(ACJC) 2024 J2 H2 Physics Prelim P2 Solution

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2024 Page 1 of 7 Qn Suggested Answer 1 (a) Precision is defined as a measure of how close the experimental values are to each other. Accuracy is defined as a measure of how close the experimental values are to the true value of the physical quantity. (b) ( ) 222 -2 22 4 50 0 104 9 78933 m s1 42 Lg T . ..  − = = = 2g L T g L T    = +  0 2 0 02 9 78933 0 1770450 0 1 42g  = +  = .. .... = ± 0.2 m s−2 (1 s.f.) g = 9.8 ± 0.2 m s−2 (c) There is error due to human reaction time . Taking a large number of oscillations will reduce the fractional/percentage uncertainty of the measurement of time. The absolute uncertainty is the same but taking more oscillations reduces the effect in the calculation of T. (ΔT = Δt / n where Δt = ± 0.2 – 0.4 s (human reaction time) and n is the number of oscillations.) 2 (a)(i) Taking upwards as positive, 22 22 2 0 2( 9.81)(27) v u as u =+ = + − -1 23.02 23 m s (2 s.f.) (shown) u = =

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2024 Page 2 of 7 (a)(ii) Straight line with negative gradient. Initial velocity of 23 m s−1. Max height at 2.34 s. Graph stops at 6.0 s with velocity of -35.9 m s−1. max height -1 0 ( 23) 2.34 s9.81 At 6.0 s 23 (9.81)(6.0) 35.9 m s vu gt vut g t v u at − = − − −= = = = = + = − =− (a)(iii) Steeper slope before v = 0. Gentler slope after v = 0. Gradient at v = 0 should be parallel to graph in (a)(ii). (b) GPE linear with negative gradient. EPE parabolic shape starting from x0. KE shape and correct KE value at x0 and xs (TE is constant). 3 (a) The resultant force acting on the object must be zero. The resultant torque about any axis is zero. (b) Taking moments about the hinge, ( ) ( ) ( ) 3 1.32 0.80052 3 1.32 47.5 0.80052 TW W     =           =       70.2 NW = 70.2 7.2 kg (2 s.f.)9.81m==

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2024 Page 3 of 7 (c) The tension in the cables has a horizontal component, while the weight has only vertical component. For the canopy to be in equilibrium, there must be a horizontal component by the hinge away from the hinge. The sum of the vertical component of the tensions in both cables is less than the weight, hence the force by the hinge has a vertical upward component. 4 (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. (b)(i) Gravitational force provides the centripetal force. 2 2 21Kinetic energy 2 2 GMm mv rr mv GMm r = = = (b)(ii) Total energy Kinetic energy Gravitational potential energy ( )2 GMm GMm rr =+ = + − 2 GMm r=− (c)(i) Period of rotation of satellite, T = 24.6 x 3 = 73.8 hours Gravitational force of Mars on satellite provides the centripetal force. 2 2 11 23 2 3 2 2() (6.67 10 )(6.39 10 )(73.8 60 60) 4 GMm mr Tr r   − =    = 74.24 10 mr = (3 s.f.) (c)(ii) 11 23 7 Work done ( ) 2(2 ) 2 1()22 (6.67 10 )(6.39 10 )(470) 4(4.24 10 ) GMm GMm rr GMm r − = − − − = =  118 MJ= (3 s.f.) (d) Total energy will be reduced. This means that the radius of circular orbit will gradually decrease. Decreasing circular orbit means that the kinetic energy increases and thus the speed of the satellite would increase.

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2024 Page 4 of 7 5 (a)(i) The volt is defined as the potential difference between two points in a circuit where 1 J of electrical energy is converted to other forms of energy when 1 C of charge passes from one point to the other. (a)(ii) e.m.f. is the amount of non -electrical energy converted into electrical energy per unit charge passing through the terminals of the cell (source). p.d. is the amount of electrical energy converted to other forms of energy per unit charge passing from one point to the other. (b)(i) When current in P is 0.15 A, the p.d. across P is 2.70 V. The p.d. across Q is also 2.70 V. From the I-V graph of Q, the current through Q is 0.0900 A. Total current in battery = current in P + current in Q = 0.15 + 0.0900 = 0.24 A (b)(ii) P.d. across R = 4.5 – 2.70 = 1.8 V Resistance of R = Ω1.8 7.5 0.24 = (b)(iii) In the given circuit, Resistance of P = Ω2.70 18 0.15 = Resistance of Q = Ω2.70 30.0 0.0900 = Since resistance ,L RARL A  == Since 22, constantA d L Rd   = 2 2 18 2 30 1 2.4 P P P Q Q Q L R d L R d   =         =     = (b)(iv) When Q stops conducting, effective resistance of the two lamps increases. By potential divider principle, the p.d. across lamp P increases. The current through P also increases as no current passes through Q. From the graph, as p.d. across P (or current in P) increases, its resistance increases. 6 (a) Random: Impossible to predict when a particular nucleus in a sample is going to decay. Spontaneous: The decay of a nucleus is not affected by by the presence of other nuclei, chemical reactions or physical conditions such as temperature and pressure. (b)(i) 210 206 4 84 82 2Po Pb He→+

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 2 Guide H2 (9749) Physics JC2 2024 Page 5 of 7 (b)(ii) 2 2 Since magnitude of is the same, 206 4 Pb Pb pKE m p KE m KE m   = == 51.5 Pb KE KE  = (b)(iii) ( ) ( ) 11 4 2 ln2 4.2 10 2.44 10 Bq 138 24 60 ooAN = =  =   12 ln2 ln2 600138 tt ooA A e A e  −  −   == 1200 Bq (2 s.f.)A= (b)(iv) The reading will be lower. Particles are emitted in all directions and only a proportion will be detected by the counter. (c)(i) By conservation of momentum, for a two particle system, the ratio of the speeds of the particles is fixed (inverse ratio of masses). By conservation of energy, energy released per decay is constant, the speed of the beta particle emitted should be constant. However, the range of KE suggest that there must be a third particle. (c)(ii) Half-life of platinum is much longer than that of gold. Formation of gold is slower than its decay, hence gold will be the nuclei of the smallest percentage. 7 (a)(i) 1030 9.81 400 P gh= =   4.04 MPa (3 s.f.)= (a)(ii) 0 6 0 6 2 6.30 4.04 10 2 2 550 10 dP t dPt   =   0.0231m 2.4 cm (round up to 2 s.f.) (shown) t  = (a)(iii) The thickness of the hull is much sma