ACSI 2020 Y3 IP Physics Paper1+Paper2 Answers

1 2020 Year3 IP EOY Answer Scheme 1 2 3 4 5 6 7 8 9 10 C C D B D B C B B C 11 12 13 14 15 16 17 18 19 20 A A A C C B B A D D 21 22 23 24 25 26 27 28 29 30 B B D D B C B C C C 31 32 33 34 35 36 37 38 39 40 C D C C D B C C B D Qns Answer Marks A1(a) Speed is a scalar quantity. Velocity is a vector quantity A1 (b)(i) W = mg = 75 × 10 = 750 N A1 (b)(ii) 750 N A1 (c)(i) 54 m s-1 A1 (c)(ii) Area under graph = 54 × 12 = 648 m or 650 m A1 (d)(i) Air resistance increases A1 (d)(ii) Deceleration or velocity decreases A1 (e) Air resistance decrease A1 A2(a) Net (unbalanced) force is equal to the product of its mass and acceleration B1 2(b)(i) a = (13 – 9) / (10 – 0) = 0.40 m s-2 M1 A1 2(b)(ii) F = ma = 900 × 0.40 = 360 N C1 2(b)(iii) Fnet = 360 N = Fhorizontal – air resistance & friction – tension Fhorizontal = 360 + 520 + 420 = 1300 N B1 C1

2 A3(a) mass = density × volume = 690 × 3.0 × 1.0 × 0.02 = 41.4 or 41 kg M1 A1 3(b) The centre of gravity of an object is defined as the point through which its whole weight appears to act for any orientation of the object. B1 3(c) Clockwise B1 3(d) No net moment caused by the CG about the pivot. B1 A4(a) Friction B1 4(b)(i) dperpendicular = 3.0 × Cos 30  = 2.59 = 2.6 m M1 C1 4(b)(ii) 90 × 6.0 = W × 2.6 W = 208 N M1 A1 A5(a) ½ mv2 = mgh v = (2gh) = (2*10**360) 7200 = 84.85 v = 84.9 m s-1 M2 A1 5(b)(i) Only one force acting on them. Gravitational force which exerts the same acceleration on both coins. B1 5(b)(ii) A heavier object needs a larger air resistance to balance its weight and hence it will have a larger falling speed. The converse is also true B1 B1 A6a Air particles are contionusly in random motion and collision with surface of mercury. As P = F/ A , force acting on the surface of the mercury exerts pressure. B1 B1 bi 730 +18 = 748 mm Hg M1 ii P = hpg = 0.748*13600*10 P = 101728 = 1.02 * 105 Pa 3sf M1 A1 c P = h pg = {(0.83-0.10-0.30)* 13600*10] + [(0.018*13600*10)] P = 60928 = 60900 Pa or 610 00 Pa {2 or 3 sf] M1 A1 d Remians same A1

3 A7a Thermometer A Amount of expansion is more per unit change in temperature, so change in mercury column per unit change in temperature is more than B. A1 B1 b Wider bore / longer bore A1 ci Conduction A1 ii Glass paticles near the hot end gains the thermal energy, vibrates more vigrously about their fixed position and makes collision with neighbouring particles and transfer thermal energy. B1 B1 A8 a Black suraface with steeper curve than white For both black and white shows same final temperature as room temperature. A1 A1 b In first few seconds, the temperature difference between the surroundings and the subtance is greater than the last few seconds. As the rate of thermal enegy loss is depends on the temperature differnce , at the first few seconds temp drops is greater than last few seconds. B1 B1 c Particles with high kinetic energy evaporate first and the average kinetic of the remaining paticles lesser. As average KE of the particles dereases, temperature decreases and causes cooling. B1 B1 X Y

4 B9(a)(i) magnitude indicated. M1 (a)(ii) vector length shoule be the same with magnitudea indicated M1 M1 (a)(iii) vector length shoule be the significantly longer with magnitudes indicated M1 M1 (b)(i) 1 mark for every 2 correct. No ½ mark. Distance fallen / m 20 45 80 Gravitational Potential Energy / J decreasing decreasing zero Kinetic Energy / J Increasing decreasing zero Elastic Potential Energy / J Zero increasing maximum A1 A1 (b)(ii) Net force is upwards. B1 (b)(iii) The forms of energy change will be converted to work done against friction in the bungee cord and eventually it will be converted to heat & sound which will be dissipated into the surroundings. So energy is continually changing its forms & is never lost. B1 B1 900 N 900 N 900 N 900 N 2400 N

5 B10 a Liquid A1 b 2.95 *105 J of thermal energy is needed to change unit mass of a substance X from liquid state to gaseous state without any change in temprature. B1 c E = 0.100 * 140* 37.0 E= 518 J [2 or 3 sf] M1 A1 d E = ( 0.100* 140* 332) +(0.100*2.95 *105) E= 4648 + 29500 E = 34148 J 34000 J or 34100 J [2sf /3sf] M1 M1 A1 e Ave KE decreases gas particles decreases as the temerature of gas decreases Rate of collison decreases As P = F/A , average force decreases, the pressure of gas decreases B1 B1 B1 B11a It is the angle of incidence in the optically denser medium, for which the angle of refraction is 90. B1 aii Critical angle is inversly proportional to refractive index B1 b 30 º c n=1 / sin c n=1/sin 30 º n= 2.0 [2sf] M1 A1 d The angle incidence in the optically denser medium is greater than the critical angle, TIR takes place. B1 e n=sin i/sin r r=sin-1(sin20 º /2.0) = 9.8º M1 A1 f PQ will refracted, no TIR As the ray travels from optically less dense medium to denser medium B1 B1