NYGH 2020-S3EOY-IP Physics P1 and P2 Ans

2020 Sec 3 IP Physics EOY Exams Mark Scheme Paper 1: 1….….5 6…….10 11…..15 16…..20 21..…25 26..…30 CAABB DAADD CDCBA CDDCB BCBCA ADCBD 3. density = m / V = 5.94 x 1024 / (1.08 x 1012 x 109) since 1 km3 = (1000 m)3 = 109 m3 9. v2 = u2 + 2as, find a where a is negative. Same braking force means same deceleration. Find s. 10. During reaction time, distance already covered, d = speed x time = 100 km h-1 x 0.50 s v2 = u2 + 2as where s = 100 m – d u = 100 km h-1 = 27.78 m s-1 11. u = 0, t = 1.0 s, s = 6.0 m ➔ s = ut + ½ at2, find a. 13. Constant resultant force ➔ constant acceleration = constant gradient on v-t graph 3 graphs are for different objects. 15. Horizontally, no force acts between P and Q Q: Fnet = ma ➔ 15 N = (10.0 kg) a P: no horizontal force, a = 0 initially 18. Pushing X slightly causes it rotate and stay in its new position. Pushing Y slightly can cause it to topple easily. 21. bubble of gas obeys Boyle’s law: p1V1 = p2V2 At depth h: p1 = Patm + hdg, V1 = 15.0 cm3 At surface: p2 = Patm, V2 = 22.5 cm3 29. If mirror moves to the right from the stationary man at a speed u, it is equivalent to the man moving to the left from the stationary mirror at speed u. this will result in the man’s image moving to the right from the mirror also at speed u. Relative to moving man, his image is moving away from him at twice the speed or 2u. 2u = 3.0 m s-1, so u = 1.5 m s-1.

2020 End-of-Year Examination Secondary Three 2 Nanyang Girls’ High School Setters: AJL & TPY Physics II Paper 2: Section A 1(a) Taking the average of the time for 10 oscillations will reduce the random error in the measurement of the period T. (b) The first measurement could have a miscount of the number of oscillations. Or Taking another measurement will improve the reliability of the measurements. (c) The graph does not support the hypothesis. [No mark given for this alone!] The hypothesis indicates that a straight line graph passing through the origin is expected. The graph plotted is a curve, and the graph does not pass through (or very close to) the origin. 2(a) v = u + at ➔ t = (v-u) / a = (0 – 44.1) m s-1 / (-10 m s-2) = 4.41 s (b) Assume that there is negligible air resistance. (c) v2 = u2 + 2as ➔ s = (v2 – u2) / (2a) = (0 - (44.1)2) / 2(-10) = 97.2 m (d) Time taken to fall from maximum height = 6.0 – 4.41 s = 1.59 s Distance travelled when falling down s = ut + ½ at2 = 0 + ½ (10 m s-2)(1.59 s)2 = 12.64 m Total distance travelled = 97.2 m + 12.64 m = 109.8  110 m 3(a) v = u + at ➔ u = v – at = 10 m s-1 – (2.0 m s-2)(3.0 s) = 4.0 m s-1 (b) v = u + at = 10 + (2.0)(2.5) = 15 m s-1 (c) Straight line from (0,4) to (5.5, 15) : labelling of values not required! velocity / m s-1 time / s 0 2 4 5.5 6 15 10 5 4

2020 End-of-Year Examination Secondary Three 3 Nanyang Girls’ High School Setters: AJL & TPY Physics II [Turn over (d) s = ut + ½ at2 = (4.0 m s-1)(5.5 s) + ½ (2.0 m s-2)(5.5 s)2 = 52.25  52 m OR s = vave t = ½ (u+v)t = ½ (4.0 + 15)(5.5) = 52 m OR use area under the graph 4(a) Comments: • Normal contact force and weight are equal in magnitude; hence both should be drawn to the same length. • Forces must be labelled with words spelt in full. (b) m = 1.10 kg, M = 0.20 kg Fnet = m a ➔ T = m a  M g – T = M a  Combining  + , M g = (m + M)a ➔ a = [M/(m + M)] g = [0.20 / (0.20 + 1.10)] (10) = 1.538  1.5 m s-2 (c) T = m a = (1.10 kg)(1.538 m s-2) = 1.6918  1.7 N (d) Fnet = M a = (0.20)(1.538) = 0.3076  0.31 N Allow for e.c.f. from (b) OR M g – T (e) When the string breaks, tension T becomes zero / There is no net force on B Or By Newton’s 1st law of motion, B will continue moving to right with a constant velocity. 5 (a) This is to ensure the weight of the lifting bar would not produce any moment to affect the balancing of the system. (b) Taking moments about the pivot P Clockwise moment = anticlockwise moment F × 2.0 m = 500 N × 0.5 m F = 125 N = 130 N (2 s.f.) (c) The (effort) force F applied at Y will be much less than the weight to be lifted. B tension weight normal contact force

2020 End-of-Year Examination Secondary Three 4 Nanyang Girls’ High School Setters: AJL & TPY Physics II 6 (a) Note: • The vertical line-of-action of weight must pass exactly at the end of the foot. (b) The (vertical) line-of-action of weight falls outside the base of his feet when he tries to touch his toes. OR acts to the right of P (his toes) The weight of the man produces a net clockwise moment about P (his toes) which rotates the man forward causing him to topple. 7(a)(i) Atmospheric pressure = 0.76 m x 13600 kg m-3 x 10 m s-2 = 1.03 x 105 Pa (ii) Excess pressure = 45.0 cm Hg – 25.0 cm Hg = 20.0 cm Hg Gas pressure = 76.0 cm Hg + 20.0 cm Hg = 96.0 cm Hg (b) Mark X in mercury and label a suitable distance on Fig. 7.1 (c)(i) 76.0 cm Hg (c)(ii) P1V1 = P2V2 96.0 cm Hg x 10.0 cm3 = 76.0 cm Hg x V2 V2 = 12.63 cm3 ≈12.6 cm3 x P B 25.0 cm 45.0 cm 10.0 cm3 of gas A mercury 5.0 cm X 15.0 cm 30.0 cm 7b) ‘X’ must be in mercury with correct suitable distance indicated on Fig. 7.1

2020 End-of-Year Examination Secondary Three 5 Nanyang Girls’ High School Setters: AJL & TPY Physics II [Turn over 8 (a) Total energy = GPE at the start = 0.200 kg x 10 m s-2 x 6.0 m = 12 J GPE at D = 0.200 kg x 10 m s-2 x 1.5 m = 3.0 J K.E. = Total energy – GPE at D = 12 – 3.0 = 9.0 J OR Increase in KE = Decrease in GPE = mgΔh = 0.200 kg x 10 m s-2 x (6.0 - 1.5)m = 9.0 J (b) K.E = 9 J = ½ m v2 ➔ v = √(9 x 2)/0.200 = 9.487 ms-1 ≈ 9.5 ms-1 Section B 9(a)(i) This will shorten the time taken for the rover to reach Mars. Accept: Less fuel is needed for the journey to Mars. (a)(ii) average speed = d / t = 55 x 106 km / (6 x 30 x 24) h = 12731  13 000 km h-1 (b) (c)(i) v = u + at ➔ v = -9.0 m s-1 + (3.0 m s-1)(2.0 s) = - 3.0 m s-1 Or velocity is 3.0 m s-1 downwards. (c)(ii) From t = 4.5 to 5.0s, the module is falling freely, a = -2.0 m s-2 Acceleration due to free fall g is 2.0 m s-2 on this planet (c)(iii) F – mg = ma ➔ F = m(g + a) = 250 kg (2 + 3) m s-2 = 1250 N 10 (a) Any 3: Image formed is [2] • Virtual • Upright • Same size as object • As far behind the mirror as the object is in front of the mirror • Laterally inverted landing module thrust weight

2020 End-of-Year Examination Secondary Three 6 Nanyang Girls’ High School Setters: AJL & TPY Physics II (c) 80 cm – 20 cm = 60 cm Rays drawn to indicated the bottom point of image that the eye can see OR 5.0 m + 5.0 m = 10.0 m x 30 cm = 10.0 m 5.0 m ➔ x = 60 cm OR Length of image = 2 x size of mirror = 2 x 30.0 cm = 60.0 cm or 0.600 m (d) image distance from the reflective window = 5.000 m horizontal distance from the man’s eye to the image of the rule = 5.000 m + 5.000 m = 10.000 m 11 Either (a) The ice block has a large mass and hence a large inertia,