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 bottom left So, resultant field strength towards top left 21 (B) Positive charge experience upward electric force to balance the weight. The lower plate at higher potential than the top plate. When V is decreased, P falls towards the lower plate. – Gravitational potential energy will decrease. – P gains electric potential moving towards lower plate, so gain electric potential energy When V is increased, P rises towards the upper plate. – Gravitational potential energy will increase. – P losses electric potential moving towards upper plate, so loses electric potential energy

6 22 (D) For closed circuits, terminal p.d. is lower when there is internal resistance in the source (non-ideal). 23 (C) At X, voltmeter measure emf of battery. At Y, voltmeter measure emf of battery. Voltmeter and lamp is connected in series. At Z, it is a open circuit. Voltmeter reads 0 V. 24 (A) N-pole of magnet experience a force in the same direction of field line. S-pole of magnet experience a force in the opposite direction of field line. Magnet will turn anticlockwise. Field line of closer spacing indicates higher magnetic field strength. Force at S > force at N. So component of resultant force to the left and magnet accelerates left. 25 (C) Average induced e.m.f = 2.1 (610 -120) x 10-4 / (16 x10-3) = 6.4 V 26 (C) T = 4 x 5 x 10-3 = 20 ms 3 22 20 10Tx  −== = 314 =310 27 (C) energy of electron 𝐸 = 𝑒𝑉 energy of electron converted to a single photon 𝐸 = ℎ𝑓 = ℎ𝑐 𝜆0 ⇒ 𝜆0 = ℎ𝑐 𝑒𝑉 ⇒ 𝜆0 ∝ 1 𝑉 ratio = (𝜆0)6 𝑘𝑉 (𝜆0)8 𝑘𝑉 = 8 6 = 1.3 28 (A) 𝛥𝑝𝛥𝑥 = ℎ ⇒ 𝛥𝑝 = 6.63×10−34 1.00×10−10 = 6.63 × 10−24 𝑘𝑔 𝑚 𝑠−1

7 29 (C) ratio of diameter of nucleus to atom should be small for most of the alpha particles to pass through undeflected. 30 (A) alpha: few cm of air, about 1 mm of aluminium beta: few mm of aluminium (about 5 mm) gamma: a few cm of lead, thus several more cm of aluminium