2019 9749 H2 Physics MS EJC Suggested ver Aug2022

©EJC 2021 9749/H2 PHYSICS 2019 PHYSICS SUGGESTED MARK SCHEME Maximum Mark: 190 9749 October/November 2019 Paper 1 Multiple Choice Question Key Question Key Question Key 1 D 6 B 11 D 2 D 7 A 12 A 3 D 8 D 13 C 4 C 9 D 14 C 5 A 10 B 15 B 16 B 21 B 26 C 17 B 22 B 27 B 18 C 23 C 28 C 19 D 24 D 29 B 20 C 25 B 30 A Notes: Q5, 13, 23, 24 and questions on parallel D.C. circuits were not well done. Question 5 As the total momentum of the system is zero before the explosion, the centre of gravity cannot move throughout. If it does, this would mean that an external resultant force must have been applied to the system. Question 8 Those who incorrectly selected option A thought that KE was dropping, and missed out on the fact that there was a force exerted to the body continually.

2 ©EJC 2021 9749/H2 PHYSICS 2019 Question 13 Those who chose option A (43 m s−1) did not convert the mass of the helium into kilograms. Since the r.m.s. speed of a typical air molecule is about 500 m s−1, a helium atom with the same KE but smaller mass should have a higher speed. Question 15 Those who chose option C did not consider the direction of heat transfer. Question 17 Option A - the times of minimum and maximum disturbance Option C – the times of maximum disturbance Question 18 Question 21 A charged particle in a uniform electric field undergoes constant acceleration, and the gradient of the v-t graph is constant. Question 23 Possible misconception – if two components are connected in parallel and one of them is not conducting, then the other will not conduct. Question 24 Those who chose C did not realise that when the value of the variable resistor is 0, the resistance of the left-hand side of the circuit is 0, not 1000 Ω.

©EJC 2021 9749/H2 PHYSICS 2019 Paper 2 Structured Questions Notes: Singaporean candidates have to be good at the maths and the presentation of maths. Weaker answers often did not make use of scientific terms appropriately. Weaker answers for definitions were often imprecise. Better answers tend to be short and direct. Qns Marks 1(a) resultant force (in any direction) is zero B1 resultant torque/moment (about any point) on the body is zero Notes: must refer to resultant/net/sum of… B1 1(b) 12 sum of forces along vertical: sin sin 10 7 020 00TT °+ °− = M1 12 sum of forces along horizon cos20 tal: cos10 0TT °− °= M1 1 2 solving simultaneousl 0 y N N 1400 13 0 T T = = OR 1 1 70 4 0 80 30 1 00 sin sin N T T = =°° 2 2 700 70 30 1300 sin sin N T T = =°° A1 A1 (M1) (A1) (M1) (A1) 1(c) let angle that T makes with pole be θ 16 tan 12 53 1o . . . θ θ = = B1 take moments about base of pole, sum of clockwise moments = sum anti-clockwise moments ( ) ( )( ) ( ) ( )1 2 sin53.1 1 2 0 60 150 cos 10o.T . .= + ° M1 T = 280 N A1 2(a)(i) oscillations perpendicular to the direction of energy transfer B1 T1 T2 700 30o 70 o 80 o

4 ©EJC 2021 9749/H2 PHYSICS 2019 Qns Marks visible light / mechanical waves in string under tension etc Notes: use words like vibrations/oscillations B1 2(a)(ii) Wave’s oscillations are made to occur in one direction, in a plane normal to the direction of energy transfer. B1 2(b)(i) ( ) ( ) ( ) ( ) 2 0 22 T0 4 0 0 cos 30 cos intensity after 1st fi 30 cos 0 lter 6 30 cos 3 56 0 0. = °  ° °− °=  = ° = I II I I C1 C1 A1 2(b)(ii) intensity is directly proportional to square of amplitude, 2A∝I M1 0TT 00 0 05 06 75.A A .= = = II II A1 2(b)(iii) B1 – shape of cosine-squared B1 – passes through important points such as 0 at 45 135 225 3152 ,,,°°°°I 2(c)(i) sound waves are longitudinal where oscillations of air particles are parallel to direction of energy transfer B1 2(c)(ii) 1period 4 1 70T f= = C1 6.8 cm represents 1 740 s; 1 cm represents 1 1 1 6 8 s cm740 ms0 c20 m . . − − ÷ = C1 A1 3(a) work done per unit positive charge in moving a small positive test charge from infinity to that point B1

5 ©EJC 2021 9749/H2 PHYSICS 2019 Qns Marks 3(b)(i) ( )( ) ( ) 10 00 0 12 10 19 10 m, 780 V 10 10 10 54 2 10 44 4 4 8 85 1 0 780 1 60 r. Q ne rr rV e . V . . V n . εε ε ππ π π − −− − ×= = = = ×× = = = × 1 proton has a positive charge of + e. As charge in nuclear is 54e, it has 54 protons. Notes: examiners required full substitution B1 C1 A1 3(b)(ii) distance between proton and centre of nucleus much larger than proton radius and nuclear radius Notes: precision of terms required. “size” was not accepted; use “radius”/”diameter” B1 3(b)(iii) ( )( ) ( ) ( )( ) ( ) 22 00 219 2 2712 8 27 16 2 11 1 0 108 85 10 10 10 kg 1 9 10 N 54 1 44 54 1 6 1 s 3 07 7 6 0 164 20 1 67 m pp p F ma a . eeF Qq m . . . . rm r m . . πε πε π − −−− − − = = = = × = ××× = × +   = × + C1 M1 A1 3(b)(iv) Gain in kinetic energy = loss in electric potential energy B1 ( ) ( )( ) K intial at infinity nucleus 0 219 12 8 19 0 10 10 10 6 2 10 J 4 54 1 60 4 88 5 20 E qV qV Qe r . . . . πε π − −− − = − =   × ×× = × = ∆ − C1 A1

6 ©EJC 2021 9749/H2 PHYSICS 2019 Qns Marks 4(a)(i) Using Fleming’s left hand rule, the proton will experience a magnetic force that is vertically upwards. Electric force on proton must act vertically downwards to balance the magnetic force. Since Q is grounded and potential of P must be higher and so is positive polarity B1 B1 Notes: type of force and direction was required 4(a)(ii) ( ) ( ) ( ) ( ) ( )( ) 5 K 1 2 K 1 3 7 1 9 2 2 64 1 60 10 45 10 16 22 2 64 7 1 2 5 1 0 0 1074 10 ms 6 0 N 10 C 4 pp eB p E mm F eE Bv B m m eE . qE B v q F . v v v . − − − − − × × × × = = = = = = = = = = Notes: common mistakes included not converting the energy from eV to Joule, and using the mass of electron instead of mass of proton. C1 C1 C1 C1 A1 4(a)(iii) eBF F> curve downwards [not parabola, so the acceleration is not constant] Notes: attention to details in drawing are demanded. the particle experiences vertical force once inside the region so the path cannot continue straight for too long. the path must stretch to include the straight exit of the proton. see below for the dotted lines illustrating a sample construction: B1 straight straight once exit smoothly curving once inside field this is the angle associated with change in velocity θ

7 ©EJC 2021 9749/H2 PHYSICS 2019 Qns Marks 4(b) magnetic field surrounding X due to current in Y ( ) ( ) ( ) 7 0 5 4 10 0 12 1 4 10 22 T 84Y . . B r. πµ ππ − − × = × = = I ( ) ( ) 5 51 1 4 10 7 7 10 55 N m XX X X F BL F L . B . . − −− = = × = = × I I To the right C1 C1 A1 A1 5(a) energy transformed from chemical to electrical per unit charge that is driven round a complete circuit B1 5(b)(i) when switch is open, voltmeter reads e.m.f. of cell E = 12.0 V A1 effective resistance of S in parallel with T ( )( ) eff 200 300 200 300 120 R = = + Ω C1 C1 by potential divider rule terminal eff eff eff terminal 1 121 10 8 13 01 3 2 V r R E Rr ER V . . = = − − Ω +    =  = C1 C1 A1 OR When switch is closed, voltmeter reads terminal p.d. which is also p.d. across S-in-parallel-with-T terminal eff terminal eff 0 0900 A 10 8 120 VR V . R . = = = = I I Consider p.d. across internal resistance 1 13 3 1 20 08 0 09 r r Vr V ..r . . = = Ω = −= I I C1 C1 A1

8 ©EJC 2021 9749/H2 PHYSICS 2019 Qns Marks 5(b)(ii) ( ) ( ) 2 2 J 12 1 5 0 60 3 5 0 60 2 3 3 rVP . . t r . . × = × = = C1 A1 OR ( ) ( )( ) 2 2 5 60 0 0900 5 0 60 32 J 13 3 P r . t . .