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 = n