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