2023 HCI H2 PH Prelim P2 SS

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 1 2023 H2 Physics Preliminary Examination Paper 2 Suggested Solutions Q1 Suggested Solutions Mark 1 (a) In the direction normal to the slope, the block is in equilibrium. net force on the block = 0 N thus, Normal contact force by the scale on the block, N = weight component of the block normal to the slope = 1.6 g cos 30 The scale measures the normal contact force exerted by the block on the scale, that is equal and opposite to the normal contact force by the scale on the block. (Newton’s 3rd Law of Motion) Thus, the reading on the scale = normal contact force by the block on the scale/ g = 1.6 g cos 30 / g = 1.39 or 1.4 kg. B1 B1 B1 (b) (i) Consider the block moving from the bottom to maximum height: Let the velocity of the block immediately after collision be v. By conservation of energy, Loss in KE = Gain in GPE ( )( ) 2 -1 1 02 2 2 9.81 1.3 1.3cos37 2.3 m s mv mgh v gh −= = = −  = B1 B1 (ii) applying N2L on the block: ( ) ( )1.6 2.3 0 0.2 18.4 N m v uF t −−==  = B1 A1 30 wooden block weighing scale normal contact force by scale on block, N weight of block, W = 1.6 g W⊥ = 1.6 g cos 30 Normal contact force by block on scale, N’ 37 1.3 m h

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 2 OR variations: ( )J F t m v u=  = − ; ( )vuF ma m t −==  (iii) By conservation of linear momentum, mballuball = mblockvblock + mballvball uball = (mblockvblock / mball ) + vball = ( ) -11.6 2.3 19 44.4 m s0.058 −= Or ( ) ( ) on ball -1 -1 0.058 1918.4 0.2 44 m s i.e. v = 44.4 m s opposite to the final velocity. m v uF t u u −=  −= =− B1 A1 (c) Relative speed of approach = 44 m s-1 -1 h R 1elat ive s speed o c f separ 2.3 19.1 2 .4 m rel p ati e on ative s e d of approa = + =  Thus, the collision is not elastic. OR Initial KE = ½ (0.058)(44)2 = 56 J Final KE = ½ (0.058)(19)2 + ½ (1.6)(2.3)2 = 14.7 J Since total KE s not conserved/ initial KE is greater than final KE, the collision is not elastic. B1 A1 1.6 kg before after 58 g u 2.3 m s-1 1.6 kg 58 g 19 m s-1 0.058 kg u 19 m s -1 F

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 3 Q2 Suggested Solutions Mark 2 (a) Newton’s Law of Gravitation states that every point mass attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. B1 (b) (i) The gravitational force on the satellite is always directed towards the centre of the Earth and so any circular orbit must have its centre at the centre of the Earth. The orbit must be in the equatorial plane, otherwise the satellite will sometimes be over the northern hemisphere and sometimes over the southern hemisphere. B1 B1 (ii) 1. Gravitational force provides centripetal force (full statement required) Let G be the gravitational constant, M the mass of the Earth, m the mass of the satellite, r the radius of the satellite’s orbit, ω the satellite’s angular velocity, and T the orbital period, T = 24 x 60 x 60 = 86 400 s. ( ) 2 2 2 2GMm m r mrrT ==   (show equation) 2 3 24 GMTr =  The radius of the orbit of the satellite about the centre of the Earth is ( )( )( ) 211 24 3 2 6.67 10 6.0 10 24 60 60 4r −    =  (values entered explicitly) 74.230 10 m= Let h be the altitude of the satellite above the surface of the Earth h = 4.23 x 107 - 6.4 x 106 (show values) = 3.59 x 107 m B1 B1 B1 B1 A0 2. Linear speed = rω ( ) 7 -1 2 24.230 10 24 60 60 3080 m s v r r T ==  =   =   B1 A1

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 4 Q3 Suggested Solutions Mark 3 (a) The incident wave from the oscillator is reflected at Q. The incident and reflected waves have the same speed, wavelength (or frequency) and amplitude, travelling in opposite directions overlap/meet/superpose and form a stationary wave. B1 B1 (b) Q is a fixed end or node . Destructive interference must occur, so that the net displacement is always zero. Hence there must be a phase change of 180° (or π radians) at Q. B1 (c) (i) λ/2 = 0.30 m λ = 0.60 m v = f λ = (120) (0.60) = 72 m s-1 M1 A1 (ii) v0 = ω x0 =2 (120) (0.0080) = 6.03 m s-1 M1 A1

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 5 Qn 4 Suggested Solutions Mark 4(a) The electric potential at a point is the work done per unit positive charge, in bringing a small test charge from infinity to that point. B1 4(b)(i) At r = 1.0 x 10-10 m, the potential is 780 V. ( ) ( ) 0 0 12 10 0 19 4 , where is number of protons4 4 780 4 (8.85 10 )(1.0 10 ) 1.60 10 54 QV r ne nr Vrn e     −− − = = ==  = M1 M1 M1 4(b)(ii) The distance 82.0 10 m− is considerably greater than the diameters or radii of the nucleus and the proton. B1 (b)(iii) 1. The magnitude of the electric field strength at a point is equal to the gradient of the potential-distance curve at that point. B1 2. Approach 1 determine E by finding gradient of tangent at r = 2.6 x10-10 m ( ) ( )( ) 12 -1 10 19 12 7 550 100 1.15 10 NC0.40 4.30 10 1.60 10 1.1538 10 1.85 10 N electric VE r F qE − − − −=− =− =  − = =   = Direction: radially away from the nucleus gradient – 1 mark substitution of values into force equation – 1 mark value of force – 1 mark direction (radially must be present) – 1 mark M1 M1 A1 A1

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 6 Approach 2 From Fig. 4.1, when r = 2.6 x10-10 m, V = 300 V ( ) 0 19 10 7 54 1 4 3001.60 10 2.6 10 1.85 10 N electric dVF qE q dr Veqe r r r − − − ==     ==         =   =  radially away from the nucleus

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 7 M1 M1 A1 A1

2023 HWA CHONG INSTITUTION (COLLEGE SECTION) C2 8 Q5 Suggested Solutions Mark 5 (a) e.m.f. is the energy per unit charge converted from other forms of energy to electrical energy as charge is driven through a whole circuit. Potential difference between two points in a circuit is the energy per unit charge converted from electrical energy to other forms of energy when charge passes between the two points. B1 B1 (b) Since this is a resistance wire of uniform resistivity and cross-sectional area (by the formula R =  l /A), the resistance R is proportional to the distance l between the two mentioned points. Since a constant current flows through the whole circuit , the potential difference between two points on the resistance wire is proportional to resistance R and proportional to the distance l between the two mentioned points. B1 B1 (c) (i) When CK is 60.0 cm, AJ is 54.0 cm. Since ammeter reading is zero, p.d. across CK = p.d. across AJ = VAB (AJ/ AB) = 12.0 (54.0/120.0) = 5.4 V B1 B1 (ii) When CK = 60.0 cm, resistance across CK = 4.5  . When AJ = 54 cm , p.d. across AJ = 5.4 V p.d. across AJ = p.d. across CK 4.55.4   4.5 Er= + ……………………(1) When CK = 20.0 cm, resistance across CK = 1.5  . When AJ = 45 cm , p.d. across AJ = 4.5 V p.d. across AJ = p.d. across CK 1.54.5   1.5 Er= + ……………………….(2) Eqns [1]/[2] , ( )3 1.51.2   4.5 5.4 1.2 4.5 3   0.5  r r rr r += + + = + = M1 M1

2023 HWA CHONG INS