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(a) Using \( v = u + at \) with \( u = 0 \), \( a = 4.0\text{ m/s}^2 \), and \( t = 3.0\text{ s} \): \( v = 0 + (4.0 \times 3.0) = 12.0\text{ m/s} \). (b) After fuel is exhausted, the rocket is in free fall. Using \( v = u + at \) where final velocity \( v = 0 \), initial velocity \( u = 12.0\text{ m/s} \), and deceleration \( a = -9.8\text{ m/s}^2 \): \( 0 = 12.0 - 9.8t_{up} \) which gives \( t_{up} = \frac{12.0}{9.8} \approx 1.22\text{ s} \). Total time from launch = \( 3.0\text{ s} + 1.22\text{ s} = 4.22\text{ s} \). (c) Distance during first phase: \( s_1 = \frac{1}{2}at^2 = 0.5 \times 4.0 \times 3.0^2 = 18.0\text{ m} \). Distance during second phase: \( s_2 = \frac{v^2 - u^2}{2a} = \frac{0 - 12.0^2}{2 \times (-9.8)} = \frac{-144}{-19.6} \approx 7.35\text{ m} \). Maximum height = \( s_1 + s_2 = 18.0 + 7.35 = 25.35\text{ m} \).

(a) [2 marks]: 1 mark for formula \( v = u + at \) or substitution, 1 mark for correct value with unit (12 m/s). (b) [3 marks]: 1 mark for calculating time during deceleration phase (1.22 s), 1 mark for adding the initial 3.0 s, 1 mark for final correct value (4.22 s). (c) [2.27 marks]: 1 mark for calculating first height (18 m) or second height (7.35 m), 1.27 marks for correct total height (25.35 m or 25.4 m with unit).

(a) Work done \( W = mgh = 2.5\text{ kg} \times 9.8\text{ N/kg} \times 3.0\text{ m} = 73.5\text{ J} \). (b) Total electrical energy input to the motor: \( E_{in} = \text{Power input} \times t = 15\text{ W} \times 6.5\text{ s} = 97.5\text{ J} \). Useful energy output: \( E_{out} = 73.5\text{ J} \). Efficiency = \( \frac{E_{out}}{E_{in}} \times 100\% = \frac{73.5}{97.5} \times 100\% \approx 75.4\% \). (c) The motor loses thermal energy due to electrical resistance in its coils, and work must be done against friction in its moving parts, converting some mechanical energy into internal (thermal) energy.

(a) [2 marks]: 1 mark for formula \( W = mgh \) or substitution, 1 mark for correct value (73.5 J). (b) [3 marks]: 1 mark for calculating total input energy (97.5 J), 1 mark for the efficiency formula, 1 mark for correct final value (75.4% or 0.75). (c) [2.27 marks]: 1.13 marks for stating thermal loss in coils, 1.14 marks for stating work done against friction.

(a) The center of gravity is defined as the single point on an object through which its entire weight can be assumed to act. (b) For a uniform beam of length \( 2.0\text{ m} \), the center of gravity is at its midpoint, \( 1.0\text{ m} \) from pivot A. Taking moments about pivot A: Clockwise moment = \( W \times 1.0\text{ m} = 50\text{ N} \times 1.0\text{ m} = 50\text{ N}\cdot\text{m} \). Anticlockwise moment = \( T \times 1.5\text{ m} \). Since the system is in equilibrium: \( T \times 1.5 = 50 \implies T = 33.3\text{ N} \). (c) For vertical equilibrium, total upward force equals total downward force. Let \( F \) be the upward force at the pivot: \( F + T = W \implies F + 33.3 = 50 \implies F = 16.7\text{ N} \). The force acts vertically upwards.

(a) [2 marks]: 1 mark for identifying it as a single point, 1 mark for saying weight of the body acts through it. (b) [3 marks]: 1 mark for stating that the weight acts at the midpoint (1.0 m), 1 mark for equating clockwise and anticlockwise moments, 1 mark for correct tension (33.3 N). (c) [2.27 marks]: 1.27 marks for calculating the magnitude (16.7 N), 1 mark for specifying the upward direction.

(a) The combined resistance \( R_p \) of parallel resistors \( R_2 \) and \( R_3 \) is given by: \( \frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{3}{6.0} = \frac{1}{2.0} \), so \( R_p = 2.0\ \Omega \). (b) The total resistance of the circuit is \( R_{\text{total}} = R_1 + R_p = 4.0\ \Omega + 2.0\ \Omega = 6.0\ \Omega \). The total current is \( I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A} \). (c) The potential difference across the parallel pair is \( V_p = I \times R_p = 2.0\text{ A} \times 2.0\ \Omega = 4.0\text{ V} \). The current through the \( 6.0\ \Omega \) resistor is \( I_2 = \frac{V_p}{R_2} = \frac{4.0\text{ V}}{6.0\ \Omega} = 0.67\text{ A} \).

(a) [2 marks]: 1 mark for parallel formula or correct substitution, 1 mark for correct value (2.0 ohms). (b) [2 marks]: 1 mark for total resistance (6.0 ohms), 1 mark for correct current (2.0 A). (c) [3.27 marks]: 1 mark for formula \( V_p = I R_p \), 1 mark for calculating PD (4.0 V), 1.27 marks for calculating correct current (0.67 A).

(a) Using Snell's law: \( n = \frac{\sin(i)}{\sin(r)} \) where \( n = 1.52 \) and \( i = 45.0^\circ \). Thus, \( \sin(r) = \frac{\sin(45.0^\circ)}{1.52} = \frac{0.7071}{1.52} = 0.4652 \). This gives \( r = \arcsin(0.4652) = 27.7^\circ \). (b) The critical angle \( c \) is given by \( \sin(c) = \frac{1}{n} = \frac{1}{1.52} = 0.6579 \). This yields \( c = \arcsin(0.6579) = 41.1^\circ \). Since the angle of incidence \( 45.0^\circ \) is greater than the critical angle \( 41.1^\circ \), the ray of light will undergo total internal reflection. (c) The critical angle is defined as the angle of incidence in the optically denser medium at which the angle of refraction in the less dense medium is \( 90^\circ \).

(a) [2.27 marks]: 1 mark for Snell's Law equation, 1.27 marks for correct angle of refraction (27.7 degrees). (b) [3 marks]: 1 mark for formula for critical angle, 1 mark for correct critical angle (41.1 degrees), 1 mark for correct conclusion based on comparing the two angles. (c) [2 marks]: 1 mark for referencing angle of incidence in the denser medium, 1 mark for specifying the angle of refraction is 90 degrees.

(a) Sound is a longitudinal wave. In a longitudinal wave, the particles of the medium vibrate back and forth parallel to the direction of energy propagation. In a transverse wave, the particles of the medium vibrate perpendicular to the direction of energy propagation. (b) The time interval \( T \) between claps is the reciprocal of the frequency of claps: \( T = \frac{1}{f} = \frac{1}{2.4\text{ claps/s}} = 0.417\text{ s} \). (c) The sound travels from the student to the wall and back, which is a distance of \( 2d \), during the time interval \( T \). Thus, \( 2d = v \times T \), so \( 2d = 340\text{ m/s} \times 0.417\text{ s} = 141.7\text{ m} \). This gives \( d = \frac{141.7}{2} = 70.8\text{ m} \) (or \( 70.83\text{ m} \)).

(a) [3 marks]: 1 mark for stating longitudinal, 1 mark for explaining longitudinal vibrations are parallel, 1 mark for explaining transverse vibrations are perpendicular. (b) [2 marks]: 1 mark for using \( T = 1/f \), 1 mark for correct time interval (0.417 s). (c) [2.27 marks]: 1 mark for using the double distance formula \( 2d = v t \), 1.27 marks for correct distance calculation (70.8 m or 71 m).

(a) Alternating current in the primary coil generates a continuously changing magnetic field in the iron core. This changing magnetic field cuts the turns of the secondary coil, inducing an alternating electromotive force (e.m.f.) across its terminals. (b) Using the transformer equation: \( \frac{V_p}{V_s} = \frac{N_p}{N_s} \), we get \( V_s = V_p \times \frac{N_s}{N_p} = 240\text{ V} \times \frac{60}{1200} = 12\text{ V} \). (c) First, find the current in the secondary coil: \( I_s = \frac{V_s}{R} = \frac{12\text{ V}}{8.0\ \Omega} = 1.5\text{ A} \). For a 100% efficient transformer, input power equals output power: \( V_p I_p = V_s I_s \). Thus, \( 240 \times I_p = 12 \times 1.5 \), which gives \( I_p = \frac{18}{240} = 0.075\text{ A} \) (or \( 75\text{ mA} \)).

(a) [2 marks]: 1 mark for mentioning changing magnetic field, 1 mark for stating that it induces an e.m.f. in the secondary coil. (b) [2 marks]: 1 mark for the transformer equation, 1 mark for correct voltage (12 V). (c) [3.27 marks]: 1 mark for finding secondary current (1.5 A), 1 mark for power conservation equation or equivalent turns ratio current relation, 1.27 marks for correct primary current value (0.075 A).

(a) Redshift shows that the light from distant galaxies is shifted to longer wavelengths, which means that these galaxies are moving away from us. Furthermore, more distant galaxies show greater redshift, meaning they are moving faster. This suggests that the Universe is expanding. Extrapolating back in time, all matter must have been concentrated at a single, incredibly hot and dense point, which supports the Big Bang theory. (b) (i) The change in wavelength is \( \Delta\lambda = 672.5\text{ nm} - 656.3\text{ nm} = 16.2\text{ nm} \). (ii) Using the redshift formula \( \frac{\Delta\lambda}{\lambda_0} = \frac{v}{c} \), we find \( v = c \times \frac{\Delta\lambda}{\lambda_0} = 3.0 \times 10^8\text{ m/s} \times \frac{16.2\text{ nm}}{656.3\text{ nm}} \approx 7.405 \times 10^6\text{ m/s} \), which is approximately \( 7.4 \times 10^6\text{ m/s} \).

(a) [3 marks]: 1 mark for stating redshift indicates galaxies are moving away, 1 mark for stating that speed of recession increases with distance (Hubble's observations), 1 mark for linking the expanding universe back to a common starting point (Big Bang). (b)(i) [1.27 marks]: 1.27 marks for correct subtraction (16.2 nm). (b)(ii) [3 marks]: 1 mark for rearraging the formula for \( v \), 1 mark for substituting values, 1 mark for correct calculation to 2 significant figures (7.4 x 10^6 m/s).

(a) Redshift is the increase in the observed wavelength of electromagnetic radiation received from a receding celestial body compared to that emitted by the source. (b) Light from all distant galaxies is redshifted, showing that they are moving away from Earth. More distant galaxies show a greater redshift, which indicates they are moving away faster. This linear relationship (Hubble's Law) implies that the Universe is expanding from a single point of origin, supporting the Big Bang theory. (c) First, calculate the change in wavelength: \(\Delta \lambda = 681\text{ nm} - 656\text{ nm} = 25\text{ nm}\). Using the formula: \(\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}\). Substitute the values to find \(v\): \(v = c \times \frac{\Delta \lambda}{\lambda_0} = (3.0 \times 10^8\text{ m/s}) \times \frac{25\text{ nm}}{656\text{ nm}}\), which gives \(v \approx 1.14 \times 10^7\text{ m/s}\).

(a) [2 marks] Correct definition: increase in observed wavelength / light shifted towards the red end of spectrum (1 mark); Due to the source moving away from the observer (1 mark). (b) [2 marks] Redshift shows galaxies are moving away / receding (1 mark); The larger redshift for more distant galaxies indicates faster recession, showing expansion of space / universe originating from a single point (1 mark). (c) [3.27 marks] Calculation of wavelength shift \(\Delta \lambda = 25\text{ nm}\) (1 mark); Correct formula statement or rearrangement: \(v = c \frac{\Delta \lambda}{\lambda_0}\) (1 mark); Correct substitution: \(3.0 \times 10^8 \times \frac{25}{656}\) (0.27 marks); Correct final answer with units: \(1.14 \times 10^7\text{ m/s}\) (accept \(1.1 \times 10^7\text{ m/s}\)) (1 mark).

(a)(i) Total resistance of a series circuit is the sum of individual resistances: \(R = R_1 + R_{\text{LDR}} = 400\ \Omega + 200\ \Omega = 600\ \Omega\). (a)(ii) Use Ohm's Law for the whole circuit: \(I = \frac{V}{R} = \frac{12.0\text{ V}}{600\ \Omega} = 0.020\text{ A}\) (or \(20\text{ mA}\)). (a)(iii) Potential difference across the LDR: \(V_{\text{LDR}} = I \times R_{\text{LDR}} = 0.020\text{ A} \times 200\ \Omega = 4.0\text{ V}\). (b) When the light intensity decreases, the resistance of the LDR increases. Because the LDR now represents a larger fraction of the total resistance, a larger proportion of the supply voltage is dropped across the LDR. Consequently, the potential difference across the fixed resistor \(R_1\) decreases.

(a)(i) [1 mark] \(600\ \Omega\) (1 mark). (a)(ii) [2 marks] Use of \(I = V/R\) (1 mark); Correct answer: \(0.020\text{ A}\) (or \(20\text{ mA}\)) (1 mark). (a)(iii) [2 marks] Use of \(V = I \times R\) or ratio method (1 mark); Correct answer: \(4.0\text{ V}\) (1 mark). (b) [2.27 marks] Resistance of LDR increases when light intensity decreases (1 mark); Total resistance increases, so current in the circuit decreases (1 mark); Therefore, potential difference across the fixed resistor decreases (or reference to sharing of voltage) (0.27 marks).

(a) Using the transformer turn ratio equation: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\). Rearranging for secondary voltage \(V_s\): \(V_s = V_p \times \frac{N_s}{N_p} = 240\text{ V} \times \frac{60}{1200} = 12\text{ V}\). (b)(i) Power of the lamp is given by \(P = V_s \times I_s\). Rearranging for current in the secondary coil \(I_s\): \(I_s = \frac{P}{V_s} = \frac{24\text{ W}}{12\text{ V}} = 2.0\text{ A}\). (b)(ii) For a 100% efficient transformer, input power equals output power: \(P_{\text{primary}} = P_{\text{secondary}} = 24\text{ W}\). Therefore, \(I_p = \frac{P}{V_p} = \frac{24\text{ W}}{240\text{ V}} = 0.10\text{ A}\). (Alternatively: \(I_p = I_s \times \frac{N_s}{N_p} = 2.0 \times \frac{60}{1200} = 0.10\text{ A}\)).

(a) [2.27 marks] State or use formula: \(\frac{V_p}{V_s} = \frac{N_p}{N_s}\) (1 mark); Correct substitution: \(V_s = 240 \times \frac{60}{1200}\) (0.27 marks); Correct final answer with unit: \(12\text{ V}\) (1 mark). (b)(i) [2 marks] Use of \(P = V I\) rearranged to \(I = P / V\) (1 mark); Correct calculation: \(2.0\text{ A}\) (1 mark). (b)(ii) [3 marks] Equating primary power to secondary power (24 W) OR stating formula \(I_p V_p = I_s V_s\) (or turn-current ratio) (1 mark); Substitution of values: \(I_p = \frac{24}{240}\) or equivalent (1 mark); Correct calculation of primary current: \(0.10\text{ A}\) with unit (1 mark).