2023 ACSBR Phy Prelims Ans

1 Anglo-Chinese School (Barker Road) 2023 Sec 4 Exp Physics (6091) Preliminary Examination Marking Scheme Paper 1 Paper 2 Overall remarks: ● Students who give answers in fractions will get 1 mark deducted off the whole paper. ● 1 mark is deducted for every different incorrect unit. Answers Marks Examiner’s Comments 1 a Velocity changes from positive to negative. Line goes below x-axis. after time t, velocity becomes negative. B1 Do not accept: line is negative (line cannot be negative) velocity is below x-axis (velocity can be negative but cannot be below x-axis) 1 C 11 D 21 C 31 C 2 C 12 A 22 C 32 A 3 A 13 A 23 A 33 A 4 C 14 D 24 B 34 D 5 A 15 A 25 B 35 A 6 C 16 C 26 D 36 B 7 D 17 B 27 A 37 C 8 D 18 B 28 A 38 C 9 C 19 C 29 A 39 B 10 A 20 C 30 B 40 D

2 1 b a = (v-u) / t -10 = (0 – 20) / t time = 20 / 10 = 2 s A1 1 c a = (v-u) / t -10 = (v – 0) / (6-2) v = - 40 m/s B1 must have -ve sign 1 d Height = ½ x 4 x 40 – ½ x 2 x 20 = 60 m M1 A1 2 a Resultant force is zero. Resultant moment is zero OR Sum of clockwise moments about a pivot is equal to sum of anticlockwise moments about the same pivot. (pivot must be mentioned) B1 B1 Common misconception: resultant force is upwards “Word” pivot must be present 2 b Taking moments about rear wheel, T2 x 4 = 18 000 x (4 - 0.8) T2 = 14 400 N T1 = 18 000 – 14 400 = 3 600 N M1 A1 A1 3 a F = pA = 140 000 x 0.012 = 1680 N M1 A1 3 b Liquid pressure + atmospheric pressure = gas pressure hpg + 100 000 = 140 000 hpg = 40 000 (0.80 – 0.30) x p x 10 = 40 000 p = 8000 kg/m3 OR 8 g/cm3 M1 A1 Need to convert 50 cm to 0.5 m to work in SI units 3 c KE increase and speed of molecules increase Frequency of collison of molecules with piston walls increases Total force of collision increases Force per unit area increases B1 B1 B1

3 Gas pressure increases Resultant pressure outwards (and pushes piston to the left) *1 mark for every 2 points 4 a Energy cannot be created or destroyed but can only be converted from one form to another form (total amount of energy in an isolated system remains constant) B1 4 bi GPE = mgh = 0.5 x 10 x 13 = 65 J M1 A1 4 bii Total energy at end + energy converted to heat due to slope friction = Total energy at start GPE at C + 10.7 = GPE at A + KE at A mgh + 10.7 = mgh + ½ mv2 0.5 x 10 x 13 + 10.7 = (0.5 x 10 x 7.5) + (½ x 0.5 x v2) ½ x 0.5 x v2 = 38.2 v = 12.4 m/s M1 A1 4 c No energy to overcome air resistance or converted to sound energy B1 Do not accept if answer is not specific - no energy loss (must be specific to which form of energy is energy lost to) 5 ai Q = mcΔƟ Pt = mcΔƟ 450 x t = 2 x 900 x (100 – 25) t = 300 s OR 5 mins M1 A1 5 aii D. Highest melting point so it can withstand the heat/wont melt so easily

4 Smallest specific heat capacity so temperature will rise fastest for the same amount of thermal energy supplied Any 1 point – 1 mark B1 B1 5 b Q = mlv 5 190 000 = 2.3 x lv lv = 5 190 000 / 2.3 =2 256 522 J / kg M1 A1 Units must be present. Common mistake was J/ kg / deg celsius which is for specific heat capacity 6 a wavelength = 35 / 7 = 5 cm A1 6 b period = 2.5 / 5 T = 0.5 Frequency = 1/ period = 1/ T = 1 / 0.5 = 2Hz OR v = 25 / 2.5 = 10 cm/s v = frequency x wavelength 10 = f x 5 f = 2 Hz OR 2.5 s → 5 complete waves 1 s → 5/2.5 = 2 complete waves (frequency → no. of complete waves in 1 s) f = 2 Hz M1 A1 7 a Using graph For P, When current = 0.15 A in P, V across P = 2.7 V

5 Since P and Q are in parallel, V across Q = V across P = 2.7 V Using graph, When V across Q = 2.7 V, current in Q = 0.09 A Total current = 0.15 + 0.09 = 0.24 A M1 A1 7 b V across resistor = 4.5 – 2.7 = 1.8 A1 7 c ((pL/A) of P )/ ((pL/A) of Q) = 18/3 21 4AD = Since diameter of P is twice diameter of Q, AP = 4AQ Since resistivity p is the same, (LP / 4A) / ((LQ / A) = 6 (LP / LQ )= 6 x 4 = 24 M1 A1 8 ai 1 / R = 1/ 1800 + 1 / 9000 R = 1500 Ω B1 Must see working 8 aii I = V / R = 4.5 / 1500 = 0.003 A B1 Must see working 8 b Light intensity increases, resistance of LDR decreases Effective resistance of circuit decreases [Current increases OR ammeter reading increases] Either one B1 B1 The word “effective” is important because the LDR’s resistance is one of two resistors present in the circuit. The ammeter reading measures the circuit’s current. 9 a A : North B: South B1

6 9 b Magnetic field of current interacts with magnetic field of magnet Downward force acting on PQ Upward force acting on RS [Forces act at a distance from the axis of coil] optional Coil rotates anticlockwise (ECF based on magnetic poles in 9(a) B2 3 Pts – 2 marks 2 Pts – 1 mark 1 Pt – 0 mark Need to state clearly direction of force on each side of coil (use Flemming’s left hand rule to determine direction of force) 9 c Ensure the current in the circuit reverses for every half revolution B1 9 d Due to inertia / momentum, the coil continues to turn though current is not flowing in coil in vertical position. B1 10 a Ratio of turns = Ratio of voltage 3300 / 220 = 15 *1500 / 100 = 15 Select M as Primary coil K as Secondary coil M1 A1 10 b Power in Secondary = 0.75 x Power in Primary 15 000 = 0.75 x I x V 15 000 = 0.75 x I x 3300 I = 6.06 A M1 A1 75% efficient means output power is 75% of input power 10 c To ensure maximum magnetic flux linkage of primary coil with secondary coil by concentrating magnetic field lines. Accept if a student is able to mention one of the two bolded ideas. B1 11 ai The refractive index is the ratio of the speed of light in vacuum to the speed of light in the medium B1 aii Total internal reflection A1

7 aiii For total internal reflection to occur in the core, light has to travel from an optically denser to a less dense medium (optically must be present) Hence, the cladding has a lower refractive index. B1 aiv 1. The speed of light is lower in the optical fibre than in the air since the optical density of the optical fibre is higher than that of air. 2. The path of green light is shorter (straight path) than that of violet light (jagged path) for the same distance between the transmitter and receiver. B1 B1 bi When r = 90, i = 40, this is the critical angle Total internal reflection will occur and the light ray is reflected back into the glass or water. B1 B1 bii n = 1/sin c n = 1/sin 50 n = 1.31 Award 1 mark to students who use i = 40 and r = 30 as data and substituted correctly. (No other angles are allowed) M1 A1 Remarks: Students should examine graphs carefully when selecting data points. i = 40 and r = 30 is not an exact point and hence only 1 mark is awarded. biii The graph(gradient) for glass will be less steep since the amount of bending will be lesser hence angle of refraction will be lower for each angle of incidence. Marking point: State how the graph for glass would change: gradient less steep, gentler slope Explain: bending less (from normal), angle of refraction is smaller as compared to when light emerges to air. B1 12 a Sound wave of above 20 kHz. (frequency is implied) OR Sound wave of high frequency above human audibility range B1

8 b 6 divisions on time base. T = 6 x 1 x 10-6 = 6 x 10-6 s A1 c Not all ultrasound is reflected from