbulent).
- The sphere must be moving slowly.
- The container must be wide compared to the sphere's radius (no wall/boundary effects).

(a) [2 Marks]
- Calculates volume of the sphere (\(V = 1.41 \times 10^{-8}\text{ m}^3\)) [1M]
- Uses \(U = \rho V g\) to get \(1.75 \times 10^{-4}\text{ N}\) (shown with at least 3 s.f.) [1M]

(b) [4 Marks]
- Identifies force balance equation: \(W = U + F_{\text{drag}}\) [1M]
- Calculates weight of the sphere (\(W = 1.08 \times 10^{-3}\text{ N}\)) [1M]
- Calculates required viscous drag force (\(F_{\text{drag}} = 9.07 \times 10^{-4}\text{ N}\)) [1M]
- Rearranges and calculates terminal velocity \(v = 2.28 \times 10^{-2}\text{ m s}^{-1}\) [1M]

(c) [1 Mark]
- States that the flow must be laminar OR the sphere must be small/moving slowly OR no wall effects [1M]

(a) By the principle of conservation of linear momentum:
Total initial momentum = Total final momentum
\(m_A u_A + m_B u_B = (m_A + m_B) v\)

Substitute the given values:
\((12000 \times 3.5) + (18000 \times 0) = (12000 + 18000) v\)
\(42000 = 30000 v\)
\(v = \frac{42000}{30000} = 1.4\text{ m s}^{-1}\)

(b) Total initial kinetic energy (\(E_{k,\text{initial}}\)):
\(E_{k,\text{initial}} = \frac{1}{2} m_A u_A^2 = \frac{1}{2} \times 12000 \times (3.5)^2 = 73500\text{ J}\)

Total final kinetic energy (\(E_{k,\text{final}}\)):
\(E_{k,\text{final}} = \frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2} \times 30000 \times (1.4)^2 = 29400\text{ J}\)

Kinetic energy lost (\(\Delta E_k\)):
\(\Delta E_k = 73500 - 29400 = 44100\text{ J}\) (or \(44.1\text{ kJ}\)).

(c) The collision is inelastic because kinetic energy is not conserved (there is a loss of kinetic energy). The lost kinetic energy is transferred/converted into thermal energy (heating up the coupling mechanism and wheels) and sound energy.

(a) [2 Marks]
- Uses conservation of momentum equation: \(m_A u_A = (m_A + m_B) v\) [1M]
- Correct calculation of \(v = 1.4\text{ m s}^{-1}\) [1M]

(b) [3 Marks]
- Calculates initial kinetic energy as \(73.5\text{ kJ}\) [1M]
- Calculates final kinetic energy as \(29.4\text{ kJ}\) [1M]
- Subtracts values to get \(44.1\text{ kJ}\) (or \(44100\text{ J}\)) [1M]

(c) [2 Marks]
- Explains that 'inelastic' means kinetic energy is not conserved (or decreases) [1M]
- Identifies that the lost KE is converted to thermal energy / sound energy [1M]

(a) Since the plank is uniform, its weight (\(W_p\)) acts at its center of gravity, which is at the midpoint of the plank:
\(W_p = m g = 18.0 \times 9.81 = 176.58\text{ N}\) acting at \(1.80\text{ m}\) from either end.

Take moments about the left-hand end (where Rope P is attached):
- Clockwise moment due to weight of plank = \(176.58\text{ N} \times 1.80\text{ m} = 317.84\text{ N m}\)
- Counter-clockwise moment due to tension in Rope Q (\(T_Q\)) = \(T_Q \times (3.60 - 0.80)\text{ m} = T_Q \times 2.80\text{ m}\)

For rotational equilibrium:
\(2.80 \times T_Q = 317.84\)
\(T_Q = 113.5\text{ N}\)
This is approximately \(110\text{ N}\).

(b) Rope P goes slack when \(T_P = 0\). At this point, the plank is on the verge of rotating about the pivot point at Rope Q (\(x = 2.80\text{ m}\) from the left end).

Take moments about Rope Q:
- The weight of the plank acts at \(1.80\text{ m}\) from the left, which is \(2.80 - 1.80 = 1.00\text{ m}\) to the left of Q. This creates a counter-clockwise moment:
\(M_{\text{plank}} = 176.58\text{ N} \times 1.00\text{ m} = 176.58\text{ N m}\)

- The builder of weight \(W_b = 75.0 \times 9.81 = 735.75\text{ N}\) must be at a distance \(d\) to the right of Q to create an equal clockwise moment:
\(M_{\text{builder}} = 735.75 \times d\)

Equating the moments:
\(735.75 \times d = 176.58\)
\(d = \frac{176.58}{735.75} = 0.240\text{ m}\) to the right of Rope Q.

Therefore, the total distance from the left-hand end is:
\(x = 2.80\text{ m} + 0.240\text{ m} = 3.04\text{ m}\).

(c) The principle of moments states that for a system in rotational equilibrium, the sum of the clockwise moments about any pivot point is equal to the sum of the anticlockwise moments about that same pivot point.

(a) [3 Marks]
- Identifies weight of plank acts at \(1.80\text{ m}\) from left and calculates \(W_p = 176.6\text{ N}\) [1M]
- Takes moments about Rope P: \(T_Q \times 2.8 = W_p \times 1.8\) [1M]
- Calculates \(T_Q = 113.5\text{ N}\) (must show unrounded value to show it is about \(110\text{ N}\)) [1M]

(b) [3 Marks]
- Identifies that when Rope P goes slack, pivot is at Rope Q (or sets up moment equation about Q with \(T_P = 0\)) [1M]
- Formulates moment equation: \(735.75 \times d = 176.58 \times 1.0\) [1M]
- Solves for total distance from left end as \(3.04\text{ m}\) (accept range \(3.0 - 3.1\text{ m}\)) [1M]

(c) [1 Mark]
- States that clockwise moments equal anticlockwise moments about a point for a body in equilibrium [1M]

(a) Using the equation of motion:
\(v^2 = u^2 + 2 a s\)

Substitute the values:
\((8.0)^2 = (24.0)^2 + 2 \times a \times 64.0\)
\(64.0 = 576.0 + 128.0 a\)
\(128.0 a = 64.0 - 576.0 = -512.0\)
\(a = \frac{-512.0}{128.0} = -4.0\text{ m s}^{-2}\)

Thus, the deceleration of the car is \(4.0\text{ m s}^{-2}\).

(b) Using the equation of motion:
\(v = u + a t\)
\(8.0 = 24.0 + (-4.0) t\)
\(4.0 t = 16.0\)
\(t = 4.0\text{ s}\)

(c) First, calculate the loss in kinetic energy (\(\Delta E_k\)):
\(\Delta E_k = \frac{1}{2} m u^2 - \frac{1}{2} m v^2 = \frac{1}{2} m (u^2 - v^2)\)
\(\Delta E_k = \frac{1}{2} \times 1100 \times (24.0^2 - 8.0^2) = 550 \times (576 - 64) = 550 \times 512 = 281600\text{ J}\)

The average rate of energy dissipation (power \(P\)) is:
\(P = \frac{\Delta E_k}{t} = \frac{281600\text{ J}}{4.0\text{ s}} = 70400\text{ W} = 70.4\text{ kW}\)

(Alternative method:
Average braking force \(F = m a = 1100 \times 4.0 = 4400\text{ N}\).
Average power \(P = F \times v_{\text{average}} = 4400 \times \frac{24.0 + 8.0}{2} = 4400 \times 16.0 = 70400\text{ W} = 70.4\text{ kW}\).)

(a) [2 Marks]
- Selects \(v^2 = u^2 + 2 a s\) [1M]
- Substitutes correctly and obtains \(a = -4.0\text{ m s}^{-2}\) (showing clearly the calculation steps) [1M]

(b) [2 Marks]
- Selects \(v = u + a t\) or \(s = \frac{1}{2}(u + v)t\) [1M]
- Correct calculation to give \(t = 4.0\text{ s}\) [1M]

(c) [3 Marks]
- Calculates change in kinetic energy (\(281600\text{ J}\)) OR calculates braking force (\(4400\text{ N}\)) [1M]
- Divides change in energy by time (\(4.0\text{ s}\)) OR multiplies force by average speed (\(16\text{ m s}^{-1}\)) [1M]
- Correct calculation of rate to give \(70.4\text{ kW}\) (or \(70400\text{ W}\)) [1M]

(a) The student should set up the apparatus as follows:
- Clamp one end of the copper wire securely to the bench.
- Pass the other end of the wire over a pulley clamped at the edge of the bench, and attach a mass hanger for adding loads.
- Measure the original length \(L\) from the clamp to a reference marker taped on the wire using a tape measure.
- Place a metre rule on the bench close to the reference marker to measure its displacement (extension \(\Delta x\)) as masses are added.
- Use a micrometer screw gauge to measure the diameter of the wire at several points and orientations to calculate an average cross-sectional area \(A = \pi \frac{d^2}{4}\).

(b) The formula for extension is \(\Delta x = \frac{F L}{A E}\).
- A long wire (large \(L\)) and thin wire (small \(A\)) will produce a much larger extension \(\Delta x\) for any given load \(F\).
- Larger extensions can be measured with a lower percentage uncertainty, improving the accuracy of the calculated Young modulus.

(c) Possible systematic error:
- The support or clamp may slip as heavier loads are added, which adds an extra 'apparent' extension that is not due to the wire stretching. This makes the measured extension larger than the true extension, leading to a calculated Young modulus that is too low (\(E \propto 1/\Delta x\)).
- To minimize this: Ensure the wire is clamped tightly between two wooden blocks, or use a Searle's apparatus which has a control wire attached to the same support to automatically compensate for support yielding.

(a) [3 Marks]
- Describes securing one end of the wire and passing it over a pulley with loads [1M]
- Explains measuring original length with a tape measure and extension using a reference marker and rule/scale [1M]
- Mentions measuring diameter using a micrometer at multiple places/orientations to find average area [1M]

(b) [2 Marks]
- States that long and thin wires increase the extension for a given force [1M]
- Explains that larger extensions reduce the percentage uncertainty in the extension measurement [1M]

(c) [3 Marks]
- Identifies a valid systematic error (e.g., clamp slipping / wire not initially straight) [1M]
- Explains how this error leads to an overestimated extension (which underestimates the Young modulus) [1M]
- Suggests a suitable remedy (e.g., use of wooden clamping blocks / applying a pre-tension load before taking zero readings) [1M]

(a) To find the time of flight \( t \), we use the vertical motion equation: \( s_y = u_y t + \frac{1}{2} a_y t^2 \). Since the motorcycle launches horizontally, the initial vertical velocity \( u_y = 0 \). Therefore, \( 5.20\text{ m} = \frac{1}{2} \times 9.81\text{ m s}^{-2} \times t^2 \), which gives \( t^2 = \frac{2 \times 5.20}{9.81} = 1.0599\text{ s}^2 \), so \( t = 1.030\text{ s} \). For the horizontal motion, with zero horizontal acceleration: \( v_x = \frac{s_x}{t} = \frac{14.0\text{ m}}{1.030\text{ s}} = 13.59\text{ m s}^{-1} \), which is approximately \( 14\text{ m s}^{-1} \).

(b) Air resistance acts in the direction opposite to the motorcycle's velocity. This introduces a horizontal retarding force (drag), causing a horizontal deceleration. As a result, the horizontal component of velocity decreases throughout the flight, meaning the motorcycle covers less horizontal distance in the same time of flight. Thus, the horizontal distance traveled will be less than \( 14.0\text{ m} \).

(c) The horizontal velocity remains constant at: \( v_x = 13.6\text{ m s}^{-1} \). The vertical velocity just before impact is given by: \( v_y = u_y + a_y t = 0 + (9.81\text{ m s}^{-2} \times 1.030\text{ s}) = 10.10\text{ m s}^{-1} \). The magnitude of the velocity \( v \) is: \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{13.6^2 + 10.10^2} = \sqrt{184.96 + 102.01} = 16.94\text{ m s}^{-1} \approx 16.9\text{ m s}^{-1} \). The angle \( \theta \) below the horizontal is: \( \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{10.10}{13.6}\right) = 36.6^\circ \).

Part (a) [3 Marks]:
- Use of \( s = \frac{1}{2} g t^2 \) with vertical values to find time of flight (1)
- Correct calculation of time of flight \( t = 1.03\text{ s} \) (1)
- Use of \( v = d/t \) to show value of \( 13.6\text{ m s}^{-1} \) (1)

Part (b) [2 Marks]:
- Identifies that air resistance opposes horizontal motion or causes deceleration (1)
- Explains that horizontal velocity decreases, resulting in a smaller horizontal distance (1)

Part (c) [3 Marks]:
- Correct calculation of vertical velocity component \( v_y = 10.1\text{ m s}^{-1} \) (1)
- Correct calculation of magnitude of velocity \( v = 16.9\text{ m s}^{-1} \) (allow 17.0 if using rounded values) (1)
- Correct angle calculated as \( 36.6^\circ \) (or \( 37^\circ \)) below the horizontal (1)

(a) To find the time of flight \( t \), we use the vertical motion equation: \( s_y = u_y t + \frac{1}{2} a_y t^2 \). Since the motorcycle launches horizontally, the initial vertical velocity \( u_y = 0 \). Therefore, \( 5.20\text{ m} = \frac{1}{2} \times 9.81\text{ m s}^{-2} \times t^2 \), which gives \( t^2 = \frac{2 \times 5.20}{9.81} = 1.0599\text{ s}^2 \), so \( t = 1.030\text{ s} \). For the horizontal motion, with zero horizontal acceleration: \( v_x = \frac{s_x}{t} = \frac{14.0\text{ m}}{1.030\text{ s}} = 13.59\text{ m s}^{-1} \), which is approximately \( 14\text{ m s}^{-1} \).

(b) Air resistance acts in the direction opposite to the motorcycle's velocity. This introduces a horizontal retarding force (drag), causing a horizontal deceleration. As a result, the horizontal component of velocity decreases throughout the flight, meaning the motorcycle covers less horizontal distance in the same time of flight. Thus, the horizontal distance traveled will be less than \( 14.0\text{ m} \).

(c) The horizontal velocity remains constant at: \( v_x = 13.6\text{ m s}^{-1} \). The vertical velocity just before impact is given by: \( v_y = u_y + a_y t = 0 + (9.81\text{ m s}^{-2} \times 1.030\text{ s}) = 10.10\text{ m s}^{-1} \). The magnitude of the velocity \( v \) is: \( v = \sqrt{v_x^2 + v_y^2} = \sqrt{13.6^2 + 10.10^2} = \sqrt{184.96 + 102.01} = 16.94\text{ m s}^{-1} \approx 16.9\text{ m s}^{-1} \). The angle \( \theta \) below the horizontal is: \( \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{10.10}{13.6}\right) = 36.6^\circ \).

Part (a) [3 Marks]:
- Use of \( s = \frac{1}{2} g t^2 \) with vertical values to find time of flight (1)
- Correct calculation of time of flight \( t = 1.03\text{ s} \) (1)
- Use of \( v = d/t \) to show value of \( 13.6\text{ m s}^{-1} \) (1)

Part (b) [2 Marks]:
- Identifies that air resistance opposes horizontal motion or causes deceleration (1)
- Explains that horizontal velocity decreases, resulting in a smaller horizontal distance (1)

Part (c) [3 Marks]:
- Correct calculation of vertical velocity component \( v_y = 10.1\text{ m s}^{-1} \) (1)
- Correct calculation of magnitude of velocity \( v = 16.9\text{ m s}^{-1} \) (allow 17.0 if using rounded values) (1)
- Correct angle calculated as \( 36.6^\circ \) (or \( 37^\circ \)) below the horizontal (1)

Since the wires are connected in series, they carry the same current \(I\).

The formula for current is \(I = n A v e\), where \(n\) is the number density of conduction electrons, \(A\) is the cross-sectional area of the wire, \(v\) is the drift velocity, and \(e\) is the elementary charge.

Since both wires are made of copper, \(n\) is identical for both. Therefore, \(v \propto \frac{1}{A}\).

Since \(A = \frac{\pi d^2}{4}\), the area is proportional to the square of the diameter: \(A \propto d^2\).

Thus, the drift velocity is inversely proportional to the square of the diameter: \(v \propto \frac{1}{d^2}\).

The ratio of the drift velocities is:
\frac{v_{\text{X}}}{v_{\text{Y}}} = \frac{d_{\text{Y}}^2}{d_{\text{X}}^2} = \frac{(2d)^2}{d^2} = 4

1 mark for the correct answer.

- Correct option: A
- Rejects other options based on incorrect application of the relationship between drift velocity, current, and cross-sectional area.

When unpolarized light of intensity \(I_0\) passes through the first polarizer, it becomes plane-polarized, and its intensity is reduced by half:
\(I_1 = \frac{1}{2} I_0\)

When this light passes through the second polarizer, Malus's law applies:
\(I_2 = I_1 \cos^2\theta\)

Substituting \(\theta = 60^\circ\) and \(I_1 = \frac{1}{2} I_0\):
\(I_2 = \left(\frac{1}{2} I_0\right) \cos^2(60^\circ)\)

Since \(\cos(60^\circ) = 0.5\):
\(I_2 = \frac{1}{2} I_0 \times (0.5)^2 = \frac{1}{2} I_0 \times \frac{1}{4} = \frac{1}{8} I_0\)

1 mark for the correct answer.

- Correct option: A
- Rejects B, C, D due to failure to apply the unpolarized light halving rule and/or incorrect calculation of Malus's Law.

1. For a negative temperature coefficient (NTC) thermistor, its resistance increases as the temperature decreases.
2. In a series potential divider, the output voltage across the thermistor is given by:
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{thermistor}}}{R + R_{\text{thermistor}}}\)

As the thermistor's resistance increases relative to the fixed resistor, it takes a greater share of the input voltage. Thus, \(V_{\text{out}}\) increases.

1 mark for the correct answer.

- Correct option: A
- Rejects other options because NTC behavior causes resistance to increase, which increases its share of the potential difference.

For a string fixed at both ends, the boundaries require nodes at each end. The wavelengths of the harmonics are given by:
\(L = n \frac{\lambda_n}{2}\)

For the third harmonic (\(n = 3\)):
\(L = \frac{3\lambda_3}{2} \implies \lambda_3 = \frac{2L}{3}\)

Using the wave equation \(v = f \lambda\):
\(f_3 = \frac{v}{\lambda_3} = \frac{v}{\frac{2L}{3}} = \frac{3v}{2L}\)

1 mark for the correct answer.

- Correct option: C
- Rejects other options due to incorrect harmonic wavelength relationships or wave equation application.

The volume of a cylindrical wire is \(V = A \times l\).
Since the volume remains constant when the wire is stretched to a new length of \(l' = 2l\), the new area \(A'\) must satisfy:
\(A \times l = A' \times (2l) \implies A' = \frac{A}{2}\)

The resistance of a wire is given by:
\(R = \rho \frac{l}{A}\)

The new resistance \(R'\) is:
\(R' = \rho \frac{l'}{A'} = \rho \frac{2l}{\frac{A}{2}} = 4 \rho \frac{l}{A} = 4R\)

1 mark for the correct answer.

- Correct option: C
- Rejects A, B, D because they fail to account for the simultaneous halving of the cross-sectional area when the length is doubled.

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

If the frequency is doubled to \(2f\):
\(E_{\text{k,max}}' = h(2f) - \Phi = 2hf - \Phi\)

We can substitute \(hf = E_{\text{k,max}} + \Phi\) into the equation:
\(E_{\text{k,max}}' = 2(E_{\text{k,max}} + \Phi) - \Phi = 2E_{\text{k,max}} + \Phi\)

Since the work function \(\Phi\) of the metal is a positive constant (\(\Phi > 0\)), the new maximum kinetic energy must be greater than \(2E_{\text{k,max}}\).

1 mark for the correct answer.

- Correct option: B
- Rejects other options because they do not correctly mathematically relate the change in photon energy to the constant work function.

The voltmeter connected across the terminals of the cell measures the terminal potential difference \(V\).
The terminal potential difference is given by:
\(V = \varepsilon - I r\)
where \(I\) is the current in the circuit, given by:
\(I = \frac{\varepsilon}{R + r}\)

When the variable resistance \(R\) is decreased, the total circuit resistance \((R + r)\) decreases, causing the current \(I\) to increase.
Since \(I\) increases, the lost volts \(I r\) across the internal resistance increase.
Consequently, the terminal potential difference \(V = \varepsilon - I r\) decreases, which means the voltmeter reading also decreases.

1 mark for the correct answer.

- Correct option: A
- Rejects options suggesting that terminal potential difference or voltmeter reading increase or differ from each other.

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

For the first-order maximum (\(n = 1\)):
\(d \sin(30^\circ) = 1 \cdot \lambda\)

Since \(\sin(30^\circ) = 0.5\):
\(0.5 d = \lambda \implies d = 2\lambda\)

For the second-order maximum (\(n = 2\)):
\(d \sin\theta_2 = 2 \lambda\)

Substitute \(d = 2\lambda\) into this equation:
\((2\lambda) \sin\theta_2 = 2\lambda\)
\(\sin\theta_2 = 1\)

Therefore, the angle is:
\\theta_2 = \sin^{-1}(1) = 90^\circ\)

1 mark for the correct answer.

- Correct option: D
- Rejects other options based on incorrect algebraic substitution into the diffraction grating formula.

According to Einstein's photoelectric equation:
\(E_{\text{k}} = hf - \phi\)
where \(\phi\) is the work function of the metal.

When the frequency is doubled to \(2f\), the new maximum kinetic energy \(E_{\text{k}}'\) is:
\(E_{\text{k}}' = h(2f) - \phi = 2hf - \phi\)

This can be rewritten as:
\(E_{\text{k}}' = hf + (hf - \phi)\)

Substituting \(E_{\text{k}} = hf - \phi\) into this equation gives:
\(E_{\text{k}}' = E_{\text{k}} + hf\)

Therefore, B is the correct option.

B - \(E_{\text{k}} + hf\)

[1 mark] Correctly identifies the option B.

Since the wires are connected in series, the same current \(I\) flows through both wires.

The relationship between current and mean drift velocity \(v\) is given by:
\(I = nAvq\)
where \(n\) is the number density of conduction electrons, \(A\) is the cross-sectional area, and \(q\) is the elementary charge.

Since both wires are made of the same metal, \(n\) is constant. Thus:
\(A_{\text{X}} v_{\text{X}} = A_{\text{Y}} v_{\text{Y}}\)

This gives the ratio of drift velocities as:
\(\frac{v_{\text{X}}}{v_{\text{Y}}} = \frac{A_{\text{Y}}}{A_{\text{X}}}\)

The cross-sectional area of a wire of diameter \(d\) is proportional to \(d^2\):
\(\frac{A_{\text{Y}}}{A_{\text{X}}} = \frac{(2d)^2}{d^2} = 4\)

So, the ratio is \(4:1\).

A - \(4 : 1\)

[1 mark] Correctly identifies the option A.

(a) Coherent sources have a constant phase difference and the same frequency (or wavelength).

(b) The distance to the 4th fringe is \( y_4 = 1.88 \text{ cm} = 0.0188 \text{ m} \). The fringe spacing is \( x = \frac{y_4}{4} = \frac{0.0188}{4} = 4.70 \times 10^{-3} \text{ m} \). Using the double-slit formula \( \lambda = \frac{d x}{D} \):\
\( \lambda = \frac{0.25 \times 10^{-3} \text{ m} \times 4.70 \times 10^{-3} \text{ m}}{1.80 \text{ m}} = 6.53 \times 10^{-7} \text{ m} \) (or \( 653 \text{ nm} \)).

(c) If the slit separation \( d \) is halved, the fringe width \( x \) will double because \( x \propto \frac{1}{d} \). The fringes will be more widely spaced.

(a) [1 mark] Constant phase difference. [1 mark] Same frequency/wavelength.
(b) [1 mark] Identification of fringe spacing \( x = 4.70 \times 10^{-3} \text{ m} \). [1 mark] Correct substitution into \( \lambda = \frac{d x}{D} \). [1 mark] Correct calculation of wavelength with units (\( 6.53 \times 10^{-7} \text{ m} \)).
(c) [1 mark] State that the fringe spacing increases / doubles. [1 mark] State the algebraic relationship \( x \propto \frac{1}{d} \) or equivalent. [0.78 mark] Clear explanation of the inverse relationship.

(a) Cross-sectional area \( A = \pi r^2 = \pi \left(0.20 \times 10^{-3}\right)^2 = 1.257 \times 10^{-7} \text{ m}^2 \). Using \( R = \frac{\rho L}{A} = \frac{1.10 \times 10^{-6} \times 1.50}{1.257 \times 10^{-7}} = 13.13 \\ \Omega \) (approx. \( 13.1 \\ \Omega \)).

(b) As temperature increases, the lattice ions in the metal vibrate with greater amplitude. This increases the frequency of collisions between the conducting free electrons and the vibrating ions, hindering the flow of charge (current) and thus increasing the resistance.

(c) Use a micrometer screw gauge (or digital calliper). Take measurements at several points along the wire and at different orientations, then calculate an average diameter to account for any non-uniformity.

(a) [1 mark] Calculation of area \( A = 1.26 \times 10^{-7} \text{ m}^2 \). [1 mark] Use of formula \( R = \frac{\rho L}{A} \). [1 mark] Answer \( 13.1 \\ \Omega \) (accept \( 13.0 \\ \Omega \) to \( 13.1 \\ \Omega \)).
(b) [1 mark] Lattice ions vibrate with greater amplitude/kinetic energy. [1 mark] More frequent collisions between conduction electrons and ions. [1 mark] This reduces current / increases resistance.
(c) [0.78 mark] Identify micrometer screw gauge / digital calipers. [1 mark] Describe taking measurements at multiple positions and orientations and averaging.

(a) Using Snell's law: \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \) with \( n_1 = 1.52 \), \( \theta_1 = 35.0^\circ \), \( n_2 = 1.00 \).\
\( 1.52 \times \sin(35.0^\circ) = 1.00 \times \sin \theta_2 \) which yields \( \sin \theta_2 = 0.8718 \) and \( \theta_2 = 60.7^\circ \).

(b) Critical angle \( \theta_c \) in air: \( \sin \theta_c = \frac{1}{n} = \frac{1}{1.52} = 0.6579 \) which gives \( \theta_c = 41.1^\circ \).

(c) When submerged in water (\( n_{\text{water}} = 1.33 \)), the critical angle is given by:\
\( \sin \theta_{c,\text{water}} = \frac{n_{\text{water}}}{n_{\text{glass}}} = \frac{1.33}{1.52} = 0.8750 \) which gives \( \theta_{c,\text{water}} = 61.0^\circ \). This is larger than the critical angle in air because the boundary has a smaller relative refractive index, requiring a larger angle of incidence for total internal reflection to occur.

(a) [1 mark] State or use \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \). [1 mark] Correct calculation of angle of refraction: \( 60.7^\circ \) (allow \( 60.5^\circ \) to \( 61^\circ \)).
(b) [1 mark] State or use \( \sin \theta_c = \frac{1}{n} \). [1 mark] Correct critical angle \( 41.1^\circ \).
(c) [1 mark] Use of \( \sin \theta_c = \frac{n_2}{n_1} \) with correct values (\( 1.33 \) and \( 1.52 \)). [1 mark] Correct calculation of critical angle in water: \( 61.0^\circ \). [1.78 marks] Explains that the critical angle is larger because the difference in refractive index between the two media is smaller.

(a) Total resistance \( R_{\text{total}} = 3.3 \text{ k}\Omega + 1.8 \text{ k}\Omega = 5.1 \text{ k}\Omega \). Using the potential divider formula:\
\( V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{fixed}}}{R_{\text{total}}} = 9.0 \times \frac{3.3}{5.1} = 5.82 \text{ V} \) (or \( 5.8 \text{ V} \)).

(b) When the light level decreases, the LDR's resistance increases. This increases the total resistance of the circuit, which reduces the current. Since \( V_{\text{fixed}} = I R_{\text{fixed}} \) and \( R_{\text{fixed}} \) is constant, a smaller current means the potential difference across the fixed resistor decreases.

(c) Under bright conditions, total resistance \( R_{\text{total}} = 3300 \\ \Omega + 220 \\ \Omega = 3520 \\ \Omega \). The current is \( I = \frac{V}{R_{\text{total}}} = \frac{9.0 \text{ V}}{3520 \\ \Omega} = 2.557 \times 10^{-3} \text{ A} \). The power dissipated by the fixed resistor is:\
\( P = I^2 R_{\text{fixed}} = \left(2.557 \times 10^{-3}\right)^2 \times 3300 = 0.0216 \text{ W} \) (or \( 21.6 \text{ mW} \)).

(a) [1 mark] Correct total resistance \( 5.1 \text{ k}\Omega \). [1 mark] Correct application of potential divider formula. [1 mark] Correct potential difference \( 5.8 \text{ V} \).
(b) [1 mark] Identification that LDR resistance increases when dark. [1 mark] Identification that total circuit current decreases. [1 mark] Conclusion that voltage across the fixed resistor decreases.
(c) [0.78 mark] Calculation of current under bright conditions. [1 mark] Correct power calculation (\( 0.0216 \text{ W} \) or \( 21.6 \text{ mW} \)).

(a) Unpolarised light contains electric field oscillations in all planes perpendicular to the direction of wave propagation. A polarising filter only transmits components of the oscillations parallel to its transmission axis and absorbs the perpendicular components. The emerging light oscillates in a single plane.

(b) After passing through the first filter, the light is polarised and its intensity is halved: \( I_1 = 0.5 I_0 \). Applying Malus's law for the second filter: \( I_2 = I_1 \cos^2 \theta = (0.5 I_0) \cos^2(40.0^\circ) \). Since \( \cos(40.0^\circ) \approx 0.7660 \), we get \( \cos^2(40.0^\circ) \approx 0.5868 \). Thus, \( I_2 = 0.5 I_0 \times 0.5868 = 0.293 I_0 \), which is approximately \( 0.29 I_0 \).

(c) As the angle increases to \( 90.0^\circ \), the transmission axes of the two filters become perpendicular. Since \( \cos(90.0^\circ) = 0 \), the transmitted intensity falls to zero.

(a) [1 mark] Explains unpolarised light oscillations. [1 mark] Explains filter action (only transmitting parallel component).
(b) [1 mark] Identifies intensity after first filter is \( 0.5 I_0 \). [1 mark] Recalls and uses Malus's Law \( I = I_1 \cos^2 \theta \). [1 mark] Correct mathematical derivation of \( 0.29 I_0 \).
(c) [1 mark] States that the intensity decreases to zero. [0.78 mark] Explains that the transmission axes are crossed/perpendicular.

(a) The circuit diagram must show: a cell connected in series with an ammeter and a variable resistor. A voltmeter must be connected in parallel across the cell (or across the combination of the cell and its internal resistance).

(b) The terminal potential difference is given by \( V = \mathcal{E} - Ir \), which matches the straight-line equation form \( y = mx + c \) rewritten as \( V = -rI + \mathcal{E} \). Comparing with \( V = -0.65 I + 1.45 \):\
- The intercept on the vertical axis represents the EMF, so \( \mathcal{E} = 1.45 \text{ V} \).\
- The magnitude of the gradient represents the internal resistance, so \( r = 0.65 \\ \Omega \).

(c) EMF is the total work done per unit charge inside the cell. When a current flows, charge carriers move through the cell's internal resistance. Work must be done against this internal resistance, which dissipates electrical energy as thermal energy within the cell. This voltage drop (called 'lost volts', \( I r \)) results in a terminal potential difference that is less than the EMF.

(a) [1 mark] Cell and variable resistor in series. [1 mark] Ammeter in series. [1 mark] Voltmeter across the cell.
(b) [1 mark] Correctly identifies EMF as \( 1.45 \text{ V} \). [1 mark] Correctly identifies internal resistance as \( 0.65 \\ \Omega \).
(c) [1 mark] Defines EMF in terms of energy supplied per unit charge. [1 mark] Explains that work is done against internal resistance. [0.78 mark] Identifies that energy is dissipated as thermal energy ('lost volts').

(a) According to wave theory, the energy carried by a wave is proportional to its intensity and is transferred continuously. If low-frequency light is incident, energy should eventually accumulate over time to allow emission. However, no emission occurs below the threshold frequency, showing that light acts as discrete packets (photons) where energy is dependent on frequency (\( E = hf \)).

(b) First, convert the work function to joules: \( \Phi = 4.30 \text{ eV} \times 1.60 \times 10^{-19} \text{ J eV}^{-1} = 6.88 \times 10^{-19} \text{ J} \).\
Then, calculate the photon energy: \( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \text{ J s} \times 3.00 \times 10^8 \text{ m s}^{-1}}{2.40 \times 10^{-7} \text{ m}} = 8.288 \times 10^{-19} \text{ J} \).\
Using Einstein's photoelectric equation: \( E_k = E - \Phi = 8.288 \times 10^{-19} \text{ J} - 6.88 \times 10^{-19} \text{ J} = 1.408 \times 10^{-19} \text{ J} \) (approx. \( 1.41 \times 10^{-19} \text{ J} \)).

(c) The maximum kinetic energy of the photoelectrons would remain unchanged. Doubling the intensity only increases the number of photons incident per second, which increases the rate of emission of photoelectrons, but does not alter the energy of individual photons.

(a) [1 mark] Explains wave theory suggestion of continuous energy accumulation. [1 mark] Contrast with photon theory (discrete packet energy transfer). [1 mark] Relates photon energy directly to frequency (\( E = hf \)) explaining threshold requirement.
(b) [1 mark] Correct conversion of work function to Joules. [1 mark] Correct calculation of photon energy. [1 mark] Correct calculation of maximum kinetic energy (\( 1.41 \times 10^{-19} \text{ J} \)).
(c) [1 mark] States that maximum kinetic energy is unchanged. [0.78 mark] Explains that intensity only increases photon/electron count per unit time.

(a) A wave travels down the wire and is reflected at the fixed end. The outgoing wave and the reflected wave, which have the same frequency, amplitude, and speed, travel in opposite directions and superpose. When they meet in phase, constructive interference occurs (nodes are formed where they are completely out of phase, and antinodes where they are in phase).

(b) First, find the mass per unit length \( \mu = \frac{\text{mass}}{\text{length}} = \frac{1.20 \times 10^{-3} \text{ kg}}{0.75 \text{ m}} = 1.60 \times 10^{-3} \text{ kg m}^{-1} \).\
Using the wave speed formula: \( v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{60.0 \text{ N}}{1.60 \times 10^{-3} \text{ kg m}^{-1}}} = 193.6 \text{ m s}^{-1} \) (approx. \( 194 \text{ m s}^{-1} \)).

(c) For the first harmonic, the wavelength of the standing wave is \( \lambda = 2L = 2 \times 0.75 \text{ m} = 1.50 \text{ m} \).\
Using the wave equation \( v = f \lambda \):\
\( f = \frac{v}{\lambda} = \frac{193.6 \text{ m s}^{-1}}{1.50 \text{ m}} = 129.1 \text{ Hz} \) (approx. \( 129 \text{ Hz} \)).

(a) [1 mark] Identifies reflection of waves from boundaries. [1 mark] Mentions superposition of two waves traveling in opposite directions with identical frequency and amplitude. [1 mark] Explains nodes (phase opposition/destructive) and antinodes (in phase/constructive).
(b) [1 mark] Calculation of mass per unit length \( \mu = 1.60 \times 10^{-3} \text{ kg m}^{-1} \). [1 mark] Correct calculation of wave speed (\( 194 \text{ m s}^{-1} \)).
(c) [1 mark] Correct wavelength for the first harmonic (\( 1.50 \text{ m} \)). [1.78 marks] Correct calculation of frequency (\( 129 \text{ Hz} \)).

**(a)**

First, find the distance \( x \) from the central maximum to one of the second-order maxima:

\[ x = \frac{1.28 \text{ m}}{2} = 0.64 \text{ m} \]

Now, use trigonometry to find the angle of diffraction \( \theta \):

\[ \tan\theta = \frac{x}{D} = \frac{0.64 \text{ m}}{1.45 \text{ m}} = 0.44138 \]

\[ \theta = \arctan(0.44138) = 23.81^{\circ} \approx 23.8^{\circ} \text{ (or } 0.416 \text{ rad)} \]


**(b)**

Use the diffraction grating equation:

\[ d \sin\theta = n\lambda \]

Rearrange to solve for the grating spacing \( d \) with \( n = 2 \):

\[ d = \frac{n\lambda}{\sin\theta} = \frac{2 \times (6.33 \times 10^{-7} \text{ m})}{\sin(23.81^{\circ})} \]

\[ d = \frac{1.266 \times 10^{-6} \text{ m}}{0.4037} = 3.136 \times 10^{-6} \text{ m} \]

Calculate the number of lines per metre \( N \):

\[ N = \frac{1}{d} = \frac{1}{3.136 \times 10^{-6} \text{ m}} = 3.189 \times 10^{5} \text{ lines m}^{-1} \]

Convert this to lines per millimetre:

\[ N_{\text{mm}} = \frac{3.189 \times 10^{5}}{1000} \approx 319 \text{ lines mm}^{-1} \]

*(Accept values between 318 and 320 due to rounding variations)*


**(c)**

- The wavelength \( \lambda \) of green light is smaller than that of red light.
- From \( d \sin\theta = n\lambda \), since the grating spacing \( d \) is constant, a smaller wavelength \( \lambda \) results in a smaller value for \( \sin\theta \) (and thus a smaller diffraction angle \( \theta \)).
- Therefore, the bright fringes will be closer together, meaning the distance between adjacent bright fringes on the screen decreases.

**(a)**
* **MP1:** Use of \( x = 0.64 \text{ m} \) in a valid trigonometric relationship to find \( \theta \) (e.g., \( \tan\theta = 0.64 / 1.45 \)). (1)
* **MP2:** Correct value of \( \theta = 23.8^{\circ} \) or \( 24^{\circ} \) (or \( 0.416 \text{ rad} \)). (1)

**(b)**
* **MP1:** Correct use of the grating formula \( d \sin\theta = n\lambda \) with \( n = 2 \). (1)
* **MP2:** Calculation of grating spacing \( d = 3.14 \times 10^{-6} \text{ m} \) (allow error carried forward from (a)). (1)
* **MP3:** Correct division of 1 by \( d \) and conversion to yield \( 319 \text{ lines mm}^{-1} \) (accept range 318 to 320). (1)

**(c)**
* **MP1:** State that green light has a shorter wavelength than red light. (1)
* **MP2:** Reference to \( d \sin\theta = n\lambda \) to show that a smaller wavelength results in a smaller angle of diffraction \( \theta \) (as \( d \) is constant). (1)
* **MP3:** Conclude that the distance between adjacent bright fringes on the screen decreases. (1)

(a) Releasing the sphere above the light gate ensures that the sphere is already moving at a steady, measurable speed when it passes the first light gate, allowing the timer to start cleanly without any delay or contact effects from the release mechanism.

(b) Starting with \(d = ut + \frac{1}{2}gt^2\):
Divide both sides by \(t\):
\(\frac{d}{t} = u + \frac{1}{2}gt\)
Multiply both sides by 2:
\(\frac{2d}{t} = gt + 2u\)
Comparing this to \(y = mx + c\):
- \(y = \frac{2d}{t}\)
- \(x = t\)
- \(m = g\) (gradient)
- \(c = 2u\) (y-intercept)
Thus, a graph of \(\frac{2d}{t}\) against \(t\) yields a straight line with a gradient equal to \(g\).

(c) Calculation:
\(\frac{2d}{t} = \frac{2 \times 0.800}{0.324} = 4.938\text{ m s}^{-1} \approx 4.94\text{ m s}^{-1}\)

Percentage uncertainty calculation:
\(\% \text{ uncertainty in } d = \frac{0.002}{0.800} \times 100 = 0.25\%\)
\(\% \text{ uncertainty in } t = \frac{0.005}{0.324} \times 100 \approx 1.54\%\)
Total percentage uncertainty in \(\frac{2d}{t}\) = \(\% \Delta d + \% \Delta t = 0.25\% + 1.54\% = 1.79\% \approx 1.8\%\).

(d) Use a set square against the metre rule and the light gates, and ensure the line of sight is perpendicular to the scale of the rule at the point of measurement.

(e) A steel sphere has a much higher density than a tennis ball, so air resistance has a negligible effect on its motion compared to its weight. This ensures the acceleration is very close to \(g\).

(f) Measure the card length using a digital vernier caliper.

- (a) To ensure the sphere has a non-zero initial velocity / avoids release-mechanism lag/errors (1 mark)
- (b) Rearranging correctly to \(\frac{d}{t} = u + \frac{1}{2}gt\) (1 mark); multiplying by 2 and identifying that gradient \(m = g\) (1 mark)
- (c) Value of 4.94 m/s (or 4.9) (1 mark); % uncertainty in d = 0.25% AND % uncertainty in t = 1.54% (1 mark); Sum of uncertainties = 1.8% (or 1.79%) (1 mark)
- (d) Use of set square (1 mark); viewing horizontally / perpendicularly to the scale (1 mark)
- (e) Steel sphere has less air resistance relative to weight / is more dense (1 mark); therefore acceleration is closer to free fall acceleration / g is more accurate (1 mark)
- (f) Use a vernier caliper (or micrometer) (1 mark)

(a) The circuit must include:
- A power supply connected to a variable length of test wire (using a crocodile clip or sliding contact).
- An ammeter in series with the test wire.
- A voltmeter in parallel across the measured length of the wire.
- A switch/variable resistor to control current.

(b) Two precautions:
1. Check and correct for zero error on the micrometer before taking readings.
2. Measure the diameter at multiple different positions and orientations along the wire, and calculate a mean.

(c) Cross-sectional area:
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\)
Percentage uncertainty in \(d\):
\(\% \Delta d = \frac{0.01}{0.38} \times 100 = 2.63\%\)
Percentage uncertainty in \(A\):
\(\% \Delta A = 2 \times \% \Delta d = 2 \times 2.63\% = 5.26\% \approx 5.3\%\)

(d) From the formula \(R = \frac{\rho L}{A}\), we have \(\frac{R}{L} = \frac{\rho}{A}\). Thus:
\(\text{gradient} = \frac{\rho}{A} \implies \rho = \text{gradient} \times A\)
\ ho = 4.15 \times 1.134 \times 10^{-7} = 4.71 \times 10^{-7}\ \Omega\text{ m}

(e) A large current would cause the wire to heat up. An increase in temperature would increase the resistivity of the wire, making the resistance measurements inconsistent and invalidating the constant-temperature assumption.

- (a) Correct series circuit with cell/power supply, ammeter, and test wire (1 mark); voltmeter connected in parallel across the variable length of wire (1 mark); variable resistor or switch included (1 mark)
- (b) Check/adjust for zero error (1 mark); take readings at different orientations/positions along the wire and average (1 mark)
- (c) Correct calculation of area as \(1.13 \times 10^{-7}\text{ m}^2\) (1 mark); % uncertainty in d calculated as 2.63% (1 mark); % uncertainty in area correctly doubled to 5.3% (or 5.26%) (1 mark)
- (d) Recall of \(\rho = \text{gradient} \times A\) (1 mark); value of \(4.71 \times 10^{-7}\ \Omega\text{ m}\) (allow ecf from c) (1 mark)
- (e) Current causes heating/temperature rise (1 mark); resistance/resistivity varies with temperature (1 mark)

(a) A long, thin wire produces a much larger extension for a given load since \(\Delta x = \frac{F L_0}{A E}\). This larger extension reduces the percentage uncertainty in the extension measurement.

(b) Attach a marker (e.g., a piece of tape or paper) to the wire near a fixed metre rule, or use a vernier scale arrangement. Read the position of the marker against the scale before and after adding each load, keeping the scale close to the wire to avoid parallax errors.

(c) Area \(A = \frac{\pi d^2}{4} = \frac{\pi (0.56 \times 10^{-3}\text{ m})^2}{4} = 2.463 \times 10^{-7}\text{ m}^2\).

Young Modulus:
\(E = \frac{F L_0}{A \Delta x} = \frac{35.0 \times 2.450}{2.463 \times 10^{-7} \times 1.8 \times 10^{-3}} = 1.933 \times 10^{11}\text{ Pa} \approx 1.9 \times 10^{11}\text{ Pa}\).

Percentage uncertainties:
- \(\% \Delta L_0 = \frac{0.002}{2.450} \times 100\% = 0.08\%\)
- \(\% \Delta d = \frac{0.01}{0.56} \times 100\% = 1.79\%\)
- \(\% \Delta A = 2 \times 1.79\% = 3.58\%\)
- \(\% \Delta F = \frac{0.5}{35.0} \times 100\% = 1.43\%\)
- \(\% \Delta(\Delta x) = \frac{0.1}{1.8} \times 100\% = 5.56\%\)

Total percentage uncertainty in \(E\):
\(\% \Delta E = \% \Delta L_0 + \% \Delta A + \% \Delta F + \% \Delta(\Delta x) = 0.08\% + 3.58\% + 1.43\% + 5.56\% = 10.65\% \approx 11\%\).

(d) The student can check if the elastic limit has been exceeded by removing the load and checking if the wire returns to its original length. If there is permanent deformation, the limit has been exceeded.

- (a) Larger extension for same force (1 mark); reduces the percentage uncertainty in extension (1 mark)
- (b) Use a marker/reference point on the wire (1 mark); read against a scale/metre rule (1 mark); read at eye level / use a vernier setup to improve precision (1 mark)
- (c) Calculate cross-sectional area correctly (1 mark); calculate Young Modulus as \(1.9 \times 10^{11}\text{ Pa}\) (1 mark); uncertainty in area = 3.6% (1 mark); sum of percentage uncertainties correctly calculated as 11% (or 10.7%) (1 mark)
- (d) Remove the masses/load (1 mark); check if the wire returns to its original length / has a permanent extension (1 mark)

(a) The ray enters along the normal to the curved boundary (at \(90^\circ\) to the tangent), so the angle of incidence at this first boundary is zero, resulting in zero refraction.

(b) Draw around the block on paper, mark the entry and exit points of the ray, remove the block, draw the normal at the boundary point, and use a protractor to measure the angles of the rays relative to this normal.

(c) Increase the angle of incidence \(i\) until the refracted ray travels along the flat glass-air interface (at \(90^\circ\) to the normal). The angle of incidence at this position is the critical angle \(\theta_c\).
This method has high percentage uncertainty because the refracted ray becomes very faint, spreads out, or disappears near \(90^\circ\), making it difficult to pinpoint the exact transition angle.

(d) Refractive index:
\(n = \frac{\sin r}{\sin i} = \frac{\sin 58^\circ}{\sin 35^\circ} = \frac{0.8480}{0.5736} = 1.478 \approx 1.48\)

Maximum possible value of \(n\):
To maximise \(n\), we need the maximum possible value of \(r\) and the minimum possible value of \(i\).
\(r_{\text{max}} = 59^\circ\)
\(i_{\text{min}} = 34^\circ\)
\(n_{\text{max}} = \frac{\sin 59^\circ}{\sin 34^\circ} = \frac{0.8572}{0.5592} = 1.533 \approx 1.53\)

- (a) Ray enters along the normal / perpendicular to surface (1 mark); so angle of incidence is 0 degrees and no deviation occurs (1 mark)
- (b) Draw around the block and mark ray paths (1 mark); draw normal at the point of incidence and measure angles to the normal (1 mark)
- (c) Vary angle until refracted ray is along flat boundary / 90 degrees (1 mark); ray becomes dim / hard to see exact point of total internal reflection (1 mark)
- (d) Correct formula used: \(n = \frac{\sin r}{\sin i}\) (1 mark); \(n = 1.48\) (1 mark); use of \(r = 59^\circ\) and \(i = 34^\circ\) to find maximum (1 mark); maximum value \(n = 1.53\) (1 mark)