2024 AQA IAS-Level Physics (9630) Practice Paper with Answers

Using Snell's Law at the upper boundary: \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\). Since the light travels from air into glass: \(1.00 \times \sin(38.0^\circ) = 1.55 \times \sin(\theta_2)\). This gives \(\sin(\theta_2) = \frac{\sin(38.0^\circ)}{1.55} = 0.3972\), so \(\theta_2 = 23.4^\circ\). Since the top surface and the vertical side face of the rectangular block are at \(90.0^\circ\) to each other, the angle of incidence at the vertical face is \(\theta_3 = 90.0^\circ - \theta_2 = 90.0^\circ - 23.4^\circ = 66.6^\circ\).

1 mark for using Snell's Law to find the angle of refraction: \(\sin(\theta_2) = \frac{\sin(38.0^\circ)}{1.55}\). 1 mark for calculating \(\theta_2 = 23.4^\circ\). 1 mark for realizing that the angle at the vertical face is the complement of the refracted angle: \(\theta_3 = 90.0^\circ - \theta_2\). 1 mark for the correct final answer of \(66.6^\circ\).

The maximum angle at which light can escape is the critical angle, \(\theta_c = 43.5^\circ\). The relationship between critical angle and refractive index of the liquid \(n\) is given by: \(\sin(\theta_c) = \frac{1}{n}\). Therefore, \(n = \frac{1}{\sin(43.5^\circ)} = 1.453\). The speed of light in the liquid \(v\) is given by: \(v = \frac{c}{n} = \frac{3.00 \times 10^8}{1.453} = 2.07 \times 10^8\text{ m s}^{-1}\).

1 mark for identifying that the critical angle is \(43.5^\circ\). 1 mark for stating or using \(\sin(\theta_c) = \frac{1}{n}\). 1 mark for calculating \(n = 1.45\). 1 mark for using \(v = \frac{c}{n}\) with their value of \(n\). 1 mark for the final speed of \(2.07 \times 10^8\text{ m s}^{-1}\).

First, resolve the initial velocity into horizontal and vertical components: \(u_x = 18.0 \cos(25.0^\circ) = 16.31\text{ m s}^{-1}\) and \(u_y = 18.0 \sin(25.0^\circ) = 7.61\text{ m s}^{-1}\). Next, find the time of flight \(t\) using the vertical displacement equation: \(s_y = u_y t - \frac{1}{2} g t^2\). Substituting the values: \(-45.0 = 7.61 t - 4.905 t^2\). Rearranging gives the quadratic equation: \(4.905 t^2 - 7.61 t - 45.0 = 0\). Solving this equation using the quadratic formula: \(t = \frac{7.61 + \sqrt{(-7.61)^2 - 4(4.905)(-45.0)}}{2(4.905)} = 3.90\text{ s}\). Finally, calculate the horizontal distance: \(x = u_x t = 16.31 \times 3.90 = 63.7\text{ m}\).

1 mark for calculating vertical component of initial velocity \(u_y = 7.61\text{ m s}^{-1}\). 1 mark for setting up the vertical displacement equation with correct signs: \(-45.0 = 7.61 t - 4.905 t^2\). 1 mark for solving the quadratic equation to find the time of flight \(t = 3.90\text{ s}\). 1 mark for calculating horizontal component of initial velocity \(u_x = 16.31\text{ m s}^{-1}\). 1 mark for multiplying the horizontal component of velocity by the time of flight. 1 mark for the correct horizontal distance of \(63.7\text{ m}\) (or \(64\text{ m}\)).

For the vertical motion: \(s_y = \frac{1}{2} g t^2\), so \(1.65 = 4.905 t^2\), which gives the time of flight \(t = \sqrt{\frac{1.65}{4.905}} = 0.580\text{ s}\). For the horizontal motion: \(v_x = \frac{s_x}{t} = \frac{24.0}{0.580} = 41.38\text{ m s}^{-1}\). The vertical component of velocity just before impact is \(v_y = g t = 9.81 \times 0.580 = 5.69\text{ m s}^{-1}\). The magnitude of the final velocity is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{41.38^2 + 5.69^2} = 41.8\text{ m s}^{-1}\).

1 mark for calculating time of flight: \(t = 0.580\text{ s}\). 1 mark for calculating horizontal velocity: \(v_x = 41.4\text{ m s}^{-1}\). 1 mark for calculating vertical velocity at impact: \(v_y = 5.69\text{ m s}^{-1}\). 1 mark for using Pythagoras' theorem: \(v = \sqrt{v_x^2 + v_y^2}\). 1 mark for the correct final magnitude: \(41.8\text{ m s}^{-1}\).

First, find the radius \(r\) of the circular path: \(r = L \sin(28.0^\circ) = 0.850 \times \sin(28.0^\circ) = 0.399\text{ m}\). The tension \(T\) in the string provides both the upward vertical force supporting the sphere's weight and the inward horizontal centripetal force: \(T \cos(28.0^\circ) = m g\) and \(T \sin(28.0^\circ) = \frac{m v^2}{r}\). Dividing these two equations gives: \(\tan(28.0^\circ) = \frac{v^2}{r g}\). Rearranging for velocity \(v\): \(v = \sqrt{r g \tan(28.0^\circ)} = \sqrt{0.399 \times 9.81 \times \tan(28.0^\circ)} = 1.44\text{ m s}^{-1}\).

1 mark for calculating the radius of the circle: \(r = 0.399\text{ m}\). 1 mark for resolving the tension vertically: \(T \cos(28.0^\circ) = m g\). 1 mark for resolving the tension horizontally: \(T \sin(28.0^\circ) = \frac{m v^2}{r}\). 1 mark for obtaining the relationship \(\tan(28.0^\circ) = \frac{v^2}{r g}\). 1 mark for substituting values correctly and calculating speed: \(1.44\text{ m s}^{-1}\).

In the fundamental mode (first harmonic), the wavelength \(\lambda\) of the wave is twice the length of the wire: \(\lambda = 2 L = 2 \times 0.650 = 1.30\text{ m}\). The speed of the wave on the wire is: \(v = f \lambda = 120 \times 1.30 = 156\text{ m s}^{-1}\). The speed is also given by: \(v = \sqrt{\frac{T}{\mu}}\), where \(T\) is the tension and \(\mu\) is the mass per unit length. Rearranging gives: \(T = v^2 \mu = (156)^2 \times 4.20 \times 10^{-3} = 102\text{ N}\).

1 mark for calculating the wavelength: \(\lambda = 1.30\text{ m}\). 1 mark for calculating the wave speed: \(v = 156\text{ m s}^{-1}\). 1 mark for stating or using \(v = \sqrt{\frac{T}{\mu}}\). 1 mark for rearranging the formula: \(T = v^2 \mu\). 1 mark for calculating the final tension: \(102\text{ N}\).

First, resolve the weight into components parallel and perpendicular to the inclined plane: \(W_{\parallel} = m g \sin(30.0^\circ) = 6.50 \times 9.81 \times \sin(30.0^\circ) = 31.88\text{ N}\) acting down the plane. \(W_{\perp} = m g \cos(30.0^\circ) = 6.50 \times 9.81 \times \cos(30.0^\circ) = 55.22\text{ N}\). The normal contact force is \(N = W_{\perp} = 55.22\text{ N}\). The maximum frictional force is \(F_f = \mu N = 0.240 \times 55.22 = 13.25\text{ N}\) acting down the plane to oppose motion up the plane. The net force \(F_{\text{net}}\) acting up the plane is: \(F_{\text{net}} = P - W_{\parallel} - F_f = 52.0 - 31.88 - 13.25 = 6.87\text{ N}\). The acceleration \(a\) is: \(a = \frac{F_{\text{net}}}{m} = \frac{6.87}{6.50} = 1.06\text{ m s}^{-2}\).

1 mark for calculating weight component down the plane: \(W_{\parallel} = 31.9\text{ N}\). 1 mark for calculating normal contact force: \(N = 55.2\text{ N}\). 1 mark for calculating frictional force: \(F_f = 13.3\text{ N}\). 1 mark for setting up the net force equation: \(F_{\text{net}} = P - W_{\parallel} - F_f\). 1 mark for finding \(F_{\text{net}} = 6.87\text{ N}\). 1 mark for the correct final acceleration: \(1.06\text{ m s}^{-2}\).

First, calculate the time constant \(\tau\) of the circuit: \(\tau = R C = 82.0 \times 10^3 \times 47.0 \times 10^{-6} = 3.854\text{ s}\). The formula for the potential difference during discharging is: \(V = V_0 e^{-t/\tau}\). Substituting the values: \(V = 12.0 \times e^{-2.50 / 3.854} = 12.0 \times e^{-0.6487} = 12.0 \times 0.5227 = 6.27\text{ V}\).

1 mark for calculating the time constant: \(\tau = 3.85\text{ s}\). 1 mark for stating or using the discharging formula: \(V = V_0 e^{-t/\tau}\). 1 mark for calculating the ratio \(\frac{t}{\tau} = 0.649\). 1 mark for the correct final voltage: \(6.27\text{ V}\).

To find the maximum angle of incidence \(\theta\) at the entrance:

1. Calculate the critical angle \(\theta_c\) at the core-cladding boundary:
\[ \sin\theta_c = \frac{n_{\text{cladding}}}{n_{\text{core}}} = \frac{1.41}{1.48} \]
\[ \theta_c = \arcsin(0.9527) \approx 72.31^{\circ} \]

2. For total internal reflection to occur, the angle of incidence on the boundary must be greater than or equal to \(\theta_c\). Under limiting conditions, the angle of refraction \(r\) at the first interface is:
\[ r = 90^{\circ} - \theta_c = 90^{\circ} - 72.31^{\circ} = 17.69^{\circ} \]

3. Apply Snell's Law at the air-core interface:
\[ n_{\text{air}} \sin\theta = n_{\text{core}} \sin r \]
\[ 1.00 \times \sin\theta = 1.48 \times \sin(17.69^{\circ}) \]
\[ \sin\theta = 1.48 \times 0.3039 = 0.4498 \]
\[ \theta = \arcsin(0.4498) \approx 26.7^{\circ} \]

Award [2 marks] for finding the critical angle \(\theta_c = 72.31^{\circ}\).
Award [1 mark] for relating the angle of refraction \(r\) to the critical angle: \(r = 90^{\circ} - \theta_c\).
Award [1 mark] for calculating the angle of refraction \(r = 17.69^{\circ}\).
Award [1.6 marks] for the application of Snell's law at the air-core boundary to get \(\theta = 26.7^{\circ}\) (accept range \(26.5^{\circ}\) to \(27.0^{\circ}\)).

1. Resolve the initial velocity into horizontal and vertical components:
\[ u_x = u \cos\theta = 22.0 \cos(35.0^{\circ}) \approx 18.02\text{ m s}^{-1} \]
\[ u_y = u \sin\theta = 22.0 \sin(35.0^{\circ}) \approx 12.62\text{ m s}^{-1} \]

2. Determine the time \(t\) taken to travel the horizontal distance of \(32.0\text{ m}\) to the wall:
\[ t = \frac{s_x}{u_x} = \frac{32.0}{18.02} \approx 1.776\text{ s} \]

3. Calculate the vertical displacement \(s_y\) at time \(t\):
\[ s_y = u_y t - \frac{1}{2}gt^2 \]
\[ s_y = (12.62 \times 1.776) - (0.5 \times 9.81 \times 1.776^2) \]
\[ s_y = 22.41 - 15.47 = 6.94\text{ m} \]

Award [1 mark] for correctly resolving the horizontal velocity component (\(u_x = 18.02\text{ m s}^{-1}\)).
Award [2 marks] for calculating the time of flight to the wall (\(t = 1.78\text{ s}\)).
Award [1 mark] for resolving the vertical velocity component (\(u_y = 12.62\text{ m s}^{-1}\)).
Award [1.6 marks] for using the equations of motion to solve for the final vertical height (\(s_y = 6.94\text{ m}\); accept range \(6.90\text{ m}\) to \(7.00\text{ m}\)).

The critical angle \(\theta_c\) at the boundary between two media of refractive indices \(n_1\) (where light is travelling) and \(n_2\) (the adjacent medium) is given by: \(\sin \theta_c = \frac{n_2}{n_1}\) where \(n_1 > n_2\). Here, \(n_1 = n_g = 1.52\) and \(n_2 = n_{\text{oil}} = 1.35\). Therefore: \(\sin \theta_c = \frac{1.35}{1.52} \approx 0.8882\). Taking the inverse sine: \(\theta_c = \sin^{-1}(0.8882) \approx 62.7^\circ\).

- 1 mark for the correct answer (C).

At maximum height, the vertical component of the velocity of the projectile is zero (\(v_y = 0\)). The horizontal component of the velocity remains constant throughout the motion because there is no horizontal acceleration: \(v_x = u \cos\theta = 25.0 \times \cos(35.0^\circ) \approx 20.5\text{ m s}^{-1}\). The speed at this point is therefore equal to this horizontal component: \(v = \sqrt{v_x^2 + v_y^2} = 20.5\text{ m s}^{-1}\).

- 1 mark for the correct answer (C).

At the lowest point of the circular path, the upward tension \(T\) in the string and the downward force of gravity \(mg\) combine to provide the necessary centripetal force: \(T - mg = \frac{m v^2}{r}\). Rearranging this gives: \(T = \frac{m v^2}{r} + mg = \frac{0.15 \times 5.0^2}{0.80} + 0.15 \times 9.81 = 4.6875 + 1.4715 = 6.159\text{ N}\), which rounds to \(6.2\text{ N}\).

- 1 mark for the correct answer (C).

For a wire fixed at both ends, the second harmonic consists of two loops, meaning the length of the wire \(L\) is exactly equal to one wavelength \(\lambda\). Thus: \(\lambda = L = 0.60\text{ m\). Using the wave equation: \(v = f \lambda = 240 \times 0.60 = 144\text{ m s}^{-1}\).

- 1 mark for the correct answer (B).

The force parallel to the incline trying to pull the box down is the component of its weight: \(F_{\text{weight}} = mg \sin\theta = 8.0 \times 9.81 \times \sin(30^\circ) = 39.24\text{ N}\). Applying Newton's second law along the incline (with the direction up the incline defined as positive): \(F_{\text{net}} = F - F_{\text{weight}} = 60 - 39.24 = 20.76\text{ N}\). The acceleration is: \(a = \frac{F_{\text{net}}}{m} = \frac{20.76}{8.0} \approx 2.6\text{ m s}^{-2}\).

- 1 mark for the correct answer (A).

First, calculate the time constant \(\tau\) of the circuit: \(\tau = RC = (15 \times 10^3\text{ }\Omega) \times (470 \times 10^{-6}\text{ F}) = 7.05\text{ s}\). The potential difference \(V\) at time \(t\) during discharge is: \(V = V_0 e^{-t/\tau} = 12.0 \times e^{-10.0 / 7.05} = 12.0 \times e^{-1.418} \approx 12.0 \times 0.242 = 2.90\text{ V}\).

- 1 mark for the correct answer (A).

The refractive index \(n\) of the plastic block is given by the ratio of the wavelength in vacuum to the wavelength in the medium: \(n = \frac{\lambda_{\text{vacuum}}}{\lambda_{\text{medium}}} = \frac{633}{422} = 1.50\). The speed of light in the medium is: \(v = \frac{c}{n} = \frac{3.00 \times 10^8}{1.50} = 2.00 \times 10^8\text{ m s}^{-1}\).

- 1 mark for the correct answer (B).

First, find the time of flight \(t\) using the vertical motion: \(s_y = \frac{1}{2} g t^2 \implies 15.0 = \frac{1}{2} \times 9.81 \times t^2 \implies t^2 \approx 3.058\text{ s}^2 \implies t \approx 1.749\text{ s}\). The horizontal distance is the horizontal velocity multiplied by the time of flight: \(s_x = v_x \times t = 18.0 \times 1.749 \approx 31.5\text{ m}\).

- 1 mark for the correct answer (B).

Using Snell's Law:
\(n_1 \sin \theta_1 = n_2 \sin \theta_2\)

Substitute the given values to find \(n_2\):
\(1.60 \times \sin(35.0^\circ) = n_2 \times \sin(48.0^\circ)\)
\(1.60 \times 0.5736 = n_2 \times 0.7431\)
\(n_2 = \frac{0.9177}{0.7431} \approx 1.235\)

Now, calculate the speed of light in medium 2:
\(v_2 = \frac{c}{n_2} = \frac{3.00 \times 10^8}{1.235} \approx 2.43 \times 10^8\text{ m s}^{-1}\).

1 mark for correct calculation and choosing option C.

First, find the initial velocity components:
\(u_x = 25.0 \cos(30.0^\circ) \approx 21.65\text{ m s}^{-1}\)
\(u_y = 25.0 \sin(30.0^\circ) = 12.50\text{ m s}^{-1}\)

Next, calculate the total time of flight \(T\) using the vertical motion:
\(s_y = u_y T - \frac{1}{2}g T^2 = 0\)
\(T = \frac{2 u_y}{g} = \frac{2 \times 12.50}{9.81} \approx 2.548\text{ s}\)

Finally, calculate the horizontal range \(R\):
\(R = u_x T = 21.65 \times 2.548 \approx 55.2\text{ m}\).

1 mark for correct calculation and choosing option C.

The centripetal force required to keep the car in circular motion is provided by the static frictional force:
\(F_{\text{centripetal}} \le F_{\text{friction, max}}\)
\(\frac{m v^2}{r} \le \mu_s m g\)

Rearranging for \(v\):
\(v \le \sqrt{\mu_s g r}\)
\(v_{\text{max}} = \sqrt{0.45 \times 9.81 \times 0.80} = \sqrt{3.5316} \approx 1.88\text{ m s}^{-1}\).

1 mark for correct calculation and choosing option B.

For a pipe closed at one end and open at the other, only odd harmonics exist. The frequencies are given by:
\(f_n = n f_1\) where \(n = 1, 3, 5, \dots\)

We are given the frequency of the third harmonic (\(n = 3\)):
\(f_3 = 3 f_1 = 450\text{ Hz} \implies f_1 = 150\text{ Hz}\)

The fifth harmonic (\(n = 5\)) has a frequency of:
\(f_5 = 5 f_1 = 5 \times 150\text{ Hz} = 750\text{ Hz}\).

1 mark for correct calculation and choosing option B.

Let \(a\) be the acceleration and \(T\) be the tension in the string.
For the hanging mass \(m_1\):
\(m_1 g - T = m_1 a\)

For the mass on the horizontal table \(m_2\):
\(T - f_k = m_2 a\)
where the frictional force is \(f_k = \mu_k m_2 g\).

Adding these two equations gives:
\(m_1 g - \mu_k m_2 g = (m_1 + m_2) a\)

Rearranging to solve for \(a\):
\(a = \frac{m_1 - \mu_k m_2}{m_1 + m_2} g\)
\(a = \frac{3.0 - (0.20 \times 5.0)}{3.0 + 5.0} \times 9.81 = \frac{2.0}{8.0} \times 9.81 = 2.45\text{ m s}^{-2}\).

1 mark for correct calculation and choosing option B.

First, calculate the initial charge on the capacitor \(Q_0\):
\(Q_0 = C V = 47 \times 10^{-6}\text{ F} \times 12\text{ V} = 5.64 \times 10^{-4}\text{ C}\)

Next, calculate the time constant \(\tau\) of the circuit:
\(\tau = R C = (150 \times 10^3\text{ }\Omega) \times (47 \times 10^{-6}\text{ F}) = 7.05\text{ s}\)

Now, find the charge at \(t = 5.0\text{ s}\) using the exponential discharge formula:
\(Q(t) = Q_0 e^{-t/\tau}\)
\(Q(5.0) = 5.64 \times 10^{-4} \times e^{-5.0 / 7.05}\)
\(Q(5.0) \approx 5.64 \times 10^{-4} \times 0.4920 \approx 2.77 \times 10^{-4}\text{ C} \approx 0.28\text{ mC}\).

1 mark for correct calculation and choosing option B.

(a) From Snell's law:
\(n_1 \sin \theta_1 = n_2 \sin \theta_2\)
Substituting the values:
\(1.58 \sin(42.0^\circ) = n_{\text{liquid}} \sin(53.2^\circ)\)
\(n_{\text{liquid}} = \frac{1.58 \times 0.6691}{0.8007} = 1.32\)

(b) The critical angle \(\theta_c\) is given by:
\(\sin \theta_c = \frac{n_{\text{liquid}}}{n_{\text{glass}}}\)
\(\sin \theta_c = \frac{1.32}{1.58} = 0.8354\)
\(\theta_c = \sin^{-1}(0.8354) = 56.7^\circ\) (or using unrounded figures: \(\theta_c = 56.7^\circ\))

(a)
- Recall and use of Snell's law: \(n_1 \sin \theta_1 = n_2 \sin \theta_2\) [1 mark]
- Substitution of correct values: \(1.58 \sin(42.0^\circ) = n_{\text{liquid}} \sin(53.2^\circ)\) [1 mark]
- Correct calculation of \(n_{\text{liquid}} = 1.32\) [1 mark]

(b)
- Recall of critical angle equation: \(\sin \theta_c = \frac{n_2}{n_1}\) [1 mark]
- Substitution of values: \(\sin \theta_c = \frac{1.32}{1.58}\) [1 mark]
- Correct final angle: \(\theta_c = 56.7^\circ\) (accept range \(56.6^\circ\) to \(56.7^\circ\)) [1 mark]

(a) Because the ray is incident normally on the flat surface, it enters without bending and travels parallel to the central axis. At the curved surface, the normal to the interface lies along the radius of the hemisphere. The geometry forms a right triangle where:
\(\sin \theta_1 = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3.0}{5.0} = 0.60\)
Therefore, the angle of incidence \(\theta_1\) is:
\(\theta_1 = \sin^{-1}(0.60) = 36.87^\circ \approx 37^\circ\).

(b) At the boundary with air, we apply Snell's law:
\(n_{\text{glass}} \sin \theta_1 = n_{\text{air}} \sin \theta_2\)
\(1.52 \times 0.60 = 1.00 \times
\sin \theta_2\)
\\sin \theta_2 = 0.912\)
\(\theta_2 = \sin^{-1}(0.912) = 65.78^\circ\)
The angle of deviation \(\delta\) is the change in direction of the ray:
\(\delta = \theta_2 - \theta_1 = 65.78^\circ - 36.87^\circ = 28.91^\circ \approx 28.9^\circ\).

(a)
- Recognition that the normal at the curved surface lies along the radius line [1 mark]
- Correct trigonometric equation for the angle of incidence: \(\sin \theta_1 = \frac{3.0}{5.0}\) [1 mark]
- Correct calculation showing \(\theta_1 \approx 36.9^\circ\) or \(37^\circ\) [1 mark]

(b)
- Recall and substitution into Snell's law: \(1.52 \times \sin(36.9^\circ) = 1.00 \times \sin \theta_2\) [1 mark]
- Correct calculation of angle of refraction: \(\theta_2 = 65.8^\circ\) (accept range \(65.7^\circ - 65.9^\circ\)) [1 mark]
- Correct calculation of deviation: \(\delta = 65.8^\circ - 36.9^\circ = 28.9^\circ\) (accept range \(28.8^\circ - 29.0^\circ\)) [1 mark]

(a) Resolving the initial velocity into vertical and horizontal components:
\(u_y = 22.0 \sin(30.0^\circ) = 11.0\text{ m s}^{-1}\)
\(u_x = 22.0 \cos(30.0^\circ) = 19.05\text{ m s}^{-1}\)
Using the vertical displacement formula with \(s_y = -45.0\text{ m}\) and \(g = 9.81\text{ m s}^{-2}\):
\(s_y = u_y t - \frac{1}{2}g t^2 \implies -45.0 = 11.0 t - 4.905 t^2\)
Rearranging into standard quadratic form:
\(4.905 t^2 - 11.0 t - 45.0 = 0\)
Solving for \(t\) using the quadratic formula:
\(t = \frac{11.0 + \sqrt{(-11.0)^2 - 4(4.905)(-45.0)}}{2 \times 4.905} = \frac{11.0 + \sqrt{121 + 882.9}}{9.81} = \frac{11.0 + 31.68}{9.81} = 4.35\text{ s}\)
Which is approximately \(4.4\text{ s}\).

(b) The horizontal component of velocity is constant:
\(x = u_x \times t = 19.05\text{ m s}^{-1} \times 4.35\text{ s} = 82.9\text{ m}\) (or \(83.8\text{ m}\) using rounded \(4.4\text{ s}\)).

(a)
- Correct vertical component of initial velocity: \(u_y = 11.0\text{ m s}^{-1}\) [1 mark]
- Correct setup of vertical motion equation with appropriate signs: \(-45.0 = 11.0 t - 4.905 t^2\) [1 mark]
- Correct calculation yielding \(t \approx 4.35\text{ s}\) or \(4.4\text{ s}\) [1 mark]

(b)
- Correct horizontal component of initial velocity: \(u_x = 19.05\text{ m s}^{-1}\) [1 mark]
- Use of horizontal distance equation: \(x = u_x t\) [1 mark]
- Correct final horizontal distance: \(82.9\text{ m}\) (accept range \(82.5\text{ m} - 84.0\text{ m}\) depending on rounding of time of flight) [1 mark]

(a) The horizontal motion is described by:
\(x = (u \cos \theta) t \implies t = \frac{x}{u \cos \theta}\)
The vertical displacement \(y\) is:
\(y = (u \sin \theta) t - \frac{1}{2}g t^2\)
Substituting \(t\) and using \(\theta = 45.0^\circ\):
\(y = x \tan(45.0^\circ) - \frac{g x^2}{2 u^2 \cos^2(45.0^\circ)}\)
Since \(\tan(45.0^\circ) = 1\) and \(\cos^2(45.0^\circ) = 0.5\):
\(y = x - \frac{g x^2}{u^2}\).

(b) The horizontal displacement from release to basket is \(x = 8.00\text{ m}\).
The vertical displacement is \(y = 3.05 - 2.00 = 1.05\text{ m}\).
Using the derived equation:
\(1.05 = 8.00 - \frac{9.81 \times 8.00^2}{u^2}\)
\(\frac{627.84}{u^2} = 8.00 - 1.05 = 6.95\)
\(u^2 = \frac{627.84}{6.95} = 90.337\)
\(u = \sqrt{90.337} = 9.50\text{ m s}^{-1}\).

(a)
- Correct expression for horizontal motion and time: \(t = \frac{x}{u \cos 45^\circ}\) [1 mark]
- Correct vertical displacement equation: \(y = u t \sin 45^\circ - \frac{1}{2}g t^2\) [1 mark]
- Use of \(\tan 45^\circ = 1\) and \(\cos^2 45^\circ = 0.5\) to derive the final expression [1 mark]

(b)
- Correct identification of vertical displacement: \(y = 1.05\text{ m}\) [1 mark]
- Correct substitution of values into equation: \(1.05 = 8.00 - \frac{9.81 \times 64.0}{u^2}\) [1 mark]
- Correct final velocity calculation: \(9.50\text{ m s}^{-1}\) (accept \(9.5\text{ m s}^{-1}\)) [1 mark]

(a) The centripetal acceleration is:
\(a_c = \frac{v^2}{r} = \frac{12.0^2}{22.0} = \frac{144}{22.0} = 6.55\text{ m s}^{-2}\).

(b) At the highest point of the vertical circle, the forces acting on the car are the weight \(mg\) downwards and the normal reaction \(N\) upwards. The resultant force towards the centre of the circle is the centripetal force:
\(F_c = mg - N = ma_c\)
Rearranging for \(N\):
\(N = m(g - a_c) = 1400 \times (9.81 - 6.545) = 1400 \times 3.265 = 4571\text{ N}\) (or \(4570\text{ N}\) to 3 significant figures).

(a)
- Recall of centripetal acceleration formula: \(a = \frac{v^2}{r}\) [1 mark]
- Correct substitution of values: \(a = \frac{12.0^2}{22.0}\) [1 mark]
- Correct acceleration: \(6.55\text{ m s}^{-2}\) (accept range \(6.54 - 6.55\)) [1 mark]

(b)
- Formulation of equations of motion: \(mg - N = m a_c\) [1 mark]
- Correct substitution of mass and values: \(N = 1400(9.81 - 6.55)\) [1 mark]
- Correct value for normal contact force: \(4570\text{ N}\) (accept range \(4560\text{ N} - 4580\text{ N}\)) [1 mark]

(a) For a string of length \(L\) vibrating in its third harmonic, there are 3 loops. Therefore:
\(L = \frac{3}{2}\lambda \implies \lambda = \frac{2}{3}L = \frac{2}{3} \times 1.20\text{ m} = 0.80\text{ m}\).
The wave speed \(v\) is:
\(v = f \lambda = 150\text{ Hz} \times 0.80\text{ m} = 120\text{ m s}^{-1}\).

(b) The speed of transverse waves on a stretched string is given by:
\(v = \sqrt{\frac{T}{\mu}}\)
Rearranging for tension \(T\):
\(T = \mu v^2 = (3.50 \times 10^{-3}\text{ kg m}^{-1}) \times (120\text{ m s}^{-1})^2\)
\(T = 3.50 \times 10^{-3} \times 14400 = 50.4\text{ N}\).

(a)
- Relate wavelength and length for the third harmonic: \(\lambda = 0.80\text{ m}\) [1 mark]
- Use wave speed formula: \(v = f \lambda\) [1 mark]
- Correctly calculate wave speed: \(120\text{ m s}^{-1}\) [1 mark]

(b)
- Recall equation for wave speed on a string: \(v = \sqrt{\frac{T}{\mu}}\) [1 mark]
- Correct rearrangement: \(T = \mu v^2\) [1 mark]
- Correct calculation of tension: \(50.4\text{ N}\) (accept \(50\text{ N}\) if done to 2 s.f.) [1 mark]

(a) The normal reaction force \(N\) is:
\(N = mg \cos \theta = 8.00 \times 9.81 \times \cos(25.0^\circ) = 71.13\text{ N}\)
The frictional force \(f\) is:
\(f = \mu N = 0.300 \times 71.13 = 21.34\text{ N} \approx 21.3\text{ N\}.

(b) The component of the box's weight acting down the slope is:
\)W_{\parallel} = mg \sin \theta = 8.00 \times 9.81 \times \sin(25.0^\circ) = 33.17\text{ N}\)
Applying Newton's second law along the slope (taking up the slope as positive):
\(F_{\text{net}} = F - f - W_{\parallel} = 65.0 - 21.34 - 33.17 = 10.49\text{ N}\)
The acceleration is:
\(a = \frac{F_{\text{net}}}{m} = \frac{10.49}{8.00} = 1.31\text{ m s}^{-2}\).

(a)
- Recall normal reaction force equation: \(N = mg \cos \theta\) [1 mark]
- Substitution of values: \(8.00 \times 9.81 \times \cos(25.0^\circ)\) [1 mark]
- Calculate friction force: \(f = 21.3\text{ N}\) (accept \(21.34\text{ N}\)) [1 mark]

(b)
- Resolve weight component down the slope: \(W_{\parallel} = 33.2\text{ N}\) (or \(33.17\text{ N}\)) [1 mark]
- Formulate Newton's second law equation along slope: \(65.0 - 21.34 - 33.17 = 8.00 a\) [1 mark]
- Calculate acceleration: \(a = 1.31\text{ m s}^{-2}\) (accept range \(1.30 - 1.32\)) [1 mark]

(a) The time constant \(\tau\) is:
\(\tau = RC = 150 \times 10^3\ \Omega \times 47.0 \times 10^{-6}\text{ F} = 7.05\text{ s}\).

(b) The formula for potential difference decay is:
\(V = V_0 e^{-\frac{t}{\tau}}\)
Substituting the values:
\(V = 12.0 \times e^{-\frac{5.00}{7.05}} = 12.0 \times e^{-0.7092} = 12.0 \times 0.492 = 5.90\text{ V}\).

(a)
- Recall formula for time constant: \(\tau = RC\) [1 mark]
- Correct unit conversions for resistance and capacitance: [1 mark]
- Correct calculation of time constant: \(7.05\text{ s}\) [1 mark]

(b)
- Recall exponential discharge equation: \(V = V_0 e^{-\frac{t}{RC}}\) [1 mark]
- Substitution of correct values: \(V = 12.0 \times e^{-\frac{5.00}{7.05}}\) [1 mark]
- Correct final voltage: \(5.90\text{ V}\) (accept range \(5.90 - 5.91\text{ V}\)) [1 mark]

1. Identify the angle of incidence at the glass-liquid hypotenuse boundary:
Since the ray enters normally to the short side, it travels undeflected and meets the hypotenuse. From the geometry of a \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle, the angle of incidence \(\theta_i = 45.0^\circ\).

2. Calculate the critical angle \(\theta_c\) at the glass-liquid boundary:
\(\sin(\theta_c) = \frac{n_{\text{liquid}}}{n_{\text{glass}}} = \frac{1.30}{1.55} = 0.8387\)
\(\theta_c = \sin^{-1}(0.8387) = 57.0^\circ\).

3. Determine if total internal reflection (TIR) occurs:
Since the angle of incidence \(\theta_i = 45.0^\circ\) is less than the critical angle \(\theta_c = 57.0^\circ\), TIR does not occur. The ray will refract into the liquid.

4. Calculate the angle of refraction \(\theta_r\) using Snell's Law:
\(n_{\text{glass}} \sin(\theta_i) = n_{\text{liquid}} \sin(\theta_r)\)
\(1.55 \sin(45.0^\circ) = 1.30 \sin(\theta_r)\)
\(\sin(\theta_r) = \frac{1.55 \times 0.7071}{1.30} = 0.8431\)
\(\theta_r = \sin^{-1}(0.8431) = 57.5^\circ\).

- State or show that the angle of incidence on the hypotenuse is \(45.0^\circ\) [1 mark]
- State critical angle equation \(\sin \theta_c = \frac{n_2}{n_1}\) and substitute values correctly [1 mark]
- Calculate critical angle as \(57.0^\circ\) (or \(57^\circ\)) [1 mark]
- Compare angle of incidence with critical angle and conclude that refraction occurs because \(45.0^\circ < 57.0^\circ\) [1 mark]
- Apply Snell's Law: \(1.55 \sin(45.0^\circ) = 1.30 \sin(\theta_r)\) [1 mark]
- Calculate final angle of refraction as \(57.5^\circ\) (accept \(57^\circ\) to \(58^\circ\) depending on intermediate rounding) [1 mark]

1. Resolve the initial velocity into horizontal and vertical components:
Horizontal component: \(u_x = 24.0 \cos(35.0^\circ) = 19.66\text{ m s}^{-1}\)
Vertical component: \(u_y = 24.0 \sin(35.0^\circ) = 13.77\text{ m s}^{-1}\)

2. Calculate the time \(t\) taken to travel the horizontal distance of \(32.0\text{ m}\):
\(t = \frac{x}{u_x} = \frac{32.0}{19.66} = 1.628\text{ s}\)

3. Calculate the vertical height \(y\) at this time \(t\) using the equation of motion:
\(y = u_y t - \frac{1}{2}g t^2\)
\(y = (13.77 \times 1.628) - (0.5 \times 9.81 \times 1.628^2)\)
\(y = 22.42 - 13.00 = 9.42\text{ m}\) (using precise unrounded numbers: \(22.407 - 12.995 = 9.41\text{ m}\)).

- Calculate horizontal component of initial velocity \(u_x = 19.7\text{ m s}^{-1}\) [1 mark]
- Calculate vertical component of initial velocity \(u_y = 13.8\text{ m s}^{-1}\) [1 mark]
- Calculate time of flight to the wall \(t = 1.63\text{ s}\) [1 mark]
- State or use correct vertical displacement equation \(y = u_y t - \frac{1}{2}g t^2\) [1 mark]
- Substitute calculated values of \(u_y\) and \(t\) correctly into the equation [1 mark]
- Obtain final height as \(9.41\text{ m}\) (accept range \(9.40\text{ m}\) to \(9.43\text{ m}\)) [1 mark]

1. State the discharge equation for a capacitor:
\(V = V_0 e^{-\frac{t}{RC}}\)

2. Substitute the given values into the equation:
\(3.00 = 12.0 e^{-\frac{15.0}{R \times 470 \times 10^{-6}}}\)

3. Simplify the fraction:
\(\frac{3.00}{12.0} = 0.250 = e^{-\frac{15.0}{RC}}\)

4. Take the natural logarithm of both sides:
\(\ln(0.250) = -1.386\)
\(-1.386 = -\frac{15.0}{RC}\)

5. Solve for the time constant \(\tau = RC\):
\(RC = \frac{15.0}{1.386} = 10.82\text{ s}\)

6. Calculate the resistance \(R\):
\(R = \frac{10.82}{470 \times 10^{-6}} = 23021\ \Omega = 23.0\text{ k}\Omega\).

- Use or state the exponential decay equation \(V = V_0 e^{-t/RC}\) [1 mark]
- Substitute \(V = 3.00\text{ V}\) and \(V_0 = 12.0\text{ V}\) to get ratio \(0.250\) [1 mark]
- Take natural logarithms correctly to show \(\ln(0.250) = -1.39\) (or equivalent log form) [1 mark]
- Find the time constant \(\tau = RC = 10.8\text{ s}\) (or \(10.82\text{ s}\)) [1 mark]
- Rearrange equation to make \(R\) the subject: \(R = \frac{\tau}{C}\) [1 mark]
- Calculate final value of resistance as \(23.0\text{ k}\Omega\) (or \(2.30 \times 10^4\ \Omega\); accept \(2.3\text{ k}\Omega\) to \(2.31\text{ k}\Omega\)) with appropriate units [1 mark]

Using Snell's Law across boundaries: \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2) = n_3 \sin(\theta_3)\). Thus, for the air-to-liquid boundary, the intermediate glass layer does not alter the final angle of refraction: \(n_{\text{air}} \sin(\theta_{\text{air}}) = n_{\text{liquid}} \sin(\theta_{\text{liquid}})\). Substituting the values: \(1.00 \times \sin(40.0^\circ) = 1.33 \times \sin(\theta_{\text{liquid}})\). \\ \sin(\theta_{\text{liquid}}) = \frac{\sin(40.0^\circ)}{1.33} = 0.4833 \implies \theta_{\text{liquid}} = \arcsin(0.4833) \approx 28.9^\circ\).

[1 mark] Correctly applies Snell's Law boundary relation and calculates \(28.9^\circ\).

The initial vertical component of velocity is \(v_{y0} = 25.0 \sin(35.0^\circ) = 14.34\text{ m s}^{-1}\). At maximum height, the vertical component of velocity \(v_y = 0\). Using the SUVAT equation \(v_y^2 = v_{y0}^2 - 2gH\), we get: \(0 = 14.34^2 - 2(9.81)H \implies 19.62H = 205.62 \implies H \approx 10.5\text{ m}\).

[1 mark] Correctly determines the initial vertical velocity component and applies the SUVAT equation to calculate \(10.5\text{ m}\).

At the highest point of the vertical circle, both the tension \(T\) and the weight \(mg\) act downwards towards the centre of the circle, providing the centripetal force: \(T + mg = \frac{m v^2}{r}\). Substituting the values: \(1.5 + (0.20 \times 9.81) = \frac{0.20 v^2}{0.80}\). \\ \(1.5 + 1.962 = 0.25 v^2 \implies 3.462 = 0.25 v^2 \implies v^2 = 13.848 \implies v \approx 3.72\text{ m s}^{-1}\).

[1 mark] Correctly writes the dynamic equation at the top of the circle and solves for velocity \(3.72\text{ m s}^{-1}\).

For a string fixed at both ends, the wavelength of the \(n\)-th harmonic is given by \(\lambda = \frac{2L}{n}\). For the third harmonic (\(n=3\)): \[\lambda = \frac{2 \times 1.2}{3} = 0.8\text{ m}\] The wave speed \(v\) is found using the wave equation: \[v = f \lambda = 240 \times 0.8 = 192\text{ m s}^{-1}\]

[1 mark] Finds wavelength for the third harmonic and uses wave equation to determine \(192\text{ m s}^{-1}\).

The forces acting down the slope are the component of gravity, \(F_g = mg \sin(30.0^\circ) = 4.0 \times 9.81 \times 0.5 = 19.62\text{ N}\), and the friction force \(f_k = \mu_k R = \mu_k mg \cos(30.0^\circ) = 0.15 \times 4.0 \times 9.81 \times \cos(30.0^\circ) = 5.10\text{ N}\). The net force up the slope is \(F_{\text{net}} = 35 - 19.62 - 5.10 = 10.28\text{ N}\). The acceleration is \(a = \frac{F_{\text{net}}}{m} = \frac{10.28}{4.0} \approx 2.57\text{ m s}^{-2}\).

[1 mark] Resolves forces along the incline, calculates net force, and finds acceleration \(2.57\text{ m s}^{-2}\).

The time constant \(\tau = RC = 150 \times 10^3 \times 47 \times 10^{-6} = 7.05\text{ s}\). The discharge equation for potential difference is \(V = V_0 e^{-t/\tau}\). Substituting the values: \(V = 12.0 \times e^{-5.0 / 7.05} = 12.0 \times e^{-0.7092} = 12.0 \times 0.4920 \approx 5.90\text{ V}\).

[1 mark] Correctly evaluates the time constant and exponential discharge formula to obtain \(5.90\text{ V}\).

The refractive index of the plastic is \(n = \frac{c}{v} = \frac{3.00 \times 10^8}{2.05 \times 10^8} \approx 1.463\). The wavelength of the light in the plastic is \(\lambda_n = \frac{\lambda_0}{n} = \frac{632.8\text{ nm}}{1.463} \approx 433\text{ nm}\).

[1 mark] Computes the refractive index and uses it to find the scaled wavelength of \(433\text{ nm}\).

First, find the time of flight using vertical motion: \(s_y = \frac{1}{2}gt^2 \implies 45.0 = 0.5 \times 9.81 \times t^2 \implies t^2 = 9.174 \implies t \approx 3.029\text{ s}\). Next, find the horizontal speed: \(s_x = v_x t \implies 60.0 = v_x \times 3.029 \implies v_x = 19.8\text{ m s}^{-1}\).

[1 mark] Determines the correct time of flight and horizontal speed of \(19.8\text{ m s}^{-1}\).

Total internal reflection occurs when light travels from a more optically dense medium to a less optically dense medium, and the angle of incidence exceeds the critical angle \(\theta_c\).

The critical angle is given by the formula:
\(\sin\theta_c = \frac{n_2}{n_1}\)

Substituting the given refractive indices:
\(\sin\theta_c = \frac{1.25}{1.50} = 0.8333\)

\(\theta_c = \arcsin(0.8333) = 56.4^\circ\)

1 mark for the correct calculation of the critical angle, leading to option C.

At the maximum height of its path, the vertical component of the projectile's velocity is zero (\(v_y = 0\)).

The horizontal component of the velocity remains constant throughout the flight because there is no horizontal acceleration:
\(v_x = u \cos\theta\)

The initial kinetic energy is:
\(E_k = \frac{1}{2} m u^2 = 240\text{ J}\)

The kinetic energy at the maximum height is:
\(E_{\text{top}} = \frac{1}{2} m v_x^2 = \frac{1}{2} m (u \cos\theta)^2 = \left(\frac{1}{2} m u^2\right) \cos^2\theta\)

Substituting the values:
\(E_{\text{top}} = 240 \times \cos^2(60.0^\circ) = 240 \times (0.500)^2 = 240 \times 0.250 = 60\text{ J}\)

1 mark for identifying that only the horizontal component of velocity contributes to kinetic energy at the peak and correctly calculating it to be 60 J (Option B).

At the top of the bridge, the forces acting on the car are its weight \(mg\) downwards and the normal contact force \(R\) upwards.

The resultant downward force provides the centripetal acceleration:
\(mg - R = \frac{m v^2}{r}\)

For the car to just lose contact with the road, the normal contact force \(R\) becomes zero:
\(mg = \frac{m v^2}{r}\)

Solving for the maximum speed \(v\):
\(v = \sqrt{gr}\)

Substituting the values:
\(v = \sqrt{9.81 \times 25} \approx 15.7\text{ m s}^{-1}\)

1 mark for equating the gravitational force to the required centripetal force and correctly solving for the maximum speed of 15.7 m s⁻¹ (Option C).

For a pipe closed at one end, the boundary conditions require a node at the closed end and an antinode at the open end. This means only odd harmonics can form.

The fundamental frequency (first harmonic) is:
\(f_1 = 180\text{ Hz}\)

The next possible harmonic is the third harmonic (\(n = 3\)):
\(f_3 = 3 f_1 = 3 \times 180\text{ Hz} = 540\text{ Hz}\)

Even harmonics (such as \(2 \times 180 = 360\text{ Hz}\)) cannot exist in a pipe closed at one end.

1 mark for identifying that a pipe closed at one end only supports odd harmonics and calculating the next harmonic frequency as 540 Hz (Option C).

Let \(m_1 = 3.0\text{ kg}\) (on the table) and \(m_2 = 2.0\text{ kg}\) (hanging).

Applying Newton's second law to each mass:
For the hanging mass \(m_2\):
\(m_2 g - T = m_2 a\)

For the mass on the table \(m_1\):
\(T = m_1 a\)

Adding these equations to find the acceleration \(a\):
\(m_2 g = (m_1 + m_2) a\)

\(a = \frac{m_2 g}{m_1 + m_2} = \frac{2.0 \times 9.81}{3.0 + 2.0} = 3.924\text{ m s}^{-2}\)

Now, substitute \(a\) back to find the tension \(T\):
\(T = m_1 a = 3.0 \times 3.924 = 11.77\text{ N} \approx 11.8\text{ N}\)

1 mark for setting up the equations of motion, finding the acceleration, and solving for the tension to obtain 11.8 N (Option B).

The discharging of a capacitor's voltage is given by the formula:
\(V = V_0 e^{-t/RC}\)

We are looking for the time \(t\) when \(V = 0.25 V_0\):
\(0.25 V_0 = V_0 e^{-t/RC}\)

\(0.25 = e^{-t/RC}\)

Taking the natural logarithm of both sides:
\(\ln(0.25) = -\frac{t}{RC} \implies t = RC \ln(4)\)

Calculate the time constant \(\tau = RC\):
\(RC = 47 \times 10^3\ \Omega \times 220 \times 10^{-6}\text{ F} = 10.34\text{ s}\)

Now, calculate \(t\):
\(t = 10.34 \times \ln(4) \approx 10.34 \times 1.3863 \approx 14.3\text{ s}\)

1 mark for calculating the correct time constant and using the exponential decay equation to find the time of 14.3 s (Option C).

(a) Using Snell's Law at the air-oil interface:
\(n_{\text{air}} \sin \theta_i = n_1 \sin \theta_{\text{oil}}\)
\(1.00 \times \sin(48.0^\circ) = 1.48 \times \sin \theta_{\text{oil}}\)
\(\sin \theta_{\text{oil}} = \frac{\sin(48.0^\circ)}{1.48} = \frac{0.7431}{1.48} = 0.5021\)
\ heta_{\text{oil}} = \arcsin(0.5021) = 30.14^\circ \approx 30.1^\circ\)

(b) The lateral displacement \(x\) of the beam within the oil slab of thickness \(d\) is given by:
\(x = d \frac{\sin(\theta_i - \theta_{\text{oil}})}{\cos \theta_{\text{oil}}}\)
\(\theta_i - \theta_{\text{oil}} = 48.0^\circ - 30.14^\circ = 17.86^\circ\)
\(x = 2.50\text{ cm} \times \frac{\sin(17.86^\circ)}{\cos(30.14^\circ)}
= 2.50 \times \frac{0.3067}{0.8648} = 2.50 \times 0.3547 = 0.887\text{ cm} = 8.87 \times 10^{-3}\text{ m}\)

(c) The maximum angle of incidence in air is \(\theta_i = 90.0^\circ\) (grazing incidence).
Under this condition, the maximum possible angle of refraction inside the oil layer is:
\(1.00 \times \sin(90.0^\circ) = 1.48 \times \sin \theta_{\text{oil, max}}\)
\(\sin \theta_{\text{oil, max}} = \frac{1}{1.48} = 0.6757 \implies \theta_{\text{oil, max}} = 42.51^\circ\)

The critical angle \(\theta_c\) at the oil-water interface is:
\(\sin \theta_c = \frac{n_2}{n_1} = \frac{1.33}{1.48} = 0.8986\)
\(\theta_c = \arcsin(0.8986) = 63.98^\circ \approx 64.0^\circ\)

Since \(\theta_{\text{oil, max}} < \theta_c\) (\(42.51^\circ < 63.98^\circ\)), total internal reflection cannot occur.

Part (a) [3.1 marks]:
- [1 mark] State or apply Snell's law: \(n_1 \sin \theta_1 = n_2 \sin \theta_2\).
- [1 mark] Correct substitution of values: \(1.00 \times \sin(48.0^\circ) = 1.48 \times \sin \theta_{\text{oil}}\).
- [1.1 marks] Correct final angle: \(30.1^\circ\) (accept \(30^\circ\) to \(30.14^\circ\)).

Part (b) [3.1 marks]:
- [1 mark] State or apply formula for lateral displacement: \(x = d \frac{\sin(\theta_i - \theta_{\text{oil}})}{\cos \theta_{\text{oil}}}\) (or alternative geometric breakdown).
- [1 mark] Substitution of values: \(2.50 \times \frac{\sin(17.86^\circ)}{\cos(30.14^\circ)}\).
- [1.1 marks] Correct evaluation: \(0.887\text{ cm}\) or \(8.87\text{ mm}\) (allow ecf from (a)).

Part (c) [3.1 marks]:
- [1 mark] Correct calculation of the maximum angle of refraction in oil (\(42.5^\circ\)).
- [1 mark] Correct calculation of the critical angle at the oil-water interface (\(64.0^\circ\)).
- [1.1 marks] Correct comparison and concluding statement that TIR is not possible.

(a) Resolving vertically with upwards as the positive direction:
\(u_y = u \sin \theta = 15.0 \sin(35.0^\circ) = 8.604\text{ m s}^{-1}\)
\(a_y = -g = -9.81\text{ m s}^{-2}\)
\(s_y = -18.0\text{ m}\)

Using the kinematic equation:
\(s_y = u_y t + \frac{1}{2} a_y t^2\)
\(-18.0 = 8.604 t - 4.905 t^2\)
\(4.905 t^2 - 8.604 t - 18.0 = 0\)

Solving using the quadratic formula:
\(t = \frac{-(-8.604) \pm \sqrt{(-8.604)^2 - 4 \times 4.905 \times (-18.0)}}{2 \times 4.905}\)
\(t = \frac{8.604 \pm \sqrt{74.03 + 353.16}}{9.81} = \frac{8.604 \pm \sqrt{427.19}}{9.81} = \frac{8.604 \pm 20.669}{9.81}\)
Taking the positive root:
\(t = \frac{29.273}{9.81} = 2.984\text{ s} \approx 2.98\text{ s}\)

(b) Resolving horizontally:
\(u_x = u \cos \theta = 15.0 \cos(35.0^\circ) = 12.29\text{ m s}^{-1}\)
\(R = u_x \times t = 12.29 \times 2.984 = 36.67\text{ m} \approx 36.7\text{ m}\)

(c) The horizontal velocity remains constant:
\(v_x = 12.29\text{ m s}^{-1}\)
The vertical velocity just before impact:
\(v_y = u_y - g t = 8.604 - 9.81 \times 2.984 = 8.604 - 29.273 = -20.67\text{ m s}^{-1}\)

The magnitude of the velocity is:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12.29^2 + (-20.67)^2} = \sqrt{151.04 + 427.25} = \sqrt{578.29} = 24.05\text{ m s}^{-1} \approx 24.1\text{ m s}^{-1}\)
(Alternatively, using conservation of energy: \(\frac{1}{2} m u^2 + m g h = \frac{1}{2} m v^2 \implies v = \sqrt{u^2 + 2 g h} = \sqrt{15.0^2 + 2 \times 9.81 \times 18.0} = \sqrt{225 + 353.16} = \sqrt{578.16} = 24.04\text{ m s}^{-1}\).)

Part (a) [3.1 marks]:
- [1 mark] Correct vertical component of initial velocity \(u_y = 8.60\text{ m s}^{-1}\).
- [1 mark] Setting up the correct quadratic equation: \(4.905 t^2 - 8.604 t - 18.0 = 0\).
- [1.1 marks] Finding the correct positive time: \(2.98\text{ s}\) (accept \(3.0\text{ s}\)).

Part (b) [3.1 marks]:
- [1 mark] Correct horizontal component of velocity \(u_x = 12.3\text{ m s}^{-1}\).
- [1 mark] State or apply \(R = u_x t\).
- [1.1 marks] Correct final distance: \(36.7\text{ m}\) (accept range \(36.5\text{ m}\) to \(37.0\text{ m}\) based on rounding).

Part (c) [3.1 marks]:
- [1 mark] Correctly stated or calculated component velocities at landing, or statement of conservation of energy principle.
- [1 mark] Correct substitution into energy conservation or Pythagoras formula.
- [1.1 marks] Correct final velocity magnitude: \(24.0\text{ m s}^{-1}\) to \(24.1\text{ m s}^{-1}\).

(a) Since there is no vertical motion, the vertical component of the normal reaction force balances the weight:
\(N \cos \theta = m g\)
\(N \cos(22.0^\circ) = 0.250 \times 9.81 = 2.4525\text{ N}\)
\(N = \frac{2.4525}{\cos(22.0^\circ)} = \frac{2.4525}{0.92718} = 2.645\text{ N} \approx 2.65\text{ N}\)

(b) The horizontal component of the normal force provides the centripetal force:
\(N \sin \theta = \frac{m v^2}{r}\)
Substituting \(N = \frac{mg}{\cos \theta}\) into the equation:
\(m g \tan \theta = \frac{m v^2}{r} \implies v = \sqrt{r g \tan \theta}\)
\(v = \sqrt{1.80 \times 9.81 \times \tan(22.0^\circ)}\)
\(v = \sqrt{17.658 \times 0.4040} = \sqrt{7.134} = 2.671\text{ m s}^{-1} \approx 2.67\text{ m s}^{-1}\)

(c) The relationship between angular speed \(\omega\) and linear speed \(v\) is:
\(\omega = \frac{v}{r}\)
\(\omega = \frac{2.671}{1.80} = 1.484\text{ rad s}^{-1} \approx 1.48\text{ rad s}^{-1}\)

Part (a) [3.1 marks]:
- [1 mark] Resolving forces vertically to obtain \(N \cos \theta = mg\).
- [1 mark] Substitution of values: \(0.250 \times 9.81 / \cos(22.0^\circ)\).
- [1.1 marks] Correct evaluation of \(N = 2.65\text{ N}\).

Part (b) [3.1 marks]:
- [1 mark] State centripetal force relation: \(N \sin \theta = \frac{mv^2}{r}\).
- [1 mark] Derive or state the expression \(v = \sqrt{rg \tan \theta}\).
- [1.1 marks] Correct evaluation of speed: \(2.67\text{ m s}^{-1}\) (allow ecf from (a)).

Part (c) [3.1 marks]:
- [1 mark] State expression relating angular and linear speed: \(\omega = \frac{v}{r}\) or \(\omega = \sqrt{\frac{g \tan \theta}{r}}\).
- [1 mark] Substitution: \(2.671 / 1.80\).
- [1.1 marks] Correct final angular speed: \(1.48\text{ rad s}^{-1}\) (allow ecf from (b)).

(a) The mass \(m\) of the wire of length \(L\) and cross-sectional area \(A\) is given by:
\(m = \rho V = \rho A L\)
The mass per unit length \(\mu\) is:
\(\mu = \frac{m}{L} = \rho A\)
\(\mu = 7850\text{ kg m}^{-3} \times 4.50 \times 10^{-7}\text{ m}^2 = 3.5325 \times 10^{-3}\text{ kg m}^{-1} \approx 3.53 \times 10^{-3}\text{ kg m}^{-1}\)

(b) The speed \(v\) of transverse waves on the wire is given by:
\(v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{120}{3.5325 \times 10^{-3}}} = \sqrt{33970} = 184.31\text{ m s}^{-1}\)

For the fundamental frequency (1st harmonic) on a fixed string:
\(\lambda_1 = 2L = 2 \times 0.850 = 1.70\text{ m}\)
\(f_1 = \frac{v}{\lambda_1} = \frac{184.31}{1.70} = 108.42\text{ Hz} \approx 108\text{ Hz}\)

(c) For the third harmonic:
\(f_3 = 3 \times f_1 = 3 \times 108.42 = 325.26\text{ Hz} \approx 325\text{ Hz}\)

The distance between adjacent nodes is half a wavelength:
\(d_{\text{node}} = \frac{\lambda_3}{2}\)
Since \(\lambda_3 = \frac{2L}{3} = \frac{2 \times 0.850}{3} = 0.5667\text{ m}\),
\(d_{\text{node}} = \frac{0.5667}{2} = 0.283\text{ m}\).

Part (a) [3.1 marks]:
- [1 mark] State the formula \(\mu = \rho A\).
- [1 mark] Substitution: \(7850 \times 4.50 \times 10^{-7}\).
- [1.1 marks] Correct value: \(3.53 \times 10^{-3}\text{ kg m}^{-1}\).

Part (b) [3.1 marks]:
- [1 mark] Calculate the wave speed \(v = 184\text{ m s}^{-1}\) using \(v = \sqrt{\frac{T}{\mu}}\).
- [1 mark] State the fundamental wavelength relationship: \(\lambda = 2L\).
- [1.1 marks] Correct fundamental frequency: \(108\text{ Hz}\) (accept range \(108\text{ Hz}\) to \(109\text{ Hz}\)).

Part (c) [3.1 marks]:
- [1 mark] Correctly calculate third harmonic frequency: \(325\text{ Hz}\) (allow ecf from (b)).
- [1 mark] Express node spacing as half of the wavelength: \(d = \frac{\lambda}{2}\).
- [1.1 marks] Correct node-to-node distance: \(0.283\text{ m}\).

(a) Resolving perpendicular to the incline:
\(N = m g \cos \theta\)
\(N = 45.0 \times 9.81 \times \cos(25.0^\circ)\)
\(N = 441.45 \times 0.9063 = 400.09\text{ N} \approx 400\text{ N}\)

(b) The friction force is given by:
\(F_f = \mu_k N\)
\(F_f = 0.350 \times 400.09 = 140.03\text{ N} \approx 140\text{ N}\)

(c) The net force parallel to the incline is:
\(F_{\text{net}} = P - m g \sin \theta - F_f\)
First, calculate the component of gravity parallel to the incline:
\(F_{\parallel} = m g \sin \theta = 45.0 \times 9.81 \times \sin(25.0^\circ) = 441.45 \times 0.4226 = 186.56\text{ N}\)

Now, substitute the forces:
\(F_{\text{net}} = 380 - 186.56 - 140.03 = 53.41\text{ N}\)

According to Newton's Second Law:
\(a = \frac{F_{\text{net}}}{m} = \frac{53.41}{45.0} = 1.187\text{ m s}^{-2} \approx 1.19\text{ m s}^{-2}\)

Part (a) [3.1 marks]:
- [1 mark] State expression: \(N = mg \cos\theta\).
- [1 mark] Substitution: \(45.0 \times 9.81 \times \cos(25.0^\circ)\).
- [1.1 marks] Correct normal force: \(400\text{ N}\) (accept range \(400\text{ N}\) to \(400.1\text{ N}\)).

Part (b) [3.1 marks]:
- [1 mark] State definition of kinetic friction: \(F_f = \mu_k N\).
- [1 mark] Substitution: \(0.350 \times 400\).
- [1.1 marks] Correct frictional force: \(140\text{ N}\) (allow ecf from (a)).

Part (c) [3.1 marks]:
- [1 mark] Set up equations of motion: \(P - mg \sin\theta - F_f = ma\).
- [1 mark] Correctly determine gravity component parallel to incline: \(187\text{ N}\).
- [1.1 marks] Correct acceleration: \(1.19\text{ m s}^{-2}\) (allow range \(1.18\text{ m s}^{-2}\) to \(1.20\text{ m s}^{-2}\)).

(a) The time constant \(\tau\) is:
\(\tau = R C = 15.0 \times 10^3\ \Omega \times 470 \times 10^{-6}\text{ F} = 7.05\text{ s}\)

(b) The energy stored at \(t = 0\) is:
\(E_0 = \frac{1}{2} C V_0^2\)
\(E_0 = 0.5 \times 470 \times 10^{-6}\text{ F} \times (12.0\text{ V})^2 = 2.35 \times 10^{-4} \times 144 = 0.03384\text{ J} = 3.38 \times 10^{-2}\text{ J}\ (33.8\text{ mJ})\)

(c) The voltage across the capacitor during discharge decreases according to:
\(V = V_0 e^{-t/\tau}\)
\(2.50 = 12.0 e^{-t/7.05}\)
\(e^{-t/7.05} = \frac{2.50}{12.0} = 0.2083\)

Taking the natural logarithm of both sides:
\(-\frac{t}{7.05} = \ln(0.2083) = -1.5686\)
\(t = 1.5686 \times 7.05 = 11.06\text{ s} \approx 11.1\text{ s}\)

Part (a) [3.1 marks]:
- [1 mark] State formula: \(\tau = RC\).
- [1 mark] Correct unit conversion and substitution: \(15.0 \times 10^3 \times 470 \times 10^{-6}\).
- [1.1 marks] Correct time constant: \(7.05\text{ s}\).

Part (b) [3.1 marks]:
- [1 mark] State energy formula: \(E = \frac{1}{2}CV^2\).
- [1 mark] Substitution of values: \(0.5 \times 470 \times 10^{-6} \times 12.0^2\).
- [1.1 marks] Correct energy value with SI units: \(3.38 \times 10^{-2}\text{ J}\) or \(33.8\text{ mJ}\).

Part (c) [3.1 marks]:
- [1 mark] State or apply exponential discharge formula: \(V = V_0 e^{-t/\tau}\).
- [1 mark] Correct algebraic manipulation involving taking logarithms.
- [1.1 marks] Correct evaluation of time: \(11.1\text{ s}\) (allow range \(11.0\text{ s}\) to \(11.1\text{ s}\)).

(a) First, calculate the cross-sectional area \(A\) of the wire:
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.380 \times 10^{-3}\text{ m})^2}{4} = 1.1341 \times 10^{-7}\text{ m}^2\)

Now, calculate the resistivity \(\rho\):
\(R = \rho \frac{L}{A} \implies \rho = \frac{R A}{L}\)
\(\rho = \frac{5.20 \times 1.1341 \times 10^{-7}}{2.40} = 2.4572 \times 10^{-7}\ \Omega\text{ m} \approx 2.46 \times 10^{-7}\ \Omega\text{ m}\n
(b) Using Ohm's Law:
\)I = \frac{V}{R} = \frac{3.00}{5.20} = 0.5769\text{ A} \approx 0.577\text{ A}\)

(c) Current is the rate of flow of charge:
\(I = \frac{Q}{t} = \frac{N e}{t}\)
Let \(n = \frac{N}{t}\) be the number of electrons per second:
\(n = \frac{I}{e} = \frac{0.5769}{1.60 \times 10^{-19}} = 3.606 \times 10^{18}\text{ s}^{-1} \approx 3.61 \times 10^{18}\text{ s}^{-1}\)

Part (a) [3.1 marks]:
- [1 mark] Correctly calculate area \(A = 1.13 \times 10^{-7}\text{ m}^2\).
- [1 mark] State resistivity formula \(\rho = \frac{RA}{L}\).
- [1.1 marks] Correct final resistivity value with correct units: \(2.46 \times 10^{-7}\ \Omega\text{ m}\).

Part (b) [3.1 marks]:
- [1 mark] State Ohm's law relation: \(I = \frac{V}{R}\).
- [1 mark] Substitution of values: \(3.00 / 5.20\).
- [1.1 marks] Correct current: \(0.577\text{ A}\).

Part (c) [3.1 marks]:
- [1 mark] State the definition of current: \(I = \frac{\Delta Q}{\Delta t}\).
- [1 mark] Relate total charge to number of electrons: \(Q = Ne\).
- [1.1 marks] Correct rate of electrons: \(3.61 \times 10^{18}\text{ s}^{-1}\) (allow ecf from (b)).

The relationship between the critical angle \(\theta_c\) and the refractive indices of the two media is given by \(\sin(\theta_c) = \frac{n_2}{n_1}\), where \(n_1\) is the refractive index of the denser medium (glass, 1.52) and \(n_2\) is the refractive index of the less dense medium (liquid). Rearranging the formula: \(n_2 = n_1 \sin(\theta_c) = 1.52 \times \sin(58.0^\circ) = 1.52 \times 0.8480 = 1.29\).

1 mark for the correct calculation of the refractive index of the liquid, matching option A.

The maximum height \(H\) achieved by the projectile is given by \(H = \frac{u^2 \sin^2\theta}{2g}\). The horizontal range \(R\) is given by \(R = \frac{u^2 \sin(2\theta)}{g} = \frac{2 u^2 \sin\theta \cos\theta}{g}\). We are given that \(R = 3H\). Substituting the equations: \(\frac{2 u^2 \sin\theta \cos\theta}{g} = 3 \left(\frac{u^2 \sin^2\theta}{2g}\right)\). Cancelling out \(u^2 / g\) and dividing both sides by \(\sin\theta \cos\theta\) gives \(2 = 1.5 \tan\theta\). Therefore, \(\tan\theta = \frac{2}{1.5} = 1.333\), which yields \(\theta = \arctan(1.333) = 53.1^\circ\).

1 mark for the correct angle calculated from equating the range and three times the height, matching option C.

The maximum centripetal force is provided by the maximum frictional force, so \(F_{\text{max}} = \frac{m v^2}{r}\). Rearranging to solve for the maximum speed \(v\): \(v = \sqrt{\frac{F_{\text{max}} \times r}{m}} = \sqrt{\frac{1.6 \times 0.80}{0.20}} = \sqrt{6.4} = 2.53\text{ m s}^{-1} \approx 2.5\text{ m s}^{-1}\).

1 mark for the correct calculation of maximum speed using the centripetal force equation, matching option B.

For a stretched string fixed at both ends, the frequency of the \(n\)-th harmonic is given by \(f_n = n f_1\), where \(f_1\) is the fundamental frequency (first harmonic). We are given \(f_3 = 240\text{ Hz}\), so \(3 f_1 = 240\text{ Hz} \implies f_1 = 80\text{ Hz}\). The frequency of the fifth harmonic is therefore \(f_5 = 5 f_1 = 5 \times 80\text{ Hz} = 400\text{ Hz}\).

1 mark for identifying the fundamental frequency and calculating the fifth harmonic frequency, matching option B.

According to Newton's second law, the net force along the inclined plane is \(F_{\text{net}} = F - m g \sin\theta = m a\). Rearranging for \(F\) gives \(F = m a + m g \sin\theta = m(a + g \sin\theta)\). Substituting the values: \(F = 4.0 \times (2.0 + 9.81 \times \sin(30.0^\circ)) = 4.0 \times (2.0 + 4.905) = 4.0 \times 6.905 = 27.62\text{ N} \approx 27.6\text{ N}\).

1 mark for setting up the equation of motion along the plane and calculating the force, matching option C.

The electrical energy stored in a capacitor is given by \(E = \frac{1}{2} C V^2\). Since the potential difference during discharge is \(V = V_0 e^{-t/\tau}\), the energy at time \(t\) is \(E = \frac{1}{2} C (V_0 e^{-t/\tau})^2 = E_0 e^{-2t/\tau}\), where \(E_0\) is the initial energy. We require \(E = 0.10 E_0\), so \(0.10 E_0 = E_0 e^{-2t/\tau} \implies e^{-2t/\tau} = 0.10\). Taking the natural logarithm of both sides: \(-2t/\tau = \ln(0.10) \approx -2.303 \implies 2t = 2.303 \tau \implies t \approx 1.15 \tau\).

1 mark for relating energy decay to the time constant and solving for time, matching option C.

Using Snell's Law: \(n_{\text{air}} \sin(\theta_{\text{air}}) = n_{\text{block}} \sin(\theta_{\text{block}})\). Given \(n_{\text{air}} = 1.00\), we find \(n_{\text{block}} = \frac{\sin(45^\circ)}{\sin(28^\circ)} = \frac{0.7071}{0.4695} \approx 1.506\). The speed of light in the block \(v\) is given by \(v = \frac{c}{n_{\text{block}}} = \frac{3.00 \times 10^8}{1.506} \approx 1.99 \times 10^8\text{ m s}^{-1}\).

1 mark for calculating the refractive index first and then using it to determine the speed of light, matching option B.

Resolving the forces acting on the bob vertically: since there is no vertical acceleration, the vertical component of the tension must balance the weight of the bob. This is represented by \(T \cos\theta = m g\). Solving for tension \(T\) gives \(T = \frac{m g}{\cos\theta}\).

1 mark for resolving the vertical forces and identifying the correct expression for tension, matching option B.

Using Snell's Law at the boundary:

\(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\)

Substitute the given values:

\(1.45 \sin(32.0^\circ) = 1.62 \sin(\theta_2)\)

\(\sin(\theta_2) = \frac{1.45 \times 0.5299}{1.62} \approx 0.4743\)

\(\theta_2 = \arcsin(0.4743) \approx 28.3^\circ\)

For a ray travelling from glass to liquid, the critical angle \(\theta_c\) is defined when the angle of refraction in the liquid is \(90^\circ\):

\(n_2 \sin(\theta_c) = n_1 \sin(90^\circ)\)

\(\sin(\theta_c) = \frac{n_1}{n_2} = \frac{1.45}{1.62} \approx 0.8951\)

\(\theta_c = \arcsin(0.8951) \approx 63.5^\circ\).

Award 1 mark for the correct answer.
- Incorrect choices: B, C, D.

First, resolve the initial velocity into vertical and horizontal components:

\(u_y = 24 \sin(35^\circ) \approx 13.77\text{ m s}^{-1}\)

\(u_x = 24 \cos(35^\circ) \approx 19.66\text{ m s}^{-1}\)

Next, calculate the total time of flight \(t\) by setting the vertical displacement \(s_y = 0\):

\(s_y = u_y t - \frac{1}{2} g t^2 = 0\)

\(t = \frac{2 u_y}{g} = \frac{2 \times 13.77}{9.81} \approx 2.81\text{ s}\)

Finally, calculate the horizontal range \(R\):

\(R = u_x t = 19.66 \times 2.81 \approx 55.2\text{ m}\), which rounds to \(55\text{ m}\).

Award 1 mark for the correct answer.
- Incorrect choices: A, B, D.

For the rider not to slide, the frictional force \(f\) must balance the weight of the rider:

\(f = mg\)

The maximum force of static friction is given by:

\(f_{\text{max}} = \mu_s N\)

where \(N\) is the normal force. In this case, the normal force provides the required centripetal force:

\(N = m \omega^2 r\)

To prevent sliding, we require:

\(mg \le \mu_s N \implies mg \le \mu_s m \omega^2 r\)

Cancelling mass \(m\) and solving for \(\omega\):

\(\omega^2 \ge \frac{g}{\mu_s r}\)

\(\omega_{\text{min}} = \sqrt{\frac{9.81}{0.40 \times 3.5}} = \sqrt{\frac{9.81}{1.40}} \approx 2.65\text{ rad s}^{-1}\).

This rounds to \(2.6\text{ rad s}^{-1}\).

Award 1 mark for the correct answer.
- Incorrect choices: A, C, D.

For a tube closed at one end and open at the other, the boundary conditions require a node at the closed end and an antinode at the open end. The possible wavelengths \(\lambda\) are given by:

\(L = (2n - 1) \frac{\lambda}{4}\) where \(n = 1, 2, 3, \dots\)

The resonant frequencies are therefore:

\(f = \frac{v}{\lambda} = \frac{(2n - 1) v}{4 L}\)

Calculate the fundamental frequency (for \(n = 1\)):

\(f_1 = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100\text{ Hz}\)

The permitted harmonic frequencies are odd multiples of the fundamental frequency:

\(f_n = (2n - 1) \times 100\text{ Hz}\)

This yields allowed frequencies of \(100\text{ Hz}, 300\text{ Hz}, 500\text{ Hz}, 700\text{ Hz}, \dots\)

Thus, \(200\text{ Hz}\) is not a possible resonant frequency.

Award 1 mark for the correct answer.
- Incorrect choices: A, C, D.

Resolve the forces acting parallel to the inclined plane. The component of the weight pulling the block down the slope is:

\(W_{\parallel} = m g \sin(30^\circ)\)

The frictional force \(F_f\) acts up the slope opposing motion. Applying Newton's second law along the incline:

\(F_{\text{net}} = m g \sin(30^\circ) - F_f = m a\)

Rearranging to solve for the frictional force \(F_f\):

\(F_f = m g \sin(30^\circ) - m a\)

\(F_f = (4.0 \times 9.81 \times 0.5) - (4.0 \times 1.5)\)

\(F_f = 19.62 - 6.0 = 13.62\text{ N}\), which rounds to \(13.6\text{ N}\).

Award 1 mark for the correct answer.
- Incorrect choices: A, C, D.

The initial energy stored in the fully charged capacitor is given by:

\(E_0 = \frac{1}{2} C V_0^2 = \frac{1}{2} \times (470 \times 10^{-6}\text{ F}) \times (12\text{ V})^2 = 0.03384\text{ J} = 33.84\text{ mJ}\)

During discharging, the potential difference across the capacitor after time \(t\) is:

\(V = V_0 e^{-t/RC}\)

The energy stored in the capacitor at time \(t\) is proportional to the square of the potential difference:

\(E = \frac{1}{2} C V^2 = \frac{1}{2} C \left(V_0 e^{-t/RC}\right)^2 = E_0 e^{-2t/RC}\)

After one time constant (\(t = RC\)):

\(E = E_0 e^{-2}\)

\(E = 33.84\text{ mJ} \times e^{-2} \approx 33.84\text{ mJ} \times 0.1353 \approx 4.58\text{ mJ}\), which rounds to \(4.6\text{ mJ}\).

Award 1 mark for the correct answer.
- Incorrect choices: B, C, D.

The pump must do work to increase both the gravitational potential energy and the kinetic energy of the water.

In each second (\(\Delta t = 1.0\text{ s}\)), the mass of water moved is \(m = 15\text{ kg}\).

1. Gravitational potential energy gained per second:

\(\frac{\Delta E_p}{\Delta t} = \frac{m g h}{\Delta t} = 15 \times 9.81 \times 6.5 = 956.5\text{ W}\)

2. Kinetic energy gained per second:

\(\frac{\Delta E_k}{\Delta t} = \frac{\frac{1}{2} m v^2}{\Delta t} = 0.5 \times 15 \times (8.0)^2 = 480.0\text{ W}\)

3. Minimum total power output:

\(P = 956.5\text{ W} + 480.0\text{ W} = 1436.5\text{ W} \approx 1.44\text{ kW}\).

Award 1 mark for the correct answer.
- Incorrect choices: A, B, D.