ASRJC 2024 JC2 H2 Physics Prelim P3 MS

1 9749/03/ASRJC/2024PRELIM [Turn Over Anderson Serangoon Junior College 2024 H2 Physics Preliminary Examination Mark Scheme Paper 3 (80 marks) 1ai Any two of time, temperature, current, (luminous intensity) B2 1aii Any derived quantity, e.g. energy, force, power, velocity, acceleration, pressure, density, etc B1 1bi Percentage uncertainty = 2 + (3x2) = 8% A1 1bii 2 2 2 (4 1.50) 2.48 9.63 m s g − = = Absolute uncertainty = 0.08 x 9.63 = 0.8 m s–2 g = 9.6 ± 0.8 m s-2 (Note: g must have same place value as Δg) C1 C1 A1 2a As the sky-diver picks up speed, air resistance increases, the resultant force decreases and hence the acceleration decreases. B1 B1 2b (Since the sky-diver starts from rest), there is no air resistance initially / the only force he experiences is his weight, hence his initial acceleration is equal to 9.81 m s–2. M1 A1 2c Terminal velocity is the area under the a-t graph (by using trapezium rule/counting squares) A1 2d Correct shape start with zero gradient and ends with constant gradient from about t = 24.0 s M1 A1

2 9749/03/ASRJC/2024PRELIM 3a Resultant/net force (in any direction) on the object must be zero and Resultant/net moment / torque on the object about any point / axis must be zero. B1 B1 3b Taking moments about end A, (W × 0.25) + (12 × 0.35) = (17 sin 50° × 0.50) W = 9.246 = 9.2 N C1 A1 3c Consider vertical equilibrium, taking upwards as positive Sum of forces in vertical direction = 0 Fy + 17 sin 50° − 9.2 − 12 = 0 Fy = 8.177 N Consider horizontal equilibrium, taking rightwards as positive Sum of forces in horizontal direction = 0 17 cos 50° − Fx = 0 Fx = 10.93 N F = 22(8.177 10.93 )+ = 13.7N C1 C1 A1 3d By taking moments about end A, The moment due to the force by the block on the beam decreases, , the tension in the string decreases. When the tension in the string deceases at the same angle, by considering horizontal equilibrium, the horizontal component of the force exerted on the beam by the hinge decreases. M1 A1

3 9749/03/ASRJC/2024PRELIM [Turn Over 4ai reduction in energy / amplitude (of the oscillations) due to force opposing motion / resistive forces / dissipative forces B1 B1 4aii amplitude is decreasing (very) gradually / oscillations would continue for a long time / many complete oscillations hence light damping M1 A1 4bi frequency = 1 / 0.3 = 3.3 Hz A1 4bii energy = ½ mv2 and v = ωxo = ½ × 0.065 × (2π/0.3)2 × (1.5 × 10–2)2 = 0.00321 J = 3.2 mJ B1 B1 A0 4c amplitude reduces exponentially / amplitude decreases by a smaller amount (for each oscillation) / does not decrease linearly / decrease at a decreasing rate so amplitude will not be 0.7 cm M1 A1 4d Relevant examples e.g. pointer in ammeter, car suspension system etc. where critical damping is used A1 5ai Using potential divider rule, 2.01.0 RV R = + BD BD BD 4.0 2.0 1.64.0 1.0V =  = + BD V C1 A1 5aii Since current in Cell Y is zero, VBC = e.m.f. of cell Y 1.5 100 941.6 =  = cml C1 A1 5b Replace cell X with another cell with lower internal resistance OR Use a resistance wire of higher resistivity, keeping cross-sectional area and length unchanged OR Use a resistance wire of smaller cross-sectional area, keeping resistivity and length unchanged. OR Use resistance wire with a higher resistance per unit length By potential divider rule, this increases the p.d. across BD and allows a smaller value of l (while keeping current in cell Z zero). M1 A1

4 9749/03/ASRJC/2024PRELIM 5c (When current is zero,) VBC = e.m.f. of cell Z p.d. across internal resistance is zero hence terminal p.d. = e.m.f. Distance l remains unchanged. M1 A1 6a electric and magnetic fields normal to each other in the same region either charged particles enters region normal to both fields or correct B direction wrt E for zero deflection (in drawing) For no deflection, v = E/B or no net force. M1 A1 A1 6bi magnetic force on ion in path B provides for centripetal force By N2L, Bqv = m v2 r m = rBq v = 12.3 2 ×10−2 × 640×10−3 × 1.6×10−19 9.6×104 = 6.56 × 10−26 kg = 6.56×10−26 1.66×10−27 = 40 u (or 39.5 u) B1 C1 A1 6bii Since the ions are of the same isotope, they all have the same mass regardless of the paths undertaken. Using the equation in answer to (b)(i), the radius of the path is inversely proportional to q (or state equation for r) Hence, the ions in path A have thrice the charge compared to ions in path B. B1 B1 B1 7a energy from 1 nucleus = (1.77 × 1013) / (6.02 × 1023) ( = 2.94 × 10-11 J) Energy released = Binding energy of products – Binding energy of reactants binding energy of Z = [(1.25 + 1.81) × 10-10 ] – 2.94 × 10-11 ( = 2.77 × 10-10 J) nucleon number of Z = 93 + 139 + 2 – 1 (= 233) Binding energy per nucleon of Z = (2.77 × 10-10) / (233 × 1.60 × 10-13) = 7.43 MeV (3 s.f.) C1 C1 C1 A1 7bi N0 = 0.874 / (238 × 1.66 × 10-27) = 2.212 × 1024 =2.21 × 1024 A1 7bii 1 2 24 14 ln2 ln2 2.21 1087.7 365 24 3600 = 5.54 10 Bq A N N t== =      C1

5 9749/03/ASRJC/2024PRELIM [Turn Over A1 7biii 0 ln2ln0.653 ( ) 87.7 53.9 years tA A e t t −= =− = C1 A1 7biv half-life shorter, will not provide power for long enough / require frequent replacement of probe B1 8ai Horizontal component of tension / spring force provides centripetal force Weight of sphere is (now) equal to the vertical component of tension / spring force OR horizontal and vertical components of tension / spring force combine to give a greater tension in spring Greater tension/spring force so greater extension / since extension is proportional to spring force B1 M1 A1 8aii1 Radius, r = 10.8 sin 27° = = 4.903 cm ≈ 4.9 cm A1 8aii2 Fspring cos θ = mg OR sum of vertical forces = 0 0.29 9.81 3.19cos cos27 spring mgF  = = =  N Fspring ≈ 3.2 N (shown) B1 A1 A0 8aii3 ( ) ( ) 1 3.2 0.29 9.81 10.8 8.5 0.15 springFk x k −  −== − = N cm C1 A1 8aiii 1 sin 3.2 sin27 0.29 spring c Fa m  == ac = 5.0 m s−2 C1 A1 8aiii 2 2 2 2 ca r r T  ==  ( )2 0.049 / 5.0 0.62T =  = s C1 A1 8bi From dg dr =− , when the gravitational field strength is zero, the potential gradient at that point would be zero. (Thus the point will be a turning point, which is a maximum point.) Thus, x is 0.52 x 1012 m Accepted range of x : 0.50 x 1012 m − 0.54 x 1012 m M1 A1

6 9749/03/ASRJC/2024PRELIM 8bii1 ( ) 8180 14 10 10Um   =  =  − − −  = −7.2 x 1010 J Change = 7.2 x 1010 J The change in kinetic energy is an increase. C1 A1 A1 8bii2 energy required (to reach maximum point) = 180 × (10 – 4.4) × 108 or energy per unit mass (to reach maximum point) = (10 – 4.4) × 108 ½ × 180 × v2 = 180 × (10 – 4.4) × 108 or ½ × v2 = (10 – 4.4) × 108 v = 3.3 × 104 m s−1 C1 C1 A1 9a += 22 . . . 2 (0.002) 1 (0.002) 0.01 r m sV