2022 FHSS Phy Prelims Ans

Paper 1 1 2 3 4 5 6 7 8 9 10 C A C B C D A B D D 11 12 13 14 15 16 17 18 19 20 A C B D D B C A A C 21 22 23 24 25 26 27 28 29 30 D D A D A A D B A A 31 32 33 34 35 36 37 38 39 40 D B D B C C B C D B

Section A 1 (a) [1] (a) W = mg = (0.900 kg)(10 N/kg) = 9.00 N (3sf) or 9.0 N (2sf) [1] (b) 1m – appropriate scale (1.0 cm to 2.0 N) 1m – correct vector diagram (right angle triangle, direction of arrows) 1m – correct magnitude of friction force Allow for e.c.f. from (a) [1] [1] [1] (c) Newton’s first law. Forces are balanced. Resultant force of friction, weight and normal contact force = 0 N. (Any one) [1] 2 (a) moment by brake pad = moment by master piston (220)(32) = (F)(8) F = 880 N [1] [1] Weight of box Normal Contact Force Friction

(b) Pmaster piston = Pslave piston F1/A1 = F2/A2 880 / 1.5 = F2/5.0 F2 = 2930 N or 2900 N (allow for e.c.f.) [1] [1] 3 (a) The product of force applied and distance moved in the direction of the force [1] (b) KE = ½ x m x v2 = 0.5 x 0.20 x 202 = 40 J [1] [1] (c) G.P.E at max height = m x g x h = 0.200 x 10 x 12 = 24 J (2sf) [1] [1] (d) At max height, KE remaining = 40 J – 24 J = 16 J (allow for e.c.f.) 0.5 x 0.200 x v2 = 16 v = 12.6 m / s or 13 m / s [1] [1] 4 (a) The focal length is the distance along the principal axis, between the principal focus and the optical centre of the lens. OR Distance between the optical centre of the lens and the principal focus (focal point). [1] (b) Accept either position of principal focus of (iii) Rays and arrowheads should be in solid lines [1] [1] [1] (c)(i) 60.0 cm [1] (ii) 2f = 60.0 cm f = 30.0 cm [1] (iii) 1. Virtual 2. Upright 3. Magnified [1] 5 (a) Negatively charged [1] (b) The negative charges will be transferred from rod Y to sphere B. [1] I

The negative charges are attracted by positively charged rod X as unlike charges attract. [1] Hence, there is more negative charges than positive charges in sphere B upon removal of the rods. (Sphere B has a net negative charge) [1] (c) Sphere A will become positively charged (and B is negatively charged). [1] They will attract each other. [1] Since unlike charges attract. [1] 6 (a)(i) I = V/R = 4.8 / 32 = 0.15 A [1] [1] (ii) Q = I(t) = (0.15)(25) = 3.75 C or 3.8 C (allow for e.c.f.) [1] [1] (iii) Pd across parallel branch = 6.0 – 4.8 V = 1.2 V R = V/I = 1.2 / 0.15 = 8.0 Ω Alternative method: Potential Divider method [32 / (32 + R)] x 6.0 = 4.8 (32 + R) / 32 = 6.0 / 4.8 R = 8.0 Ω [1] [1] (iv) Method 1 (1/R1 + 1/R2 = 1/Rtotal) 1/R + 1/24 = 1/8 R = 12 Ω allow for e.c.f. Method 2 Current flowing through Y = 1.2 / 24 = 0.050 A Current flowing through X =