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In \(\beta^+\) decay, a proton in the fluorine nucleus decays into a neutron, a positron (\(e^+\)), and an electron neutrino (\(\nu_e\)). The quark composition of a proton is \(uud\), and that of a neutron is \(udd\). Thus, the decay of a proton to a neutron involves an up quark changing into a down quark (\(u \to d\)). To conserve lepton number, the antiparticle positron (lepton number \(-1\)) is emitted together with an electron neutrino (lepton number \(+1\)).

1 mark for the correct option B. Reject options with incorrect quark changes (A and D) or incorrect lepton pair (C).

Electron capture is mediated by the weak interaction because it involves a change in quark flavour (one up quark in the proton changes into a down quark in the neutron) and involves leptons. None of the particles involved (proton, electron, neutron, electron neutrino) possess any strange quarks, so the initial and final total strangeness are both \(0\). Thus, the change in the total strangeness is \(0\).

1 mark for the correct option B. Reject options with incorrect force (A and C) or incorrect change in strangeness (D).

The change in momentum \(\Delta p\) is equal to the area under the force-time graph. The area under the graph from \(t = 0\) to \(t = 5.0\text{ s}\) consists of: 1. A triangular area from \(t = 0\) to \(t = 3.0\text{ s}\): \(\text{Area}_1 = \frac{1}{2} \times 3.0\text{ s} \times 12.0\text{ N} = 18.0\text{ N s}\). 2. A rectangular area from \(t = 3.0\text{ s}\) to \(t = 5.0\text{ s}\): \(\text{Area}_2 = (5.0 - 3.0)\text{ s} \times 12.0\text{ N} = 24.0\text{ N s}\). The total change in momentum is: \(\Delta p = 18.0 + 24.0 = 42.0\text{ N s}\). The change in velocity is: \(\Delta v = \frac{\Delta p}{m} = \frac{42.0\text{ N s}}{0.50\text{ kg}} = 84.0\text{ m s}^{-1}\). The final velocity \(v\) is: \(v = u + \Delta v = 4.0\text{ m s}^{-1} + 84.0\text{ m s}^{-1} = 88.0\text{ m s}^{-1}\).

1 mark for the correct option C. Incorrect option B neglects the initial velocity. Incorrect option A is the value of the impulse itself without dividing by mass.

Let the direction to the right be positive. The initial velocities are \(u_{\text{X}} = +5.0\text{ m s}^{-1}\) and \(u_{\text{Y}} = -2.0\text{ m s}^{-1}\). Since the collision is perfectly elastic, the relative speed of approach is equal to the relative speed of separation: \(u_{\text{X}} - u_{\text{Y}} = v_{\text{Y}} - v_{\text{X}} \implies 5.0 - (-2.0) = v_{\text{Y}} - v_{\text{X}} \implies v_{\text{Y}} - v_{\text{X}} = 7.0 \implies v_{\text{Y}} = v_{\text{X}} + 7.0\). By conservation of linear momentum: \(m_{\text{X}} u_{\text{X}} + m_{\text{Y}} u_{\text{Y}} = m_{\text{X}} v_{\text{X}} + m_{\text{Y}} v_{\text{Y}} \implies (2.0)(5.0) + (3.0)(-2.0) = 2.0 v_{\text{X}} + 3.0 v_{\text{Y}} \implies 10.0 - 6.0 = 2.0 v_{\text{X}} + 3.0 (v_{\text{X}} + 7.0) \implies 4.0 = 5.0 v_{\text{X}} + 21.0 \implies 5.0 v_{\text{X}} = -17.0 \implies v_{\text{X}} = -3.4\text{ m s}^{-1}\). A negative sign indicates motion to the left, so the velocity of glider X after the collision is \(3.4\text{ m s}^{-1}\) to the left.

1 mark for the correct option A. Reject B and C which are incorrect velocity values. Reject D which has the incorrect direction.

The extension \(e\) of a wire is given by \(e = \frac{F L}{A E} = \frac{4 F L}{\pi d^2 E}\), where \(E\) is the Young modulus and \(d\) is the diameter. For wire P: \(x = \frac{4 F L}{\pi d^2 E_{\text{P}}}\). For wire Q: \(e_{\text{Q}} = \frac{4 (3F) (2L)}{\pi (2d)^2 (2E_{\text{P}})} = \frac{4 \times 6 F L}{\pi \times 4 d^2 \times 2 E_{\text{P}}} = \frac{6}{8} \left( \frac{4 F L}{\pi d^2 E_{\text{P}}} \right) = 0.75 x\).

1 mark for the correct option B. Reject A (if they forgot to square the diameter factor), C, and D (other algebraic errors).

The elastic strain energy recovered when the wire is unloaded is the area under the unloading curve. Since the unloading curve is a straight line from \((5.0\text{ mm}, 200\text{ N})\) to \((2.0\text{ mm}, 0\text{ N})\), this area is a triangle of height \(200\text{ N}\) and base \(5.0\text{ mm} - 2.0\text{ mm} = 3.0\text{ mm} = 3.0 \times 10^{-3}\text{ m}\). Elastic strain energy recovered \(= \frac{1}{2} \times 3.0 \times 10^{-3}\text{ m} \times 200\text{ N} = 0.30\text{ J}\). The work done during loading is \(0.45\text{ J}\). The energy retained as thermal energy due to plastic deformation is the difference: \(\text{Thermal energy} = \text{Work done during loading} - \text{Elastic strain energy recovered} = 0.45\text{ J} - 0.30\text{ J} = 0.15\text{ J}\).

1 mark for the correct option A. Reject C (which is the recovered elastic energy) and B (incorrect calculation).

At \(20^\circ\text{C}\), the potential difference across the thermistor is \(V_{\text{th}} = 4.5\text{ V}\). The potential difference across the fixed resistor \(R = 4.0\ \Omega\) is \(V_R = 6.0 - 4.5 = 1.5\text{ V}\). Since \(R\) and the thermistor are in series, the ratio of their resistances equals the ratio of their potential differences: \(\frac{R_{\text{th}}}{R} = \frac{V_{\text{th}}}{V_R} = \frac{4.5}{1.5} = 3 \implies R_{\text{th}} = 3 \times 4.0 = 12.0\ \Omega\). At \(60^\circ\text{C}\), the resistance of the thermistor becomes \(R_{\text{th}}' = \frac{12.0}{3} = 4.0\ \Omega\). Now, the potential difference across the thermistor is: \(V_{\text{th}}' = \frac{R_{\text{th}}'}{R + R_{\text{th}}'} \times \text{e.m.f.} = \frac{4.0}{4.0 + 4.0} \times 6.0 = 3.0\text{ V}\).

1 mark for the correct option C. Reject A (if they assumed the potential difference itself scaled by \(\frac{1}{3}\)) and D.

The fringe width \(w\) is given by: \(w = \frac{\lambda D}{a}\). Here, \(\lambda = 630\text{ nm} = 6.30 \times 10^{-7}\text{ m}\), \(D = 1.8\text{ m}\), and \(a = 0.30\text{ mm} = 3.0 \times 10^{-4}\text{ m}\). Therefore, \(w = \frac{6.30 \times 10^{-7} \times 1.8}{3.0 \times 10^{-4}} = 3.78 \times 10^{-3}\text{ m} = 3.78\text{ mm}\). The distance from the central bright fringe to the third dark fringe (third minimum) is \(2.5 w\) (since the first minimum is at \(0.5 w\), the second is at \(1.5 w\), and the third is at \(2.5 w\)). Distance \(= 2.5 \times 3.78\text{ mm} = 9.45\text{ mm} \approx 9.5\text{ mm}\).

1 mark for the correct option B. Reject C (if calculated to the third bright fringe, \(3.0 w = 11.3\text{ mm}\)), A (if calculated to the second dark fringe, \(1.5 w = 5.7\text{ mm}\)), and D (if calculated to the fourth dark fringe, \(3.5 w = 13.2\text{ mm}\)).

In the decay, the initial state contains quarks \(u\), \(s\), and \(s\). The final state contains a \(\Lambda^0\) (quarks \(u\), \(d\), \(s\)) and a \(\pi^0\) (quarks \(u\), \(\bar{u}\)). Since one \(s\) quark remains in the final state baryon and a \(u\bar{u}\) pair is created from energy, the fundamental change is that the other strange (\(s\)) quark decays into a down (\(d\)) quark.

Correct answer is A (1 mark).

Using conservation of linear momentum, taking right as positive: \(m(3v) + 2m(-v) = m(-v) + 2m(v_Q)\). This simplifies to \(3mv - 2mv = -mv + 2mv_Q\), which gives \(mv = -mv + 2mv_Q\) and thus \(v_Q = +v\) (to the right). To check if the collision is elastic, we compare kinetic energies. Initial kinetic energy is \(0.5m(3v)^2 + 0.5(2m)(-v)^2 = 5.5mv^2\). Final kinetic energy is \(0.5m(-v)^2 + 0.5(2m)(v)^2 = 1.5mv^2\). Since kinetic energy is lost, the collision is inelastic.

Correct answer is B (1 mark).

The current in the circuit is given by \(I = \frac{E}{R+r}\). As the resistance \(R\) is decreased, the total resistance of the circuit decreases, so the current \(I\) increases. The potential difference across the internal resistance is \(V_r = Ir\); since \(I\) increases, \(V_r\) increases. The voltmeter reading (terminal potential difference) is \(V = E - Ir\); since \(I\) increases, the voltmeter reading \(V\) decreases.

Correct answer is B (1 mark).

The extension is given by \(\Delta L = \frac{FL}{AE}\) where \(A = \frac{\pi d^2}{4}\). This means \(\Delta L \propto \frac{L}{d^2}\) because the force \(F\) and Young Modulus \(E\) are the same. For wire X, \(\Delta L_X \propto \frac{L}{d^2}\). For wire Y, \(\Delta L_Y \propto \frac{2L}{(2d)^2} = \frac{L}{2d^2}\). Therefore, the ratio \(\frac{\Delta L_X}{\Delta L_Y} = \frac{L/d^2}{L/(2d^2)} = 2\).

Correct answer is D (1 mark).

The original fringe spacing is given by \(x = \frac{\lambda D}{a}\). Under the new conditions, the new fringe spacing is \(x' = \frac{\lambda' D'}{a'} = \frac{(1.5\lambda)(2D)}{(a/2)} = \frac{3.0\lambda D}{0.5 a} = 6 \frac{\lambda D}{a} = 6x\).

Correct answer is D (1 mark).

The motion has three phases: 1) Acceleration: \(s_1 = ut + 0.5at^2 = 0 + 0.5(2.0)(5.0)^2 = 25\text{ m}\). The velocity reached is \(v = u + at = 0 + 2.0(5.0) = 10\text{ m s}^{-1}\). 2) Constant velocity: \(s_2 = vt = 10 \times 10 = 100\text{ m}\). 3) Deceleration: \(s_3 = \frac{u+v}{2} t = \frac{10+0}{2} \times 4.0 = 20\text{ m}\). The total distance is \(s_1 + s_2 + s_3 = 25 + 100 + 20 = 145\text{ m}\).

Correct answer is B (1 mark).

Electric field strength \(E\) is force per unit charge (\(E = F/q\)). The SI unit is \(\text{N C}^{-1}\). Since force has base units of \(\text{kg m s}^{-2}\) and charge has base units of \(\text{A s}\), the base units of electric field strength are \(\frac{\text{kg m s}^{-2}}{\text{A s}} = \text{kg m s}^{-3}\text{ A}^{-1}\).

Correct answer is B (1 mark).

The useful power output consists of raising the potential energy of the water plus giving it kinetic energy. Rate of potential energy increase = \(\frac{dm}{dt} g h = 8.0 \times 9.81 \times 15 = 1177.2\text{ W}\). Rate of kinetic energy increase = \(0.5 \frac{dm}{dt} v^2 = 0.5 \times 8.0 \times 6.0^2 = 144.0\text{ W}\). Total useful power output = \(1177.2 + 144.0 = 1321.2\text{ W}\). Power input is \(1.8\text{ kW} = 1800\text{ W}\). Efficiency = \(\frac{1321.2}{1800} \times 100\% \approx 73\%\).

Correct answer is C (1 mark).

The baryon consists of three quarks. Since each quark has a baryon number of \(+\frac{1}{3}\), the total baryon number is 3 \(\times\) (\(+\frac{1}{3}\)) = \(+1\). The charge is the sum of the constituent quark charges: \(2 \times (+\frac{2}{3}e) + (-\frac{1}{3}e) = +e\). Since there are no strange quarks, the strangeness of the baryon is 0. This corresponds to the values in option A.

1 mark for correctly determining the charge as \(+e\), baryon number as \(+1\), and strangeness as 0 to identify option A as correct.

Let upwards be the positive direction. The initial velocity is \(-v\) and the final velocity is \(+0.8v\). The change in momentum is \(\Delta p = m(0.8v) - m(-v) = 1.8mv\). The net upward force on the ball during contact is \(F_{\text{net}} = \frac{\Delta p}{\Delta t} = \frac{1.8mv}{\Delta t}\). The actual forces acting on the ball are the upward force from the floor \(F_{\text{floor}}\) and the downward weight \(mg\). Thus, \(F_{\text{net}} = F_{\text{floor}} - mg\), which gives \(F_{\text{floor}} = \frac{1.8mv}{\Delta t} + mg\).

1 mark for correctly applying the impulse-momentum equation taking direction into account, and correctly including the weight term to find the force from the floor.

The Young modulus is given by \(E = \frac{F L}{A x}\), so extension \(x = \frac{F L}{A E}\). Since they are connected in series, the tension \(F\) is the same in both. The material is the same, so \(E\) is the same. The ratio of cross-sectional areas is \(A_Y / A_X = (2d/d)^2 = 4\). Thus, \(x_X = \frac{F L}{A_X E}\) and \(x_Y = \frac{F (2L)}{4 A_X E} = 0.5 x_X\). The ratio \(x_X / x_Y = 2.0\).

1 mark for expressing extension in terms of area and length, correctly applying the area ratio of 4, and finding the final ratio of 2.0.

For an NTC thermistor, a decrease in temperature causes its resistance to increase. In the potential divider, the output voltage across the thermistor is given by \(V_{\text{out}} = E \times \frac{R_{\text{th}}}{R + R_{\text{th}}}\). As \(R_{\text{th}}\) increases, the proportion of the total resistance represented by the thermistor increases, causing \(V_{\text{out}}\) to increase.

1 mark for identifying that thermistor resistance increases when temperature decreases, and that this increases the output voltage across the thermistor.

Fringe separation is given by \(x = \frac{\lambda D}{a}\). For the new setup, \(\lambda' = 1.2\lambda\), \(D' = 2D\), and \(a' = 0.5a\). The new fringe separation is \(x' = \frac{(1.2\lambda)(2D)}{0.5a} = \frac{2.4}{0.5} \frac{\lambda D}{a} = 4.8x\).

1 mark for using the double-slit formula with the updated parameters to obtain a factor of 4.8.

Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with upwards as the positive direction: initial velocity is \(+u\), acceleration is \(-g\), and the final displacement is \(-h\) (where \(h\) is the height of the cliff). This gives \(-h = uT - \frac{1}{2}gT^2\). Rearranging for \(h\) yields \(h = \frac{1}{2}gT^2 - uT\).

1 mark for correctly applying the sign convention to displacement, initial velocity, and acceleration to solve for the height of the cliff.

The relationship between terminal potential difference and current is \(V = E - Ir\), which can be rewritten as \(V = -rI + E\). Comparing this with the equation for a straight line \(y = mx + c\), a plot of \(V\) against \(I\) is a straight line with a vertical intercept equal to \(E\) and a negative gradient equal to \(-r\).

1 mark for using the terminal potential difference equation to identify the straight-line relationship with negative gradient of magnitude r and intercept E.

During alpha decay, the nucleon number \(A\) decreases by 4 and the proton number \(Z\) decreases by 2. During each beta-minus decay, \(A\) remains constant and \(Z\) increases by 1. For one alpha emission followed by two beta-minus emissions: new \(A = 212 - 4 = 208\), and new \(Z = 83 - 2 + (2 \times 1) = 83\). Because the proton number remains 83, the isotope is bismuth-208 (\(^{208}_{83}\text{Bi}\)).

1 mark for applying conservation of nucleon and proton numbers to determine the final state with A = 208 and Z = 83.

The baryon number of a proton is +1 and that of an antiproton is -1, so the total baryon number before the interaction is 0. Pions are mesons, which have a baryon number of 0, so the total baryon number after the interaction is 0. Thus, baryon number is conserved. Charge is also conserved because it is 0 before and 0 after. Lepton number is conserved because no leptons are present before or after.

1 mark for identifying that the total baryon number before and after the interaction is 0, demonstrating conservation of baryon number.

Let the direction to the right be positive. Total initial momentum = \(m_1 u_1 + m_2 u_2 = (3.0 \times 4.0) + (5.0 \times -2.0) = 12.0 - 10.0 = +2.0\text{ kg m s}^{-1}\). Total final momentum = \(m_1 v_1 + m_2 v_2 = (3.0 \times v_1) + (5.0 \times 1.0) = 3.0 v_1 + 5.0\). By conservation of momentum: \(3.0 v_1 + 5.0 = 2.0\), which gives \(3.0 v_1 = -3.0\), so \(v_1 = -1.0\text{ m s}^{-1}\). The negative sign indicates the direction is to the left.

1 mark for using conservation of momentum with correct signs to find the final velocity of the 3.0 kg block.

Using the potential divider formula in the dark: \(V_{\text{out}} = V_s \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\). Thus, \(8.0 = 12 \times \frac{R_{\text{dark}}}{4.0 + R_{\text{dark}}}\), which gives \(8.0(4.0 + R_{\text{dark}}) = 12 R_{\text{dark}}\), leading to \(32 = 4 R_{\text{dark}}\) and \(R_{\text{dark}} = 8.0\text{ k}\Omega\). When light is shone, the resistance of the LDR becomes \(1.0\text{ k}\Omega\). The new voltage reading is \(V_{\text{new}} = 12 \times \frac{1.0}{4.0 + 1.0} = 12 / 5.0 = 2.4\text{ V}\).

1 mark for calculating the dark resistance of the LDR and then using it to find the new voltmeter reading.

Young's modulus E is given by \(E = \frac{F L}{A x}\). Since both wires are made of the same material, they have the same Young's modulus. For the second wire, \(E = \frac{(2F) (2L)}{(3A) x_2} = \frac{4 F L}{3 A x_2}\). Equating the two expressions for E: \( \frac{F L}{A x} = \frac{4 F L}{3 A x_2} \). This simplifies to \(\frac{1}{x} = \frac{4}{3 x_2}\), which gives \(x_2 = \frac{4}{3}x\).

1 mark for setting up the ratio of Young's modulus and correctly solving for the new extension in terms of x.

The fringe spacing w is given by \(w = \frac{\lambda D}{a}\). Substituting the given values: \(w = \frac{600 \times 10^{-9} \times 2.0}{0.15 \times 10^{-3}} = 8.0 \times 10^{-3}\text{ m} = 8.0\text{ mm}\). The central bright fringe is at y = 0. The first dark fringe is at \(0.5w\), the second at \(1.5w\), and the third dark fringe is at \(2.5w\). Thus, the distance from the central bright fringe to the third dark fringe is \(2.5 \times 8.0\text{ mm} = 20.0\text{ mm} = 2.0 \times 10^{-2}\text{ m}\).

1 mark for calculating the fringe spacing and multiplying by 2.5 to find the distance to the third-order dark fringe.

During beta-minus decay, a neutron (udd) turns into a proton (uud). This means a down quark changes into an up quark. To conserve charge at the quark vertex, the down quark (charge -1/3) emits a virtual \(W^-\) boson (charge -1) to become an up quark (charge +2/3). The \(W^-\) boson then decays into an electron and an electron antineutrino.

1 mark for identifying the correct quark flavor change (down to up) and the correct exchange particle (W- boson).

Average power is defined as the work done divided by the time taken. The work done by the constant force F in moving the body a distance d is \(W = F \times d\). Therefore, the average power delivered is \(P = \frac{W}{t} = \frac{Fd}{t}\).

1 mark for identifying that average power is the total work done divided by time, which simplifies to Fd/t.

When a wire is stretched beyond its elastic limit and then unloaded, the unloading curve is a straight line parallel to the initial elastic loading line. The area under the loading curve represents the total work done in stretching the wire. The area under the unloading curve represents the elastic energy recovered. The difference between these two areas, which is the closed area enclosed by the loading curve, the unloading line, and the extension axis, represents the work done that is not recovered but is instead dissipated as heat (work done in plastic deformation).

1 mark for identifying the enclosed area between the loading and unloading curves as the net work done in permanent deformation.

The charge of a down quark \(d\) is \(-\frac{1}{3}e\) and its baryon number is \(+\frac{1}{3}\). The charge of an anti-strange quark \(\bar{s}\) is \(+\frac{1}{3}e\), its baryon number is \(-\frac{1}{3}\), and its strangeness is \(+1\) (since the strange quark \(s\) has strangeness \(-1\)). Summing these values gives a total charge of \(0\), a total baryon number of \(0\), and a total strangeness of \(+1\). Therefore, the correct option is A.

1 mark for identifying the correct combination of charge, baryon number, and strangeness.

A neutron consists of \(udd\) quarks and a proton consists of \(uud\) quarks. In beta-minus (\(\beta^-\)) decay, one of the down quarks (\(d\)) in the neutron changes into an up quark (\(u\)), transforming the neutron into a proton. This weak interaction is mediated by the emission of a virtual \(W^-\)
boson, which subsequently decays into an electron and an electron antineutrino. Therefore, the change in flavour is \(d \rightarrow u\) and the boson is \(W^-\).

1 mark for identifying both the correct quark flavor change and the correct exchange boson.

In a time interval \(\Delta t\), the spacecraft travels a distance \(v \Delta t\). The volume of dust swept out is \(V = A v \Delta t\). The mass of dust collected in this time is \(\Delta m = \rho V = \rho A v \Delta t\). These dust particles are initially at rest and are accelerated to the speed of the spacecraft, \(v\). The change in momentum of this dust mass is \(\Delta p = \Delta m \cdot v = \rho A v^2 \Delta t\). The rate of change of momentum is the force required to accelerate the dust: \(F = \frac{\Delta p}{\Delta t} = \rho A v^2\). By Newton's third law, the dust exerts an equal and opposite force of this magnitude on the spacecraft. To maintain a constant speed, the engines must exert an equal forward force of \(F = \rho A v^2\).

1 mark for deriving the correct expression for force based on the rate of change of momentum.

Let \(v_1\) be the final velocity of the ball of mass \(m\) and \(v_2\) be the final velocity of the ball of mass \(3m\). Using conservation of linear momentum: \(u = v_1 + 3 v_2\) (Equation 1). For an elastic collision, the relative velocity of approach is equal to the relative velocity of separation: \(u = v_2 - v_1\) (Equation 2). Adding Equations 1 and 2 gives \(2u = 4 v_2\), which simplifies to \(v_2 = 0.5 u\). The initial kinetic energy of the moving ball is \(E_k = \frac{1}{2} m u^2\). The kinetic energy transferred to the stationary ball is \(E_{k2} = \frac{1}{2} (3m) v_2^2 = \frac{1}{2} (3m) (0.5u)^2 = \frac{3}{8} m u^2\). The fraction of kinetic energy transferred is \(\frac{E_{k2}}{E_k} = \frac{3/8}{1/2} = 0.75\).

1 mark for correctly calculating the fraction of energy transferred.

In bright light, the voltmeter reading across the LDR is: \(V_{\text{bright}} = 12.0 \times \frac{2.0}{6.0 + 2.0} = 3.0\text{ V}\). In darkness, the voltmeter reading across the LDR is: \(V_{\text{dark}} = 12.0 \times \frac{18}{6.0 + 18} = 9.0\text{ V}\). The change in the voltmeter reading is \(\Delta V = 9.0\text{ V} - 3.0\text{ V} = 6.0\text{ V}\).

1 mark for calculating the potential difference in both states and finding the correct difference of 6.0 V.

The extension of the first wire under force \(F\) is \(x_1 = \frac{F L}{A E}\), and the work done is \(W_1 = \frac{1}{2} F x_1 = W\). For the second wire, the extension is \(x_2 = \frac{F (2L)}{(2A) E} = \frac{F L}{A E} = x_1\). Since both the applied force and the resulting extension are identical to those of the first wire, the work done in stretching the second wire is also \(W_2 = \frac{1}{2} F x_2 = W\).

1 mark for establishing that the extensions and thus the strain energies of both wires are equal.

The fringe spacing is given by \(x = \frac{\lambda D}{a}\). With the new values \(\lambda' = 2\lambda\), \(D' = 2D\), and \(a' = 0.5 a\), the new fringe spacing is \(x' = \frac{(2\lambda)(2D)}{0.5 a} = 8 \left( \frac{\lambda D}{a} \right) = 8x\).

1 mark for correctly using the double-slit formula to find the ratio of the new fringe spacing to the old one.

The grating spacing is \(d = \frac{1 \times 10^{-3}\text{ m}}{500} = 2.0 \times 10^{-6}\text{ m} = 2000\text{ nm}\). Using the grating equation \(d \sin\theta = n\lambda\), the maximum order occurs when \(\sin\theta \le 1\), which yields \(n \le \frac{2000}{480} \approx 4.17\). Thus, the maximum integer order is \(n = 4\). The total number of bright fringes includes the central maximum and 4 orders on each side, which is \(2(4) + 1 = 9\).

1 mark for correctly determining the maximum integer order as 4, and calculating the total number of fringes as 9.

Part (a)
- A hadron is a composite particle made of quarks.
- The two main classes of hadrons are baryons (made of three quarks) and mesons (made of a quark-antiquark pair).

Part (b)
(i) The decay is a beta-minus decay:
\(^{14}_{6}\text{C} \rightarrow {^{14}_{7}\text{N}} + {^{0}_{-1}\text{e}} + \bar{\nu}_e\)
where particle \(Y\) is an electron antineutrino (\(\bar{\nu}_e\)).
(ii) A neutron in the carbon nucleus turns into a proton. In terms of quarks, a down quark (\(d\)) changes into an up quark (\(u\)), changing the nucleon from \(udd\) (neutron) to \(uud\) (proton).
(iii) The fundamental interaction responsible is the weak interaction (or weak nuclear force).

Part (c)
- Charge conservation:
LHS charge = \(+6e\) (or \(+6\))
RHS charge = \(+7e\) (nitrogen nucleus) \(- 1e\) (beta particle) \(+ 0\) (antineutrino) = \(+6e\)
Since \(+6e = +6e\), charge is conserved.
- Lepton number conservation:
LHS lepton number = \(0\) (baryons only)
RHS lepton number = \(0\) (nitrogen nucleus) \(+ 1\) (electron) \(- 1\) (electron antineutrino) = \(0\)
Since \(0 = 0\), lepton number is conserved.

(a)
- Hadron: particle composed of quarks (or affected by the strong nuclear force) [1]
- Classes: baryons and mesons [1]

(b)(i)
- Correct equation including both beta-minus and antineutrino: \(^{14}_{6}\text{C} \rightarrow {^{14}_{7}\text{N}} + {^{0}_{-1}\text{e}} + \bar{\nu}_e\) [1]
- Correct identification of \(Y\) as an electron antineutrino (or antineutrino) [1]

(b)(ii)
- Identifies neutron decaying to proton (or \(udd \rightarrow uud\)) [1]
- Explicitly states a down quark changes to an up quark [1]

(b)(iii)
- Weak nuclear force / weak interaction [1]

(c)
- Charge conservation calculation: shows LHS is \(+6\) and RHS is \(7 - 1 + 0 = 6\) [1]
- Lepton number conservation calculation: shows LHS is \(0\) and RHS is \(0 + 1 - 1 = 0\) [1]
- Explicit statement confirming conservation for both quantities [1]

Part (a)
- The principle of conservation of momentum states that the total momentum of a system of interacting bodies remains constant, provided no external forces act on the system.

Part (b)
(i) Let \(m_A = 0.45\text{ kg}\), \(u_A = 1.20\text{ m s}^{-1}\), \(m_B = 0.25\text{ kg}\), and \(u_B = 0\).
Let \(v_A = 0.35\text{ m s}^{-1}\) and \(v_B\) be the final velocity of glider \(B\).
Using conservation of momentum:
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
\((0.45 \times 1.20) + 0 = (0.45 \times 0.35) + 0.25 v_B\)
\(0.540 = 0.1575 + 0.25 v_B\)
\(0.25 v_B = 0.3825\)
\(v_B = 1.53\text{ m s}^{-1}\)

(ii) Calculate total initial kinetic energy (\(E_{k,i}\)) and total final kinetic energy (\(E_{k,f}\)):
\(E_{k,i} = \frac{1}{2} m_A u_A^2 = 0.5 \times 0.45 \times 1.20^2 = 0.324\text{ J}\)
\(E_{k,f} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = (0.5 \times 0.45 \times 0.35^2) + (0.5 \times 0.25 \times 1.53^2)\)
\(E_{k,f} = 0.02756 + 0.29261 = 0.32017\text{ J} \approx 0.320\text{ J}\)
Since \(E_{k,f} < E_{k,i}\) (a loss of about \(0.004\text{ J}\)), kinetic energy is not conserved, meaning the collision is inelastic.

Part (c)
Using Newton's second law, average force \(F\) is equal to the rate of change of momentum of glider \(B\):
\(F = \frac{\Delta p}{\Delta t} = \frac{m_B v_B - m_B u_B}{\Delta t}\)
\(F = \frac{0.25 \times 1.53}{0.085} = \frac{0.3825}{0.085} = 4.5\text{ N}\)

(a)
- Total momentum before a collision equals total momentum after [1]
- Provided no external force acts on the system (closed system) [1]

(b)(i)
- Uses \(m_A u_A + m_B u_B = m_A v_A + m_B v_B\) [1]
- Correct substitution: \(0.45(1.20) = 0.45(0.35) + 0.25 v_B\) [1]
- Correct calculation to give \(1.53\text{ m s}^{-1}\) (or \(1.5\text{ m s}^{-1}\)) [1]

(b)(ii)
- Calculates initial kinetic energy: \(0.324\text{ J}\) [1]
- Calculates final kinetic energy: \(0.320\text{ J}\) (using \(1.53\text{ m s}^{-1}\)) [1]
- Compares energies and concludes collision is inelastic since kinetic energy is not conserved [1]

(c)
- Uses \(F = \Delta p / \Delta t\) [1]
- Obtains \(4.5\text{ N}\) (allow ECF from b(i)) [1]

Part (a)
- Electromotive force (e.m.f.) is defined as the work done (or energy transformed) per unit charge from non-electrical to electrical forms of energy.
- Potential difference (p.d.) is the work done (or energy transformed) per unit charge from electrical to other forms of energy (such as heat or light).

Part (b)
(i) The diagram must show a single loop consisting of:
- A battery (represented by cells, possibly with a separate small resistor labeled \(r\) in a dashed box to show internal resistance)
- A fixed resistor \(R_1\)
- A thermistor (symbol: a rectangular box with a diagonal line through it, with a flat line at the bottom left end)
- A voltmeter connected in parallel across only the thermistor.

(ii) Total resistance of the circuit:
\(R_{\text{total}} = R_1 + R_T + r = 12.0 + 18.5 + 1.50 = 32.0\ \Omega\)
The current \(I\) in the circuit is:
\(I = \frac{E}{R_{\text{total}}} = \frac{9.00}{32.0} = 0.28125\text{ A}\)
The reading on the voltmeter across the thermistor is:
\(V_T = I \times R_T = 0.28125 \times 18.5 = 5.2031\text{ V} \approx 5.20\text{ V}\)
Alternatively, using the potential divider formula:
\(V_T = \left( \frac{R_T}{R_1 + R_T + r} \right) \times E = \left( \frac{18.5}{32.0} \right) \times 9.00 = 5.20\text{ V}\)

(iii) As the temperature of the thermistor increases:
- The resistance \(R_T\) of the NTC thermistor decreases.
- The total resistance of the circuit decreases, which increases the current \(I\).
- However, because \(R_T\) decreases, the ratio of its resistance to the total resistance decreases, meaning the voltage share across the thermistor decreases.
- Therefore, the reading on the voltmeter decreases.

(a)
- e.m.f. is non-electrical to electrical energy conversion per unit charge AND p.d. is electrical to other forms of energy conversion per unit charge [2]

(b)(i)
- Single complete series loop with battery, fixed resistor, and correct thermistor symbol [1]
- Voltmeter connected in parallel across the thermistor [1]

(b)(ii)
- Calculates total resistance of loop: \(32.0\ \Omega\) (explicit or implicit) [1]
- Shows correct substitution into divider formula or current method: \(I = 0.281\text{ A}\) or \(\frac{18.5}{32.0} \times 9.00\) [1]
- Gives final correct value of \(5.20\text{ V}\) (allow 5.2 V) [1]

(b)(iii)
- States that resistance of the thermistor decreases as temperature increases [1]
- Explains that this decreases its fraction of the total circuit resistance (or increased current increases p.d. across other components) [1]
- Concludes that the voltmeter reading decreases [1]

Part (a)
- Elastic deformation occurs when a material returns to its original shape and size after the deforming force (or load) is removed.
- Plastic deformation occurs when the material is permanently deformed and does not return to its original shape and size after the load is removed.

Part (b)
(i) Stress is defined as force per unit cross-sectional area:
\(\text{Stress} = \frac{\text{Force}}{\text{Area}} = \frac{75.0}{3.50 \times 10^{-7}} = 2.143 \times 10^8\text{ Pa} \approx 2.14 \times 10^8\text{ Pa}\) (or \(\text{N m}^{-2}\)).

(ii) The Young modulus \(E\) is given by:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\)
Rearranging for extension \(x\):
\(x = \frac{F L}{A E} = \frac{75.0 \times 2.40}{3.50 \times 10^{-7} \times 2.00 \times 10^{11}}\)
\(x = \frac{180.0}{70000} = 2.5714 \times 10^{-3}\text{ m} \approx 2.57 \times 10^{-3}\text{ m}\).

(iii) Assuming Hooke's law is obeyed, the elastic potential energy \(E_p\) stored is equal to the work done in stretching the wire:
\(E_p = \frac{1}{2} F x\)
\(E_p = 0.5 \times 75.0 \times 2.5714 \times 10^{-3} = 0.09643\text{ J} \approx 0.0964\text{ J}\) (or \(9.64 \times 10^{-2}\text{ J}\)).

(a)
- Elastic: returns to original shape/length when force/load is removed [1]
- Plastic: permanent deformation / does not return to original shape [1]

(b)(i)
- Formula: \(\text{Stress} = F/A\) [1]
- Correct calculation with SI units: \(2.14 \times 10^8\text{ Pa}\) (or \(\text{N m}^{-2}\)) [1]

(b)(ii)
- Stated Young modulus formula: \(E = \frac{F L}{A x}\) or \(E = \frac{\text{Stress}}{\text{Strain}}\) [1]
- Correct substitution: \(x = \frac{75.0 \times 2.40}{3.50 \times 10^{-7} \times 2.00 \times 10^{11}}\) [1]
- Shows intermediate working reaching \(2.57 \times 10^{-3}\text{ m}\) (minimum 3 sig figs) [1]

(b)(iii)
- Stated energy formula: \(E_p = \frac{1}{2} F x\) [1]
- Correct substitution of values [1]
- Final correct value of \(0.0964\text{ J}\) (or \(0.096\text{ J}\) with correct unit) [1]

Part (a)
- Coherent light sources have a constant phase difference between them and have the same frequency (or wavelength).

Part (b)
(i) The distance from the 1st to the 7th bright fringe spans exactly 6 fringe spacings (\(7 - 1 = 6\)).
Therefore, the fringe separation \(x\) is:
\(x = \frac{25.4\text{ mm}}{6} = 4.2333\text{ mm} \approx 4.23 \times 10^{-3}\text{ m}\).

(ii) Using the double-slit formula:
\(\lambda = \frac{a x}{D}\)
Substitute the values in SI units:
\(\lambda = \frac{0.280 \times 10^{-3} \times 4.2333 \times 10^{-3}}{1.85}\)
\(\lambda = 6.407 \times 10^{-7}\text{ m} \approx 6.41 \times 10^{-7}\text{ m}\) (or \(641\text{ nm}\)).

Part (c)
For a diffraction grating, the formula is:
\(d \sin\theta = n\lambda\)
First, calculate the grating spacing \(d\):
\(d = \frac{1}{500\text{ lines mm}^{-1}} = 2.00 \times 10^{-3}\text{ mm} = 2.00 \times 10^{-6}\text{ m}\)
For the second-order maximum, \(n = 2\).
\(\sin\theta = \frac{2 \lambda}{d} = \frac{2 \times 6.407 \times 10^{-7}}{2.00 \times 10^{-6}} = 0.6407\)
\(\theta = \sin^{-1}(0.6407) = 39.84^{\circ} \approx 39.9^{\circ}\) (or \(40^{\circ}\) to 2 significant figures).

(a)
- Constant phase difference [1]
- Same frequency / wavelength [1]

(b)(i)
- Identifies that there are 6 fringe intervals [1]
- Correctly calculates \(x = 4.23\text{ mm}\) (or \(4.23 \times 10^{-3}\text{ m}\)) [1]

(b)(ii)
- Recalls and uses \(\lambda = \frac{a x}{D}\) [1]
- Substitutes values correctly in SI units [1]
- Obtains wavelength in range \(6.40 \times 10^{-7}\text{ m}\) to \(6.42 \times 10^{-7}\text{ m}\) [1]

(c)
- Calculates \(d = 2.00 \times 10^{-6}\text{ m}\) [1]
- Uses \(d \sin\theta = n\lambda\) with \(n = 2\) [1]
- Correctly calculates \(\theta = 39.9^{\circ}\) (or \(40^{\circ}\)) [1]

Part (a)
- Velocity is the rate of change of displacement.
- Acceleration is the rate of change of velocity.

Part (b)
(i) The initial vertical component of velocity \(u_y\) is:
\(u_y = u \sin\theta = 15.0 \sin(35.0^{\circ}) = 8.603\text{ m s}^{-1} \approx 8.60\text{ m s}^{-1}\)
The initial horizontal component of velocity \(u_x\) is:
\(u_x = u \cos\theta = 15.0 \cos(35.0^{\circ}) = 12.287\text{ m s}^{-1} \approx 12.3\text{ m s}^{-1}\)

(ii) Take upwards as positive. The vertical displacement is \(s = -25.0\text{ m}\).
The vertical acceleration is \(a = -9.81\text{ m s}^{-2}\).
Using the equation of motion:
\(s = u_y t + \frac{1}{2} a t^2\)
\(-25.0 = 8.603 t - 0.5 \times 9.81 t^2\)
\(4.905 t^2 - 8.603 t - 25.0 = 0\)
Using the quadratic formula:
\(t = \frac{-(-8.603) \pm \sqrt{(-8.603)^2 - 4 \times 4.905 \times (-25.0)}}{2 \times 4.905}\)
\(t = \frac{8.603 \pm \sqrt{74.01 + 490.5}}{9.81} = \frac{8.603 \pm \sqrt{564.51}}{9.81}\)
\(t = \frac{8.603 \pm 23.759}{9.81}\)
Taking the positive root:
\(t = \frac{32.362}{9.81} = 3.299\text{ s} \approx 3.30\text{ s}\).

(iii) The horizontal component of the velocity is constant since there is no air resistance.
\(x = u_x \times t\)
\(x = 12.287 \times 3.299 = 40.53\text{ m} \approx 40.6\text{ m}\) (or \(40.5\text{ m}\) if rounded values are used).

(a)
- Velocity: change in displacement / time [1]
- Acceleration: change in velocity / time [1]

(b)(i)
- Vertical component: shows \(15.0 \sin(35.0^{\circ}) = 8.60\text{ m s}^{-1}\) [1]
- Horizontal component: calculates \(15.0 \cos(35.0^{\circ}) = 12.3\text{ m s}^{-1}\) [1]

(b)(ii)
- Uses \(s = ut + \frac{1}{2}at^2\) with correct vertical values (\(s = -25\), \(u_y = 8.60\), \(a = -9.81\)) [1]
- Correctly sets up the quadratic equation or calculates final vertical velocity (\(v_y = -23.8\text{ m s}^{-1}\)) first [1]
- Obtains time \(t = 3.30\text{ s}\) (allow 3.3 s) [1]

(b)(iii)
- Recalls horizontal distance formula: \(x = u_x \times t\) [1]
- Substitutes horizontal component and time from (ii) [1]
- Correctly calculates distance in the range \(40.5\text{ m}\) to \(40.6\text{ m}\) [1]