EJC Physics 2020 J2 H1 JCT P1 MS

©EJC 2020 9749/J2H2JCT/2020 EUNOIA JUNIOR COLLEGE JC2 June Common Test 2020 General Certificate of Education Advanced Level Higher 1 PHYSICS MARK SCHEME 8867 June/July 2020 Paper 1 Multiple Choice Question Key Question Key Question Key 1 B 6 B 11 D 2 A 7 D 12 B 3 D 8 B 13 B 4 B 9 D 14 B 5 A 10 D 15 B 16 D 21 D 26 C 17 A 22 C 27 B 18 D 23 C 28 D 19 D 24 C 29 C 20 A 25 B 30 C 1 Estimate the kinetic energy using t = 10 s E k = ½ (80)(10)2 = 4000 J 2 units of 21 2 xy is same as units of pressure 2 2 22 33 2 1 units of units of units of N m kg m s m kg m kg m ms m s Py x − −− −− − − = = = = = 3 4 F = mA Direction of acceleration is always the same as the direction of the net force 5 At terminal velocity, height is dropping at a constant rate. The velocity is the gradient of the graph 6 Area under F- t graph gives change in momentum. Considering from t = 2 s: ( )( ) 1 1 1 5 1 2 5 N s2 25 1 25 m s2 5 12 5 63 m s p. p.v v .. .m − − ∆= = + ∆∆= = = + = += 7 Y will always have a lower velocity than X. Using the eqn: v = u + at As a result the distance between X and Y will always be increasing. ( ) ( ) π ππ π ⇒   = ± ∆∆= = ∆∆= = ∆ = =  3 32 2 1 70 0 005 cm 4 33 4 343 4 1 70 0 005 0 2 (1 s.f.) r. . VrVr Vr rV r rrr .. .

2 ©EJC 2020 9749/J2H2JCT/2020 8 System has common acceleration, a ( ) − = °− = °= = = = −= = BB down ramp, A A 2 sin 25 2 4 4 sin 2 4 g 25 2 8 m6 s A F ma a FT T aa ga aa m . m 9 Since it is an totally-inelastic collision, there is bound to be less total KE after collision so must avoid conserving KE of bullet as work done against friction. By PCLM, bullet total bullet total uMv uv M m m = = By conserving energy, 2 total 1 2 Mv F .s= ( ) ( ) ( )( ) ( ) ( )( ) 2 2 bullet total t otal total 2 bullet total 23 3 230 06 11 22 1 2 20 10 2 30 20 500 10 8 m uMv MF FM uFM ms m . − −    × +× = = = = = 10 When man accelerates upwards, N1 > mg When man accelerates downwards, N2 < mg When man is not accelerating, N3 = mg 11 The 2 objects will experience the same force by virtue of Newton’s 3rd Law. 2 x a = 5 x 10 a = 25 m s-2 12 Eliminate C and D as the sum of the three forces ≠ 0 Eliminate A because the line of action of the 3 forces do not intersect 13 If the two forces act about the same point there will be no torque 14 Pivot about where beam attaches to wall. By POM, ( ) ( ) beam boy beam boy moments 0 22 moments sin 60 1 022 sin 60 1 200 0 2 5002 sin 60 230 9 N L W . LW LT W .W T . . = + = ° + = ° + = ° = ∑ ∑   15 Elastic Potential Energy = ½ k x2 ½ k (60 x 10-3)2 = 15 J k = 8333 ½ k (90 x 10-3)2 = 33.75 J ∆ EPE = 33.75 – 15 = 18.75 J 16 d sin 30° = 1.5 d = 3 m Work done against friction = 150 x 3 Gain in GPE = 200 x 1.5 Total work done = 150 x 3 + 200 x 1.5 = 750 J 17 On level ground at constant speed, driving force is same magnitude as resistive force 3 r driving 370 14 7 10P v P Fv FF . = = = ×= On slope, component of weight downramp is in same direction as resistive force driving r slope sin sinr F mg P Fgv F m θ θ = + = + θ

3 ©EJC 2020 9749/J2H2MCT/2020 ( ) ( ) 1 slope 33 1 3 1 sin 3 10 70 10 370 10 sin 26 5 10 9 8 14 7 6 1 2 r P Fv mg . . .. θ − − = −   ×× −    ×    = = ° 18 By conserving energy, loss in GPE = gain in KE ( ) 2 B 2 B2 1 2 60 cos m gL L v g h mv − °= ∆= Considering circular motion at B, ( ) ( ) c 2 1 cos 62 01 11 2 F T mg mg mg m mg mg vT r L L = − = + = − °+ = += 19 GMm r2 =mrω2 T= 2π � r3 GM When 4 r, 2π�(4r)3 GM =8 ×2π�(r)3 GM = 8 T 20 T = ke 𝑘𝑘𝑘𝑘 = 𝑚𝑚1𝑣𝑣2 (𝐿𝐿+𝑒𝑒) ------------- (1) 𝑘𝑘(2𝑘𝑘 + 𝐿𝐿) = 𝑚𝑚2𝑣𝑣2 (2𝐿𝐿+2𝑒𝑒) -----------(2) (1)/(2): 𝑚𝑚2 = 𝑚𝑚1 × 2 × (2𝑒𝑒+𝐿𝐿) 𝑒𝑒 21 The elevation of a mountain is insignificant compared to the radius of Earth. Hence there is no change. 22 E = Ir + IR E = 3r + 3 ---------- (1) E = 2r + 4 ---------- (2) Solve (1) & (2): r = 1 Ω and E = 6 V 23 R = ρ L/A Maximum R corresponds to larger length and smallest area. 24 P = I V 100 k W = I x 10 kV I = 10 A Power = I2R = 102 x 5 = 500 Ω 25 Resistance of ammeter = 0 Ω Resistance of voltmeter = ∞ Ω 1 Ω // 2 Ω = 0.66667 Ω 1 Ω // 1.66667 Ω = 0.625 Ω = 0.63 Ω (2 SF)

4 ©EJC 2020 9749/J2H2JCT/2020 26 27 Eliminate A and C which are electrically equivalent. In parallel circuits, the effective resistance will be smaller than the branch with the least resistance so eliminate D: QS (can also be used to eliminate A and C as well); all 3 options involve one branch with only 1 resistor. 28 Using Right Hand Grip Rule, the induced magnetic field direction is going into the paper. 29 For velocity selector, 1 20000 0.25 80000 m s− = = = = Bqv qE Ev B For charged particle in circular motion within magnetic field ( )( ) ( ) 2 27 19 116 1.66 10 80000 0.25 1.6 10 0.385 m − − = = × = × = mvBqv r mvr Bq Distance = Diameter = 2r = 0.770 m 30 In a current balance, the sum of clockwise moment = sum anticlockwise moment about pivot. BI L x d 1 = mg x d2 0.022 x I x 0.4 x 0.8 = 2 x 10-3 x 9.81 x 0.9 I = 2.5 A