EJC 2023 J2 H2 PRELIM P1 MS

©EJC 2023 9749/J2H2PRELIM/2023 EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2023 9749 PHYSICS MARK SCHEME Paper 1 – Multiple Choice Questions 1 Answer: A relative velocity of bicycle to car = velocity of bicycle – velocity of car = velocity of bicycle + (– velocity of car) 2 Answer: A Horizontally, 3.0cos30 3.0 cos30u t t u =  =  Vertically, taking upwards as positive, 2 2 1sin30 0.252 Sub-in the expression for above, sin30 1 3.03.0 0.25cos30 2 cos30 u t gt t g u  − =−  − =−  Solving, 2 29.7u = , 15.4 m su −= Option C: forgot to take square root Option B: took vertical displacement to be 0.25 m Option D: took vertical displacement to be 0.25 m and forgot to take square root 15 m s -1 relative velocity of bicycle to car

2 ©EJC 2023 9749/J2H2PRELIM/2023 3 Answer: B From FBD of block A: Fnet = ma 12 + 2.0× 9.81× sin35° - = 2.0a 23.25 - = 2.0 Eqn(1) T T a - -- > From FBD of block B: Fnet = ma - 20 - 5.0×9.81× sin20° - 20 = 5.0 - 36.78 = 5.0 - -- > Eqn(2) Ta Ta From Eqn(1), 58.125-2.5T=5.0a -→ Eqn(3) (3) – (2): 94.905-3.5T=0 T=27 N 4 Answer: B 12 12 12 1 Horizontally, 50 cos60 50 cos60 0 Vertically, 50 sin60 50 sin60 100(8) 0 50 sin60 50 sin60 100(8) 0 9.2 m s oo oo oo vv vv vv vv v − −= = + − = + − = =

3 ©EJC 2023 9749/J2H2PRELIM/2023 5 Answer: A When 3 nonparallel forces act on a object in equilibrium, the must intersect at a point. 6 Answer: B Let d be the initial height from ground, h be the height at t. For EP-s graph, EP = mgh = mg(d-s) => Sketch y = b − cx (where b, c are constants) For EP-t graph, EP = mgd − mgs = mgd − mg(ut+1/2gt2) = mgd − ½ mg2t2 => Sketch y = p − qx2 (where p, q are constants) W CG

4 ©EJC 2023 9749/J2H2PRELIM/2023 7 Answer: C 2 3 33 (20) 1000(0.50) 500 400 40 10 (500 400 )(20) 3.75 (3.75)(25) 59 kW driving driving driving new driving new friction new F ma Fk Fk P F v k k P F v P F v P kv = −= =+ =  = + = = = = = = 8 Answer: C Horizontal component of Normal contact force provides centripetal force ( ) 2 sinθ .... 1 net cF ma mvN r = = For vertical equilibrium: cosN mg =θ …. (2) ( ) ( ) 2 1 1 : tan θ2 50 9.81 tan25 15.1m s v rg v − = =    = N W

5 ©EJC 2023 9749/J2H2PRELIM/2023 9 Answer: D Gain in GPE = Loss in KE 221(1 cos ) ( ) 2 ifmgr m v v + = − 221(1 cos ) ( ) 2 ifgr v v + = − 221(9.81)(7.7)(1 cos30 ) (25 )2 o fv+ = − Vf = 18.5 m s−1 Assume rod in tension 2mvmg F r+= 2(3.5)(18.5)(3.5)(9.81) 7.7F+= F = 121 N (tension) 10 Answer: B 3 3 3 (273.15) (1) (2) (1) 273.15:(2) 273.15 pV nRT z nR x nRT z xT xT z = = −−− = −−− = = 11 Answer: C pxVx = nxRT pyVy = nyRT pV = nRT assuming no gas particles escaped out of the flasks, n = nx + ny Therefore, pV = (nx + ny)RT = pxVx + pyVy

6 ©EJC 2023 9749/J2H2PRELIM/2023 12 Answer: D 21 3 Nmpc V= 52 1(1.00 10 ) (1.60)3 c= crms = 433 ms−1 13 Answer: A supplied rate of heat loss rate of heat loss rate of heat loss rate of heat loss 1 (3.0)650 2.0 60 (3.0)1200 1.0 60 (3.0) (3.0)1200 650 1.0 60 2.0 60 22 kJ kg f f f f f f mP l h t mP l ht lh lh l l − =+ =+ =+  =+  − = −   = 14 Answer: D ∆U is zero as there is no change in temperature. W is zero as the volume of the refrigerator is assumed to be constant. Since, Q = ∆U – W, Q is also zero. 15 Answer: B At t = 0 s, the projection is at maximum amplitude. Hence displacement-time graph is cosine. Differentiating will give velocity-time and acceleration-time.

7 ©EJC 2023 9749/J2H2PRELIM/2023 16 Answer: C The pulse undergoes 1800 phase change upon hitting the fixed point O, options reduce to either C or D. However, the smaller amplitude portion will travel first before the larger amplitude portion as it is the portion that encounters fixed point O first. Thus, answer is C. 17 Answer: A 18 Answer: A The wavelength being 5.0 m, the path difference (147.5 − 135 = 12.5 m) corresponds to 2.5 wavelengths. Hence, the two waves meet in anti-phase. Intensity = k(amplitude)2, k is a constant 2 11 2 22 * 1 42 * k A A AkA I I =  = , or A2 = 2A1 When meet in anti-phase, the two vectors are in opposite direction. The resultant vector is the difference between them: Atot = A2 − A1 = 2A1 − A1 = A1 Hence, the resulting intensity is 22 resultant tot 1I kA kA I= = = Displacement: upwards Velocity: upwards Displacement: upwards Velocity: downwards

8 ©EJC 2023 9749/J2H2PRELIM/2023 19 Answer: B 11sin sin if angles are small2 2 2b   = =  b is doubled, double of the wave energy is allowed to pass through the slit. Since double of the energy is now spread over half the area, the maximum intensity is increased to 4 times. 20 Answer: D Dx a = ( )( ) 9600 10 1 00 . y a − = ( ) 9400 10 D y a − = D = 1.50 m 21 Answer: A At equilibrium, mg = FE = qV/d When V increases to 2V, Resultant force, ma = q(2V)/d – mg = 2mg – mg = mg Therefore, a = g FE mg

9 ©EJC 2023 9749/J2H2PRELIM/2023 22 Answer: A I = nAvq I, n and q are constant. Therefore, A ∝ 1/v But A is linearly related to x, Hence v = k/(x+constant) 23 Answer: A Current in circuit, I = Psupply/Vsupply = 2400/240 = 10 A Pkettle = Psupply – Ploss = 2400 – (102)(0.5+0.5) = 2300 W Potential difference across kettle = 240 – (10)(0.5+0.5) = 230 V 24 Answer: B Effective resistance between X and Y = 4.0 Ω Effective resistance between A and B = ½ (8.0) = 4.0 Ω By potential divider rule, potential difference between A and B = 20 × 4.0 /(6.0+4.0) = 8.0 V Therefore potential difference between X and Y = ½ (8.0) = 4.0 V B A X Y 4.0 Ω 4.0 Ω 8.0 Ω 6.0 Ω 20 V

10 ©EJC 2023 9749/J2H2PRELIM/2023 25 Answer: A F will always be perpendicular to the component of velocity normal to the B field. Hence electron will move in circular motion. At the same time, the component of velocity parallel to the B field remains constant. Hence, the electron will move in a helix. F = Bqv sin θ v = 7.3 ×10−16 (0.084)(1.6 ×10−19) sin 230 = 1.4 x 105 ms-1. 26 Answer: C Magnetic flux,  = BA (where A is the area exposed to perpendicular B-field) At 4s, Q and R are entirely within the field while coil P has length of 4.0 m within. P = 8B; Q =6B; R = 12B (largest) Magnetic flux linkage, = N P = 8B; Q =18B(largest); R = 12B At 4s, coils Q and R would be experiencing maximum flux linkage. Hence no change in flux linkage => no induced e.m.f. and no induced current. Only coil P has changing flux linkage, induced e.m.f. and current. 27 Answer: A The brightness depends on the power delivered to the bulb. The peak power is 2Vp R= , which does not depend on the frequency f of the alternating voltage. v 230 B field (0.084 T) Conventional current direction due to electron going downwards at P From Fleming’s Left Hand Rule, electromagnetic force, F on electron is into the plane