HCI Sample A answer key

1 Sample A Sec 3 EOY Exam Answer Key Whole paper: Subtract 1 mark for improper presentations. Subtract 1 mark for any number of sf mistakes. Subtract 1 mark for any number of unit mistakes. Paper 1 1 C 11 D 21 A 2 D 12 D 22 B 3 C 13 D 23 C 4 D 14 A 24 A 5 D 15 A 25 D 6 C 16 B 26 D 7 D 17 A 27 B 8 C 18 C 28 A 9 C 19 A 29 D 10 A 20 C 30 D Paper 2 1(a) Zero error = - 0.02 mm [1] Corrected reading = 3.77 mm – (- 0.02 mm) = 3.79 mm [1] (b) Mass = 0.0152 / 10 = 0.00152 kg [1] Vol = 3.14 x (3.79 / 2 x 1000)2 x 3.00 = 3.38 x 10-5 m3 [1] Density = 0.00152 kg / 3.38 x 10-5 m3 = 45.0 kg / m3 [1 – ecf for calculation of density] 2(a) θ = sin-1 (1/1.47) = 42.9º [1] Diameter = 2 x (80 x tan 42.9º) = 1.5 x 102 mm [1] (b) The ‘halo’ diameter would be larger [1] as the critical angle increased [1]. Award 1 mark (process mark) if only observed reading is correct (3.77 mm)

2 3) (a) Magnification factor = 3.0 cm / 1.2 cm = 2.5 (2.3 to 2.7) [1] (b) Focal length = 4.0 cm (+ / - 0.4 cm) [1] 4(a) v2 = u2 + 2as 0 = 152 + 2(-10)s [1] s = 11.25 m = 11 m [1] (b) t1 = (0 – 15) / (-10) = 1.5 s s = ut + 0.5at2 t2 = √[(11.25 + 80) / 5] = 4.3 s [1] for correct calculation for t1 and t2 t = t1 + t2 = 5.8 s [1] or s = ut + 0.5at2 80 = −15t + 5t2 or −80 =15t − 5t2 [1] t = 5.8 s [1] 5) Scale: Let 1.0 cm rep. 50 N Object Image F 120˚ 30˚ T1 = 10.0 cm T2 = 4.9 cm T3 = 8.7 cm [1] for light ray that cut through principle axis and lens. [1] for light ray from top of image to lens and from top of image to principle axis. Subtract 1 mark if arrows are not drawn.

3 Team 2 tension = 4.9 x 50 = 245 N (+ / - 10 N) [1] Team 3 tension = 8.7 x 50 = 435 N (+ / - 10 N) [1] Correct vector diagram drawn [1] Appropriate scale used – diagram occupy at least half of given space [1] 6a(i)(ii) (b) The direction of the satellite is changing all the time [1] and this means that its velocity is changing. A change in velocity [1] denotes that the satellite is accelerating. (c) The direction of the moving satellite is always normal / at right angle to the force [1]. Therefore, no work is done [1]. 7(a) The water is changing phase / boiling / kinetic energy of molecules is constant while potential energy is changing. [1] (b) Energy supplied = 300 W x 420 s = 126 kJ 126 kJ = 0.40 x 4.2 x 103 x ∆T [1] ∆T = 75˚C therefore, boiling temperature = 75 + 20 = 95˚C [1] 8(a) Temp = (804-802) x 100 / (815- 802) [1] = 15.4˚C [1] (b) Y [1]. Thermal energy will take a longer time to pass through the thicker glass bulb wall of Y, compared to X [1]. F V [1] – F drawn as a straight line towards centre of the Earth. [1] – V drawn as a tangent to the position of the satellite.

4 9(a)(b) (c) Since the wavelength is larger in medium I [1], the speed of light is higher in medium I [1]; or the angle of incidence is larger than the angle of refraction [1], the speed of light is higher in medium I [1]. (d) vI / vR = λI / λR = 2.9 cm / 1.5 cm [1 awarded: 2.9 cm for numerator and denominator is based on their wavelength measured in medium R] = 1.9 (+ / - 0.1) [1] 10(a)(i) S (a) 1 mark: parallel to D. (b) 1 mark: bend towards normal and 90˚ to (a). Ecf awarded as long as wavefront is at right angle to direction of refracted water wave. Tension exerted by thread on object OR Tension Weight exerted by object on thread or Weight Object [1] [1] Subtract 1 mark if tension is not drawn longer than weight.

5 (a)(ii) s = ut + 0.5at2 s = 0.5at2 a = 2s/t2 = (2 x 0.84) / (2.2)2 = 0.35 m s-2 [1] T – mg = ma T = m (g + a) = 0.015 (10 + 0.35) [1] = 0.16 N [1] (b)(i) Increase in potential energy = (0.015 kg)(10 m s-2)(0.84 m) = 0.13 J [1] Power = Work done / time = 0.126 J / 3.4 s = 0.037 W [1] (ii) Power input to motor = VI = (6.0 V)(0.045 A) = 0.27W [1 - ecf] Efficiency = Pout / Pin = (0.03706 W / 0.27 W) x 100% = 14% [1 - ecf] (iii) The efficiency will be higher as the frictional force will be reduced / less energy will be lost [1 mark for both points]. 11(a)(i) The pressure rises [1] above atmospheric pressure as the temperature rises due to increase in kinetic energy of the particles. As the pressure increases, the boiling point of water rises hence the water can reach a higher temperature [1]. (ii) The weighted vent was stuck / fail to open / hole was stuck with food. [1] * Either one reason is accepted. (iii) The high temperature caused the alloy plug to melt [1]. Some gas molecules inside the pressure cooker escape into the air / the rate at which the gas molecules hit the wall of the cooker decreases [1]. (b)(i) Gas molecules are moving randomly all the time, bombarding the walls of the cylinder with a force [1]. Since pressure is defined as force per unit area [1], the gas exerts a pressure on P. (ii) The piston moved to the right [1] when it was heated up as the particles gained average kinetic energy [1] and hit the wall with a greater force / pressure [1]. 12A(a) Transverse wave / Travel at speed of light in vacuum / are electromagnetic radiation / travel through vacuum / can be reflected, refracted, polarized, e t c- any one of the mentioned. [1] (b) Power = 0.36 / (3.0 x 60) = 2.0 x 10-3 W [1] SAR = 2.0 x 10-3 W / 0.010 kg = 0.20 W/kg [1] (c) It is converted into thermal energy [1] before being carried away by the body. Award 1 mark only if a is calculated wrongly but the process for T is correct. If calculation was made without following the correct units, award only 1 mark.

6 (d) Radio waves have low frequency, long wavelength or low energy and this makes it not so penetrating. (e) Sound energy / thermal energy / light energy – Any one. [1] (f) Keep phone at a distance; use headphone or switch off hand set; make brief calls only; direct antenna away from head, etc. (Any one) [1] (g) λ = c / f = 3.0 x 108 m s-1 / 1200 x 106 Hz = 0.25 m [1] Length of antenna = 0.25 / 4 = 0.0625 = 0.063 m [1] (h) Gamma ray I X-rays I UV (any one) that damage/mutate/ionize body cells. [1] 12B(a)(i) Polystyrene is a poor thermal conductor so less thermal energy is conducted away from the hot water to the outside environment [1]. The lid reduces the heat loss as it cuts down the evaporation of hot water [1]. The lid reduces heat loss as it does not encourage the setting up of convect