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