2024 HCI H2 Physics Paper 3 Solutions

2024 HCI Preliminary Examination Paper 3 Suggested Solutions Q1 (a) Random errors are deviations of the measured value from the mean value, with varying signs and magnitudes. Systematic errors are deviations of the mean value from the true value, with same sign and similar magnitude. B1 B1 (b) The units of P are (kg m s−2)(m) s = kg m2 s-3 The units of pqk Av are 2 1 3 2 3 s kg( )(m ) (m ) kg () m1 m p q p q q s− − + + −= For the equation to be homogeneous, the units of P must be equal to the units of pqk Av . Comparing power: seconds: -q = -3  q = 3 metres: -3 + 2p + q = 2  p =1 M1 M1 A1

Q2 (a) The acceleration of an object is its rate of change of velocity with respect to time. A1 (b)(i) Taking upward to be positive and using 22 2y y y yv u a s=+ , where vy is the vertical component of the final velocity, uy is the vertical component of the initial velocity, ay is the vertical component of the acceleration and sy is the vertical component of the displacement. At the highest point, the vertical component of the velocity is zero. Furthermore, ignoring air resistance, the vertical component of the acceleration is ay = -g = -9.81 m s-2. Hence, 0 = (210 sin (30⁰))2 + 2(-9.81) h h = 561.9 m = 560 m (1, 2 or 3 s.f.) M1 A1 (b)(ii) Taking upwards as positive and using y y yv u a t=+ 0 = (210 sin (30⁰)) + (-9.81) t t = 10.703 s = 11 s (1, 2 or 3 s.f.) M1 A1 (b)(iii) From (b)(i), we know that the highest point for P is 561.9 m. Using 21 2 y y ys u t a t=+ , 561.9 = (210 sin (60⁰))tQ + (0.5)(-9.81) tQ2 t = 3.4 s or t = 33.7 s Since t must be less than 10.7 [from part (b)(ii)], t = 33.7 s should be rejected. t = 3.4 s (shown) B1 A0 (b)(ii) B1 – parallel to and above original line. (P starts at -1210sin30 105 m s= while Q starts at -1210sin60 182 m s= . Values not needed in sketch) B1 – starts after mid-point of original line and ends at the same time (3.4 s not needed in sketch) vertical velocity / m s -1 time / s 3.4 s 105 182

Q3 (a)(i) Fluid pressure increases with depth. The upward forces due to the fluid pressure acting on the lower surface of the material are larger than the downward forces due to the fluid pressure acting on the upper surface of the material, resulting in a net upward force called the upthrust. B1 B1 (a)(ii) Upthrust on cup = weight of water displaced = V ρwater g = (6.8 × 10-5) × (1000) × (9.81) = 0.66708 N Weight of cup = V ρcup g = (6.8 × 10-5) × (2200) × (9.81) = 1.4676 N Since the weight of the cup is larger than the upthrust acting on the cup, an upwards external force F is required to keep the cup stationary. Since the cup is held in equilibrium, the net force acting on the cup is zero. Hence, F + U = Wcup F = Wcup – U = 1.4676 – 0.66708 = 0.800496 = 0.80 N M1 M1 A0 (a)(iii) Since the cup was already fully submerged, there is no difference in the volume and hence weight of fluid displaced, the upthrust stays constant. or The difference in pressure between the upper and lower surface remains the same and therefore the upthrust stays constant. B1 B1 (b) Pressure of compressed air = pressure at depth d = (atmospheric pressure) + (hydrostatic pressure) = patm + ρ g d = (1.0 × 105) + (1000) × (9.81) × (0.30) = 1.02943 × 105 Pa Applying p V = constant, since temperature is unchanged, (1.0 × 105) (5.50 x 10-4) = (1.02943 × 105) V V = 5.3429 × 10-4 m3 = 5.3 × 10-4 m3 B1 M1 A1

Q4 (a) An oscillatory motion where the acceleration is directly proportional to displacement from equilibrium, and where acceleration is always opposite to displacement / acceleration is always directed toward equilibrium. B1 B1 (b) Since ay− By comparing with 2ax =− , 2  = Ag M 2 2     = = T M Ag 4 0.012 0.0252 1000 6.0 10 9.81 0.498 s 0.50 s  − +=    = = B1 B1 (c)(i) Energy of oscillation is the (maximum) kinetic energy the test-tube possesses, which decreases with time due to damping. Energy of oscillation = ( ) 22 2 2 0 1 1 1 2 2 2mv m A m A== . A reduction of 75 % would mean that the energy of oscillation remaining is 25 % of its original. 2 2 '' 1.0 EA E = 2 2 1' 4 1.0 A= ' 0.50 cmA = From the graph, this happens at 1.0 s. M1 A1 (c)(ii) Natural frequency of the system is 2.0 Hz (since period is 0.5 s). However, the driving frequency is only 1.0 Hz. Energy transfer from the (external forcing agent) water waves to the test-tube is not optimal / does not result in resonance. B1 B1

Q5 (a)(i) Q = mass x specific latent heat of vaporisation = 0.37 × 2.3 × 106 = 8.5 × 105 J A1 (a)(ii) pV = nRT T = (100 + 273) = 373 K Number of moles, n = 0.37 × 1000 g / 18 g ( ) ( ) 5 3 0.37 1000 8.31 373 18 1.0 10 0.63714 0.64 m V   =  = = B1 B1 B1 (a)(iii) Work done by the water = (atmospheric pressure)(increase in volume) = (1.0 x 105)(0.64) = 6.4 x 104 J A1 (a)(iv) Work done on water is negative. From the first law of thermodynamics, increase in internal energy = heat supplied + work done on water = (8.5 – 0.64) x 105 = 7.9 x 105 J M1 A1 (b) Kinetic energy of the molecules remains unchanged because there is no temperature change. Potential energy of the molecules increases, because molecular bonds are broken and the molecules are further apart. Hence, the internal energy of the system increases. B1 B1 A0

Q6 (a)(i) The resistance is infinite. B1 (a)(ii) The resistance decreases as V increases. B1 (b)(i) Method 1: When the galvanometer reads zero, 1600 (9.0) 5.143 V1600 1200 LDR XZ LDR LDR RV V E RR= = = = ++ For the wire, 5.143 (1.2) 0.68579.0 0.69 m XZ XZ XZ XZ YZ XY YZ XY V L V LLV L V=  = = = = Method 2: 1 4 3 600 1200 XZ R XZ ZY LDR ZY V V kL I V V kL I=  = =   1.2 0.69 m 4 7 XZL  = = M1 A1 M1 A1 (b)(ii)1 As intensity of light is increased, the resistance of the LDR decreases and there is a smaller potential difference across the LDR. The length of XZ decreases. M1 A1 (b)(ii)2 The total resistance of the circuit decreases and more current is drawn from the battery. Hence power produced by the battery increased. OR The total resistance of the circuit decreases. Since power produced by battery = V2/Rtotal, power produced by battery increased. M1 A1 M1 A1

Q7 (a) The magnetic flux (linkage) is given by ()NBA B wx = = where x is the distance AB has moved past P. Hence the induced emf is given using Faraday’s Law by d dxE Bwdt dt == dxBw Bwvdt = B2 (b)(i) As AB moves from P towards Q, magnetic flux linkage over the area ABCD enclosed by the frame increases resulting in an induced e.m.f. generated in the frame by Faraday’s Law. By Lenz’s Law , an induced current will flow such that it opposes the increase in magnetic flux linkage. (current flows in anticlockwise direction) Consequently, a magnetic force acts on AB towards the left, causing it to slow. (Using Fleming’s Left Hand Rule) Alternative: AB cuts the magnetic flux as it moves through PQ, resulting in an induced e.m.f . generated in the frame by Faraday’s Law. There is induced current in the full circuit in the anticlockwise direction. (or electrons in the clockwise direction) Consequently, a magnetic force acts on AB towards the left, causing it to slow. B1 – why there is an induced emf B1 – how is the direction of induced current determined B1 – effect on AB/frame B1 B1 B1 B1 B1 B1 (b)(ii) B1 – same slope where the speed is the same as the original graph B1 – plateau B1 – shorter time to pass throug