NJC 2022 Prelim P1 Guide

National Junior College 2022 Preliminary Examination 9749 H2 Physics Paper 1 Answer Key 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 B D C B C B D C D C C D B D B 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A B A C A B D C A C C C A C A Suggested Solutions 1 (B) Mass of sprinter = 65 kg Speed of sprinter = 24 km h-1 = 6.7 m s-1 KE = 1500 J 2 (D) 3% of 327.66 = 9.8… = 10 m s-1 3 (C) [Method 1] ratio = 3 parts / 1 part = 3 [Method 2] Between 0 and 8 s, distance travelled = 0 × 8 + 1 2 𝑎(8)2 = 32𝑎 and speed = 0 + 𝑎(8) = 8𝑎 Between 8 s and 16 s, distance travelled = 8𝑎 × 8 + 1 2 𝑎(8)2 = 96𝑎 Ratio = 96𝑎/32𝑎 = 3

2 4 (B) horizontal velocity after 4 s = 35 𝑐𝑜𝑠 40° = 26.8 vertical velocity after 4 s = 35 𝑠𝑖𝑛 40° + (−9.81)(4.0) = −16.7 speed after 4 s = √(35 𝑐𝑜𝑠 40° )2 + (35 𝑠𝑖𝑛 40° − 9.81 × 4)2 = 32 𝑚 𝑠−1 5 (C) F – f = ma mg sin 15o – f = ma, thus mg sin 20o – f = 2ma f = mg (sin 20o – 2 sin 15o) f = 0.86 N 6 (B) Using conservation of momentum, vmumum f=+ 2211 )(0.15)0(0.10)0.15(00.5 v=+ 00.5=v m s−1 Loss of kinetic energy = 1 2 𝑚1𝑢12 − 1 2 𝑚𝑓𝑣2 22 )00.5)(0.15(2 1)0.15)(00.5(2 1 −= 375= J 7 (D) Pressure of oxygen = atm pressure + pressure of mercury = 1.01 x 105 + 1.52(13.6x103)9.81 = 3.038 x 105 Pa pressure of oxygen pressure of atmosphere = 3.038 x 105/1.01 x 105 = 3

3 8 (C) 9 (D) Power = Work done/time = Fs/t = Fv 10 (C) F = mrw2 = (2)(5)(2π/3)2 = 40𝜋2 9 N 11 (C) Let third particle be x from 𝑀1. 𝐺𝑀1𝑚 𝑥2 = 𝐺𝑀2𝑚 (𝑑 − 𝑥)2 ⇒ 𝑥 √𝑀1 = 𝑑 − 𝑥 √𝑀2 ⇒ 𝑥 = 𝑑√𝑀1 √𝑀1 + √𝑀2

4 12 (D) Gravitational potential at Y = 𝐺𝑀 2𝐿 = 1 2 × 𝐺𝑀 𝐿 = 1 2 × (−8) = −4 𝑘𝐽 𝑘𝑔−1 Change in gravitational potential from X to Y = (−4) − (−8) = 4 𝑘𝐽 𝑘𝑔−1 Change in gravitational potential energy of 2 kg moved from X to Y = 2 × 4 = 8 𝑘𝐽 13 (B) There is still heat flow, but the net heat flow is zero. 14 (D) A: 𝑝𝑉 = 𝑛𝑅𝑇 ⇒ 𝑇 ∝ 𝑝𝑉, (𝑝𝑉)𝐴 = 8000, (𝑝𝑉)𝐵 = 18000, so temp of B > A, so internal energy changes B: gas expands, so work is done by gas C: From A, average k.e. for B > A D: 𝑞 = 𝛥𝑈 − 𝑤 = +𝑣𝑒 − (−𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑔𝑎𝑠) > 0 15 (B) A: only true for ideal gas system C: During change of phase, i.e boiling, internal energy increased but temperature remained constant. D: U = KE + PE 16 (A) 𝐹 = − 𝛥𝐸𝑝 𝛥𝑟 = −𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝐸𝑝 − 𝑟 𝑔𝑟𝑎𝑝ℎ 17 (B) 𝑣 = 𝑓𝜆 𝜆 = 320 400 = 0.80 𝑚 𝛥𝜙 2𝜋 = 0.2 0.8 𝛥𝜙 = 𝜋 2 𝑟𝑎𝑑

5 18 (A) Path S2P is 5.0 m p.d. = 5.0 – 3.0 = 2.0 m = 1 𝜆 Since the source starts antiphase and has a path difference of 1 wavelength, they meet destructively to give 0 amplitude. 19 (C) When one end is closed, L = ¼ λ V = fλ 𝑓 = 𝑣 4𝐿 When both ends are opened, L = ½ λ V = f’λ 𝑓′ = 𝑣 2𝐿 = 2𝑓 20 (A) field strength towards –Q on top left field strength towards –Q on top right field strength towards –Q on bo