2019 9749 H2 Physics P1 MS EJC Suggested

©EJC 2021 9749/H2 PHYSICS 2019 PHYSICS SUGGESTED MARK SCHEME Maximum Mark: 190 9749 October/November 2019 Paper 1 Multiple Choice Question Key Question Key Question Key 1 D 6 B 11 D 2 D 7 A 12 A 3 D 8 D 13 C 4 C 9 D 14 C 5 A 10 B 15 B 16 B 21 B 26 C 17 B 22 B 27 B 18 C 23 C 28 C 19 D 24 D 29 B 20 C 25 B 30 A 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.

2 ©EJC 2021 9749/H2 PHYSICS 2019 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) 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 − − = ×= = =  ×  ×

3 ©EJC 2021 9749/H2 PHYSICS 2019 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 When the displacement (from the equilibrium position) is zero, the spring is not extended, hence the EPE is zero. The KE is thus at its maximum. The spring is not extended, hence the restoring force is zero. 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 interfere destructively 0.75 360 0.75 361.5 at 0.75 s, the 2 waves will meet and interfere destruc A B A B A B T T T T t T T t − − = = × = = × = = = = = = tively 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 ∆= × =

4 ©EJC 2021 9749/H2 PHYSICS 2019 23 C Without the diode, the current against p.d. curve would be a straight line through the origin, because the resistor is ohmic ( constant resistance). With the diode connected, the curve is the same straight line for p.d. < 2 V, since there is yet no current in the dio de. For p.d. > 2 V, there is a constant current in the diode. So the original straight line is shifted upwards by a constant amount for p.d. > 2 V. 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 Refer to the EM tutorial solution. 27 B pp s s sp NV NV= = I I 28 C 29 B 30 A A is comparatively the best answer, although the actual constant in the equation relating the rate of decay to the number of nuclei is −λ.