NYGH 2010 S3 Physics EOY Answers

2010 SEC 3 PHYSICS END OF YEAR EXAM ANSWER SCHEME PAPER 1 1~5: ABBDC 6~10: ACACD 11~15: DBCAB 16~20: ADABD 21~25: BBDAB 26~30: CABBC PAPER 2 Q1 (a) Volume of cube = length x height x breadth = 4.0 x 3.0 x 2.0 = 24 cm3 Density = mass / volume = 36 / 24 = 1.5 g/cm3 (b) Its density is greater than that of water. (c) Physical quantity Increase No change Decrease Length of sides of cube 9 Mass of cube 9 Weight of cube 9 Density of cube 9 Q2 (a) Distance = area under the graph = (½ x 10 x 8) + (15 x 8) + (½ x 8 x 2.5) OR [½(27.5+15) x 8] = 170 m (b) EITHER 25.0 s - 27.5 s OR when it is decelerating from 8 m/s to 0 m/s (c) Average speed = total distance/total time = 170 / 27.5 = 6.2 m/s Q3 (a) Both arrows point away from the mass to the ceiling in both strings. (b) Suitable scale, at least ½ of space available; no penalty for awkward scale. Vector polygon with arrows forming closed triangle. T 1 and T2 in acceptable range i.e. T1 = 10 N ± (0.1 cm x scale) & T2 = 20 N ± (0.1 cm x scale) Q4 Take moment abt P, (500 x 2.0) + (100 x 1.8) = R Q x 3.6 OR RQ = 328 N Take moment abt Q, (500 x 1.60) + (100 x 1.8) = R Q x 3.6 RQ = 272 N Then, R P = 500 + 100 – 328 = 272 N OR RQ = 500 + 100 – 272 = 328 N 1

Q5 (a) & (b) Mirror M1 Mirror M2 E Q6(a) The light needs to be passing from optically more dense to less dense media OR The incident ray is in the denser medium. The angle of incidence needs to be greater than the critical angle. (b) eye 45° glass prisms (c) sin c = 1/n sin 450 = 1/n n = 1.41 Q7 (a) The distance between the centre of lens and the principal focus. OR The distance from the centre of the lens at which parallel rays meet. 2

(b) (i) I F O (ii) 12 cm ± 0.5 cm (c) magnifying glass OR used to magnify a small object Q8 (a) Transverse (b) Upwards / Up (c) Downwards / Down (d) T = 5.0 s f = 1/T hence, f= 1/50 = 0.20 Hz (e) v = f λ = 0.2 x 8 m = 1.6 m/s (f) No change in frequency Q9 (a) softer. Amplitude decreasing (b) period / frequency is constant Q10 (a) NRF starts either from base or centre, friction along bottom. Other positions are not acceptable. normal reaction force / normal contact force frictional force weight 180 N (b) Using F = ma = (18.0) (4.0/5.0) = 14.4 N (c) (i) The inertia (or mention of Newton’s 1st Law) of the box causes it to continue moving forward but the base is prevented from moving forward by the friction, on the surface of the conveyer belt. (ii) Max angle of tile = tan-1(11/40) = 15.4° (d) Taking moments about the edge, clockwise moments = anticlockwise moments F × 80 = 180 × 40 Hence, F = 90 N 3

4 Q11 (a) Weight = 500 N (b) 500 N, downwards (c) Using v = u + at 0 = u + (-10) (0.7) u = 7.0 m/s (d) (i) Using v = u + (-10) (1.04) v = 0 + (-10) (1.04) v = 10.4 m/s (ii) Using s = ut + ½ a t 2 s = (7.0)(1.74) + ½ (-10) (1.74) 2 s = 2.96 m (e) Yes, The girl will be able to reach the same height. They will have the same initial velocity and subject to the same deceleration due to gravity. OR mass is not a factor in the appropriate equation of motion. Q12 (a) Energy cannot be created nor destroyed but can be transformed from one form to another. (b) GPE = mgh = 0.5 x 10 x 4.0 = 20 J (c) (i) KE = ½ x m x v2 = ½ x 0.50 x 4.02 = 4.0 J (ii) Work against friction = 1.0 x 3.0 = 3.0 J KE = 4.0 – 3.0 = 1.0 J OR a = F/m = -1/0.5 = -2 m/s 2 v2 = u2 + 2as = 16 + 2(-2)(3.0) = 4.0 KE = 1/2 x m x v2 = 1/2 x 0.5 x 4.0 = 1.0 J (iii) KE = 1.0 + 20 = 21 J O R Just before hitting floor, v 2 = (horizontal velocity)2 + (vertical velocity)2 = 2 . 0 2 + (u2 + 2as) = 4 . 0 + ( 0 2 + 2 x 10 x 4.0) = 8 4 KE = 1/2 x m x v 2 = 1/2 x 0.5 x 84 = 21 J (d) converted into sound energy & heat energy (& some transferred to the floor as the floor’s internal energy)