2023 9749 H2 Physics MS EJC Suggested Solutions

©EJC 2024 9749/H2 PHYSICS 2023 PHYSICS SUGGESTED MARK SCHEME Maximum Mark: 190 9749 October/November 2023 Paper 1 Multiple Choice Question Key Question Key Question Key 1 C 6 B 11 B 2 D 7 C 12 C 3 A 8 B 13 B 4 C 9 C 14 D 5 B 10 A 15 B 16 D 21 B 26 C 17 A 22 D 27 D 18 A 23 C 28 A 19 D 24 A 29 A 20 B 25 D 30 C 1 The error varies each time a measurement is take. 2 From graph: 21 2 1 2 = + −= + s ut at qppr r 2  r ( ) ( ) 1 2 1 2 = +− = + pr q p r q pr 3 Notice that all options has Ft , which is momentum. So let’s focus on momentum. 2= ∆= = ⇒∝ Ft p p mE Ft E Note: The change in momentum is identical to the initial momentum since the final momentum is zero. 4 Notice the u and v are speeds, which is the magnitude of velocity. Apply understanding of relative speed of approach equal to relative speed of separation, taking care of the direction of the bodies.

2 ©EJC 2024 9749/H2 PHYSICS 2023 5 Split the metal into top half (mass m) and bottom half (mass ½ m). Let x be the distance between the c.g. and the centre of top half. m 1 2=xm (4 ) 4 cm3 − = x x Distance of c.g. from P, which is 6 cm from P: 46 4.7 cm3= −=d 6 Read question carefully to deduce that 60% of the loss in KE of wind is converted to electrical output of turbine. ( ) ( ) ( ) ( ) 22 22 22 drop in Power 60% 1 1260% 60% 2 10.6 9.7 4.0 1.52 40 W = −   = = −     = −  KE t mu v mVI u vtt 7 One year, one orbit. Half year, half orbit. 2 360 180 π π = ° = ° 8 Based on definition. 9 Interpretation of the diagram is key. A: spacecraft is falling towards Earth (possible as with any normal object) B: path is around Earth (possible with some initial velocity, then spacecraft moves in a circular path dues to centripetal force) C: Not possible D : path is away from Earth (possible if it had an initial velocity, it would just be slowing down as it moves) 10 The average force by N molecules on the wall is dues to the collision of N molecules on the wall. By Newton’s 3 rd Law, force by molecule on wall equals force by wall on molecules. 22 2 ∆∆= = = =∆∆   ave mv v v vF N Nm Nm Nm ptt p v Note: Δv = 2v due to change in direction. Δt is taken to be the time between consecutive collision with the same wall, 2∆= pt v 11 Notice that the gas is heated to double its thermodynamic temperature in Kelvin. 33 22 '2 '2 = = ∴= ⇒= U pV nRT TT UU 12 Notice thermal energy is applied, and both P and Q have the same mass. Since P’s temperature increases faster (larger gradient) than Q’s, hence P has a smaller heat capacity. Since P takes a longer time to melt fully (from the point it started to melt), it has a larger specific latent heat of fusion than Q. 13 Note that molar mass is given but question asks for KE of one air molecule. ( ) ( ) 00 6 0 00 3 3 -1 2 223 max 0 23 31 sin sin 2 2 103 10 4.2 10 m s 1 1 2.9 10 4.2 1022 6.02 10 4.2 10 J π − − − − − − = ⇒= ∴= ⇒ = ×  × = ×  ×= = × × = × x x wt v wx wt v wx v KE mv

3 ©EJC 2024 9749/H2 PHYSICS 2023 14 Identify peak to peak timing, and take fraction of period: phase difference 360 10 sq phase difference 30 sq 360 phase difference 120 ∆ = ° = ° = ° t T 15 Apply Malus’s Law twice: 22 00cos 45 cos 45 0.25= ° °=II I 16 U sing trigonometry, 1 2sinθ = x D Apply given quantities in formula: sin 1 2 2 λθ = ⇒= ⇒= b cx f Db cDb xf 17 Apply information to formula: ( ) 3 9 sin 1 10 sin60 2 500 10 866 θλ − − = × °= × = dn N N stationary 18 D ouble slit experiment so graph & graph same shape constant, gradient of - is constant λ ⇒∝ ⇒− − = ⇒ = D a xD xt D D x t D t t Note: B is not an option because its gradient is zero, meaning distance D is constant. 19 Protons are positive and will experience an electric force in the direction of the E field, which is downwards, towards Y. Hence horizontal velocity constant; vertical velocity has constant acceleration. Path is NON-circular. 20 Note isolated charge at S is used. Next, a grid is given, indicating that distances can be found: SP = 3 sq, SR = 2 2 0 13 4 636 22 600 V πε= ⇒∝⇒ = ⇒ = = PP P RR P Vr VQVV r rVr V 21 B ased on definition 22 For e.m.f. of cell E to be balanced (current in ammeter = 0), the potential difference across the resistance wire must be identical to E. Due to the direction of the cell (positive to the left) in the main potentiometer circuit, the positive end of E (Y) must be connected to the left, and the negative end of E (X) to the right. (options A and C eliminated) If option B is chosen, then the p.d. across the resistance wire is fixed and may not match the e.m.f. of E. Hence D is the best option. 23 As intensity of light increase, resistance of LDR decreases. To get the correct graph, add a constant R to a decreasing R.

4 ©EJC 2024 9749/H2 PHYSICS 2023 24 Magnetic field lines indicates the direction of force on North pole. Based on the field line, N side will be pulled to the right and the S side will be pulled to the left. However, field lines are closer at S than at N, hence the leftward force on S is greater than the rightward force on N. 25 From diagram, the electron beam is pushed upwards and rightwards. Work on identifying the direction of forces by the various fields on the electron: F E: always opposite to direction of E field FB: use Fleming’s LHR to determine. 26 Scientific fact to be known. 27 From V-t graph, a.c. source must be used. Eliminate A and C. For option B, diode in that arrangement will result in no current and hence no p.d. across the resistor at all times. 28 Energy lost by electron results in formation of photon. ( ) ( ) ( )( ) 34 8 19 8 6.63 10 3.0 10 1.6 10 122.4 30.6 1.3 10 m λ λ λ − − − ∆= = ×× × −= = × hce V hf 29 constantλλ= ⇒= =hp ph Graph in option A shows that pλ is constant, no matter what λ is 30 Go through the “square ⇒ mean ⇒ root” for the graph. Once the graph is squared, it is a straight line. Hence, the mean- square graph is identical to the square graph. Then we apply root to get V 0.

5 ©EJC 2024 9749/H2 PHYSICS 2023 Paper 2 Structured Questions General Notes: Always consider the requirements of the questions fully before attempting answers. Avoid unnecessary repeating of wording of the question; this takes time and uses up answer space. For ‘show that’ questions, need to show the substitution of all numerical values in the formula used. For formulae derivation, need to show all the algebraic steps clearly. Question Suggested Solution Marks 1(a) Pressure exerted by water increases with depth hence pressure on bottom surface area of cylinder greater than pressure on top surface area of cylinder. Upwards force on bottom surface larger than downwards force on top surface hence net upwards force is upthrust. B1 B1 1(b)(i) Upthrust = Weight of fluid displaced 63 Upthrust (27.8 10 1 0 10 9 81 0.27 N (shown) )( . )( . )− = = ×× = Vρg Note: Need to give explanation to working B1 B1 1(b)(ii) Taking moments about the pivot in Fig. 1.1, Using Principle of Moments, Clockwise moments = Anticlockwise moments 22(8.3 10 (18.0 10 (1) Taking moments about the pivot in Fig. 1.2, Using Prin )) ∑∑ −−×× W = F --- ciple of Moments, Clockwise moments = Anticlockwise moments 2 22(7.8 10 + (0.27)(19.0 10 (19.0 10 (2) Solving (1) & (2), = 2.46 N and = 5.34 N ) )) ∑∑ − −−× ×× W = F --- FW Note: Make sure that moments due to both F an