NYGH 2017-S3MYE-Physics Ans

NANYANG GIRLS’ HIGH SCHOOL ANSWERS Mid-Year Examination 2017 Secondary Three PHYSICS 1 hour 30 minutes Thursday 11 May 2017 08:45 – 10:15 Calculators may be used Section A 1 D 6 A 11 D 16 C 2 A 7 C 12 B 17 A 3 B 8 C 13 A 18 B 4 D 9 D 14 D 19 A 5 C 10 B 15 C 20 C Section B 21 (a) Measuring cylinder A; precision = 0.1 cm3 (b) (i) Cylinder B. Total volume of the 5 marbles exceeds to maximum capacity of cylinder A. (ii) Possible Reasons: • Increases precision of average volume as the answer will have 3 s.f. (or 2 d.p.) compared to only 1 or 2 s.f. (or 1 d.p.) if measured directly from the measuring cylinders. • Increases precision as any random error in taking volume reading once will be divided by 5. • Variations in volume of individual marbles will be partially cancel each other to increase accuracy Mention of the following without elaboration is not accepted: • increase accuracy • reduce (random) errors • variations in the marbles • human judgement error (no such term as human error in this context) Following is not accepted: • Reading from larger volume is more obvious (the rise in level is smaller for the same volume change in a wider cylinder) • The chance / possibility / likelihood of error is reduced (this remains the same. It is the size of the error that is reduced) • Parallax error (not accepted as error, should be taken care of in expt) • Any general statements about precision, accuracy, errors etc without linking to the actual question.

Physics 22 (a) 100.0/2.00 = 50.0 cm s-1 (b) Acceleration is constant. Correct calculation of velocity for 3 consecutive time intervals: first 0.40 s, vel = 10.0 / 0.40 = 25 cm s-1; for second 0.40 s, vel = 15.0 / 0.40 = 37.5 cm s-1; for third 0.40 s, vel = 20.0 / 0.40 = 50.0 cm s-1; for fourth 0.40 s, vel = 25.0 / 0.40 = 67.5 cm s-1; for fifth 0.40 s, vel = 30.0 / 0.40 = 75.0 cm s-1. Calculate acceleration correctly e.g. (37.5 – 25) / 0.40 = 31.25 cm s-2 is constant. 23 (a) Constant velocity till from 1 s to 9 s. Constant deceleration to zero from 9 s to 12 s. Reverse direction at 12 s / travels in opposite direction from 12 s. (Change direction, negative direction – not accepted) Constant acceleration from 12 s to 15 s. (b) Positive displacement = area from 0 s to 12 s = ½ (12+9) x 28 = 294 m Negative displacement = area from 12 s to 15 s = 1/2 x 3 x 28 = 42 m Net displacement = 294 – 42 = 252 m 24 (a) Ave velocity = 4.0 cm / (3 x 1/50) s = 66.7 cm s-1 (b) Ave acc = (66.7 – 25) / (5 x 1/50) = 417 cm s-2 or 420 cm s-2 25 (a) W=mg = 75 x 1.6 = 120 N (b) Same. This is the weight and it is a force. 26 (a) ρ=m/V V=m/ρ =9.0/0.45 = 20 cm3 (b) At the top if the liquid (must be shown partially out / partially submerged ~50%) (c) Zero newtons 27 (a) T, tension W, weight

Section C (b) Fnet / W = tan 20° Fnet = mg tan 20° = 0.60 x 10 tan 20° = 2.183 ≈ 2.2 N (c) Fnet = ma è a = Fnet / m = 2.183 / 0.6 [1] = 3.640 ≈ 3.6 m s-2 [1] 28 (a) (i) v=u+at = 0 + (0.5x10) = 5.0 m s-1 (a) (ii) Fnet = ma =70 x 0.5 = 35 N (a) (iii) N – W = ma N = ma + mg = 35 + 700 (b) Constant velocity è zero acceleration è net force is zero (0 N) (c) v2 = u2 + 2as è a = (v2 – u2)/2s = (0 – 52) / 2(15) = -0.833 m s-2 Consider freebody diagram of man+lift upwards & decelerating. T – W = ma, where m = 730+70 = 800 kg, W = mg = 8000 N T = mg + ma è T = 8000 + 800(-0.833) = 7333 ≈ 7300 N (d) constant acceleration to 5 m/s over 10 seconds constant velocity from 10 s to 15 s constant deceleration from 15 s to 21 s v = u + at è t = (v – u)/a = (0 – 5)/ (-0.833) = 6.0 s W T Fnet 20° weight = 700 N normal contact force = 735 N time / s velocity / m s-1 0 10 15 21 5