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For a perfectly elastic head-on collision between mass \(m\) (velocity \(u\)) and mass \(3m\) (at rest): Conservation of momentum gives \(m u = m v_1 + 3m v_2\), which simplifies to \(u = v_1 + 3v_2\). For an elastic collision, the relative speed of approach equals the relative speed of separation: \(u = v_2 - v_1\). Adding these two equations yields \(2u = 4v_2\), which gives \(v_2 = \frac{1}{2}u\).

Award [1] for the correct answer B. Correctly write down the conservation of momentum and conservation of kinetic energy equations (or relative velocity relation) and solve for the final velocity of the heavier mass.

The constant power supplied to the solid is given by \(P = \frac{Q}{\Delta t} = \frac{m c_s \Delta T}{\Delta t}\). During melting, the total heat energy supplied is \(Q_m = P t_m = m L_f\). Substituting the expression for \(P\) into the melting equation gives \(m L_f = \left( \frac{m c_s \Delta T}{\Delta t} \right) t_m\), which simplifies to \(L_f = c_s \frac{\Delta T \cdot t_m}{\Delta t}\).

Award [1] for the correct answer B. Relate the constant heating power to the temperature change in the solid phase, then equate it to the energy rate required for the phase change.

The induced electromotive force (emf) as the coil enters the magnetic field is \(\varepsilon = B L v\). The resulting induced current in the coil is \(I = \frac{\varepsilon}{R} = \frac{B L v}{R}\). The magnetic force acting on the leading wire of the coil as it moves through the field is \(F = I L B = \left( \frac{B L v}{R} \right) L B = \frac{B^2 L^2 v}{R}\).

Award [1] for the correct answer C. Correctly identify the induced emf in the moving side of the coil, calculate the resulting current, and use the formula for the magnetic force on a current-carrying conductor.

At the distance of closest approach, the initial kinetic energy \(E\) of the alpha-particle is converted entirely into electrostatic potential energy: \(E = \frac{k (2e)(Ze)}{d}\), which means \(d = \frac{2 k Z e^2}{E}\). Thus, the distance of closest approach is proportional to \(\frac{Z}{E}\). When both the nuclear charge \(Z\) and the initial kinetic energy \(E\) are doubled, the ratio \(\frac{Z}{E}\) remains constant, and therefore the distance of closest approach is unchanged and remains \(d\).

Award [1] for the correct answer B. Use the conservation of energy to establish that the closest approach distance is proportional to the ratio of nuclear charge to kinetic energy, noting that doubling both keeps this ratio constant.

Upon passing through the first polarizer, the unpolarized light of intensity \(I_0\) becomes vertically polarized with intensity \(I_1 = \frac{1}{2} I_0\). When this light passes through the second polarizer whose axis is at \(30^\circ\) to the first, we apply Malus's Law: \(I_2 = I_1 \cos^2(30^\circ) = \left(\frac{1}{2} I_0\right) \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} I_0 \cdot \frac{3}{4} = \frac{3}{8} I_0\).

Award [1] for the correct answer B. Use the fact that the first polarizer halves the intensity of unpolarized light, and then apply Malus's Law using the angle between the two polarizing axes.

Let \(x\) be the distance from the charge \(+q\) where the electric potential is zero. Since the point lies on the line segment between the charges, the distance from the charge \(-4q\) is \(L - x\). The total electric potential is the sum of individual potentials: \(V = \frac{k q}{x} + \frac{k (-4q)}{L - x} = 0\). Simplifying this equation: \(\frac{1}{x} = \frac{4}{L - x} \implies L - x = 4x \implies 5x = L \implies x = \frac{L}{5}\).

Award [1] for the correct answer A. State the equation for electric potential of point charges, set the sum to zero for a point between them, and solve for the distance.

In the isothermal expansion (state 1 to state 2): \(P_1 V_1 = P_2 V_2 \implies P V = P_2 (2V) \implies P_2 = \frac{P}{2}\). The isobaric compression (state 2 to state 3) takes place at constant pressure \(P_2 = \frac{P}{2}\) from volume \(2V\) to volume \(V\). The work done by the gas is given by \(W = P_2 \Delta V = \frac{P}{2} (V - 2V) = -\frac{PV}{2}\).

Award [1] for the correct answer B. Use the isothermal relation to find the intermediate pressure, then compute the work done during the compression using \(W = P \Delta V\).

Einstein's photoelectric equation states that \(hf = \Phi + E_{\text{k,max}}\). For wavelength \(\lambda\): \(\frac{hc}{\lambda} = \Phi + E_{\max}\) (Equation 1). For wavelength \(\frac{\lambda}{2}\): \(\frac{2hc}{\lambda} = \Phi + 3E_{\max}\) (Equation 2). Multiplying Equation 1 by 2 gives \(\frac{2hc}{\lambda} = 2\Phi + 2E_{\max}\). Equating this to Equation 2 yields: \(2\Phi + 2E_{\max} = \Phi + 3E_{\max}\), which simplifies to \(\Phi = E_{\max}\).

Award [1] for the correct answer B. Express Einstein's photoelectric equation for both wavelengths, eliminate the photon energy term, and solve for the work function in terms of the maximum kinetic energy.

The induced electromotive force (emf) \(\varepsilon\) in the square loop as it is pulled out is given by Faraday's Law: \(\varepsilon = B v s\). The current \(I\) flowing through the loop of resistance \(R\) is \(I = \frac{\varepsilon}{R} = \frac{B v s}{R}\). The time interval \(\Delta t\) required to pull the loop of side length \(s\) completely out of the field at a constant speed \(v\) is \(\Delta t = \frac{s}{v}\). The total thermal energy dissipated is: \(E = I^2 R \Delta t = \left(\frac{B v s}{R}\right)^2 R \left(\frac{s}{v}\right) = \frac{B^2 s^3 v}{R}\).

Award [1] for the correct answer. Determine the emf, relate it to current, find the time duration, and compute energy.

For the \(2 \to 1\) transition, the energy of the photon is: \(\Delta E_{2 \to 1} = -\frac{E_0}{2^2} - \left(-\frac{E_0}{1^2}\right) = \frac{3}{4}E_0 = h f\). For the \(4 \to 2\) transition, the energy of the photon is: \(\Delta E_{4 \to 2} = -\frac{E_0}{4^2} - \left(-\frac{E_0}{2^2}\right) = -\frac{E_0}{16} + \frac{E_0}{4} = \frac{3}{16}E_0\). Comparing the two energy changes: \(\Delta E_{4 \to 2} = \frac{1}{4} \left(\frac{3}{4}E_0\right) = \frac{1}{4} h f\). Thus, the frequency of the second photon is \(0.25 f\).

Award [1] for the correct option. Use the energy level formula to express the photon energy for both transitions and take their ratio.

According to the Rayleigh criterion for a circular aperture, the minimum angular separation for resolution is \(\theta \approx 1.22 \frac{\lambda}{D}\). Thus, the minimum angle is directly proportional to the wavelength and inversely proportional to the diameter: \(\theta \propto \frac{\lambda}{D}\). Changing the wavelength to \(\lambda' = 2\lambda\) and diameter to \(D' = 0.5D\) gives: \(\theta' \propto \frac{2\lambda}{0.5D} = 4 \frac{\lambda}{D}\). Therefore, \(\theta' = 4\theta\).

Award [1] for the correct option. State the relationship between angle, wavelength, and diameter, then calculate the new value.

The electric potential \(V\) at the origin \((0,0)\) is the sum of the potentials due to the two charges: \(V = \frac{1}{4\pi\varepsilon_0} \frac{2Q}{d} + \frac{1}{4\pi\varepsilon_0} \frac{-Q}{d} = \frac{Q}{4\pi\varepsilon_0 d}\). Since the potential at infinity is zero, the work done by an external agent in bringing a charge \(+q\) to the origin is \(W = q \Delta V = q(V - 0) = \frac{Qq}{4\pi\varepsilon_0 d}\).

Award [1] for the correct option. Calculate the net potential at the origin and multiply by the test charge to find the work done.

During an isothermal compression (Process X), the pressure of the gas increases continuously from \(P_0\) to \(2P_0\) according to Boyle's law (\(PV = \text{constant}\)). During an isobaric compression (Process Y), the pressure remains constant at \(P_0\). On a \(P-V\) diagram, the curve for the isothermal process lies completely above the horizontal line for the isobaric process for any volume between \(0.5V_0\) and \(V_0\). Since the magnitude of work done is the area under the curve, the work done on the gas in the isothermal process is greater: \(W_X > W_Y\).

Award [1] for the correct option. Compare the paths on a PV diagram to determine which process has a larger area under the curve.

The de Broglie wavelength is given by \(\lambda = \frac{h}{p}\). The momentum \(p\) of a non-relativistic particle is related to its kinetic energy \(E_k\) by \(p = \sqrt{2m E_k}\). Therefore, \(\lambda = \frac{h}{\sqrt{2m E_k}}\). If the kinetic energy is doubled, \(E_k' = 2E_k\), then the new wavelength is \(\lambda' = \frac{h}{\sqrt{2m (2E_k)}} = \frac{\lambda}{\sqrt{2}}\).

Award [1] for the correct option. Relate de Broglie wavelength to kinetic energy and solve for the new wavelength.

The heat supplied to raise the temperature of the solid phase is: \(Q_1 = P t_1 = m c_s \Delta T\), where \(t_1 = 120\text{ s}\) and \(\Delta T = 80 - 20 = 60\text{ K}\). This gives \(c_s = \frac{P t_1}{m \Delta T}\). The heat supplied to melt the substance is: \(Q_2 = P t_{\text{melt}} = m L_f\), where \(t_{\text{melt}} = 240\text{ s}\). This gives \(L_f = \frac{P t_{\text{melt}}}{m}\). The ratio is: \(\frac{c_s}{L_f} = \frac{t_1}{t_{\text{melt}} \Delta T} = \frac{120}{240 \times 60} = \frac{1}{120}\text{ K}^{-1}\).

Award [1] for the correct option. Formulate equations for temperature change and phase change, then calculate their ratio.

The impulse \(J\) delivered by the force is represented by the area under the force-time graph. Since the force starts at \(0\), peaks at \(F_0\) at time \(\frac{T}{2}\), and returns to \(0\) at time \(T\), the graph is triangular. The area of this triangle is \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} T F_0\). Since the force acts to the left (opposing the initial velocity), the impulse is negative: \(J = -\frac{1}{2} F_0 T\). Using the impulse-momentum theorem: \(J = M v_f - M v \implies -\frac{1}{2} F_0 T = M (v_f - v)\). Solving for the final speed \(v_f\) gives \(v_f = v - \frac{F_0 T}{2M}\).

Award [1] for the correct option. Calculate the impulse as the area under the triangular F-t graph, then apply the impulse-momentum theorem.

The rate at which mass is collected by the spacecraft is given by \(\frac{dm}{dt} = \rho A v\). By conservation of momentum (or Newton's second law in the form \(F = \frac{dp}{dt} = v \frac{dm}{dt}\)), the resistive force acting on the spacecraft is \(F = \rho A v^2\). Applying Newton's second law, \(F = M a\), we get the magnitude of deceleration as \(a = \frac{\rho A v^2}{M}\).

Award [1] for the correct derivation of deceleration by equating force to rate of change of momentum and dividing by mass \(M\).

At \(t = 0\), the plane of the coil is perpendicular to the magnetic field, meaning the magnetic flux is at its maximum value. Thus, the magnetic flux linkage is given by \(\Phi = N B A \cos(\omega t)\). According to Faraday's law of induction, the induced electromotive force is \(\varepsilon = -\frac{d\Phi}{dt}\). Taking the derivative of \(\Phi\) with respect to time gives \(\frac{d\Phi}{dt} = -N B A \omega \sin(\omega t)\). Substituting this back into Faraday's law yields \(\varepsilon = -(-N B A \omega \sin(\omega t)) = N B A \omega \sin(\omega t)\).

Award [1] for identifying the correct flux linkage function and using Faraday's law with correct sign differentiation.

The energy difference between the levels is \(\Delta E = E_3 - E_2 = \frac{3^2 h^2}{8 m_e L^2} - \frac{2^2 h^2}{8 m_e L^2} = \frac{5 h^2}{8 m_e L^2}\). The energy of the emitted photon is related to its wavelength by \(\Delta E = \frac{h c}{\lambda}\). Equating the two expressions: \(\frac{h c}{\lambda} = \frac{5 h^2}{8 m_e L^2}\). Rearranging for \(\lambda\) gives \(\lambda = \frac{8 m_e L^2 c}{5 h}\).

Award [1] for calculating \(\Delta E\) and correctly equating to \(\frac{hc}{\lambda}\) to solve for \(\lambda\).

According to the observer on the space station, the spacecraft undergoes length contraction. Its measured length \(L\) is \(L = L_0 \sqrt{1 - \frac{v^2}{c^2}} = L_0 \sqrt{1 - 0.80^2} = 0.60 L_0\). The time interval \(\Delta t\) for this contracted length to pass the single point (beacon) at speed \(v = 0.80c\) is \(\Delta t = \frac{L}{v} = \frac{0.60 L_0}{0.80 c} = 0.75 \frac{L_0}{c}\).

Award [1] for applying length contraction correctly and then using time = distance/speed in the station's frame.

For a rolling body of mass \(M\), radius \(R\), and moment of inertia \(I = \beta M R^2\) starting from rest, conservation of energy gives \(Mgh = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} M v^2 (1 + \beta)\), which simplifies to \(v = \sqrt{\frac{2gh}{1+\beta}}\). For the solid cylinder, \(\beta = 1/2 \implies v_{\text{solid}} = \sqrt{\frac{4gh}{3}}\). For the hollow cylinder, \(\beta = 1 \implies v_{\text{hollow}} = \sqrt{gh}\). Taking the ratio gives \(\frac{v_{\text{solid}}}{v_{\text{hollow}}} = \sqrt{\frac{4}{3}}\).

Award [1] for using conservation of energy including rotational kinetic energy and correctly finding the ratio.

According to the Stefan-Boltzmann law, the radiated power is \(P = \sigma A T^4 = 4\pi \sigma R^2 T^4\). For the second star, the power is \(P_2 = 4\pi \sigma (2R)^2 \left(\frac{T}{2}\right)^4 = 4\pi \sigma (4R^2) \left(\frac{T^4}{16}\right) = \frac{4}{16} (4\pi \sigma R^2 T^4) = \frac{1}{4} P\).

Award [1] for applying Stefan-Boltzmann law with area and temperature scaling to find the ratio.

For exactly three spectral lines to be observed in the emission spectrum, the electron must have been excited to the \(n = 3\) level. This is because the possible emission transitions from \(n = 3\) are \(3 \to 2\), \(2 \to 1\), and \(3 \to 1\), which produces exactly three distinct lines. The energy of the absorbed photon is the energy required to excite the electron from \(n = 1\) to \(n = 3\): \(\Delta E = E_3 - E_1 = -\frac{13.6}{9} - \left(-\frac{13.6}{1}\right) \approx -1.51 + 13.6 = 12.1\text{ eV}\).

Award [1] for determining that the final state is \(n=3\) and calculating the transition energy of \(12.1\text{ eV}\).

For single-slit diffraction, the angular position of the first-order minimum is given by \(\theta \approx \frac{\lambda}{b}\). The distance from the central maximum to the first minimum on the screen is \(x = D \tan\theta \approx D \theta = \frac{\lambda D}{b}\). The distance between the two first-order minima on either side of the central maximum is \(y = 2x = \frac{2\lambda D}{b}\). Solving for \(b\) gives \(b = \frac{2\lambda D}{y}\).

Award [1] for writing the condition for the first diffraction minimum, relating it to the linear distance on the screen, and solving for \(b\).

The Stefan-Boltzmann law states that the power radiated by a black body is proportional to the fourth power of its absolute temperature: \(P = \sigma A T^4\). If the absolute temperature is increased by \(25\%\), the new temperature is \(1.25 T\). The new power \(P'\) is proportional to \((1.25 T)^4 = 1.25^4 T^4 \approx 2.44 T^4\). This represents a power of \(2.44 P\). The percentage increase in power is \((2.44 - 1) \times 100\% = 144\%\).

Award 1 mark for correct answer D. Method: Recognize that radiated power is proportional to the fourth power of thermodynamic temperature. Compute \((1.25)^4 \approx 2.44\). Find the percentage increase as \(144\%\).

From the conservation of linear momentum, the common final velocity \(v_f\) after the completely inelastic collision is given by: \(m v = (m + 3m) v_f \implies v_f = \frac{v}{4}\). The initial kinetic energy is \(E_{k,i} = \frac{1}{2} m v^2\). The final kinetic energy is \(E_{k,f} = \frac{1}{2} (4m) v_f^2 = \frac{1}{2} (4m) \left(\frac{v}{4}\right)^2 = \frac{1}{8} m v^2 = \frac{1}{4} E_{k,i}\). The kinetic energy dissipated as thermal energy is \(E_{k,i} - E_{k,f} = \frac{3}{4} E_{k,i}\), which is \(75\%\) of the initial kinetic energy.

Award 1 mark for correct answer C. Method: Calculate the final velocity using conservation of momentum. Compute the initial and final kinetic energies of the system. Determine the percentage of energy lost as \(75\%\).

The rate of change of magnetic flux through the loop is given by \(\frac{\Delta \Phi}{\Delta t} = B \cdot s \cdot v\). According to Faraday's law of induction, the induced electromotive force (emf) is \(\varepsilon = B s v\). The electrical power dissipated in the loop is given by \(P = \frac{\varepsilon^2}{R} = \frac{(B s v)^2}{R} = \frac{B^2 s^2 v^2}{R}\).

Award 1 mark for correct answer C. Method: Use Faraday's law to express the induced emf as \(\varepsilon = B s v\). Apply the power equation \(P = \frac{\varepsilon^2}{R}\) to obtain the correct expression.

The energy of a photon emitted during a transition is given by \(\Delta E = E_i - E_f\), and the wavelength \(\lambda\) is related to this energy by \(\lambda = \frac{hc}{\Delta E}\). According to the hydrogen energy levels, \(\Delta E \propto \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\). For the \(3 \to 2\) transition: \(\frac{1}{\lambda} \propto \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \frac{5}{36}\). For the \(4 \to 2\) transition: \(\frac{1}{\lambda'} \propto \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = \frac{3}{16}\). Taking the ratio of these two equations yields: \(\frac{\lambda'}{\lambda} = \frac{5/36}{3/16} = \frac{20}{27}\). Therefore, \(\lambda' = \frac{20}{27} \lambda\).

Award 1 mark for correct answer B. Method: Express the relationship between transition energy and wavelength. Set up the Rydberg proportions for both transitions, and take the ratio to find the new wavelength in terms of \(\lambda\).

The fringe spacing \(s\) in a double-slit experiment is given by the formula \(s = \frac{\lambda D}{d}\). Let the new parameters be \(\lambda' = 1.5\lambda\), \(d' = 0.5d\), and \(D' = 2D\). The new fringe spacing is \(s' = \frac{\lambda' D'}{d'} = \frac{(1.5\lambda) (2D)}{0.5d} = \frac{3\lambda D}{0.5d} = 6 \left(\frac{\lambda D}{d}\right) = 6s\).

Award 1 mark for correct answer C. Method: Use the double-slit fringe spacing formula \(s = \frac{\lambda D}{d}\). Substitute the modified values for wavelength, separation, and distance. Simplify to find the factor of 6.

Let the charge \(+2Q\) be at \(x = 0\) and the charge \(-Q\) be at \(x = d\). We look for a point at \(x > d\) where the potential is zero. Let the distance from \(-Q\) to this point be \(r\), so the point is at \(x = d + r\). The potential at this point is \(V = \frac{1}{4\pi\varepsilon_0} \left( \frac{2Q}{d + r} + \frac{-Q}{r} \right) = 0\). This simplifies to \(\frac{2}{d + r} = \frac{1}{r} \implies 2r = d + r \implies r = d\). Thus, the distance from \(-Q\) is \(d\).

Award 1 mark for correct answer B. Method: Express the potential at a point on the line outside the charges as the sum of potentials due to each charge. Set this sum to zero and solve for the distance from the charge \(-Q\).

For a complete thermodynamic cycle, the change in internal energy is zero (\(\Delta U_{\text{net}} = 0\)). From the First Law of Thermodynamics, the net heat added to the gas is equal to the net work done by the gas: \(Q_{\text{net}} = W_{\text{net}} = +150\text{ J}\). The net heat is the sum of the heat exchanges in each of the three processes: \(Q_{\text{net}} = Q_{\text{isobaric}} + Q_{\text{isochoric}} + Q_{\text{adiabatic}}\). Since the third process is adiabatic, \(Q_{\text{adiabatic}} = 0\). This gives \(+400\text{ J} + Q_{\text{isochoric}} + 0 = +150\text{ J} \implies Q_{\text{isochoric}} = -250\text{ J}\).

Award 1 mark for correct answer A. Method: Recognize that the net heat equals net work for a cyclic process. State that heat transfer during an adiabatic process is zero. Solve the summation of heat exchanges for the isochoric process to find \(-250\text{ J}\).

Einstein's photoelectric equation is \(E_{k,\max} = hf - \Phi\), where \(\Phi\) is the work function of the metal. If the frequency is doubled to \(2f\), the new maximum kinetic energy \(E'_{k,\max}\) is given by \(E'_{k,\max} = h(2f) - \Phi = 2hf - \Phi\). Since \(hf = E_{k,\max} + \Phi\), we can substitute this back: \(E'_{k,\max} = 2(E_{k,\max} + \Phi) - \Phi = 2E_{k,\max} + \Phi\). Since the work function \(\Phi\) is a positive quantity, \(E'_{k,\max} > 2E_{k,\max}\).

Award 1 mark for correct answer B. Method: Set up Einstein's photoelectric equation for both frequencies. Express the new maximum kinetic energy in terms of the original maximum kinetic energy and the work function. Conclude that it is greater than double the original value because the work function is positive.

The change in momentum of the ball is \(\Delta p = m(v - u)\). Taking the direction of rebound as positive: \(\Delta p = 0.20\text{ kg} \times (12\text{ m s}^{-1} - (-15\text{ m s}^{-1})) = 0.20 \times 27 = 5.4\text{ N s}\). The average force is related to momentum change by \(F = \frac{\Delta p}{\Delta t}\). Therefore, \(\Delta t = \frac{\Delta p}{F} = \frac{5.4\text{ N s}}{180\text{ N}} = 0.030\text{ s} = 30\text{ ms}\).

Award 1 mark for the correct calculation leading to option C.

From the photoelectric equation, \(E_{\text{max}} = \frac{hc}{\lambda} - \Phi\). For wavelength \(\frac{\lambda}{2}\), the incident photon energy is doubled, so \(3E_{\text{max}} = \frac{2hc}{\lambda} - \Phi\). Rearranging the first equation gives \(\frac{hc}{\lambda} = E_{\text{max}} + \Phi\). Substituting this into the second equation: \(3E_{\text{max}} = 2(E_{\text{max}} + \Phi) - \Phi\), which simplifies to \(3E_{\text{max}} = 2E_{\text{max}} + 2\Phi - \Phi\), resulting in \(\Phi = E_{\text{max}}\).

Award 1 mark for the correct algebraic derivation leading to option B.

According to Wien's displacement law, the temperature \(T\) of a blackbody is inversely proportional to its peak wavelength \(\lambda_{\text{max}}\). Therefore, \(\frac{T_X}{T_Y} = \frac{\lambda_Y}{\lambda_X} = 2\). The luminosity \(L\) is given by the Stefan-Boltzmann law as \(L = 4\pi R^2 \sigma T^4\). Taking the ratio: \(\frac{L_X}{L_Y} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{T_X}{T_Y}\right)^4 = \left(\frac{1}{4}\right)^2 \times (2)^4 = \frac{1}{16} \times 16 = 1.0\).

Award 1 mark for the correct application of Wien's law and the Stefan-Boltzmann law leading to option B.

As the loop enters the field, the change in magnetic flux induces an emf of magnitude \(V = BLv\). This produces an induced current \(I = \frac{V}{R} = \frac{BLv}{R}\). The magnetic force acting on the leading edge of the loop is \(F_M = BIL = B \left(\frac{BLv}{R}\right) L = \frac{B^2 L^2 v}{R}\). To maintain a constant speed, the external force must be equal in magnitude and opposite in direction to this magnetic force.

Award 1 mark for using Faraday's law, Ohm's law, and the magnetic force formula to find option C.

The angular position of the first minimum in single slit diffraction is given by \(\sin\theta_1 = \frac{\lambda}{b}\). Using the small angle approximation, the half-width is \(\theta_1 = \frac{\lambda}{b}\), and the angular width of the central maximum is \(\theta = 2\theta_1 = \frac{2\lambda}{b}\). If the new slit width is \(b' = 2b\) and the new wavelength is \(\lambda' = \frac{\lambda}{2}\), the new angular width is \(\theta' = \frac{2(\lambda/2)}{2b} = \frac{\lambda}{2b} = \frac{\theta}{4}\).

Award 1 mark for identifying the relationship between angular width, wavelength, and slit width to obtain option A.

The energy of the emitted photon is \(\Delta E = E_{\text{initial}} - E_{\text{final}} = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}\). Convert this energy to Joules: \(\Delta E = 1.89 \times 1.60 \times 10^{-19}\text{ J} = 3.024 \times 10^{-19}\text{ J}\). Using \(E = hf\), the frequency is \(f = \frac{E}{h} = \frac{3.024 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} = 4.56 \times 10^{14}\text{ Hz}\).

Award 1 mark for calculating the energy change in Joules and solving for frequency to get option B.

The Lorentz factor is \(\gamma = \frac{1}{\sqrt{1 - (v/c)^2}} = \frac{1}{\sqrt{1 - 0.64}} = 1.67\) (or \(\frac{5}{3}\)). The contracted length measured by the Earth observer is \(L = \frac{L_0}{\gamma} = 0.60 L_0\). The time interval measured by the Earth observer for this contracted length to pass a fixed point at speed \(v = 0.80c\) is \(\Delta t = \frac{L}{v} = \frac{0.60 L_0}{0.80c} = 0.75 \frac{L_0}{c}\).

Award 1 mark for correctly applying length contraction and kinematics in the observer's frame to find option A.

The root-mean-square speed of an ideal gas is proportional to the square root of its absolute temperature: \(v_{\text{rms}} \propto \sqrt{T}\). If the temperature increases by \(25\%\), the new temperature is \(T' = 1.25 T\). The new speed is \(v_{\text{rms}}' = \sqrt{1.25} v_{\text{rms}} \approx 1.118 v_{\text{rms}}\). The percentage increase in speed is \((1.118 - 1) \times 100\% = 11.8\%\), which rounds to \(12\%\).

Award 1 mark for identifying the square root relationship and calculating the percentage increase to get option A.

(a) Period of one oscillation:
\(T = \frac{t}{20} = \frac{35.8}{20} = 1.79\text{ s}\)

Absolute uncertainty in \(T\):
\(\Delta T = \frac{\Delta t}{20} = \frac{0.2}{20} = 0.01\text{ s}\)

Thus, \(T = 1.79 \pm 0.01\text{ s}\).

(b) Using the pendulum period formula:
\(T = 2\pi \sqrt{\frac{L}{g}} \implies g = \frac{4\pi^2 L}{T^2}\)

Substituting the values:
\(g = \frac{4\pi^2 \times 0.800}{1.79^2} = \frac{31.583}{3.2041} = 9.857\text{ m s}^{-2}\)

This rounds to \(9.9\text{ m s}^{-2}\) to 2 significant figures.

(c) Percentage uncertainty in \(T\):
\(\%\Delta T = \frac{\Delta T}{T} \times 100\% = \frac{0.01}{1.79} \times 100\% \approx 0.559\%\)

Since \(T^2\) is proportional to the square of \(T\), its percentage uncertainty is:
\(\%\Delta (T^2) = 2 \times \%\Delta T = 2 \times 0.559\% = 1.12\%\) (accept \(1.1\%\)).

(d) Percentage uncertainty in \(L\):
\(\%\Delta L = \frac{\Delta L}{L} \times 100\% = \frac{0.002}{0.800} \times 100\% = 0.250\%\)

Total percentage uncertainty in \(g\):
\(\%\Delta g = \%\Delta L + \%\Delta (T^2) = 0.250\% + 1.117\% = 1.367\%\) (or \(1.37\%\))

Absolute uncertainty in \(g\):
\(\Delta g = 1.367\% \times 9.857\text{ m s}^{-2} = 0.135\text{ m s}^{-2} \approx 0.1\text{ m s}^{-2}\) (or \(0.14\text{ m s}^{-2}\)).

(e) The theoretical formula \(T = 2\pi \sqrt{\frac{L}{g}}\) is derived using the small-angle approximation (\(\sin\theta \approx \theta\)). At larger release angles such as \(30^\circ\), the actual period of the pendulum is longer than predicted by this small-angle formula. Because \(g = \frac{4\pi^2 L}{T^2}\), a larger measured period \(T\) in the denominator will result in a systematically lower calculated value for \(g\).

(a)
- \(T = 1.79\text{ s}\) [1]
- \(\Delta T = 0.01\text{ s}\) [1]

(b)
- Clear algebraic rearrangement to find \(g = \frac{4\pi^2 L}{T^2}\) [1]
- Correct substitution and calculation showing \(9.86\text{ m s}^{-2}\) or \(9.9\text{ m s}^{-2}\) [1]

(c)
- Calculates percentage uncertainty in \(T\) to be \(0.56\%\) [1]
- Correctly doubles this value to find the percentage uncertainty in \(T^2\) as \(1.1\%\) or \(1.12\%\) [1]

(d)
- Sums the percentage uncertainties: \(0.25\% + 1.12\% = 1.37\%\) (or \(1.4\%\)) [1]
- Correctly calculates absolute uncertainty to be \(0.1\text{ m s}^{-2}\) or \(0.14\text{ m s}^{-2}\) [1]

(e)
- Explains that larger angles increase the actual period of oscillation relative to the small-angle formula [1]
- Correctly relates this to the formula to conclude that a larger period \(T\) leads to an underestimated (smaller) value of \(g\) [1]

(a) Cross-sectional area \(A\):
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\) (or \(1.1 \times 10^{-7}\text{ m}^2\))

Percentage uncertainty in \(d\):
\(\%\Delta d = \frac{0.02}{0.38} \times 100\% \approx 5.26\%\)

Percentage uncertainty in \(A\) is twice the percentage uncertainty in \(d\):
\(\%\Delta A = 2 \times 5.26\% = 10.53\%\) (rounds to \(11\%\) or \(10.5\%\))

(b) Resistivity \(\rho = \frac{R A}{L}\):
\(\rho = \frac{3.4 \times 1.134 \times 10^{-7}}{0.750} = 5.14 \times 10^{-7}\ \Omega\text{ m}\) (rounds to \(5.1 \times 10^{-7}\ \Omega\text{ m}\))

(c) Percentage uncertainty in \(R\):
\(\%\Delta R = \frac{0.1}{3.4} \times 100\% = 2.94\%\)

Percentage uncertainty in \(L\):
\(\%\Delta L = \frac{0.003}{0.750} \times 100\% = 0.40\%\)

Total percentage uncertainty in \(\rho\):
\(\%\Delta \rho = \%\Delta R + \%\Delta A + \%\Delta L = 2.94\% + 10.53\% + 0.40\% = 13.87\%\) (rounds to \(14\%\))

Absolute uncertainty in \(\rho\):
\(\Delta \rho = 13.87\% \times 5.14 \times 10^{-7} \approx 0.71 \times 10^{-7}\ \Omega\text{ m}\) (or \(0.7 \times 10^{-7}\ \Omega\text{ m}\))

(d) Practical source of systematic error:
- Contact resistance at the alligator clips / connection terminals of the circuit adds extra resistance. This systematically increases the measured resistance, leading to an overestimation of resistivity.
- Minimization: Use a four-probe measurement technique (Kelvin connection), or clean the terminals with sandpaper/emery paper to remove oxide layers and ensure connections are clamped very tightly.

(a)
- Correct calculation of area: \(1.13 \times 10^{-7}\text{ m}^2\) [1]
- Correct percentage uncertainty in \(d\): \(5.3\%\) [1]
- Correct percentage uncertainty in \(A\): \(11\%\) (or \(10.5\%\)) [1]

(b)
- Recalls resistivity formula: \(\rho = \frac{R A}{L}\) [1]
- Correct calculation of \(\rho = 5.1 \times 10^{-7}\ \Omega\text{ m}\) (allow error carried forward from a) [1]

(c)
- Calculates percentage uncertainty in \(R\) (\(2.9\%\)) and \(L\) (\(0.4\%\)) [1]
- Sums all three percentage uncertainties to find total (\(14\%\) or \(13.9\%\)) [1]
- Correctly calculates absolute uncertainty to be \(\pm 0.7 \times 10^{-7}\ \Omega\text{ m}\) [1]

(d)
- Identifies contact resistance as a systematic error that artificially inflates the measured resistance [1]
- Suggests a valid remedy (e.g. cleaning connection wires/plates with steel wool/sandpaper, tightening clamps, or using a four-point probe setup) [1]

(a) The friction force is \(f = \mu_d m_A g = 0.20 \times 1.5 \times 9.81 = 2.943\text{ N}\). The acceleration is \(a = -f / m_A = -\mu_d g = -1.962\text{ m s}^{-2}\). Using \(v_f^2 = u^2 + 2ad\), we find \(v_f = \sqrt{4.0^2 + 2 \times (-1.962) \times 0.90} = 3.53\text{ m s}^{-1}\). (b) Using conservation of momentum: \(m_A v_f = (m_A + m_B) v'\). Thus, \(1.5 \times 3.53 = (1.5 + 2.5) v'\), which gives \(v' = 1.32\text{ m s}^{-1}\). (c) The average net force on block B is \(F = \frac{\Delta p_B}{\Delta t} = \frac{m_B v' - 0}{\Delta t} = \frac{2.5 \times 1.32}{0.12} = 27.5\text{ N}\). (d) Total momentum is not conserved over the entire process because there is an external net force (friction) acting on the system. Total kinetic energy is also not conserved because of work done by friction and energy dissipation during the inelastic collision.

(a) 1 mark for calculating acceleration, 1 mark for using kinematic equation, 1 mark for correct final speed. (b) 1 mark for momentum conservation statement, 1 mark for substitution, 1 mark for final velocity. (c) 1 mark for rate of change of momentum formula, 1 mark for substitution, 1 mark for correct average force. (d) 1 mark for identifying that friction is an external force reducing momentum, 1.25 marks for discussing kinetic energy dissipation.

(a) The magnetic flux is given by \(\Phi = B A \cos\theta\). Since the field is perpendicular to the loop, \(\cos\theta = 1\). The area inside the field is \(A = L x\), so \(\Phi = B L x\). (b) According to Faraday's law, the magnitude of the induced emf is \(\varepsilon = \frac{d\Phi}{dt} = B L \frac{dx}{dt} = B L v = 0.45 \times 0.25 \times 2.0 = 0.225\text{ V}\). (c) The induced current is \(I = \frac{\varepsilon}{R} = \frac{0.225}{1.5} = 0.15\text{ A}\). By Lenz's law, the induced current must oppose the change in magnetic flux. Since the flux into the page is increasing, the induced magnetic field must point out of the page. By the right-hand grip rule, this corresponds to an anticlockwise current. (d) The magnetic force on the leading wire is \(F_B = I L B = 0.15 \times 0.25 \times 0.45 = 0.0169\text{ N}\). For a constant speed, the external force must be equal in magnitude and opposite in direction, so \(F_{ext} = 0.0169\text{ N}\).

(a) 1 mark for area expression, 1 mark for flux definition. (b) 1 mark for Faraday's law formula, 1 mark for substitution, 1 mark for correct calculation. (c) 1 mark for Ohm's law, 1 mark for direction (anticlockwise), 1.25 marks for explaining using Lenz's law. (d) 1 mark for magnetic force formula, 1 mark for substitution, 1 mark for final force value matching external force.

(a) The ground state is the lowest possible energy level of an electron in an atom, where it is most stable. (b) The energy difference is \(\Delta E = E_3 - E_2 = -6.04 - (-13.6) = 7.56\text{ eV}\). In Joules, \(\Delta E = 7.56 \times 1.60 \times 10^{-19} = 1.21 \times 10^{-18}\text{ J}\). Wavelength \(\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{1.21 \times 10^{-18}} = 1.64 \times 10^{-7}\text{ m}\). (c) For absorption, the photon energy must match the transition energy exactly (or be greater than the ionization energy of \(54.4\text{ eV}\)). Transition from ground state to \(E_2\) requires \(40.8\text{ eV}\), and to \(E_3\) requires \(48.36\text{ eV}\). Since \(48.0\text{ eV}\) does not match any transition and is less than \(54.4\text{ eV}\), the photons cannot be absorbed. (d) The colliding electron can transfer energy in discrete amounts corresponding to excitation energies. It can excite the atom to \(E_2\) by transferring \(40.8\text{ eV}\), leaving the electron with \(42.0 - 40.8 = 1.2\text{ eV}\). If the collision is elastic, no energy is transferred, so the electron retains \(42.0\text{ eV}\).

(a) 2 marks for defining ground state as the lowest/most stable energy state. (b) 1 mark for calculating energy difference in eV, 1 mark for converting to Joules, 1.25 marks for correct wavelength. (c) 1 mark for stating the resonance requirement, 1 mark for calculating the target transitions, 1 mark for concluding no absorption occurs. (d) 1 mark for noting the possibility of elastic collision, 1 mark for identifying the \(E_2\) transition requirement, 1 mark for calculating the \(1.2\text{ eV}\) option.

(a) The energy required to heat the ice is \(Q_1 = m c_{\text{ice}} \Delta T = 0.65 \times 2.1 \times 10^3 \times 15 = 2.0 \times 10^4\text{ J}\). (b) The energy to melt the ice is \(Q_2 = m L_f = 0.65 \times 3.3 \times 10^5 = 2.145 \times 10^5\text{ J}\). The time taken is \(t_2 = Q_2 / P = 2.145 \times 10^5 / 850 = 252\text{ s} \approx 250\text{ s}\). (c) The energy to heat the water is \(Q_3 = m c_{\text{water}} \Delta T_{water} = 0.65 \times 4.2 \times 10^3 \times 80 = 2.184 \times 10^5\text{ J}\). Total energy \(Q_{total} = Q_1 + Q_2 + Q_3 = 4.53 \times 10^5\text{ J}\). Total time \(t_{total} = Q_{total} / P = 4.53 \times 10^5 / 850 = 533\text{ s}\). (d) If the container is not perfectly insulated, heat will be lost to the surroundings. The rate of heat loss increases as the water temperature rises. Consequently, less net energy is transferred to the water, so the final temperature will be lower than \(80^\circ\text{C}\).

(a) 1 mark for formula, 1 mark for correct calculation. (b) 1 mark for phase change energy, 1 mark for using \(P = Q/t\), 1 mark for showing approximately \(250\text{ s}\). (c) 1 mark for heating water energy, 1 mark for summing the energies, 1.25 marks for total time calculation. (d) 1 mark for stating that heat is lost to the surroundings, 1 mark for explaining that the rate of loss increases with temperature, 1 mark for concluding the final temperature is lower.

(a) The first minimum is at \(\theta = \lambda/b = 632.8 \times 10^{-9} / 0.120 \times 10^{-3} = 5.27 \times 10^{-3}\text{ rad}\). The angular width of the central maximum is \(2\theta = 1.05 \times 10^{-2}\text{ rad}\). (b) The physical width is \(W = 2 D \tan\theta \approx 2 D \theta = 2 \times 2.40 \times 5.27 \times 10^{-3} = 0.0253\text{ m} = 2.53\text{ cm}\). (c) The new pattern consists of fine double-slit fringes spaced at \(\Delta y = \lambda D / d = 3.16\text{ mm}\) inside the single-slit diffraction envelope. This envelope has a central maximum of width \(2.53\text{ cm}\) containing about 7 to 8 bright fringes. (d) If the slit width is halved, the diffraction envelope becomes twice as wide. The individual fringe spacing remains unchanged, meaning twice as many interference fringes will be visible within the central maximum of the envelope.

(a) 1 mark for formula, 1 mark for calculating \(\theta\), 1 mark for doubling to find total angular width. (b) 1 mark for using \(W = 2D\theta\), 1 mark for substitution, 1 mark for correct calculation. (c) 1 mark for identifying double-slit fringes modulated by single-slit envelope, 1 mark for calculating fringe spacing, 1.25 marks for describing how the fringes fit in the envelope. (d) 1 mark for stating the envelope widens, 1 mark for stating that more fringes are visible.

(a) The PV diagram shows process AB as a horizontal line at \(1.0 \times 10^5\text{ Pa}\) from \(2.0 \times 10^{-3}\) to \(5.0 \times 10^{-3}\text{ m}^3\); process BC as a vertical downward line to \(4.0 \times 10^4\text{ Pa}\) at \(5.0 \times 10^{-3}\text{ m}^3\); and process CA as a smooth curve rising back to state A. (b) Work done \(W_{AB} = P \Delta V = 1.0 \times 10^5 \times (5.0 \times 10^{-3} - 2.0 \times 10^{-3}) = 300\text{ J}\). (c) From the first law of thermodynamics, \(Q_{AB} = \Delta U_{AB} + W_{AB} = 450 + 300 = 750\text{ J}\). (d) Work done is \(W_{CA} = P_C V_C \ln(V_A/V_C) = (4.0 \times 10^4 \times 5.0 \times 10^{-3}) \ln(2.0/5.0) = 200 \ln(0.4) \approx -183\text{ J}\). The negative sign indicates that work is done on the gas during compression.

(a) 1 mark for horizontal isobaric line, 1 mark for vertical isochoric line, 1 mark for isothermal curve. (b) 1 mark for formula, 1 mark for correct value. (c) 1 mark for first law, 1 mark for correct substitution. (d) 1 mark for substituting values into the formula, 1 mark for calculating \(\ln(0.4)\), 1 mark for showing approximately \(-183\text{ J}\), 1.25 marks for explaining the negative sign.

(a) Photon energy \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{380 \times 10^{-9}} = 5.23 \times 10^{-19}\text{ J}\). In eV, \(E = \frac{5.23 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.27\text{ eV}\). (b) The maximum kinetic energy is \(E_{k,max} = e V_s = 1.15\text{ eV}\). The work function is \(\Phi = E - E_{k,max} = 3.27 - 1.15 = 2.12\text{ eV}\). (c) The threshold frequency is \(f_0 = \frac{\Phi}{h} = \frac{2.12 \times 1.60 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.12 \times 10^{14}\text{ Hz}\). (d) The maximum momentum is \(p = \sqrt{2 m_e E_{k,max}} = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.15 \times 1.60 \times 10^{-19}} = 5.79 \times 10^{-25}\text{ kg m s}^{-1}\). The de Broglie wavelength is \(\lambda_{dB} = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.79 \times 10^{-25}} = 1.15 \times 10^{-9}\text{ m} = 1.15\text{ nm}\).

(a) 1 mark for \(E = hc/\lambda\), 1 mark for substituting values, 1 mark for converting to eV. (b) 1 mark for linking stopping potential to max KE, 1.25 marks for work function calculation. (c) 1 mark for \(\Phi = h f_0\), 1 mark for substitution, 1 mark for correct frequency. (d) 1 mark for relating KE to momentum, 1 mark for using de Broglie formula, 1 mark for correct wavelength.

(a) For the block: gravity \(mg\) acts downwards, tension \(T\) acts upwards. For the cylinder: normal support force acts upwards at the axis, gravity \(Mg\) acts downwards at the center of mass, and tension \(T\) acts downwards at the perimeter (tangential). (b) For the block: \(m g - T = m a\). For the cylinder: \(\tau = I \alpha \implies T R = \frac{1}{2} M R^2 (a / R) \implies T = \frac{1}{2} M a\). Substituting \(T\) into the block equation: \(m g - \frac{1}{2} M a = m a \implies m g = (m + M/2) a \implies a = \frac{g}{1 + \frac{M}{2m}}\). (c) Substituting values: \(a = \frac{9.81}{1 + 3.2/(2 \times 1.2)} = 4.20\text{ m s}^{-2}\). (d) Tension \(T = \frac{1}{2} M a = 0.5 \times 3.2 \times 4.20 = 6.72\text{ N}\) (or using \(T = m(g-a) = 1.2 \times (9.81 - 4.20) = 6.73\text{ N}\)).

(a) 1.5 marks for block FBD, 1.5 marks for cylinder FBD. (b) 1 mark for linear Newton's second law, 1 mark for rotational Newton's second law, 1.25 marks for algebraic substitution and derivation. (c) 1 mark for substituting values, 1 mark for calculating correct acceleration. (d) 1 mark for using tension formula (either block or cylinder), 1 mark for substituting values, 1 mark for correct final tension.