EJC Physics 2022 J2 H1 MYE P1 MS

1 ©EJC 2022 8867/J2H1MYE/2022 EUNOIA JUNIOR COLLEGE JC2 MID-YEAR EXAMINATIONS 2022 General Certificate of Education Advanced Level Higher 1 H1 PHYSICS PAPER 1 MARK SCHEME 8867 Jul 2022 Question Key Question Key Question Key 1 D 6 C 11 C 2 A 7 A 12 C 3 C 8 A 13 A 4 B 9 C 14 A 5 A 10 C 15 A 16 D 21 D 26 B 17 B 22 B 27 D 18 B 23 C 28 D 19 B 24 A 29 D 20 A 25 D 30 D 1 Ans: D 𝜌𝜌 = 𝑚𝑚 𝑤𝑤𝑤𝑤𝑤𝑤 = 51.6 (100.0)(10.0)(0.02) = 2.58 g cm–3 ∆𝜌𝜌 𝜌𝜌 = ∆𝑚𝑚 𝑚𝑚 + ∆𝑤𝑤 𝑤𝑤 + ∆𝑙𝑙 𝑙𝑙 + ∆𝑡𝑡 𝑡𝑡 ∆𝜌𝜌 2.58 = 0.1 51.6 + 0.1 10.0 + 0.1 100.0 + 0.01 0.20 ∆𝜌𝜌 2.58 = 0.06293798 ∆𝜌𝜌 = 0.16238 = 0.2 g cm–3 (to 1 s.f.) ρ = 2.58 = 2.6 g cm–3 (since ∆𝜌𝜌 is to 1 d.p.) 2 Ans: A 𝑣𝑣 = 𝑘𝑘 � ∆𝑃𝑃 𝜌𝜌 � 𝑛𝑛 LHS units: [v] = m s-1 RHS units: [𝑘𝑘 � ∆𝑃𝑃 𝜌𝜌 � 𝑛𝑛 ] = � 𝑘𝑘𝑘𝑘 𝑚𝑚 𝑠𝑠−2 𝑚𝑚−2 𝑘𝑘𝑘𝑘 𝑚𝑚−3 � 𝑛𝑛 =(𝑠𝑠−2 𝑚𝑚2)𝑛𝑛 Therefore n = 1/2

2 ©EJC 2022 8867/J2H1MYE/2022 3 Ans: C Accuracy refers to the degree of agreement between values of measurements and the actual or accepted value. Precision refers to the degree of agreement among values of measurements themselves. The measurements of 891 mm and 892 mm differ by more than 1 mm from the true value of 895 mm. Hence the measurements are not accurate to within 1 mm. As the measurements of 891 mm and 892 mm differ by 1 mm among themselves, they are precise to within 1 mm. 4 Ans: B 5 Ans: A 6 Ans: C He decelerated in one direction and accelerated in the opposite direction at the same rate. Therefore the acceleration vector should be in the same direction before and after he made the U-turn. 7 Ans: A Horizontal component: sx = vxt t = 2.35 s Vertical component: 8 Ans: A Change in momentum = Area under F-t graph = 10 x (4 + 6) / 2 = 50 N s 9 Ans: C Kinetic Energy = p2/ 2m Since both M and m have the same Kinetic Energy, p2 α m m M p p m M = ∴2 2 m M p p m M = 10 Ans: C Since the ball is momentarily stopped, the resultant force cannot be zero. The resultant force is upward since the ball is undergoing compression. 900 450cos(31.6 ) t=  21 2 yy ys ut at= + 21450sin(31.6 ) (2.348) ( 9.81 )(2.348)2 ys = +− 526 mys = (2.35) (2.35)2 527 m

3 ©EJC 2022 8867/J2H1MYE/2022 11 Ans: C For elastic collision, Relative speed of approach = relative speed of separation v2 – v1 = u1 – u2 (where the sign conventions of u1, u2, v1, v2 are to the right) Hence ux – (-uy) = vy – (-vx) ux + uy = vx + vy 12 Ans: C 13 Ans: A Στ = 0 Normal reaction from surface can be balanced by weight. Hence ΣF = 0 14 Ans: A Taking moments about A: M g cos 30o (1.0) = 4.0 g (0.60) M = 2.8 kg 15 Ans: A The 3 forces acting on the rod is weight, F and the hinge force For equilibrium, all 3 forces must pass through a common point (concurrent) and form a closed triangle. 16 Ans: D Work Done = Increase in GPE = mgh = 200 (9.81)(15) = 29 430 J Power Output = 29 430 / 60 = 490.5 W Efficiency = Power output/ Power input 0.65 = 490.5 / Power input Power Input = 490.5 / 0.65 = 750 W 17 Ans: B Lorry is accelerated from rest to a speed of 100 km h-1 . Work done by lorry's engine = increase in lorry's K.E. Friction has to be ignored as minimum time is to be considered 𝑃𝑃 𝑡𝑡 = 1 2 𝑚𝑚𝑣𝑣2 𝑡𝑡 = 𝑚𝑚𝑣𝑣2 2𝑃𝑃 where v = 100 km h-1 = 27.78 m s-1 ∴ 𝑡𝑡 = (2000)27.782 2×50000 = 15.4 s

4 ©EJC 2022 8867/J2H1MYE/2022 18 Ans: B (8 3600) (3.4 1055)Efficiency of LED = 100% 87.5%(8 3600) (15 3600) (30 1055)Efficiency of CFL = 100% 41.4%(15 3600) Difference = 87.5 % - 41.4 % 46% × −× ×=× × −× ×=× = 19 Ans: B Fnet = mv2 r ⇒ mg – R = mv2 r R = mg - mv2 r 20 Ans: A The net force is perpendicular to the velocity at all times, and therefore the instantaneous displacement. 21 Ans: D mRω2 = GMm R2 mR(2π T )2 = GMm R2 T2 ∝R3 The period of orbit of a satellite is independent of mass but dependent on the radius of orbit. As both satellites have the same period of orbit, they must have the same radius of orbit. However, just because a satellite has a 24 hour period does not mean it must be a geostationary satellite. The satellite need not be above the equator or the satellite may be orbiting in the opposite direction to the Earth’s direction of rotation. 22 Ans: B The solution can be obtained by calculating the resistance between the various junctions given in options A, B, C and D. For option B, the resistance between junctions Q and S is: 1/R Qs = 1/(2+8) +1/(4+6) 1/RQs = 1/10 +1/10 1/RQs = 2/10 RQs = 5 Ω

5 ©EJC 2022 8867/J2H1MYE/2022 23 Ans: C Potential drop across M: Looking at A-L-B-M-C, Between A and C, there is a potential drop of 20V. Since there is a 7 V drop across L, there must be a 13 V drop across M. Potential drop across Q: Looking at A-L-B-N-D-Q-C, Between A and C, there is a potential drop of 20V. Since there is a 7 V drop across L, and a 4 V drop across N, there must be 9 V drop across Q. Potential drop across P: Looking at A-P-D--Q-C, Between A and C, there is a potential drop of 20V. Since there is a 9 V drop across Q, there must be a 11 V drop across P 24 Ans: A Resistance of P is 4 times of Q thus current through Q is 4 times of P due to same pd across them. Current in P is ¼ of current in Q = 1/5 of total current (current in P + current in Q). 25 Ans: D Option A: Resistance BC = (2+2) // 2 // (2+2) = 1.0 Ω Option B: Resistance BC = 4 // 2 // 4 // 2 = 0.667 Ω Option C: Resistance = (2+2) // (2+2) = 2.0 Ω Option D: Resistance is the same as option C

6 ©EJC 2022 8867/J2H1MYE/2022 26 Ans: B Use Fleming’s LHR. 27 Ans: D According to Fleming’s left hand rule, the particle experiences a magnetic force that is perpendicular to its velocity and the magnetic field that it experiences. With magnetic field directed into the paper and current directed downwards, the magnetic forc e acting on the electrons is rightwards. 28 Ans: D Using Fleming’s left hand rule for force to be upwards, current must flow from Q to P and magnetic field must be in z direction. Magnetic force = weight , if tensions are to be zero. BIL = ρALg I = ρAg/B 29 Ans: D 30 Ans: D Majority of the α -particles managed to pass through the gold foil without being deflected since the atom consists of mostly empty space.