NYGH 2018-S3MYE-Physics Ans

2018 Mid-Year Examination Secondary Three Answer Scheme 1 …… 5 6 ……10 11 …… 15 16 …… 20 BABAA CBCDB BCDBA DABBC 2 1 light year = 3X108 m/s x 60 s/min x 60 min/h x 24 h/day x 365 days/year = 9.46 x 1015 m 1 Gm = 1 x 109 m, thus the best answer is 106 Gm. 9 Acceleration due to gravity is lower on the Moon, as compared to Earth. Thus, the rock would reach a greater height and take a longer time to fall back to the ground on the Moon. This eliminates B and C as possible answers. Since there is a net force acting on the rock, velocity (as represented by the gradient of the position-time graph) cannot be constant. Position should be increasing at a decreasing rate until the rock reaches the highest point. Thereafter, position should be decreasing at an increasing rate. 10 As the basketball is thrown downwards, the magnitude of the velocity of the ball must be increasing. This eliminates A and C as possible answers. Since the ball bounces after hitting the ground, the direction of velocity must change and part of the graph needs to be in the negative region. Thus, the answer must be B. 15 According to Newton’s Second Law of Motion, it is necessary for acceleration to act in the same direction as the net force. Consider an object moving at constant speed to the right. A net force can be applied to the left, which slows down the object. Therefore, it is not necessary for velocity to act in the same direction as the net force or acceleration. 17 The weight of the helicopter can be rewritten as: the gravitational force on the helicopter due to the Earth Therefore, the force which forms the action-reaction pair must be: the gravitational force on the Earth due to the helicopter 18 As the weight of the picture frame is the same in both diagrams, R1 = R2 Vector triangle for both diagrams: Therefore, based on the vector triangle above, as the angle θ increases, tension T would decrease. θ1 T1 T1 R1 θ2 R2 T2 T2 T1 R1 = R2 T1 T2 T2 θ1 θ2

2018 Mid-Year Examination Secondary Three 2 Nanyang Girls’ High School Setters: HWQ, STK Physics 21 (a) Micrometer screw gauge. The precision of the measurement is 0.01mm. This is the precision of the micrometer screw gauge. (b) 1) Count and measure the mass of 100 grains of rice using an electronic balance. 2) Repeat the measurement 2 more times to obtain 2 more sets of mass of 100 grains of rice. (optional) 3) Calculate the average mass of 1 grain of rice. Take the average of the 3 measurements then divide the mass by 100 to obtain the mass of 1 grain of rice. 22 (a) Anywhere less than 50 m (b) 23 (a) 0.75 s (b) 10 m s-2 (c) a = (v – u)/t 10 = (0 – u)/0.75 u = 7.5 m/s (d) (i) 0 10 20 30 40 50 60 0 s 4 s 8 s 12 s distance / m v / ms-1 t / s 7.5 0.75 1.5 - 7.5 x 4 8 12 16 0 10 20 30 40 50 x x x distance / m time / s

2018 Mid-Year Examination Secondary Three 3 Nanyang Girls’ High School Setters: HWQ, STK Physics [Turn over (d) (ii) Distance = Area under graph = ½ (7.5)(0.75) = 2.8 m 24 (a) ρ = m/V = 1.0/(0.05 × 0.05 × 0.05) = 8000 kg m-3 (b) 25 (a) Newton’s First Law of Motion states that an object at rest will remain at rest and an object in motion will continue in motion at constant speed in a straight line unless a resultant force acts on it. (b) W = mg = 5000 × 1.60 = 8000 N For the spaceship to be moving at constant velocity, the resultant force on the spaceship is zero. Therefore, thrust = weight = 8000 N (c) ma = thrust − W a = (10000 − 8000)/5000 = 0.40 m s-2 26 (a) 450 – (25 × 10) = 200 N (b) - Correct tensions: TA = 160 ± 10 N, TB = 130 ± 10 N TA cos 50° = TB cos 35° TA sin 50° + TB sin 35° = 200 Solving both equations simultaneously, TA = 160 N, TB = 130 N (c) The tensions will increase. Greater vertical components from both tensions are required to balance out the increase in the resultant force in part (a). Since the angles are constant, this can only be achieved by increasing the magnitudes of both tensions. X 35° 50° 200 N TA = 164 N TB = 129 N

2018 Mid-Year Examination Secondary Three 4 Nanyang Girls’ High School Setters: HWQ, STK Physics 27 (a) Newton’s Second Law of Motion states that the resultant force acting upon an object is equal to the product of the mass and the acceleration of the object; the direction of the force is the same as that of the object’s acceleration. (b) Weight drawn correctly on both diagrams Normal contact force drawn correctly on both diagrams Tension drawn correctly on both diagrams Note that the relative magnitudes of forces is important. (c) Let acceleration be a and tension be T Considering forces on block A, Net force = tension 2a = T Considering forces on block B, Net force = weight − tension a = 10 − T Solving them simultaneously, a = 3.3 m s-2 (d) v2 = u2 + 2as v2 = 2(3.33)(0.50) v = 1.8 m s-1 (e) weight weight normal contact force tension tension time / s velocity / m s−1 0 1.8