2023 HKDSE Physics Answers & Marking Scheme

During the melting process, the heat absorbed is \(Q_1 = P t_1 = m l_f\), which gives \(l_f = \frac{P t_1}{m}\).
During the temperature rise in the liquid state, the heat absorbed is \(Q_2 = P t_2 = m c_l \Delta T\), which gives \(c_l = \frac{P t_2}{m \Delta T}\).
Therefore, the ratio is:
\[\frac{l_f}{c_l} = \frac{P t_1 / m}{P t_2 / (m \Delta T)} = \frac{t_1 \Delta T}{t_2}\]

Award 1 mark for the correct answer B. Correct formula application is required.

First, convert temperatures to Kelvin:
\(T_1 = 27 + 273 = 300\text{ K}\)
\(T_2 = 327 + 273 = 600\text{ K}\)
Since the absolute temperature is doubled:
(1) Average kinetic energy \(E_k = \frac{3}{2} k_B T\) is proportional to \(T\). Thus, it is doubled. (Correct)
(2) Root-mean-square speed \(v_{rms} = \sqrt{\frac{3 k_B T}{m}}\) is proportional to \(\sqrt{T}\). Thus, it increases by a factor of \(\sqrt{2}\), not doubled. (Incorrect)
(3) Collision frequency is proportional to the average molecular speed since concentration is constant. It increases by a factor of \(\sqrt{2}\), not doubled. (Incorrect)
Hence, only (1) is correct.

Award 1 mark for the correct answer A. Correct conversion of Celsius to Kelvin is key.

Let \(u\) be the initial speed of the car at \(t=0\), and \(a\) be its deceleration. Since it stops at \(t=2.5\text{ s}\):
\(v(2.5) = u - 2.5a = 0 \implies u = 2.5a\).
The position of the car at any time \(t\) is given by:
\(s(t) = u t - \frac{1}{2} a t^2 = 2.5a t - 0.5a t^2\).
For the first second:
\(d_1 = s(1) - s(0) = 2.5a(1) - 0.5a(1)^2 = 2a\).
For the second second:
\(d_2 = s(2) - s(1) = [2.5a(2) - 0.5a(2)^2] - 2a = [5a - 2a] - 2a = 3a - 2a = a\).
Thus, the ratio \(d_1 / d_2 = 2a / a = 2.0\).

Award 1 mark for the correct answer B. Consistent application of kinematic equations is required.

Let the initial direction of block \(A\) be positive.
Initial momentum: \(P_i = m u\).
After the collision, the velocity of block \(A\) is \(v_A = -\frac{1}{5}u\).
Let \(v_B\) be the final velocity of block \(B\).
By conservation of linear momentum:
\[m u = m\left(-\frac{1}{5}u\right) + 3m v_B \implies \frac{6}{5}m u = 3m v_B \implies v_B = \frac{2}{5}u\]
The total kinetic energy after collision is:
\[K_f = \frac{1}{2}m v_A^2 + \frac{1}{2}(3m) v_B^2 = \frac{1}{2}m \left(-\frac{1}{5}u\right)^2 + \frac{3}{2}m \left(\frac{2}{5}u\right)^2 = \frac{1}{50}mu^2 + \frac{12}{50}mu^2 = \frac{13}{50}mu^2 = 0.26 mu^2\]

Award 1 mark for the correct answer B. Must apply conservation of momentum first to find the velocity of block B.

The mechanical energy lost due to friction is:
\(W_{\text{lost}} = m g L \sin\theta - \frac{1}{2} m v^2\).
Since acceleration is constant, the average speed of the block is \(v_{\text{avg}} = \frac{v}{2}\).
The time taken to slide down is:
\(t = \frac{L}{v_{\text{avg}}} = \frac{2L}{v}\).
The average power of friction is:
\(P_{\text{avg}} = \frac{W_{\text{lost}}}{t} = \frac{m g L \sin\theta - \frac{1}{2} m v^2}{2L / v} = \frac{v}{2L} \left(m g L \sin\theta - \frac{1}{2} m v^2\right)\).

Award 1 mark for the correct answer B. Connection between work done and average power must be applied correctly.

The critical angle \(\theta_c\) is related to the refractive indices of the two media by:
\(\sin \theta_c = \frac{n_y}{n_x}\).
Since the speed of light in a medium is inversely proportional to its refractive index, we have:
\(n_x = \frac{c}{v}\) and \(n_y = \frac{c}{v_y}\).
Substituting these into the critical angle formula:
\(\sin \theta_c = \frac{c/v_y}{c/v} = \frac{v}{v_y} \implies v_y = \frac{v}{\sin \theta_c}\).

Award 1 mark for the correct answer B. Speed of light relation to refractive index is required.

The original frequency is \(f\), and the wavelength is \(\lambda = v/f\). The path difference at \(P\) is \(\Delta = 2\lambda\).
When the frequency is increased by \(25\%\), the new frequency is \(f' = 1.25f = 1.25 \times \frac{v}{\lambda}\).
The new wavelength is \(\lambda' = \frac{v}{f'} = \frac{v}{1.25 f} = \frac{\lambda}{1.25} = 0.8\lambda\).
Thus, the path difference expressed in terms of the new wavelength is:
\(\Delta = 2\lambda = 2(1.25 \lambda') = 2.5 \lambda'\).
Since the path difference is an odd-half-integer multiple of the new wavelength, destructive interference occurs.

Award 1 mark for the correct answer B. Consistent reasoning about path difference and wavelength is required.

Using the relation \(\mathcal{E} = I(R+r)\):
For the first case: \(\mathcal{E} = 1.2(2 + r)\) --- (1)
For the second case: \(\mathcal{E} = 0.6(5 + r)\) --- (2)
Equating (1) and (2):
\(1.2(2 + r) = 0.6(5 + r)\)
\(2(2 + r) = 5 + r \implies 4 + 2r = 5 + r \implies r = 1.0\,\Omega\).
Substitute \(r = 1.0\,\Omega\) into (1):
\(\mathcal{E} = 1.2(2 + 1) = 3.6\text{ V}\).

Award 1 mark for the correct answer B. Solving the linear system of equations is required.

When the ring enters the downward magnetic field, the downward magnetic flux increases. By Lenz's law, the induced current must oppose this increase, generating an upward magnetic field. By the right-hand grip rule, the current flows anti-clockwise when viewed from above.
When the ring leaves the magnetic field, the downward magnetic flux decreases. The induced current must oppose this decrease, generating a downward magnetic field. By the right-hand grip rule, the current flows clockwise when viewed from above.

Award 1 mark for the correct answer B. Consistent application of Lenz's law and right-hand grip rule is required.

For the net electric field to be zero, the individual electric fields from \(+Q\) and \(-4Q\) must be equal in magnitude and opposite in direction. This can only occur in the region closer to the smaller charge in magnitude, i.e., to the left of \(A\) (on the side opposite to \(B\)).
Let this point be at distance \(x\) from \(A\).
Setting the field magnitudes equal:
\[\frac{k Q}{x^2} = \frac{k (4Q)}{(d + x)^2}\]
Taking the square root on both sides:
\[\frac{1}{x} = \frac{2}{d + x} \implies d + x = 2x \implies x = d\]
Thus, the position is at a distance \(d\) from \(A\) on the side opposite to \(B\).

Award 1 mark for the correct answer A. Application of Coulomb's Law and electric field superposition is required.

The melting process takes place during the constant temperature period, which lasts for \(5\text{ minutes} = 300\text{ s}\). The energy supplied during this period is \(Q = P \Delta t = 50 \times 300 = 15000\text{ J}\). Using \(Q = m L_f\), we have \(15000 = 0.2 \times L_f\), which gives \(L_f = 75000\text{ J kg}^{-1} = 75\text{ kJ kg}^{-1}\).

Award 1 mark for the correct option C. No marks for other options.

Let \(a\) be the acceleration during the first \(4\text{ s}\). Since the graph is a parabola, acceleration is uniform. The displacement in the first \(4\text{ s}\) is \(s_1 = \frac{1}{2} a (4^2) = 8a\). The velocity at \(t = 4\text{ s}\) is \(v = a (4) = 4a\). For the remaining \(2\text{ s}\) (from \(t = 4\text{ s}\) to \(6\text{ s}\)), the car moves at constant speed \(v\), so the displacement is \(s_2 = v \times 2 = 8a\). The total distance is \(s_1 + s_2 = 16a = 48\text{ m}\), which gives \(a = 3.0\text{ m s}^{-2}\).

Award 1 mark for the correct option C. No marks for other options.

By conservation of momentum: \(m v = m (-v/5) + 3m v_B \Rightarrow v_B = \frac{2}{5} v\). Initial kinetic energy \(K_i = \frac{1}{2} m v^2\). Final kinetic energy \(K_f = \frac{1}{2} m (-v/5)^2 + \frac{1}{2} (3m) (2/5 v)^2 = \frac{1}{2} m v^2 [ \frac{1}{25} + \frac{12}{25} ] = \frac{13}{25} K_i\). The fraction of kinetic energy lost is \(1 - \frac{13}{25} = \frac{12}{25}\).

Award 1 mark for the correct option A. No marks for other options.

The angle of reflection is \(\theta\). Let \(r\) be the angle of refraction in air. Since the reflected and refracted rays are perpendicular, we have \(\theta + 90^\circ + r = 180^\circ \Rightarrow r = 90^\circ - \theta\). By Snell's Law: \(n \sin \theta = 1 \cdot \sin r = \sin(90^\circ - \theta) = \cos \theta\). Thus, \\tan \theta = \frac{1}{n} = \frac{1}{1.5} = \frac{2}{3}\, giving \(\theta = \arctan(2/3) \approx 33.7^\circ\).

Award 1 mark for the correct option A. No marks for other options.

The initial fundamental frequency is \(f_1 = \frac{240}{3} = 80\text{ Hz}\). The wave speed is given by \(v = \sqrt{T/\mu}\), where \(T\) is tension and \(\mu\) is linear density. When tension is quadrupled, the wave speed doubles. Since the fundamental frequency is \(f_1 = \frac{v}{2L}\), doubling the wave speed doubles the fundamental frequency, making the new fundamental frequency \(2 \times 80 = 160\text{ Hz}\).

Award 1 mark for the correct option B. No marks for other options.

The initial force magnitude is \(F \propto |q_1 q_2| = 6 \times 2 = 12\text{ units}\). When brought into contact, the total charge of \(+4\mu\text{C}\) is shared equally, so each sphere gets \(+2\mu\text{C}\). The new force magnitude is \(F' \propto |q_1' q_2'| = 2 \times 2 = 4\text{ units}\). Hence, \(F' = \frac{4}{12} F = \frac{1}{3} F\). Since both spheres now carry positive charge, the force is repulsive.

Award 1 mark for the correct option A. No marks for other options.

As \(R\) increases, the total resistance of the circuit increases, so the current \(I = \frac{\varepsilon}{R+r}\) decreases. The terminal potential difference is \(V = \varepsilon - I r\). Since \(I\) decreases, \(V\) increases. The power dissipated in the internal resistance is \(P_{\text{int}} = I^2 r\). Since \(I\) decreases, \(P_{\text{int}}\) decreases.

Award 1 mark for the correct option B. No marks for other options.

Since the block moves at a constant velocity, its acceleration is zero, which means the net force acting on the block is zero. The forces acting on the block are gravity (\(mg\) downwards) and the force from the plane (which consists of the normal force \(N\) and friction \(f\)). For the total force to be zero, the force exerted by the plane must exactly balance the gravitational force. Therefore, the magnitude of the net force from the plane is \(mg\).

Award 1 mark for the correct option C. No marks for other options.

As the loop is pulled out, the magnetic flux decreases, inducing an emf \(\varepsilon = B L v\). The induced current is \(I = \frac{\varepsilon}{R} = \frac{B L v}{R}\). The magnetic force opposing the motion of the loop is \(F_B = I L B = \left(\frac{B L v}{R}\right) L B = \frac{B^2 L^2 v}{R}\). To keep the loop moving at a constant speed, the external pulling force must be equal in magnitude to the magnetic force, which is \(\frac{B^2 L^2 v}{R}\).

Award 1 mark for the correct option B. No marks for other options.

The acceleration of the bucket is \(a = \frac{v - u}{t} = \frac{2 - 0}{2} = 1\text{ m s}^{-2}\). The upward tension force \(T\) exerted on the bucket satisfies \(T - mg = ma \Rightarrow T = m(g+a) = 10 \times (9.81 + 1) = 108.1\text{ N}\). The average velocity of the bucket is \(v_{\text{avg}} = \frac{u+v}{2} = 1\text{ m s}^{-1}\). The average useful power output is \(P = T v_{\text{avg}} = 108.1 \times 1 = 108.1\text{ W} \approx 108\text{ W}\).

Award 1 mark for the correct option B. No marks for other options.

Using Charles' Law at constant volume, \(P_1 / T_1 = P_2 / T_2\). Note that temperature must be converted to Kelvin: \(T_1 = 27 + 273 = 300 \text{ K}\). Thus, \(T_2 = T_1 \times (P_2 / P_1) = 300 \times (3.0 \times 10^5 / 1.2 \times 10^5) = 750 \text{ K}\). Converting back to Celsius: \(750 - 273 = 477^\circ\text{C}\).

Correct answer is B. 1 mark for selecting B.

By conservation of momentum: \(m v = m v_A + 3m v_B \implies v = v_A + 3v_B\). For a perfectly elastic head-on collision, relative velocity of approach equals relative velocity of separation: \(v - 0 = v_B - v_A \implies v_A = v_B - v\). Substitute: \(v = (v_B - v) + 3v_B \implies 2v = 4v_B \implies v_B = 0.5v\). Then, \(v_A = 0.5v - v = -0.5v\).

Correct answer is A. 1 mark for selecting A.

Fringe width is given by \(\Delta y = \frac{\lambda D}{d}\). If the new distance \(D' = D/2\) and new separation \(d' = 2d\), then \(\Delta y' = \frac{\lambda (D/2)}{2d} = \frac{1}{4} \frac{\lambda D}{d} = \frac{1}{4} \Delta y\).

Correct answer is D. 1 mark for selecting D.

(1) is correct because the induced current in the ring opposes the motion of the incoming magnet, exerting an upward magnetic force, so \(a < g\). (2) is correct because as the magnet approaches, the magnetic flux increases, and as it leaves, the flux decreases, causing the induced current to reverse direction. (3) is correct because by Lenz's Law, the ring repels the approaching magnet, exerting an upward force on it; by Newton's third law, the magnet exerts an equal and opposite downward force on the ring.

Correct answer is D. 1 mark for selecting D.

Using the conservation of energy (heat lost by water = heat gained by liquid X): \(m_w c_w \Delta T_w = m_x c_x \Delta T_x\). Substituting the values: \(2.0 \times 4200 \times (80 - 50) = 3.0 \times c_x \times (50 - 20) \implies 2.0 \times 4200 \times 30 = 3.0 \times c_x \times 30 \implies 2.0 \times 4200 = 3.0 \times c_x \implies c_x = 2800 \text{ J kg}^{-1} \text{K}^{-1}\).

Correct answer is B. 1 mark for selecting B.

At maximum steady speed, the forward driving force \(F\) equals the sum of the components of gravity along the slope and the resistive force: \(F = m g \sin \theta + f = 1200 \times 10 \times 0.1 + 400 = 1200 + 400 = 1600 \text{ N}\). Since power \(P = F v\), the maximum speed is \(v = P / F = 40000 / 1600 = 25 \text{ m s}^{-1}\).

Correct answer is B. 1 mark for selecting B.

For a tube closed at one end, resonance occurs when the air column length \(L\) is an odd multiple of quarter-wavelengths: \(L = (2n-1)\lambda/4\). The first resonance occurs at \(L_1 = \lambda/4 = 16.0 \text{ cm}\). The second resonance occurs at \(L_2 = 3\lambda/4 = 3 \times 16.0 = 48.0 \text{ cm}\).

Correct answer is B. 1 mark for selecting B.

The equivalent resistance of the two parallel resistors is \(R_p = R / 2\). The total resistance of the circuit is \(R_{eq} = R/2 + R = 1.5R\). Let the total current be \(I\). The total power is \(P = I^2 R_{eq} = 1.5 I^2 R\). Since the third resistor is in series, the current passing through it is also \(I\). The power dissipated in it is \(P_3 = I^2 R\). Therefore, \(P_3 / P = (I^2 R) / (1.5 I^2 R) = 1 / 1.5 = 2/3\\, so \)P_3 = \\frac{2}{3}P\).

Correct answer is C. 1 mark for selecting C.

Let \(t\) be the time in hours. The number of remaining active nuclei of X and Y are \(N_X = N_0 (1/2)^{t/12}\) and \(N_Y = 2N_0 (1/2)^{t/4}\). Setting \(N_X = N_Y\), we get \(N_0 (1/2)^{t/12} = 2N_0 (1/2)^{t/4} \implies (1/2)^{t/12} = 2 \times (1/2)^{t/4} \implies 1 = 2 \times (1/2)^{t/4 - t/12} = 2 \times (1/2)^{2t/12} = 2 \times (1/2)^{t/6} \implies (1/2)^{t/6} = 1/2 \implies t/6 = 1 \implies t = 6 \text{ hours}\).

Correct answer is B. 1 mark for selecting B.

The gravitational force provides the centripetal force: \(\frac{G M m}{r^2} = \frac{m v^2}{r} \implies v = \sqrt{\frac{G M}{r}}\). Thus, orbital speed is inversely proportional to the square root of the orbital radius, \(v \propto \frac{1}{\sqrt{r}}\). Therefore, \(v_B / v_A = \sqrt{R / 4R} = 1/2 \implies v_B = \frac{1}{2}v\).

Correct answer is B. 1 mark for selecting B.

For the two balls to collide, they must be at the same horizontal and vertical positions at the same time \(t\).\n\nFirst, consider the horizontal motion. Ball A travels a horizontal distance \(D\) with constant speed \(u\):\n\(D = ut \implies t = \frac{D}{u}\)\n\nSecond, consider the vertical motion. For both balls, the downward acceleration is \(g\).\nVertical position of ball A: \(y_A = H - \frac{1}{2}gt^2\)\nVertical position of ball B: \(y_B = vt - \frac{1}{2}gt^2\)\n\nFor them to collide, \(y_A = y_B\):\n\(H - \frac{1}{2}gt^2 = vt - \frac{1}{2}gt^2 \implies H = vt \implies t = \frac{H}{v}\)\n\nEquating the expression for \(t\):\n\(\frac{D}{u} = \frac{H}{v} \implies u = \frac{v D}{H}\)\n\nTherefore, the correct answer is A.

1 mark for identifying the correct relationship \(u = \frac{v D}{H}\) (Option A).

Let the mass of the substance be \(m\) and the power of the heater be \(P\).\n\nDuring the heating of the solid state:\n\(P \cdot t_{\text{solid}} = m c_s \Delta T\)\nwhere \(t_{\text{solid}} = 4\text{ minutes}\) and \(\Delta T = 80 - 20 = 60\text{ K}\).\n\nDuring the melting process:\n\(P \cdot t_{\text{melt}} = m \ell_f\)\nwhere \(t_{\text{melt}} = 6\text{ minutes}\).\n\nDividing the second equation by the first equation:\n\(\frac{P \cdot t_{\text{melt}}}{P \cdot t_{\text{solid}}} = \frac{m \ell_f}{m c_s \Delta T} \implies \frac{t_{\text{melt}}}{t_{\text{solid}}} = \frac{\ell_f}{c_s \Delta T}\)\n\nRearranging the equation to find the ratio \(\frac{\ell_f}{c_s}\):\n\(\frac{\ell_f}{c_s} = \frac{t_{\text{melt}}}{t_{\text{solid}}} \cdot \Delta T = \frac{6\text{ min}}{4\text{ min}} \times 60\text{ K} = 1.5 \times 60\text{ K} = 90\text{ K}\).\n\nTherefore, the correct answer is B.

1 mark for identifying the correct ratio value of \(90\text{ K}\) (Option B).

Let the magnetic field strength be \(B\). As the triangular loop enters the magnetic field with constant velocity \(v\), the area of the loop inside the magnetic field at time \(t\) is a right-angled triangle with base and height both equal to \(vt\).\n\nThus, the area inside the field at time \(t\) is:\n\(A(t) = \frac{1}{2} (vt)(vt) = \frac{1}{2} v^2 t^2\)\n\nThe magnetic flux \(\Phi\) linking the loop is:\n\(\Phi(t) = B A(t) = \frac{1}{2} B v^2 t^2\)\n\nAccording to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. \(\varepsilon\) is:\n\(\varepsilon = \frac{d\Phi}{dt} = B v^2 t\)\n\nSince \(B\) and \(v\) are constants, \(\varepsilon \propto t\). Therefore, the magnitude of the induced e.m.f. is directly proportional to time \(t\).\n\nHence, the correct answer is C.

1 mark for identifying that \(\varepsilon \propto t\) (Option C).

(a) The graph should be a symmetric trapezium starting at the origin, rising to \(v_{\max}\) at \(t = t_1\), staying flat until \(t = t_1 + 6\), and falling to zero at \(t = 15\).
(b) Total time: \(T = 2t_1 + 6 = 15 \implies 2t_1 = 9 \implies t_1 = 4.5\text{ s}\).
The total distance is the area under the \(v\)-\(t\) graph:
\(s = \frac{1}{2} (6 + 15) v_{\max} = 120\text{ m}\)
\(10.5 v_{\max} = 120 \implies v_{\max} = \frac{120}{10.5} \approx 11.43\text{ m s}^{-1}\).
(c) The magnitude of the acceleration:
\(a = \frac{v_{\max}}{t_1} = \frac{11.43}{4.5} \approx 2.54\text{ m s}^{-2}\).

(a) 1M for a correct trapezoidal shape starting and ending at zero; 1A for correctly labeling key time intervals (\(t_1\), \(t_1+6\), \(15\)) and peak speed \(v_{\max}\).
(b) 1M for writing total time equation to find \(t_1 = 4.5\text{ s}\); 1A for correct \(t_1\); 1M for using trapezium area formula for distance; 1A for finding \(v_{\max} = 11.4\text{ m s}^{-1}\) (or \(11.43\text{ m s}^{-1}\)).
(c) 1M for using \(a = \Delta v / \Delta t\); 1.33A for correct numerical value \(2.54\text{ m s}^{-2}\) (accept \(2.5\) to \(2.6\)).

(a) Conservation of linear momentum: The total linear momentum of the two balls before collision equals the total linear momentum after collision in the absence of external forces.
Conservation of kinetic energy: The total kinetic energy of the system remains unchanged before and after the collision.
(b) Let \(v_A\) and \(v_B\) be the final velocities of A and B respectively.
From conservation of momentum:
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
\(0.5(4.0) + 0 = 0.5 v_A + 0.3 v_B \implies 0.5 v_A + 0.3 v_B = 2.0\) (Eq. 1)
Since the collision is perfectly elastic, the relative velocity of separation equals the relative velocity of approach:
\(v_B - v_A = u_A - u_B = 4.0 \implies v_B = v_A + 4.0\) (Eq. 2)
Substitute Eq. 2 into Eq. 1:
\(0.5 v_A + 0.3(v_A + 4.0) = 2.0\)
\(0.8 v_A + 1.2 = 2.0 \implies 0.8 v_A = 0.8 \implies v_A = 1.0\text{ m s}^{-1}\).
Using Eq. 2:
\(v_B = 1.0 + 4.0 = 5.0\text{ m s}^{-1}\).
(c) The average force on Ball B:
\(F_{\text{avg}} = \frac{\Delta p_B}{\Delta t} = \frac{m_B v_B - 0}{\Delta t} = \frac{0.3 \times 5.0}{0.05} = 30\text{ N}\).

(a) 1A for statement of conservation of linear momentum; 1A for conservation of kinetic energy.
(b) 1M for correct conservation of momentum equation; 1M for relative velocity relation (or kinetic energy equation); 1.33A for correct value of \(v_A\); 2A for correct value of \(v_B\).
(c) 1M for using Newton's second law/momentum-impulse relation \(F = \Delta p / \Delta t\); 1A for correct force of \(30\text{ N}\).

(a) Considering vertical motion, from \(y = u_y t + \frac{1}{2} g t^2\), since \(u_y = 0\):
\(45.0 = 0 + \frac{1}{2} (9.81) t^2\)
\(t^2 = \frac{90.0}{9.81} \approx 9.174 \implies t \approx 3.029\text{ s}\), which is approximately \(3.03\text{ s}\).
(b) Considering horizontal motion, the speed remains constant:
\(x = u_x t = 20.0 \times 3.029 \approx 60.58\text{ m} \approx 60.6\text{ m}\).
(c) Just before hitting the ground, the horizontal component of velocity is:
\(v_x = 20.0\text{ m s}^{-1}\).
The vertical component of velocity is:
\(v_y = g t = 9.81 \times 3.029 \approx 29.71\text{ m s}^{-1}\).
The speed \(v\) is:
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{20.0^2 + 29.71^2} \approx \sqrt{400 + 882.68} \approx 35.81\text{ m s}^{-1}\).
The angle \(\theta\) below the horizontal is:
\(\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{29.71}{20.0}\right) \approx 56.04^\circ \approx 56.0^\circ\).

(a) 1M for choosing the vertical displacement formula; 1.33A for substitution showing \(t = 3.03\text{ s}\).
(b) 1M for horizontal distance formula \(x = u_x t\); 2A for correct substitution and answer \(60.6\text{ m}\) (or \(60.58\text{ m}\)).
(c) 1M for calculating vertical velocity \(v_y\); 1A for finding final speed \(35.8\text{ m s}^{-1}\); 1M for using trigonometry for angle; 1A for correct angle \(56.0^\circ\) below the horizontal.

(a) Since volume is constant, by Pressure Law:
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)
\(\frac{1.0 \times 10^5}{300} = \frac{P_2}{450}\)
\(P_2 = 1.5 \times 10^5\text{ Pa}\).
(b) (i) Since pressure is constant, by Charles's Law:
\(\frac{V_2}{T_2} = \frac{V_3}{T_3}\)
\(\frac{2.0 \times 10^{-3}}{450} = \frac{3.0 \times 10^{-3}}{T_3}\)
\(T_3 = 450 \times \frac{3.0}{2.0} = 675\text{ K}\).
(ii) When volume increases at constant temperature, the average kinetic energy (and speed) of the gas molecules remains unchanged because temperature is constant. However, the number of molecules per unit volume (number density) decreases. This causes the frequency of collisions of gas molecules with the container walls per unit area to decrease, resulting in a smaller average force per unit area, and hence the pressure decreases.

(a) 1M for using \(P/T = \text{constant}\); 1M for correct substitution; 1A for finding \(1.5 \times 10^5\text{ Pa}\).
(b)(i) 1M for using \(V/T = \text{constant}\); 1M for correct substitution; 1A for finding \(675\text{ K}\).
(b)(ii) 1A for stating constant temperature means constant average kinetic energy / molecular speed; 1M for identifying that the number density of molecules decreases; 1.33A for concluding that collision frequency per unit area with walls decreases, leading to lower average force per unit area (pressure).

(a) By definition of critical angle:
\(\sin c = \frac{1}{n} = \frac{1}{1.52} \approx 0.6579\)
\(c = \sin^{-1}(0.6579) \approx 41.1^\circ\).
(b) Applying Snell's Law at the air-glass interface:
\(n_{\text{air}} \sin \theta = n \sin r\)
\(1 \cdot \sin 45.0^\circ = 1.52 \sin r\)
\(\sin r = \frac{\sin 45.0^\circ}{1.52} \approx \frac{0.7071}{1.52} \approx 0.4652\)
\(r = \sin^{-1}(0.4652) \approx 27.7^\circ\).
(c) Since the ray enters at the center of the flat face, it travels along a radius of the semicircle. It strikes the curved surface normally (i.e., the angle of incidence at the curved boundary is \(0^\circ\)). Therefore, the ray passes through the boundary into the air without changing its direction.

(a) 1M for using \(\sin c = 1/n\); 1M for correct substitution; 1A for finding \(c = 41.1^\circ\).
(b) 1M for using Snell's Law; 1M for substituting \(n = 1.52\) and \(i = 45.0^\circ\); 1.33A for finding \(r = 27.7^\circ\).
(c) 1A for stating that the ray exits without changing direction; 2A for explaining that the ray travels along the radius and strikes the boundary normally (at \(0^\circ\) incidence).

(a) Using the formula for fringe width:
\(\Delta y = \frac{\lambda D}{d}\)
Substitute the given values (converting to standard units):
\(\lambda = 650 \times 10^{-9}\text{ m}\)
\(d = 0.25 \times 10^{-3}\text{ m}\)
\(D = 1.80\text{ m}\)
\
\(\Delta y = \frac{650 \times 10^{-9} \times 1.80}{0.25 \times 10^{-3}} = 4.68 \times 10^{-3}\text{ m} = 4.68\text{ mm}\).
(b) (i) Green light has a shorter wavelength than red light (\(530\text{ nm} < 650\text{ nm}\)). Since \(\Delta y \propto \lambda\) when other variables are constant, the fringe width will decrease.
(ii) Since \(\Delta y \propto D\), doubling the distance \(D\) between the slits and the screen will double the fringe width.

(a) 1M for selecting the formula \(\Delta y = \lambda D / d\); 1M for correct conversion of units; 1.33A for finding \(4.68\text{ mm}\) (or \(4.68 \times 10^{-3}\text{ m}\)).
(b)(i) 1A for stating that the fringe width decreases; 2A for explaining that green light has a shorter wavelength and \(\Delta y \propto \lambda\).
(b)(ii) 1A for stating that the fringe width doubles; 2A for explaining that \(\Delta y\) is directly proportional to \(D\).

(a) Using Coulomb's Law:
\(F = \frac{k |Q_1 Q_2|}{r^2}\)
\(F = \frac{8.99 \times 10^9 \times (4.0 \times 10^{-6}) \times (9.0 \times 10^{-6})}{(0.30)^2}\)
\(F = \frac{0.32364}{0.09} \approx 3.596\text{ N} \approx 3.60\text{ N}\).
Since the charges have opposite signs, the force is attractive. Thus, the force on \(Q_1\) points towards \(Q_2\), which is in the positive x-direction.

(b) Let the point where the net electric field is zero be at coordinate \(x\).
Since the charges have opposite signs, the zero-field point cannot be between \(x = 0\) and \(x = 0.30\text{ m}\) (as both individual fields would point in the same direction).
Since \(|Q_1| < |Q_2|\), the point must be closer to \(Q_1\) than to \(Q_2\), which means it must lie to the left of \(Q_1\) (i.e., \(x < 0\)).
Let \(x = -d\), where \(d\) is the distance from \(Q_1\) to the left.
At this point:
\(E_1 = E_2 \implies \frac{k |Q_1|}{d^2} = \frac{k |Q_2|}{(0.30 + d)^2}\)
\(\frac{4.0 \times 10^{-6}}{d^2} = \frac{9.0 \times 10^{-6}}{(0.30 + d)^2}\)
\(\frac{2}{d} = \frac{3}{0.30 + d}\)
\(2(0.30 + d) = 3d\)
\(0.60 + 2d = 3d \implies d = 0.60\text{ m}\).
Since the point is to the left of the origin (\(x = 0\)), its coordinate is \(x = -0.60\text{ m}\).

(a) 1M for Coulomb's Law; 1M for correct substitution; 1A for force magnitude \(3.60\text{ N}\); 0.33A for correct direction (towards the right / positive x-direction).
(b) 1M for identifying that the position must be at \(x < 0\); 1M for setting up \(E_1 = E_2\); 2M for substituting the distances as \(d\) and \(0.30 + d\); 1A for finding \(d = 0.60\text{ m}\); 1A for writing the final coordinate as \(x = -0.60\text{ m}\).

(a) The equivalent resistance of the parallel combination \(R_p\) is given by:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{3}{6.0} = \frac{1}{2.0}\)
\(R_p = 2.0\text{ }\Omega\).
(b) The total external resistance of the circuit is:
\(R_{\text{ext}} = R_p + R_3 = 2.0 + 5.0 = 7.0\text{ }\Omega\).
The total resistance including the internal resistance of the battery is:
\(R_{\text{total}} = R_{\text{ext}} + r = 7.0 + 1.0 = 8.0\text{ }\Omega\).
The total current \(I\) is:
\(I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{12.0}{8.0} = 1.5\text{ A}\).
(c) The terminal voltage \(V\) is:
\(V = \mathcal{E} - I r = 12.0 - (1.5 \times 1.0) = 10.5\text{ V}\).
The potential difference across the parallel combination of \(R_1\) and \(R_2\) is:
\(V_p = I R_p = 1.5 \times 2.0 = 3.0\text{ V}\).
Therefore, the current flowing through \(R_1\) is:
\(I_1 = \frac{V_p}{R_1} = \frac{3.0}{6.0} = 0.5\text{ A}\).

(a) 1M for using parallel resistance formula; 1A for finding \(2.0\text{ }\Omega\).
(b) 1M for calculating total external resistance \(7.0\text{ }\Omega\); 1M for adding internal resistance; 1.33A for finding total current \(1.5\text{ A}\).
(c) 1M for terminal voltage equation \(V = \mathcal{E} - Ir\); 1A for terminal voltage of \(10.5\text{ V}\); 1M for finding voltage across parallel section or sharing current; 1A for finding current through \(R_1\) of \(0.5\text{ A}\).

(a) Faraday's Law states that the magnitude of the induced electromotive force (e.m.f.) in a circuit is directly proportional to the rate of change of magnetic flux linkage through the circuit.
(b) The magnetic flux \(\Phi\) through a single turn is given by:
\(\Phi = B A = B s^2\)
\(\Phi = 0.40 \times (0.10)^2 = 0.40 \times 0.01 = 4.0
\times 10^{-3}\text{ Wb}\).
(c) The rate of change of area as the coil enters the field is:
\(\frac{\Delta A}{\Delta t} = s v\)
Therefore, the induced e.m.f. \(\mathcal{E}\) in the coil of \(N\) turns is:
\(\mathcal{E} = N \frac{\Delta \Phi}{\Delta t} = N B \frac{\Delta A}{\Delta t} = N B s v\)
\(\mathcal{E} = 50 \times 0.40 \times 0.10 \times 2.0 = 4.0\text{ V}\).
(d) As the coil enters the magnetic field, the inward magnetic flux through the coil increases. According to Lenz's law, the induced current must flow in a direction that opposes this increase, which means it must produce an outward magnetic field. By the right-hand grip rule, the direction of the induced current must be counter-clockwise.

(a) 2A for a clear statement of Faraday's Law (including rate of change of magnetic flux linkage).
(b) 1M for using \(\Phi = BA\); 1.33A for obtaining \(4.0 \times 10^{-3}\text{ Wb}\).
(c) 1M for rate of change of area \(sv\); 1M for the formula \(\mathcal{E} = N B s v\) or equivalent; 1A for finding \(4.0\text{ V}\).
(d) 1A for specifying counter-clockwise direction; 1A for explaining that the induced field must point out of the page to oppose the increasing inward flux.

Since A and B form a binary star system, they rotate about their common centre of mass with the same angular velocity, hence their orbital periods are equal, so \(T_B = T\). For the center of mass, \(m_A r_A = m_B r_B\). Since \(m_A = 2m_B\), we get \(2m_B r_A = m_B r_B \Rightarrow r_A/r_B = 0.5\). Therefore, the correct option is A.

Award 1 mark for the correct option A.

Distance \(d = 1/p\). Thus \(d_X = 20\text{ pc}\) and \(d_Y = 100\text{ pc}\). Since apparent magnitudes are equal, their apparent brightness \(b\) are equal: \(\frac{L_X}{d_X^2} = \frac{L_Y}{d_Y^2} \Rightarrow \frac{L_X}{L_Y} = \left(\frac{20}{100}\right)^2 = \frac{1}{25}\). Since \(L = 4\pi R^2 \sigma T^4\), we have \(\frac{L_X}{L_Y} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{T_X}{T_Y}\right)^4 \Rightarrow \frac{1}{25} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{6000}{3000}\right)^4 = 16 \left(\frac{R_X}{R_Y}\right)^2\). Hence, \(\left(\frac{R_X}{R_Y}\right)^2 = \frac{1}{400} \Rightarrow \frac{R_X}{R_Y} = 0.05\).

Award 1 mark for the correct option A.

Redshift \(z = \frac{\Delta \lambda}{\lambda_0} = \frac{678.0 - 656.3}{656.3} \approx 0.03306\). The recession velocity \(v = z c = 0.03306 \times 3.0 \times 10^5\text{ km s}^{-1} \approx 9919\text{ km s}^{-1}\). By Hubble's Law, \(d = v / H_0 = 9919 / 70 \approx 141.7\text{ Mpc} \approx 142\text{ Mpc}\).

Award 1 mark for the correct option A.

From Einstein's equation, \(eV_1 = hf - hf_0\) and \(eV_2 = 1.5hf - hf_0\). Dividing the two: \(\frac{V_1}{V_2} = \frac{f - f_0}{1.5f - f_0}\). This gives \(1.5V_1 f - V_1 f_0 = V_2 f - V_2 f_0\), which simplifies to \(f_0(V_2 - V_1) = f(V_2 - 1.5V_1) \Rightarrow f_0 = \frac{2V_2 - 3V_1}{2(V_2 - V_1)} f\).

Award 1 mark for the correct option A.

Since \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}\), we have \(\frac{\lambda_e}{\lambda_\alpha} = \sqrt{\frac{m_\alpha q_\alpha}{m_e q_e}} = \sqrt{7300 \times 2} = \sqrt{14600} \approx 121\).

Award 1 mark for the correct option A.

In Bohr's model, the orbit radius is \(r_n = n^2 a_0\). For \(n=3\), \(r_3 = 9a_0\). According to de Broglie's wave condition for orbital stability, \(2\pi r_n = n \lambda \Rightarrow 2\pi (9a_0) = 3 \lambda_3 \Rightarrow \lambda_3 = 6\pi a_0\).

Award 1 mark for the correct option C.

Input solar power \(P_{\text{in}} = 800 \times 2.5 = 2000\text{ W}\). Output electrical power \(P_{\text{out}} = V I = 12 \times 8 = 96\text{ W}\). Efficiency \(\eta = \frac{96}{2000} \times 100\% = 4.8\%\).

Award 1 mark for the correct option A.

Electrical power input \(P_{\text{elec}} = \frac{\text{Cooling Capacity}}{\text{COP}} = \frac{3.5\text{ kW}}{2.8} = 1.25\text{ kW}\). Energy consumed \(E = P_{\text{elec}} \times t = 1.25\text{ kW} \times 8.0\text{ h} = 10.0\text{ kWh}\).

Award 1 mark for the correct option B.

For a myopic eye, the lens must form a virtual image of a distant object (\(u=\infty\)) at the far point (\(v = -50\text{ cm} = -0.5\text{ m}\)). Thus \(\frac{1}{f} = \frac{1}{\infty} + \frac{1}{-0.5} \Rightarrow f = -0.5\text{ m} = -50\text{ cm}\). The power \(P = 1/f = 1/(-0.5) = -2.0\text{ D}\).

Award 1 mark for the correct option A.

Using the attenuation equation \(I = I_0 e^{-\mu x}\), the transmitted ratio is \(\frac{I}{I_0} = e^{-0.35 \times 4.0} = e^{-1.4} \approx 0.2466\), which is \(24.7\%\).

Award 1 mark for the correct option B.

According to the Stefan-Boltzmann law, the luminosity is given by \(L = 4\pi R^2 \sigma T^4\). Therefore, \(\frac{L_A}{L_B} = \left(\frac{R_A}{R_B}\right)^2 \left(\frac{T_A}{T_B}\right)^4\). Substituting the given values: \(64 = \left(\frac{R_A}{R_B}\right)^2 (2)^4 \implies 64 = 16 \left(\frac{R_A}{R_B}\right)^2 \implies \left(\frac{R_A}{R_B}\right)^2 = 4\). Thus, \(\frac{R_A}{R_B} = 2\), which means \(R_A : R_B = 2 : 1\).

1 mark for option C.

The orbital speed of a satellite in a circular orbit of radius \(r\) is given by \(v = \sqrt{\frac{GM}{r}}\). If the radius becomes \(1.44r\), the new orbital speed \(v'\) is \(\sqrt{\frac{GM}{1.44r}} = \frac{v}{\sqrt{1.44}} = \frac{v}{1.2} \approx 0.83v\).

1 mark for option B.

According to Einstein's photoelectric equation, \(eV_s = \frac{hc}{\lambda} - \phi \implies \frac{hc}{\lambda} = eV_s + \phi\), where \(\phi\) is the work function of the metal. If the wavelength is halved, the new equation is \(eV_s' = \frac{hc}{\lambda / 2} - \phi = 2\frac{hc}{\lambda} - \phi\). Substituting the first equation into this gives \(eV_s' = 2(eV_s + \phi) - \phi = 2eV_s + \phi\). Since the work function \(\phi > 0\), we have \(eV_s' > 2eV_s \implies V_s' > 2V_s\).

1 mark for option C.

The kinetic energy gained is \(K = qV\). The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}\). Since both have the same charge magnitude \(e\) and same potential difference \(V\), we have \(\lambda \propto \frac{1}{\sqrt{m}}\). Therefore, \(\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} = \sqrt{1840} \approx 42.9\).

1 mark for option B.

The incident solar power on the panel is \(P_{\text{in}} = 800 \text{ W m}^{-2} \times 2.0 \text{ m}^2 = 1600 \text{ W}\). The output electrical power of the solar panel is \(P_{\text{out}} = P_{\text{in}} \times 15\% = 240 \text{ W} = 0.24 \text{ kW}\). The total electrical energy required to charge the battery is \(E = 12 \text{ V} \times 80 \text{ Ah} = 960 \text{ Wh} = 0.96 \text{ kWh}\). The charging time required is \(t = \frac{E}{P_{\text{out}}} = \frac{0.96 \text{ kWh}}{0.24 \text{ kW}} = 4.0 \text{ hours}\).

1 mark for option B.

The intensity reflection coefficient \(R\) is given by \(R = \frac{(Z_2 - Z_1)^2}{(Z_2 + Z_1)^2}\), where \(Z_1\) and \(Z_2\) are the acoustic impedances of muscle and bone respectively. \(R = \frac{(7.80 \times 10^6 - 1.70 \times 10^6)^2}{(7.80 \times 10^6 + 1.70 \times 10^6)^2} = \frac{(6.10 \times 10^6)^2}{(9.50 \times 10^6)^2} = \frac{37.21}{90.25} \approx 0.412 = 41.2\%\). Thus, \(41.2\%\) of the incident intensity is reflected.

1 mark for option B.

(a) Using the relation for effective half-life:\n1/T_e = 1/T_p + 1/T_b\n1/T_e = 1/6.0 + 1/24.0 = (4 + 1)/24.0 = 5/24.0\nT_e = 24.0 / 5 = 4.8 hours\n\n(b) Suitability:\n1. The effective half-life (4.8 hours) is short enough to minimize the long-term radiation dose received by the patient, yet long enough to allow the scanning procedure to be successfully completed.\n2. It emits only gamma photons (gamma decay). Gamma radiation has high penetrating power to easily escape the patient's body to be detected by external gamma cameras, and it has low ionizing power, which minimizes damage to biological tissues.\n\n(c) The remaining activity after t = 12.0 hours:\nA = A_0 * (1/2)^(t / T_e)\nA = 300 * (1/2)^(12.0 / 4.8)\nA = 300 * (1/2)^2.5 = 300 * 0.17678 = 53.03 MBq (or 53.0 MBq)

(a)\n- Formula 1/T_e = 1/T_p + 1/T_b (1M)\n- Substitution of values: 1/6.0 + 1/24.0 (1M)\n- Correct answer: 4.8 hours (1A)\n\n(b)\n- Explaining the half-life factor: Short enough to reduce patient's radiation dose (1A) but long enough to perform imaging (1A).\n- Explaining the radiation type factor: Emits gamma photons which can escape the body for detection due to high penetration (1A) and cause low tissue damage due to low ionization (1A).\n\n(c)\n- Formula A = A_0 * (1/2)^(t / T_e) or A = A_0 * e^(-lambda * t) (1M)\n- Substitution: 300 * (0.5)^(12.0 / 4.8) (1M)\n- Correct answer: 53.0 MBq (accept 53 MBq to 53.1 MBq) (1A)

(a) According to Wien's displacement law:\n\lambda_{\text{max}} * T = b\nT = b / \lambda_{\text{max}} = (2.90 * 10^{-3} \text{ m K}) / (966 * 10^{-9} \text{ m})\nT = 3002.07 \text{ K} \approx 3002 \text{ K} (or 3000 \text{ K})\n\n(b) According to Stefan-Boltzmann law, L = 4 * \pi * R^2 * \sigma * T^4.\nThus, the ratio of luminosity is given by:\nL / L_{\odot} = (R / R_{\odot})^2 * (T / T_{\odot})^4\n4000 = (R / R_{\odot})^2 * (3002 / 5780)^4\n4000 = (R / R_{\odot})^2 * (0.51939)^4\n4000 = (R / R_{\odot})^2 * 0.07277\n(R / R_{\odot})^2 = 4000 / 0.07277 = 54970\nR / R_{\odot} = \sqrt{54970} \approx 234.5 (Accept 234 to 235 depending on rounding of temperature, e.g., if T = 3000 K, R/R_{\odot} \approx 235)\n\n(c) Since the initial mass of the star is 1.5 solar masses (which is less than 8 solar masses):\n- The star will eventually shed its outer layers to form a planetary nebula, leaving behind a white dwarf as its core remnant. (2 marks)\n- The white dwarf is supported against gravitational collapse by electron degeneracy pressure. (1 mark)

(a)\n- State or use Wien's displacement law formula: \lambda_{\text{max}} * T = b (1M)\n- Substitution of values: T = (2.90 * 10^{-3}) / (966 * 10^{-9}) (1M)\n- Correct temperature: 3002 K or 3000 K (1A)\n\n(b)\n- State or use Stefan-Boltzmann ratio formula: L/L_sun = (R/R_sun)^2 * (T/T_sun)^4 (1M)\n- Substitution of values: 4000 = (R/R_sun)^2 * (3002/5780)^4 (1M)\n- Correct algebraic rearrangement for R/R_sun (1M)\n- Correct ratio: 234 or 235 (1A)\n\n(c)\n- State that the star becomes a white dwarf (and/or planetary nebula) (2A)\n- State that the support mechanism is electron degeneracy pressure (1A)