EJC Physics 2023 J2 H1 MYE PAPER 2 MS

©EJC 2021 9749/J1H2MYE/2021 EUNOIA JUNIOR COLLEGE JC2 MIDYEAR EXAMINATIONS 2023 8867 PHYSICS MARK SCHEME Paper 2 – Structured Qns Answer Marks 1ai (vernier/ digital) callipers Examiner’s comment: Common mistakes: micrometre screw gauge B1 1aii 0.0004percentage uncertainty = 100%0.0420 0.95% × = Examiner’s comment: Common mistakes: A lot of students did not leave the answer in the correct number of significant figures 2 s.f. For THIS MYE ONLY, they are awarded full mark despite wrong s.f. due to leniency in marking. B1 1bi kg m−3 = kg × mn/m −3 = n−1 n=−2 Examiner’s comment: Common mistakes: Cambridge does NOT recognise [ ] as a short cut symbol for “unit of”. Please write “unit of” in the future examinations where applicable. “Show” questions must show ALL steps. Do NOT skip steps. M1 A1 1bii 2M rL M rL ρ ρ ∆∆ ∆∆=++ 0.001 0.0004 0.0001percentage uncertainty = 2 100%1.072 0.0420 0.1242 2.1%  + +× = C1 C1 A1 1biii 2 3 3 3 1.072 0.0420 2337 (kg m )2.094 0.1242 0.021 2337 49 (kg m ) (2340 50) kg m ρ ρ ρ − − − − ×= =× ∆= × = = ± C1 C1 A1

2 ©EJC 2023 8867/J2H1MYE/2023 Qns Answer Marks 2ai By conservation of linear momentum, 5 0 5 1 5 Y X xY X XY p p mv m v mv mv m mm +=+ = + =+ ( ) ( ) ( ) 2 2 2 2 total kinetic energy of X and Y after collision total kinetic energy of X and Y before collision 1 2 1 5 2 5 1 5 0.20 XY X mm v m v v v += = = C1 C1 A1 2aii ratio = 1 B1 2b horizontal line from (0 ms, 0 squares) ending at (20 ms, 0 squares) and horizontal line from (40 ms, 4.0 squares) ending at (60 ms, 4.0 squares) straight line from (20 ms, 0 squares) ending at (40 ms, 4.0 squares [= 4.0 cm vertically]) Examiner’s comment: Note: From t = 0 to t = 20 ms, the line needs to be visible too despite on the axis. B1 B1

3 ©EJC 2023 8867/J2H1MYE/2023 Qns Answer Marks 3a Rate of change of momentum is proportional to the resultant force Takes place in the direction of the resultant force B1 B1 3b Apply N2L to each of the masses. For 8kg mass, F=ma 150 +8(9.81)sin30° − T = 8a [1] For 4kg mass, F=ma T = 4a [2] Solving the equations, T = 63 N a = 15.8 m s −2 Accept if students solving a using the whole system for this case. M1 M1 A2

4 ©EJC 2023 8867/J2H1MYE/2023 Qns Answer Marks 4a Let force in rod be 𝐹𝐹. Resultant of force in rod and component of weight towards the centre provides centripetal force Acceptable: “Component of resultant force towards centre” Not acceptable: “resultant force provides centripetal force” Assuming 𝐹𝐹 points towards the centre of the circle, 𝑚𝑚𝑚𝑚 𝑐𝑐𝑐𝑐𝑐𝑐 50° + 𝐹𝐹 = 𝑚𝑚 𝑣𝑣2 𝑅𝑅 (1.5)(9.81) cos 50° + 𝐹𝐹 = (1.5) �9.52 2.0 � 𝐹𝐹 = 58 N Since 𝐹𝐹centripetal > 𝑚𝑚𝑚𝑚 𝑐𝑐𝑐𝑐𝑐𝑐 50° , 𝐹𝐹 points towards the centre of the circle. Rod is in tension. B1 M1 A1 4b minimum magnitude of force occurs at the top of the circle. Gain in 𝐸𝐸𝑃𝑃(𝐺𝐺) = Loss in 𝐸𝐸𝐾𝐾 𝑚𝑚𝑚𝑚ℎ = 1 2 𝑚𝑚𝑣𝑣initial 2 − 1 2 𝑚𝑚𝑣𝑣top2 𝑣𝑣top2 = 𝑣𝑣initial 2 − 2𝑚𝑚ℎ 𝑣𝑣top2 = 9.52 − 2 × 9.81 × 2.0(1 − cos 50°) = 76.2 m2 s-2 Resultant of force in rod and weight provides centripetal force. 𝐹𝐹′ + 𝑊𝑊 = 𝑚𝑚 𝑣𝑣top2 𝑅𝑅 𝐹𝐹′ + (1.5)(9.81) = (1.5) �76.2 2.0 � 𝐹𝐹′ = 42.4 N B1 M1 M1 A1 Towards centre of circular path 𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚 cos 50° 𝐹𝐹 assumed W F’

5 ©EJC 2023 8867/J2H1MYE/2023 Qns Answer Marks 5a 3.6 2.1 1.5 AI =−= 4.4 VV = 4.4 2.9 1.5 VR I= = = Ω C1 C1 A1 5b E V Ir= + 12.0 4.4 3.6 r= + 2.1 r = Ω C1 A1 5c Energy loss = IVt 3(470 240) 10 (3.6)(12.0) t− ×= 5320 st = C1 A1 6a anticlockwise B1 6b towards P 6c Newton’s third law forces are equal in magnitude and opposite in direction. B1 B1 6di 31 6 5 19 sin (9.11 10 )(7.0 10 ) sin 36.9 (3.14 10 )(1.6 10 ) o mvR Bq θ − −− = ××= ×× = 0.76 m M1 M1 A1 6dii 31 5 19 2 2 (9.11 10 ) (3.14 10 )(1.6 10 ) mT Bq π π − −− = ×= ×× = 1.14 x 10-6 s M1 M1 A1 6diii Pitch = ( cos )Tv θ = (1.14 x 10-6)(7.0 x 106) cos 36.9o = 6.4 m M1 A1 6div Pitch = ( cos )Tv θ = (1.14 x 10-6)(7.0 x 106) cos 0o = 8.0 m M1 A1 7(a) In a fission reaction, the daughter nuclides should be of approximately the same mass, whereas in an alpha-decay, the alpha particle is typically much less massive than the other nuclide produced. B1 7(b)(i) From conservation of nucleon number, 235 + 1 = 144 + 89 + x ⇒ x = 3 From conservation of charge, 92 = y + 36 ⇒ y = 56 Hence, 235 1 144 89 1 92 0 56 36 0U n Ba Kr 3 n+→ + + B1

6 ©EJC 2023 8867/J2H1MYE/2023 Qns Answer Marks 7(b)(ii) 19 6 11 enery released total binding energy of pr oducts total binding energy of rea ctants 8.27 144 8.62 89 7.59 235 174.41 MeV 174.41 1.6 10 10 2.79 10 J − − = − = × + ×− × = = ×× × = × M1 M1 A1 7(b)(iii) kinetic energy of the products, gamma photons Note: Heat or sound is not accepted. B1 B1 7(b)(iv) Energy released = (total mass of reactants – total mass of products) c2 total mass of reactants = total mass of products + energy released/c 2 235.043923u + 1.008665u = 88.917633u + mass of 144 Ba + 3 × 1.008665u + energy released/c2 mass of 144 Ba = (235.043923 − 88.917633 − 2 × 1.008665)u − 2.79 × 10−11/(3.0 × 108)2 = 144.108960 × 1.66 × 10−27 − 2.79 × 10−11/(3.0 × 108)2 = 2.3891 × 10−25 kg C1 M1 A1 8(a)(i) electric force per unit positive charge on a small stationary test charge at that point B2 8(b)(ii) forcefield strength charge= work done = force × d work done chargeV = work done 1 work donefield strength d charge d charge = = ×  field strength V d= Examiner’s comment: “Show” questions must define all terms that yet to be defined in the question. For example. A lot of students just write “W” without defining “W” as Work done. This is not acceptable for “show” question because “W” can also mean Weight, Watt etc, M1 M1 M1 A0 8(b)(i) 2 41 1340 1.4 10 9.6 10 V m E − − = × = × C1 A1

7 ©EJC 2023 8867/J2H1MYE/2023 Qns Answer Marks 8(b)(ii) q = 4.6 × 10–14 / 9.6 × 104 = 4.8 × 10–19 C sign of charge: negative C1 A1 A1 8(b)(iii) 1. 2 adjacent field lines have same separation (or both patterns direction of lines changes from downwards to upwards B1 B1 8(b)(iii) 2. resultant force = 4.6 × 10–14 + (9.6 × 104 × 4.8 × 10–19) = 4.6 × 10–14 + 4.6 × 10–14 = 9.2 × 10–14 N C1 A1 8(b)(iii) 3. a = F / m or 2W / m or 2g a = 9.2 × 10–14 / (4.6 × 10–14 / 9.81) = 20 m s–2 or a = 2 × 9.81 = 20 m s–2 M1 M1 A0 8(b)(iii) 4. s = ut + ½at2 (1.4 × 10–2 / 2) = ½ × 20 × t2 t = 2.6 × 10–2 s C1 A1 8(b)(iii) 5. line from (0, 0.7 × 10–2) to a non-zero point on the t-axis magnitude of gradient of line increases Examiner’s comment: A lot of students did not answer the question to context where x is the distance from the bottom plate. Leniency in marking allows them to get partial mark despite wrong answer. M1 A1