2024 ACJC J2 H1 Physics Prelim P2 AS

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 2 Guide H1 (8867) Physics JC2 2024 Page 1 of 9 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)   2 22 -2 2 2 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, 2 2 2 2 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 H1 (8867) Physics JC2 2024 Page 2 of 9 (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 v u gt v ut 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 0.8005 2 3 1.32 47.5 0.8005 2 T W 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 H1 (8867) Physics JC2 2024 Page 3 of 9 (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) The total final momentum of a system after a collision is equal to total initial momentum of the system before the collision provided no net external force acts on the system. (b)(i) Applying the principle of conservation of momentum,   1 75.0 10.0 0 (75.0 55.0) 5.7692 5.77 m s (to 3 s.f.) (shown) v v        (b)(ii) The skaters exert forces on each other that have the same magnitude but opposite in direction. The magnitude of the force is given by  75.0 5.77 10.0 on moving skater 0.100 3172.5 3170 N (3 s.f.) F      OR  55.0 5.77 0 on stationary skater 0.100 3173.5 3170 N (3 s.f.) F    Since the force exerted by the skaters on each other is less than 4500 N, the skaters would not sustain any injury. 5 (a) 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. (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 

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 2 Guide H1 (8867) Physics JC2 2024 Page 4 of 9 (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 RAR L A    Since 2 2, constantA d L Rd     2 2 18 2 30 1 2.4 P P P Q Q Q L R d L R d                  6 (a) Force per unit length acting on a straight, current-carrying conductor, carrying unit current and placed at right angles to an external magnetic field. (b)(i) Arrow directed upwards. Must be drawn with a ruler. (b)(ii) For an undeflected beam, 3 7 electric force magnetic force sin sin90 (3.0 10 )(3.3 10 ) qE Bqv E Bv          4 19.9 10 N C (2 s.f.)  x x x x x x x x x x x x x x x x x x x x x x x x x electrons with speed v R

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 2 Guide H1 (8867) Physics JC2 2024 Page 5 of 9 (b)(iii) Magnetic force provides centripetal force. 2 2 7 2 3 sin sin90 3.3 10 (6.0 10 )(3.0 10 ) mvBqv r mvBev r e v m rB          11 -11.8 10 C kg (2 s.f.)  7 (a)(i) From graph, current from solar cell = 136.25 mA (a)(ii) At a potential difference of 550 mV, no current flows from the solar cell through the load resistor. This implies there is no potential difference across the internal resistance of the cell. Hence, the terminal p.d. is equal to the e.m.f of the solar cell. (a)(iii) 3 3 3300 10 550 10 (136.25 10 ) V E r r          I r = 1.83 Ω (3 s.f.) (a)(iv) Power dissipated in load resistorEfficiency Power generated by solar cell 300 550 V E V E     I I = 54.5% (3 s.f.) (b)(i) Rectangular shape. Extends vertically and horizontally from point P.

Anglo-Chinese Junior College 2024 J2 Preliminary Exam Paper 2 Guide H1 (8867) Physics JC2 2024 Page 6 of 9 (b)(ii)1. Increases. (b)(ii)2. Decreases. (b)(iii) Q is at (430,125.00). (c)(i) Straight line passing through origin. Ratio of I/V is a reciprocal of 4.2 Ω. (c)(ii) Locate intersection of the two graphs in Fig. 7.3. (470, 112.50) P = IV = 470  112.50  106 = 0.0529 W (3 s.f.) 8 (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   