EJC 2022 J1 H2 PROMO MS (P1 &P2)

©EJC 2022 9749/J1H2MYE/2022 EUNOIA JUNIOR COLLEGE JC1 Promo Examination 2022 9749 PHYSICS MARK SCHEME Paper 1 – Multiple Choice Questions 1 Answer: D (Left hand side): v has units of m s-1. (Right hand side): -2 -1 -2 22 -3 -1 -2 2 -2 -3 -1 units of N kg m sunits of = = = = kg m sunits of m m units of = kg m units of kg m s= = m sunits of kg m units of = m sunits of Fp A d p d p d Units on LHS = Units on RHS Therefore, γ has no units. 2 Answer: A 2 2 2 2 2 2 -1 =- = +(- ) = + - 2( )( )cos150° = 50 +80 - 2(50)(80)cos150° =126 km h 80 126=sin sin150° θ =18.5° LC L c Lc LC L C L C LC LC v v v vv v v v v v v v θ Bearing = 360°-18.5°-60° = 281.5° 𝜃 60° -Vc = 80 km h -1 150° VLC VL = 50 km h-1 N

2 ©EJC 2022 9749/J1H2MYE/2022 3 Answer: A Student A Student B Student C Student D Accuracy 0.85 0.85 0.8625 0.85 Precision (range) 0.17 0.06 0.17 0.04 4 Answer: C Minimum distance occurs when both cars have the same speed. ( ) ( ) ( ) ( )     1Car X : × 3.0+5.0 ×20 + 55×5 = 355 m2 1Car Y : 35×2.0 + × 35+55 ×3.0 = 205 m2 Minimum distance = (450+ 205)- 355 = 300 m 5 Answer: B Can be obtained by looking at the gradient of the s-t graph. 6 Answer: C 2 2 1 2 1 2 s ut at s at =+ = 2s at 2 2 2 2 31 66 3 m m m E EE m E m s a t s at t t tt = = = 7 Answer: A Action and reaction forces should act on opposite bodies. For Option A, the correct statement should look like “the gravitational forces of attraction between satellite and Earth” 8 Answer: B By Newton’s Second Law, rate of change of momentum is the resultant force acting on the body. In this case, that would be 4N.

3 ©EJC 2022 9749/J1H2MYE/2022 9 Answer: C Fnet = m/t vperpendicular = m/t (2v sin30) = 2.5 × (2 × 5.0sin30) = 13 N Option A: Forgot factor of 2 Option B: Forgot factor of 2 and considered cosine instead of sine Option C: Correct Option D: Considered consine instead of sine 10 Answer: B For Option A, on both arms at level where the two liquids intersect, pressures are the same. Since pressure decrease (by hg) as we progress upwards, pressure at Pt Q would be diff from pressure at Pt P since decrease in pressure is hg (same h but diff ). For Option B, Pt R and Pt S are at the same pressure. Consider the pressure at bottom - most of tube, the same decrease in liquid pressure by hg (same h,) For Option C, since the surface level of both arms experience the same pressure (atmospheric) and both liquids are of different densities, the surface levels cannot be changed. For Option D, since the surface level of both arms experience the same pressure (atmospheric) and both liquids are of different densities, the surface levels cannot be aligned. 11 Answer: C Given that there are 3 forces acting on the beam to bring about equilibrium, the lines of action of these forces must be at the same point (concurrent forces). Hence, either option B or C. Option B is the force action by wall on beam. Option C is the force acted by beam on wall. 12 Answer: D By principle of conservation of energy, Loss in GPE = gain in EPE of spring 1.0(9.81)(20.0) = average force of spring (5.0) Average force of spring = 39.2 N

4 ©EJC 2022 9749/J1H2MYE/2022 13 Answer: A On level road, P = Fengine v 108 000 = F(15) Fengine = 7200 N Fdrag = 7200 N when v = 15 m s-1 On slope, Since v = 10 m s-1 and Fdrag is proportional to v, New Fdrag = (10/15) x 7200 = 4800 N At constant power, Fengine = 108000/10 = 10 800 N Resultant force downslope = 4800 + 1500 (9.81) sin 30 – 10 800 = 1357.5 N Deceleration of car = 1357.5/1500 = 0.905 m s-2 14 Answer: C Using v2 = u2 + 2as Final velocity of load, v = 15.65 m s-1 Effective work done = gain in KE + gain in GPE = ½ mv2 + mgh = 23 291 J Efficiency = 23291/(8500x4.47) x 100 = 61% 15 Answer: D Centripetal force is a resultant force and should not be drawn in free-body-diagrams. The direction of friction along the slope would depend on the speed of the linear velocity. 16 Answer: C Given a=2r, and since  increases steadily with t, and rQ > rP (but constant over time for each) a t2 and aP /aQ = rP /rQ = constant at a given time.

5 ©EJC 2022 9749/J1H2MYE/2022 17 Answer: A For Q: Fnet = maQ = m2/2r TPQ = m2/2r For P: Fnet = maP = m2/r TXP − TPQ = m2/r TXP − m2/2r = m2/r TXP = 3m2/2r Option B: P and Q experience the same angular velocity Option C: Mass Q experiences a larger velocity since v=r (same , rQ>rP) Option D: Mass Q experiences a larger acc since a=2r (same , rQ>rP) 18 Answer: A Net gravitational field strength at X = gravitational field strength due to moon + gravitational field strength due to earth =(6.67 x 10-11)[ 5.97 x 1024/(387.5x106)2 + 7.35 x 1022/(2.5 x 106)2] = 0.787 N kg-1 19 Answer: C Gravitational force is the rate of change of potential energy with distance. ( ) ( )p g dE d m dF m mPdr dr dr =− =− =− =− 20 Answer: C Velocity is maximum at the equilibrium (region 3). Hence it is hardest to hit. 21 Answer: B 2 2 2 () 0 (2 ) (0.030) 9.81 2.9 Hz o o F ma W T ma T mg ma T m g x T x g f f    = −= =− =− = → = = =

6 ©EJC 2022 9749/J1H2MYE/2022 22 Answer: C 23.0 4.0cos 0.80 0.092 s t t − =−  = Time = 0.80 + 0.0920 = 0.89 s 23 Answer: D Phase difference between P and Q = phase difference between P and Q’. For Q’ , 1 =2 sin Ө Phase corresponding to x = Ө = π/6, Phase of P = π/2 (corresponding to a quarter of a wavelength) Phase difference PQ = π/2 - π/6 = π/3 x/ cm 0 t/ s 4.0 - 4.0 0.80 1.00 0.092 - 3.0 Displacement /m Distance from source / m 2 -2 P Q’ 2 4 x

7 ©EJC 2022 9749/J1H2MYE/2022 24 Answer: D Shift the graph to the left. P is instantaneously at rest. Q is moving downwards. R is instantaneously at rest. S is moving upwards and slowing down as its acceleration is towards the equilibrium position. 25 Answer: C Distance from R to S = 0.27 m = 2.5 λ Speed of sound in air ~ 330 m s-1 Frequency = 330/ (0.27/2.5) = 3.1 kHz 26 Answer: C Loudness is a measure of intensity; power of source is reduced to one third. However, intensity at new location is to be equal to intensity at 8.0 m away. 22 1 3 4 (8) 4 ( ) 8() 3 4.6  = = = new new PP r r m 27 Answer: D Tube P: ( )( ) =2 = 2 = 2 freq values = , 2 , 3 λL v f L fL fff Tube Q: ( ) 2 3 22 =4 2 1= = = 4 15freq values = , , 2 λL fLvff λL fff 28 Answer: B The wavelength of the waves before and after passing through the gap does not change.

8 ©EJC 2022 9749/J1H2MYE/2022 29 Answer: C Sources are in antiphase with each other. High intensity occurs when path difference = 0.5, 1.5λ, 2.5λ, … 30 Answer: D sin sin20 (1) 2.9 ----------(1) o dn d d    = = = Max order occurs at sin 90o sin sin90 ----------(2) o dn dn dn    = = = From (1), n = 2.9. Max n = 2 Hence max number of images = 5

9 ©EJC 2021 9749/J1H2MYE/2021 Paper 2 – Structured Questions Qns Answer Marks 1(a) Method 1   ball liquid2 ρ -ρv = gr μ 2 9 Solve using max-min method: rmax = 0.07 m, rmin = 0.05 m ρball,max = 7050 kg m-3, ρball,min = 6950 kg m-3 ρliquid,max = 1270 kg m-3, ρliquid,min = 1230 kg m-3 µmax = 3.9 Pa S, µmin = 3.7 Pa S ( ) ( )           ball,max liquid,min max max min ball,min liquid,max min max ρ -ρv gr μ ρ -ρgr μ Δv 2 2 -1 2 min 2 -1 2= 9 2 7050 -1230 = (9.81) 0.079 3.7 =16.8 m s 2v= 9 2 6950 -1270 = (9.81) 0.059 3.9 = 7.94 m s ( ) ( ) max minv - v -1 1= 2 1 = 16.8 - 7.942 = 4.43 =