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(a) The gravitational potential at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point.

(b) By definition, gravitational potential is zero at infinity. Because the gravitational force is always attractive, work is done by the gravitational field (or negative work is done by an external force) when moving a mass towards the planet from infinity. Hence, potential is negative everywhere.

(c) The potential is maximum (least negative) where the potential gradient is zero, which is where the net gravitational field strength is zero:
\( G M_X / x^2 = G M_Y / (D - x)^2 \)
\( \sqrt{M_Y / M_X} = (D - x) / x \)
\( \sqrt{7.4 \times 10^{22} / 6.0 \times 10^{24}} = 0.111 \)
\( 0.111 = (4.0 \times 10^8 - x) / x \)
\( 1.111 x = 4.0 \times 10^8 \implies x = 3.6 \times 10^8 \text{ m} \).

(d) Calculate the potentials at this distance:
\( \phi = - G ( M_X / x + M_Y / (D - x) ) \)
\( \phi = - (6.67 \times 10^{-11}) \times [ (6.0 \times 10^{24} / (3.6 \times 10^8)) + (7.4 \times 10^{22} / (0.40 \times 10^8)) ] \)
\( \phi = - (6.67 \times 10^{-11}) \times [ 1.67 \times 10^{16} + 1.85 \times 10^{15} ] \)
\( \phi = - (6.67 \times 10^{-11}) \times [ 1.855 \times 10^{16} ] = -1.24 \times 10^6 \text{ J kg}^{-1} \).

(a) Work done per unit mass [1], in bringing a test mass from infinity to that point [1].
(b) Potential is zero at infinity [1], and since the gravitational force is attractive, work is done by the field as mass approaches [1].
(c) Formula equating field strengths: \( M_X / x^2 = M_Y / (D - x)^2 \) [1]; taking square roots: \( \sqrt{M_Y / M_X} = (D - x) / x \) [1]; substitution of values [1]; final answer \( 3.6 \times 10^8 \text{ m} \) [1].
(d) Formula for total potential: \( \phi = -G(M_X/x + M_Y/(D-x)) \) [1]; calculation to give \( -1.2 \times 10^6 \text{ J kg}^{-1} \) (accept \( -1.24 \times 10^6 \text{ J kg}^{-1} \)) [1].

(a) Coulomb's law states that the electrostatic force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

(b) Using Coulomb's Law:
\( F = \frac{Q_1 Q_2}{4 \pi \varepsilon_0 d^2} \)
\( F = \frac{8.99 \times 10^9 \times 3.2 \times 10^{-9} \times 1.8 \times 10^{-9}}{0.12^2} \)
\( F = 3.60 \times 10^{-6} \text{ N} \).

(c)(i) Since A is positive and B is negative, C (positive) is repelled by A and attracted to B. For these forces to cancel out, they must be in opposite directions, which only occurs outside the segment joining A and B. Because A has a larger charge magnitude than B (\( 3.2 \text{ nC} > 1.8 \text{ nC} \)), C must be further from A than from B to balance the forces. Hence, it must be placed to the right of B.

(c)(ii) Let \( x \) be the distance from B to C.
\( F_{AC} = F_{BC} \)
\( \frac{Q_A Q_C}{4\pi\varepsilon_0 (d+x)^2} = \frac{Q_B Q_C}{4\pi\varepsilon_0 x^2} \)
\( \frac{3.2 \times 10^{-9}}{(0.12+x)^2} = \frac{1.8 \times 10^{-9}}{x^2} \)
\( \frac{3.2}{1.8} = \left(\frac{0.12+x}{x}\right)^2 \implies \frac{16}{9} = \left(\frac{0.12+x}{x}\right)^2 \)
\( \frac{4}{3} = \frac{0.12+x}{x} \implies 4x = 0.36 + 3x \implies x = 0.36 \text{ m} \).

(a) Force is proportional to product of charges [1] and inversely proportional to square of separation [1].
(b) Correct formula usage [1], substitution of values [1], final answer \( 3.6 \times 10^{-6} \text{ N} \) (accept \( 3.6 \times 10^{-6} \text{ N} \) to \( 3.61 \times 10^{-6} \text{ N} \)) [1].
(c)(i) Forces must be in opposite directions (outside the charges) [1], must be closer to the smaller charge B to balance the larger charge A [1].
(c)(ii) Correct algebraic equation set up [1], simplifying by square-rooting both sides [1], correct answer of \( 0.36 \text{ m} \) [1].

(a) Specific latent heat of fusion is the thermal energy required per unit mass to change a substance from solid to liquid phase without any change in temperature.

(b)(i) \( Q_1 = m c_{\text{ice}} \Delta T = 0.35 \times 2100 \times (0 - (-18)) = 13230 \text{ J} \approx 1.3 \times 10^4 \text{ J} \).
(b)(ii) \( Q_2 = m L_f = 0.35 \times 3.3 \times 10^5 = 115500 \text{ J} \approx 1.2 \times 10^5 \text{ J} \).
(b)(iii) \( Q_3 = m c_{\text{water}} \Delta T = 0.35 \times 4200 \times (25 - 0) = 36750 \text{ J} \approx 3.7 \times 10^4 \text{ J} \).

(c) Total thermal energy \( Q_{\text{total}} = Q_1 + Q_2 + Q_3 = 13230 + 115500 + 36750 = 165480 \text{ J} \).
Time \( t = Q_{\text{total}} / P = 165480 / 150 = 1103.2 \text{ s} \approx 1100 \text{ s} \) (or \( 1.1 \times 10^3 \text{ s} \)).

(a) Energy required per unit mass [1], to change solid to liquid at constant temperature [1].
(b)(i) Correct substitution in \( mc\Delta T \) [1], final value \( 13000 \text{ J} \) or \( 13230 \text{ J} \) [1].
(b)(ii) Correct substitution in \( mL \) [1], final value \( 116000 \text{ J} \) or \( 115500 \text{ J} \) [1].
(b)(iii) Correct substitution in \( mc\Delta T \) [1], final value \( 36800 \text{ J} \) or \( 36750 \text{ J} \) [1].
(c) Sum of energies calculated [1], divided by \( 150 \) to give \( 1100 \text{ s} \) (accept \( 1103 \text{ s} \) or answers consistent with rounded values in part b) [1].

(a) The first law of thermodynamics states that: \( \Delta U = Q + W \), where \( \Delta U \) is the increase in internal energy of the system, \( Q \) is the thermal energy supplied to the system, and \( W \) is the work done on the system.

(b)(i) \( \Delta U = 0 \). An ideal gas's internal energy depends solely on its temperature. Since the expansion is isothermal, there is no change in temperature, so internal energy remains constant.

(b)(ii) Work done ON the gas is given by:
\( W = - p \Delta V = - (1.5 \times 10^5) \times (2.0 \times 10^{-3} - 5.0 \times 10^{-3}) \)
\( W = - (1.5 \times 10^5) \times (-3.0 \times 10^{-3}) = +450 \text{ J} \).

(b)(iii) Using the first law for Y to Z:
\( \Delta U = Q + W \)
\( -1850 = Q + 450 \implies Q = -2300 \text{ J} \).
Since \( Q \) is negative, \( 2300 \text{ J} \) of thermal energy is released by the gas.

(b)(iv) For a closed cycle, the total change in internal energy is zero:
\( \Delta U_{XY} + \Delta U_{YZ} + \Delta U_{ZX} = 0 \)
\( 0 - 1850 + \Delta U_{ZX} = 0 \implies \Delta U_{ZX} = +1850 \text{ J} \).

(a) Correct equation: \( \Delta U = Q + W \) [1], with all symbols correctly defined [1].
(b)(i) \( \Delta U = 0 \) [1]; explains that for ideal gas internal energy depends on temperature, which is constant [1].
(b)(ii) Formula \( W = p \Delta V \) [1]; calculation to give \( +450 \text{ J} \) [1].
(b)(iii) Correct substitution of values into \( \Delta U = Q + W \) [1]; calculation to give \( 2300 \text{ J} \) released (negative sign or word 'released') [1].
(b)(iv) Understands total cycle \( \Delta U = 0 \) [1]; calculates \( +1850 \text{ J} \) [1].

(a) The acceleration of the object must be directly proportional to its displacement from its equilibrium position, and always directed towards that equilibrium position.

(b)(i) Comparing \( x = x_0 \cos(\omega t) \) with \( x = 4.0 \cos(15 t) \):
\( \omega = 15 \text{ rad s}^{-1} \).
\( f = \omega / (2\pi) = 15 / (2\pi) = 2.39 \text{ Hz} \approx 2.4 \text{ Hz} \).

(b)(ii) Maximum acceleration:
\( a_{\text{max}} = \omega^2 x_0 \)
\( x_0 = 4.0 \text{ cm} = 0.040 \text{ m} \)
\( a_{\text{max}} = 15^2 \times 0.040 = 225 \times 0.040 = 9.0 \text{ m s}^{-2} \).

(b)(iii) Speed \( v \) at \( x = 2.5 \text{ cm} = 0.025 \text{ m} \):
\( v = \omega \sqrt{x_0^2 - x^2} \)
\( v = 15 \times \sqrt{0.040^2 - 0.025^2} = 15 \times \sqrt{0.0016 - 0.000625} \)
\( v = 15 \times \sqrt{0.000975} = 15 \times 0.0312 = 0.468 \text{ m s}^{-1} \approx 0.47 \text{ m s}^{-1} \).

(c) Maximum kinetic energy occurs at maximum velocity (or zero displacement):
\( E_{k,\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = \frac{1}{2} m \omega^2 x_0^2 \)
\( E_{k,\text{max}} = 0.5 \times 0.12 \times 15^2 \times 0.040^2 = 0.06 \times 225 \times 0.0016 = 0.0216 \text{ J} \approx 0.022 \text{ J} \).

(a) Acceleration is proportional to displacement [1], and directed towards fixed point/equilibrium [1].
(b)(i) \( \omega = 15 \) identified [1], \( f = 2.4 \text{ Hz} \) [1].
(b)(ii) Uses \( a = \omega^2 x_0 \) with correct units conversion [1], answers \( 9.0 \text{ m s}^{-2} \) [1].
(b)(iii) Uses \( v = \omega \sqrt{x_0^2 - x^2} \) [1], answers \( 0.47 \text{ m s}^{-1} \) [1].
(c) Uses \( \frac{1}{2} m \omega^2 x_0^2 \) [1], answers \( 0.022 \text{ J} \) (or \( 0.0216 \text{ J} \)) [1].

(a) A bridge rectifier is drawn with four diodes in a diamond configuration. During each half-cycle of the AC input, two opposite diodes conduct, ensuring current always flows through the load resistor in the same direction. (Diagram: four diodes pointing in the same clockwise/counter-clockwise loop orientation around the diamond, AC connected to opposite corners, DC load connected to the other two corners).

(b)(i) The capacitor stores energy/charge when the input voltage is high (near peaks). When the input voltage falls below the capacitor's voltage, the capacitor discharges through the load resistor, maintaining a relatively steady potential difference across the load and reducing the variation (ripple).

(b)(ii) Time constant \( \tau = R C = 1500 \times 220 \times 10^{-6} = 0.33 \text{ s} \).

(b)(iii) For full-wave rectification, there are two peaks per AC cycle. Hence, the peak interval is \( t = T/2 = 1 / (2 f) = 1 / (2 \times 50) = 0.010 \text{ s} \).

(b)(iv) Since the time constant (\( 0.33 \text{ s} \)) is much larger than the discharge time (approx. \( 0.010 \text{ s} \)), we can model the exponential decay linearly:
\( V = V_0 e^{-t/RC} \approx V_0 (1 - t/RC) \)
\( \Delta V \approx V_0 \times (t / RC) \)
\( \Delta V = 12 \times (0.010 / 0.33) = 0.363 \text{ V} \approx 0.36 \text{ V} \).

(a) Sketch of bridge rectifier with correct diode orientations [1], AC input and DC output connected correctly to the bridge [1], load resistor in parallel with DC output [1].
(b)(i) Capacitor charges up to peak voltage [1], and discharges through the resistor when input voltage drops, maintaining voltage [1].
(b)(ii) Correct calculation of \( RC = 0.33 \text{ s} \) [1].
(b)(iii) Peak-to-peak interval \( t = 0.010 \text{ s} \) [1].
(b)(iv) Uses exponential decay formula or linear approximation \( \Delta V = V_0 \frac{t}{RC} \) [1], substitutes \( 12 \text{ V} \) and other values [1], final value \( 0.36 \text{ V} \) (accept range \( 0.35 \text{ V} \) to \( 0.37 \text{ V} \)) [1].

(a) Loss in electrical potential energy = Gain in kinetic energy:
\( q V = \frac{1}{2} m v^2 \implies v^2 = \frac{2 q V}{m} \implies v = \sqrt{\frac{2 q V}{m}} \).

(b)(i) The magnetic force on the moving electron is given by \( F = q v B \). By Fleming's Left Hand Rule, this force is always perpendicular to both the velocity and the magnetic field. A constant force acting perpendicular to the direction of motion provides a centripetal force, which causes circular motion.

(b)(ii) Equating centripetal force to magnetic force:
\( m v^2 / r = q v B \implies r = m v / (q B) \)
Substituting \( v = \sqrt{\frac{2 q V}{m}} \):
\( r = \frac{m}{q B} \sqrt{\frac{2 q V}{m}} = \sqrt{\frac{2 m^2 q V}{m q^2 B^2}} = \frac{\sqrt{2 m V q}}{q B} \).

(c)(i) Substituting values:
\( r = \frac{\sqrt{2 \times 9.11 \times 10^{-31} \times 850 \times 1.60 \times 10^{-19}}}{1.60 \times 10^{-19} \times 1.8 \times 10^{-3}} \)
\( r = \frac{\sqrt{2.478 \times 10^{-46}}}{2.88 \times 10^{-22}} = \frac{1.574 \times 10^{-23}}{2.88 \times 10^{-22}} = 0.0547 \text{ m} \approx 5.5 \text{ cm} \).

(c)(ii) Frequency \( f = \frac{v}{2\pi r} = \frac{q B}{2 \pi m} \)
\( f = \frac{1.60 \times 10^{-19} \times 1.8 \times 10^{-3}}{2 \pi \times 9.11 \times 10^{-31}} = 5.03 \times 10^7 \text{ Hz} \approx 5.0 \times 10^7 \text{ Hz} \) (or \( 50 \text{ MHz} \)).

(a) States \( qV = \frac{1}{2} m v^2 \) [1], rearranges correctly to get given formula [1].
(b)(i) Magnetic force is perpendicular to velocity [1], force provides a centripetal acceleration/force [1].
(b)(ii) Equates \( qvB = m v^2 / r \) [1], algebraic steps leading to final expression [1].
(c)(i) Correct values substituted into formula [1], answer of \( 5.5 \text{ cm} \) (or \( 0.055 \text{ m} \)) [1].
(c)(ii) Formula \( f = v / 2\pi r \) or equivalent [1], calculation to give \( 5.0 \times 10^7 \text{ Hz} \) [1].

(a) The de Broglie wavelength is the wavelength associated with a moving particle, given by the relation \( \lambda = h/p \), where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

(b)(i) Momentum of photon:
\( p = h / \lambda = 6.63 \times 10^{-34} / (0.28 \times 10^{-9}) = 2.37 \times 10^{-24} \text{ kg m s}^{-1} \approx 2.4 \times 10^{-24} \text{ kg m s}^{-1} \).

(b)(ii) Energy of photon:
\( E = h c / \lambda = (6.63 \times 10^{-34} \times 3.00 \times 10^8) / (0.28 \times 10^{-9}) = 7.10 \times 10^{-16} \text{ J} \approx 7.1 \times 10^{-16} \text{ J} \).

(c)(i) Momentum of electron is also \( p = 2.37 \times 10^{-24} \text{ kg m s}^{-1} \).
Kinetic energy \( E_k = p^2 / (2 m_e) = (2.368 \times 10^{-24})^2 / (2 \times 9.11 \times 10^{-31}) \)
\( E_k = 5.607 \times 10^{-48} / (1.822 \times 10^{-30}) = 3.077 \times 10^{-18} \text{ J} \).
In electron-volts:
\( E_k = 3.077 \times 10^{-18} / (1.60 \times 10^{-19}) = 19.2 \text{ eV} \approx 19 \text{ eV} \).

(c)(ii) Shorter. A neutron has a much larger mass than an electron. Since \( \lambda = h / \sqrt{2 m E_k} \), at constant kinetic energy, a larger mass \( m \) results in a smaller/shorter de Broglie wavelength.

(a) Wavelength of wave associated with a moving particle [1]; formula \( \lambda = h/p \) with symbols defined [1].
(b)(i) Correct calculation to give \( 2.4 \times 10^{-24} \text{ kg m s}^{-1} \) (accept \( 2.37 \times 10^{-24} \)) [1], correct unit [1].
(b)(ii) Uses \( E = hc/\lambda \) or \( E = pc \) [1], answers \( 7.1 \times 10^{-16} \text{ J} \) [1].
(c)(i) Uses \( E_k = p^2 / 2m \) [1], computes in Joules: \( 3.1 \times 10^{-18} \text{ J} \) [1], converts to eV: \( 19 \text{ eV} \) [1].
(c)(ii) Shorter because neutron has larger mass, so for same KE it has more momentum [1].

(a) Luminosity is defined as the total power emitted by a star (or the total electromagnetic energy emitted per unit time).

(b)(i) Redshift is the increase in the observed wavelength (or decrease in frequency) of electromagnetic radiation due to the relative motion of the source away from the observer.

(b)(ii) Using the Doppler redshift equation:
\(\frac{\Delta\lambda}{\lambda} = \frac{v}{c}\)
\
\(\Delta\lambda = 418\text{ nm} - 410\text{ nm} = 8\text{ nm}\)
\(v = c \times \frac{\Delta\lambda}{\lambda} = (3.00 \times 10^8\text{ m s}^{-1}) \times \frac{8\text{ nm}}{410\text{ nm}} = 5.85 \times 10^6\text{ m s}^{-1}\) (or \(5.9 \times 10^6\text{ m s}^{-1}\)).

(b)(iii) Using Wien's displacement law:
\(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\text{ m K}\)
\(T = \frac{2.90 \times 10^{-3}}{410 \times 10^{-9}} = 7073\text{ K}\) which is approximately \(7100\text{ K}\).

(b)(iv) 1. Relate radiant flux intensity \(F\) to luminosity \(L\):
\(F = \frac{L}{4\pi d^2} \implies L = 4\pi d^2 F\)
\(L = 4\pi \times (2.5 \times 10^{18}\text{ m})^2 \times (3.4 \times 10^{-10}\text{ W m}^{-2}) = 2.67 \times 10^{28}\text{ W}\)

2. Use Stefan-Boltzmann law:
\(L = 4\pi R^2 \sigma T^4\)
\(R^2 = \frac{L}{4\pi \sigma T^4}\)
Using \(T = 7073\text{ K}\):
\(R^2 = \frac{2.67 \times 10^{28}}{4\pi \times (5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}) \times (7073\text{ K})^4}\)
\(R^2 = \frac{2.67 \times 10^{28}}{1.78 \times 10^9} = 1.50 \times 10^{19}\text{ m}^2\)
\(R = 3.87 \times 10^9\text{ m}\) (or \(3.9 \times 10^9\text{ m}\)).

If the approximate value of \(T = 7100\text{ K}\) is used:
\(R^2 = \frac{2.67 \times 10^{28}}{4\pi \times (5.67 \times 10^{-8}) \times (7100)^4} = 1.48 \times 10^{19}\text{ m}^2\)
\(R = 3.85 \times 10^9\text{ m}\) (or \(3.8 \times 10^9\text{ m}\)).

(a) [1] For stating total power emitted (or energy per unit time) by a star.
(b)(i) [1] For explaining that the star is moving away, causing the observed wavelength to increase.
(b)(ii) [2]
- [1] Correctly calculates \(\Delta \lambda = 8\text{ nm}\) and states formula \(\frac{\Delta \lambda}{\lambda} = \frac{v}{c}\)
- [1] Correct final speed: \(5.85 \times 10^6\text{ m s}^{-1}\) (or \(5.9 \times 10^6\text{ m s}^{-1}\))
(b)(iii) [2]
- [1] For Wien's law formula \(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\)
- [1] Correct calculation leading to \(7073\text{ K}\) and stating this is \(\approx 7100\text{ K}\)
(b)(iv) [4]
- [1] States or uses \(F = \frac{L}{4\pi d^2}\)
- [1] Calculates \(L = 2.7 \times 10^{28}\text{ W}\) (accept \(2.67 \times 10^{28}\))
- [1] States or uses \(L = 4\pi R^2 \sigma T^4\)
- [1] Calculates \(R = 3.8 \times 10^9\text{ m}\) or \(3.9 \times 10^9\text{ m}\)

(a) Advantages of the continuous-flow method:
1. The thermal capacity of the calorimeter does not need to be known or measured because the apparatus reaches a steady temperature distribution and does not absorb further net energy.
2. The heat loss to the surroundings can be eliminated from the calculations by carrying out two trials with different flow rates but the same temperatures.

(b) The rate of heat loss depends on the temperature difference between the apparatus and the surroundings. Since the inlet and outlet temperatures, as well as the surrounding air temperature, are identical in both trials, the temperature profile along the tube remains the same, ensuring the rate of heat loss is constant.

(c) The continuous-flow equation is:
\(P = \frac{m}{t} c \Delta\theta + h\)

For Trial 1 (\(1.80\text{ g s}^{-1} = 1.80 \times 10^{-3}\text{ kg s}^{-1}\)):
\(45.0 = (1.80 \times 10^{-3}) c (5.4) + h\) [Equation 1]

For Trial 2 (\(3.00\text{ g s}^{-1} = 3.00 \times 10^{-3}\text{ kg s}^{-1}\)):
\(71.0 = (3.00 \times 10^{-3}) c (5.4) + h\) [Equation 2]

(d)(i) Subtract Equation 1 from Equation 2:
\(71.0 - 45.0 = (3.00 \times 10^{-3} - 1.80 \times 10^{-3}) \times c \times 5.4\)
\(26.0 = (1.20 \times 10^{-3}) \times c \times 5.4\)
\(26.0 = (6.48 \times 10^{-3}) c\)
\(c = \frac{26.0}{6.48 \times 10^{-3}} = 4012\text{ J kg}^{-1}\text{ K}^{-1}\) (or \(4010\text{ J kg}^{-1}\text{ K}^{-1}\) to 3 s.f.).

(d)(ii) Substitute \(c = 4012\text{ J kg}^{-1}\text{ K}^{-1}\) into Equation 1:
\(45.0 = (1.80 \times 10^{-3}) \times 4012.3 \times 5.4 + h\)
\(45.0 = 39.0 + h\)
\(h = 6.0\text{ W}\) (or \(6.0\text{ J s}^{-1}\)).

(a) [2] Any two distinct points from:
- No thermal capacity of the calorimeter/apparatus needs to be known/measured [1]
- Heat loss to surroundings can be eliminated from calculation [1]
- Steady-state temperatures are measured (no cooling curves/extrapolations needed) [1]
(b) [1] For stating that the temperature distribution/gradient between the apparatus and the surroundings remains identical.
(c) [2]
- [1] Correct equation for Trial 1: \(45.0 = (1.80 \times 10^{-3}) c (5.4) + h\)
- [1] Correct equation for Trial 2: \(71.0 = (3.00 \times 10^{-3}) c (5.4) + h\)
(d)(i) [3]
- [1] Correctly eliminates \(h\) by subtracting the two equations
- [1] Evaluates \((m_2 - m_1) \Delta \theta = 6.48 \times 10^{-3}\text{ kg K s}^{-1}\)
- [1] Obtains \(c = 4000\text{ J kg}^{-1}\text{ K}^{-1}\) or \(4010\text{ J kg}^{-1}\text{ K}^{-1}\)
(d)(ii) [2]
- [1] Substitution of calculated value of \(c\) into either equation
- [1] Correct calculation of \(h = 6.0\text{ W}\) (or \(\text{J s}^{-1}\))