RI 9749 2019

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 2019 A-Level H2 Physics Suggested Solutions Paper 1 1 D 2 D 0.01 4.072 0.04072 V 40.72 mV ×= = 40.72 10 50.72 mV 0.05 V (1sf) += ≈ 3 D 1 2 11 2 11 1 let ( 1.00) s be total time of fall 1at , (9.81)( )2 1at +1.00, 4 (9.81)( 1)2 solving the eqns will give 11.00 or 3 time of fall 1 1 2.00 s t ts t ts t t + = = + = − =+= 4 C Between t1 and t2 the gradient of graph is increasing. Eventually the graph becomes linear and thus acceleration tend towards zero. 5 A No net force acting on the system (X, Y and container) and thus by principle of conservation of linear momentum, no change in the combined centre of gravity of the system. 6 B 2 22 2 22 By conservation of linear momentum, 02 3 11 1KE ( ) (3 ) (9 )22 2 11 1KE ( ) (2 ) (5 )22 2 i f mu mv mv uv m u mv mv m v mv mv += + = = = = = += 7 A 22 using coordinates (10,14) 14 140010()1000 at (20,28) the elastic limit of spring is reached 1 1 20work done (1400)( ) 0.28 J2 2 1000 F kx k kx = = = = = = 8 D As the object moves towards Y, the net force on the object tends to zero and thus the rate of increase in kinetic energy tends to zero. (ie the gradient of KE - distance graph tends to zero)

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 9 D 91.3 10 9.81 2 295000 J60 60 24 GPEP t ∆ ×× ×= = =∆ ×× 10 B vR v R ω ω = = Since v is constant, graph B is correct 11 D 2 20 5.37.0 c c ar a g r ω ω = = = = 12 A 2 By CoE 1 02 2 ifEE GMmmv R GMv R = +− = = 13 C 2 20 3 23 13 22 3 1.25 10 1374 4 10 6.02 10 m c kT kTc m − − = ×= = =  ×  × 14 C By definition, total internal energy of a gas is the sum of the random distribution of KE and PE associated with the molecules of a system of gas 15 B By 1st Law of thermodynamics 16 B 17 B 3 3 1 2.0833 10480 1 2.0747 10482 0.25 120 0.25 120.5 at 0.25 s, the 2 waves will meet and int erfere destructively 0.75 360 0.75 361.5 at 0.75 s, the 2 waves will meet and int erfere destruc A B A B A B T T T T t T T t − − = = × = = × = = = = = = tively

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 18 C Between consecutive constructive intereferences, the metal sheet is moved by 60 mm. When the metal sheet is moved by 60 mm, the path difference between the 2 waves detected by the receiver is increased by 120 mm. thus wavelength is 120 mm. 19 D 8 7 14 7 1 3 2 3 sin 3 10 5 106 10 5 10sin 36.86910 4 10 36.869 2 74 ndλθ λ θ − − − − = ×= = ××   ×= =  × ×= ° 20 C b λθ = 21 B In uniform electric field, the electric force acting on electron is constant and thus acceleration is constant. By Newton’s 2 nd Law, r ate of increase of velocity is constant. 22 B 4 1000 400 V10V ∆= × = 23 C 24 D When R = 0 parallel arrangement is short circuited 1 minimum V reading when 1.0 1 12 8.01.5 R V = = ×= 25 B 32 5 By Faraday's Law of EMI, 10 10 2 10 5 104 d dtε −− −Φ × ××= −= = × 26 C 27 B pp s s sp NV NV= = I I 28 C 29 B 30 A

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 2019 A Level Paper 2 Suggested Mark Scheme and Detailed Solution 1 (a) The resultant force on the object is zero. The resultant moment (or torque) on the object about any axis is zero. [B1] [B1] (b) Considering vertical equilibrium: 12sin20 sin10 700 N°+ °=TT ---------(1) Considering horizontal equilibrium: 1 2 12 cos10cos20 cos10 cos20 °°= °⇒ = °T T TT ------------(2) Solving, T1 = 1380 N T2 = 1320 N [C1] [M1] [A1] [A1] (c) Taking pivot about base of pole. 111.2Angle between support cable and ground ta n ( ) tan (0.75)1.6 −−= = = A Considering horizontal components of forces and its moment about the base of pole: clockwise moment anti-clockwise moment ( cos )(1.2) (150)(cos10 )(1.8) 280 N = = ° = TA T [C1] [M1] [A1] 2 (a) (i) A transverse wave is one in which its particles oscillate in a direction perpendicular to the direction of energy transfer. Electromagnetic wave. [B1] [B1] (ii) Plane polarisation of a wave means that the oscillations of the transverse wave are restricted to a single plane. [B1] (b) (i) Intensity of light emerging from 1st polarising filter 2 0' cos 30II = ° Angle between 1st and 2nd transmission axis = 60° − 30° = 30° Intensity of light emerging from 2nd polarising filter. 2 22 T0 00 'cos 30 ( cos 30 )(cos 30 ) 0.5625 0.563 II I II = °= ° ° = = [C1] [M1] [A1] (ii) 0 TT 0 0T 0 and 0.5625 0.5625 0.750 II I I I = = = = = 22kA kA A A [M1] [A1]

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 5 [B1] cosine graph [B1] maximum intensity at 0, 180, 360 deg and 0 and 90 and 270 deg. (c) (i) It is because sound wave is a longitudinal wave , and the particles oscillate in the direction of energy transfer which cannot be polarised. [B1] (ii) 41 1 11period of sound wave s740 1time base of oscilloscope 6.8 cm740 1.9872 10 s cm 0.200 ms cm −− − = = = = ÷ = × = T f [M1] [A1] 3 (a) The electric potential at a point in an electric field is defined as the work done per unit positive charge by an external force in bringing a small test charge from infinity to that point. [B1] (b) (i) Since potential is positive, the charge is thus positive and provided by protons. 10 0 12 10 0 12 19 , at 6.0 10 m, 130 V4 4 4 (8.85 10 )(6.0 10 )(130) 8.675 10 (1.6 10 ) 54.2 54 (integer) X X πε πε π − −− −− == ×= = = ×× = ×= × = = qV r Vr q rV [C1] [M1] [A1] (ii) It is because the distance between the single proton and nucleus is large, in the order of 10−8 m compared to the size of nucleus which is in the order of 10−10 m. They are very far apart. [B1] (iii) p2 82 00 19 2 8 2 27 0 16 2 54 4 4 (2.0 10 ) 54(1.6 10 ) 4 (2.0 10 ) (1.67 10 ) 1.86 10 m s Qq eeF mar a πε πε πε − − −− − = = =× ×= ×× = × [C1] [M1] [A1] (iv) The proton experiences repulsion from the nucleus as both are positively charged. By conservation of energy, the decrease in potential energy leads to a corresponding increase in kinetic energy. Potential energy at infinity is zero. [B1]

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 6 19 2 r 8 0 0 19 54(1.6 10 )change in potential energy 0 0 4 4 (2.0 10 ) 6.22 10 J πε πε − ∞ − − ×=−= − = − × = −× QqUU r Hence the change in KE = + 6.22 ×10−19 J [M1] [A1] 4 (a) (i) Using Flemings Left Hand rule, the magnetic force acts upwards towards plate P. Hence electric force on the proton has to act downwards towards plate Q. Hence plate P has to be positive. [M1] [A1] (ii) For proton to not deviate, the magnetic force balances the electric force on the proton. Fm = FE 2 p 19 27 31 1KE of proton 64 eV2 2(64)(1.6 10 ) (1.67 10 ) 110.74 10 m s − − − = = ×= × = × mv v mE 33 1 (45 10 )(110.74 10 ) 4983 N C− = = ⇒= = ×× = FF Bqv qE E Bv E [B1] [C1] [C1] [M1] [A1] (ii) [B1] proton is deflected downwards within the plates (as v ↓, Fm ↓, FE > Fm) (note: upon exit, proton moves in a straight line) (b) 0y xx x 7 51 2 (4 10 )(8.4)(5.5) 2 (0.12) 7.7 10 N m III µ π π π − −− = = ×= = × FBL L d L F L The force on wire X acts towards wire Y (towards the right). [C1] [M1] [A1] [A1] 5 (a) The electromotive force (e.m.f.) of a source is defined as the amount of electrical energy that is converted from other forms of energy when the source drives a unit charge around a complete circuit. [B1] (b) (i) E = 12.0 V The equivalent resistance of S and T in parallel is [B1]

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 7 = Ω + 1 = 12011 200 300 eqR When the switch is closed, by the potential divider principle