RI 9749 2018

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 9749 H2 Physics 2018 GCE A-Level Suggested Solutions Paper 1 1 A 22 2 2 4 4 2 2 0.05100% 100%100 6.25100% 0.6%22 π π = = ∆∆ ∆= + ∆∆ −× −× ∆   ×= = = LT g Lg T gL T gL T gL gLT T 2 A Let s be displacement and x be distance: 22 1 21 2 For object P: 11 and x 22 For object Q: By symmetry, 0 and 11x area under graph 2( )( ) 2( )( )22 2 2 2 2 4 = = = −= = ×+ ×= pp Q Q s tv tv s t t t tvvv 3 B Impulse change in momentum () 0.080(18 ( 23)) 3.3 Ns = =∆ = − = −− = fi p mv v 4 C Since trolleys stick together after collision, collision is completely inelastic. Momentum is conserved and velocities of both trolleys are the same after collision. Taking right as positive 1 12 12 1 Initial momentum Final momentum () 5.0(4.0) (2.0)( 3.0) (5.0 2.0) 14 7.0 2.0 ms 14 Ns and 2. − = += + + −=+ = = ⇒= =i mu mu m m v v v v pv 10 ms−

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 5 A Since the broom is balanced at O, the CG of the broom is 1.05 m from end of X. Let m be the mass of the broom, Since the broom is in balance again, 0 0.200(1.05 0.10 0.27) (0.27) 0.504 g m m τ = −− = ⇒= ∑ 6 A arg arg Without the barge: Weight supported by bridge With the barge: Weight supported by bridge Since the barge is floating: Weight supported by brid water water b e waterdisplaced b e waterdisplaced W WWW WW = =+− = ⇒ ge There is no extra weight supported by the bridge waterW= ∴ 7 C 3 3 22 11 3 2 40 184 kW20 P Fv kv Pv Pv P ∝∝ =  = = 8 B Conventional current flows from a higher electric potenti al to lower electric potential. S is at a lower electric potential than R. An electron has a negative charge and has a higher electric potential energy at S (less negative). Point S is lower than point R. Sine GPE = mgh, GPE at S is lower than at R (independent of charge). 9 D Both P and Q are on the same rotating disc => angular velocity is the same for P and Q. Angular displacement is the same for bo th.θ 10 A When the stone is vertically above the centre of the circle: 2 2 Taking downwards as positive: T W Mr T Mr W ω ω += = −

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 11 C A: Free falling of object. No need rockets B: Gravitational force provides for centripetal force (object moves in circular path) D: initial momentum can provide it with enough energy to travel directly outwards from surface of the earth. C: The rockets need to keep firing for spacecraft to follow a straight line, if not gravitational force will cause it change direction and take on a circular/spiraling path towards earth. 12 D 11 1Since gradient, 1 4 new old PV nRT VP nRT g nT gg = = ∝ = 13 D X: Qnet < 0 i.e. Tx,initial < 30°C Y: Qnet = 0 i.e. Y is at thermal equilibrium with its surrounding Ty,initial = 30°C Z: Qnet > 0 i.e. Tz,initial > 30°C 14 C 12 21 12 12 21 21 21 0 0 0 2800 600 1000 3200 kJ bysystem bysystem bysystem U UU QW QW W = += − +− = = −+ = ∑ 15 D A: Peak-peak value is 70 cm B: KEmax occurs when velocity is max. i.e. t = T/4 C: Restoring force is proportional to acceleration. i.e. a is maximum at the amplitudes. D: The gradient of the x -t graph gives the instantaneous velocity and is a maximum at t = T/4 16 C 2 2 1, Let wave Y have amplitude and frequence y 2 4 1 12 1 2 X YY X YY A f Af A A f f λ λ λλ ∝∝ = =   = =  I I II

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 17 D 22 11 2 1 Fringe separation, 600 (2.4) 3.2 mm450 Note: 2(1.2) 2.4 mm Dxx d x x x x λ λ λ λ ∆= ⇒∆∝ ∆ =∆ ∆= = ∆= = 18 C Distance from end of tube to the first node = 0.17 m. L = λ1 = 0.68 m The next time a node is detected at the same position, (by sketching waveform) 0.17 m = 3λ2/4. Hence L = 0.68 = 3λ2 v = fλ2 340 = f (0.68/3) f = 1500 Hz 19 C Definition: The electric field strength at a point is defined as the electric force exerted per unit positive charge placed at that point 20 A dVE dx=− . When the distance between two equipotential lines are larger, the electric field strength is smaller and vice versa. As the objects are of opposite charges, the electric field strength is always towards the right. 21 A 22 332 2 32 2 (400 10 )(1.2 10 )(1.5 10 ) 0.25 m(40 10 ) (1.8 10 ) II I ρ ρ −−− −− = = ×××= = = Ω×× LPR A PA L 22 A Since cells are in parallel, 3 1.5 4.5 VV = ×= 1 11 0.3 33 eqr rr − = += Ω 23 D eq eq 6.0 6.0 12 0.5 R 12 (1.0 9.0) 2.0 1 11 R6 3.0 2 (6.0) 1.0 12 1.0 0.33 A3.0 Total VR R R V Ω = = = Ω = −+=Ω = + = Ω = = Ω = = I I

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 5 24 A Since the square coil is in equilibrium, 0τ =∑ 3 sin30 cos302 tan30 2 tan30 2 tan30 2 0.12tan30 2(50)(0.40)(60 10 ) 0.029 T I I − × °= ° °= °= °= °= × = B B LW FL WF WNB L WB NL 25 A When the bar magnet approaches the coil, magnetic flux linkage increases. Since there is a rate of change of magnetic flux linkage, there is induced emf. By Lenz’s law, the induced current flows in the direction to set up a magnetic field to oppose the increase in flux linkage. When the bar magnet is wit hin the long coil of wire, there is no rate of change of magnetic flux linkage hence there is no induced emf and the object continues to accelerate. When the bar magnet leaves the coil (now at a higher speed), magnetic flux linkage decreases. By Lenz’s law , the induced current will now flow in the opposite direction to set up a magnetic field to oppose the decrease in flux linkage. 26 A The magnetic flux density of a solenoid = 0nµ I PQBB= ⇒ 00 PQnnµµ I= I 27 C 0 2 0 2 24 4 (7.6) (9.4) 4 140 W rectified rectified P PP P R = = = = = I 28 A 34 8 9 6.63 10 (3 10 ) 1.96 eV(633 10 ) hcE λ − − ××∆= = =× 20.66 18.70 1.96 eVWXE →∆=−= 29 C Standard Nuclear Notation: Mass No Atomic No X

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 6 30 B 1/2 1/2 7 19 1/2, X 19 8 1/2, Y ln2 1 and t i.e. t t (4.60 10 )(4.00 10 ) 0.250t (2.00 10 )(3.68 10 ) λ λλ λ λ = = ∝  ××= = = = = ×× Y Y Y YX XX YX X AN A N AN A NA N

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 7 9749 H2 Physics 2018 GCE A-Level Suggested Solutions Paper 2 1 (a) The principle of moments states that, for a body to be in rotational equilibrium, the sum of the clockwise moments about any point must equal the sum of the anticlockwise moments about that same point. [1] (b) (i) Since C.G. is nearer support A, the tension at support A is greater than that in support B, i.e., 7 3ABTT= . Since ABT T mg+= , we have 107 33 BB BT T T mg+= = So 33 10 10 4.5 9.81 13 NBT mg= = ×× = (2 s.f.) and 7 3 31 NABTT= = (2 s.f.) [1] [1] (ii) Taking moments about support A, ( )×− =×0.20 0.60Bmg d T ×= += ×+= 3 10 0.60 0.20 0.60 0.20 0.38 mBTd mg (2 s.f.) [1] [1] (c) The horizontal component of the force exerted by each support is to balance the horizontal force exerted by the wind on the sign so as to maintain equilibrium. [1] 2 (a) Hooke’s Law states that the extension of a body is proportional to the applied load, provided the proportionality limit is not exceeded. [1] (b) For a compression of 85 mm, ( )( ) Elastic potential energy stored in spring Area under graph 1 0.085 6.82 0.289 J = = = [1] [1] (c) By the principle of conservation of energy, ( )( )( ) ( ) 2 1 Loss in EPE of spring Gain in GPE of ball Gain in KE of ball 10.289 0.042 9.81 0.40 0.042 2 2.43 m s v v − = + = + = [1] [1] [1] (d) (i) For the ball to move in the circular track, at P, the weight of the ball and normal contac t force on the ball provides for the centripetal force. CCN mg F F