2020 ASRJC H1 Physics Prelims P2 Answers

1 8867/02/ASRJC/2020Prelim Anderson Serangoon Junior College 2020 H1 Physics Prelim Solutions Paper 2 (80 marks) 1ai EITHER By conservation of energy, Loss in k.e. = Gain in g.p.e. ½ m (2.02 – v2) = mg∆h ½ (2.02 – v2) = 9.81 (0.13) v = 1.204 = 1.2 m s−1 OR Taking direction up the slope as positive, acceleration along slope, aslope = − g sin 30° Using v2 = u2 + 2as, we have v2 = 2.02 + 2(− g sin 30°)(0.13/sin 30°) v = 1.204 = 1.2 m s−1 1aii At the top of the ramp, horizontal component of velocity = v cos 30° = 1.2 cos 30° vertical component of velocity = v sin 30° = 1.2 sin 30° Using sy = uyt + ½ ay t2, and taking downward as positive, 0.13 = (−1.2 sin 30°)t + ½ (9.81)t2 t = 0.2351 s d = (1.2 cos 30°)(0.2351) = 0.2443 = 0.24 m 1bi Considering momentum in the direction at right−angles to the direction of the initial path of ball A: By Conservation of linear momentum, EITHER total initial momentum = total final momentum OR total initial momentum in the direction at right−angles to the direction of the initial path of ball A = 0 0 = 4.0 × 6.0 sin θ – 12 × 3.5 sin 30° θ = 61.04 = 61 ° 1bii Considering momentum along the direction of the initial path of ball A: By Conservation of linear momentum, total initial momentum = total final momentum 4.0 × v = 4.0 × 6.0 cos 61.04° + 12 × 3.5 cos 30° v = 11.998 = 12 m s−1 1biii total initial k.e. = ½ (4)(12)2 = 288 J total final k.e. = ½ (4)(6.0)2 + ½ (12)(3.5)2 = 145.5 J Since the total kinetic energy before and after collision is different, the collision is inelastic.

2 8867/02/ASRJC/2020Prelim 2ai • arrows correctly labelled and form a closed loop • angles labelled correctly 2aii EITHER use sine rule: T sin (90° − 25°) = W sin[(90° − 18°)+25°] ➔ T = 474.8 = 470 N OR considering equilibrium of forces: ∑FY = 0 ➔ R sin 25° + T cos 18° = W = 520 ∑FX = 0 ➔ R cos 25° = T sin 18° Solving, T = 474.8 = 470 N 2bi For the oil droplet to be equilibrium, it means that the electric force is upwards. Since this is a negatively charged droplet, the direction of electric field is downwards. Since electric field points from high to low potential, metal plate A is of a higher potential. 2bii Since oil droplet is in equilibrium, weight = electric force mg = qE 2.6 × 10−14(9.81) = (8 × 1.60 × 10−19)(E) E = 1.99  105 N C−1 3a elastic energy stored = ½ (1400)(0.10)2 = 7.0 J 3bi Using conservation of energy, Loss in EPE of spring = Gain in GPE of mass + WD against friction 007.0 2 9.81 3.0=   + hh Solving the equation, h0 = 0.309 = 0.31 m 3bii As the mass loses energy due to frictional forces on its way up and down, the elastic potential energy when the mass next comes to a stop will be lesser than at the start. Hence, the compression in the spring will be less than 10 cm. W T 18° R 25°

3 8867/02/ASRJC/2020Prelim 4a The direction of the velocity/momentum is always changing. Hence, there is a rate of change of velocity/momentum. Thus, by Newton’s second law, there must have a resultant force acting on it. 4bi Though all have the same angular speed, their linear speeds are proportional to their distances from the centre of the circle. No, they would not be at the same linear speed. 4bii ω = (5.5 × 2 π)/60 = 0.57596 = 0.58 rad s-1 4biii centripetal force required = m r ω2 = 32 × 3.25 × 0.575962 = 34.500 = 35 N The maximum friction is larger than the required centripetal force. Hence, he will not fall off. 5ai correct shape of graph showing resistance decreasing 5aii As V increases, temperature increases, the number of charge carriers (p ositive holes and electrons) per unit volume increases, reducing resistivity. At the same times, the amplitude of vibration of atomic cores also increases, resulting in electrons making more frequent collisions with the atomic cores, increasing resistivity. The effect due to the increase in charge carriers per unit volume i s more significant than the effect due to increased amplitude of vibration. Hence, resistance of thermistor decreases as V increases. 5bi By potential divider rule, VAB = [RAB / (RAB + 1.2)] × 9 0.444 = RAB / RAB + 1.2 RAB = 0.96 kΩ I / A V / V 0

4 8867/02/ASRJC/2020Prelim 5bii For parallel combination, 1/960 = 1/1600 + 1/RT RT = 2.4 kΩ From the graph, temperature = 11°C (accept 10.75 to 11.25 °C) 5ci Using scenario 2 (2nd row), it can be deduced that resistance of R3 = 3.0  Using scenario 3 (3rd row), resistance of R2 + resistance of R3 = 8.0  Therefore, resistance of R2 = 5.0  Using scenario 1 (1st row), resistance of R1 + resistance of R2 + resistance of R3 = 10  Therefore, resistance of R1 = 2.0  5cii effective resistance between Z and B is (1/2.0 + 1/5.0)−1 = 1.43  effective resistance between Z and Y is 1.43 + 3.0 = 4.43  Therefore, using potential divider rule, p.d. between W and Y = 3.0 /(4.43) × 12 = 8.13 V 6a At t = 0 s, the only vertical force on the parachutist is his weight, hence his vertical acceleration is 9.81 m s−2. Note: do not allow for calculation using gradient At t = 6.0 s, draw tangent and find gradient of tangent to graph. vertical acceleration = 59.0−30.0 12.4−1.0 = 2.5 m s−2

5 8867/02/ASRJC/2020Prelim 6b no. of big squares = 11 total vertical distance = 11 × (10×4) = 440 m 6c point C (vertical speed first starts to decrease) 6d The (vertical) resultant force is given by weight – (vertical) air resistance As speed increases, (vertical) air resistance increases with speed and hence (vertical) resultant force decreases. The speed continues to increase until (vertical) resultant force equals to zero at point B (where (vertical) air resistance equals to weight). The parachutist moves at constant speed from B to C. 6ei change in momentum, ∆p = m (v – u) = 85 (20 – 50) = − 2550 = − 2600 kg m s−1 6eii average vertical resultant force = ∆p/t = − 2550 / 2 = − 1275 = − 1300 N Note: allow for positive answer. 6eiii Taking downward as positive, average vertical resultant force = weight – average vertical air resistance average vertical air resistance = weight – average vertical resultant force = 85 (9.81) – (− 1275) = 2108.85 = 2100 N 6f The parachutist comes to a complete stop over a longer period reducing the force of impact. 6.0 (12.4, 59.0) (1.0, 30.0) 4 8 12 16 20 24 28

6 8867/02/ASRJC/2020Prelim The force of impact on the parachutist is also distributed across various parts of the body (instead of a single part), reducing the risk of injury. 7a Newton’s law of gravitation states that the mutual force of attraction between any two point masses is proportional to the product of the masses and inversely proportional to the square of their separation. 7b gravitational force on the satellite provides centripetal force. By Newton’s 2nd law, GMmsatellite r2 = msatelliterω2 Since ω = 2π T , the above equation becomes GM r2 = r (2π T ) 2 Rearranging the equation, T2 = 4π2 r3 GM = 39.5 r3 GM 7ci A geostationary orbit has an orbital plane that is the same as the Equator. The period of a geostationary orbit is 24 hours. The satellite in a geostationary orbit moves from west to east (in the same direction as the rotation of the Earth). 7cii Substituting T = 24 × 60 × 60 into the equation in (b), we have r3 = GMT2 39.5 = 6.67×10−11 × 5.97×1024 × (24×60×60)2 39.5 r = 4.2219×107 = 4.22×107 m (3 s.f.) 7ciii linear speed of the satellite at geostationary orbit = rω = 4.2219×107 × 2π 24×60×60 = 3070.25 = 3100 m s−1 7civ By conservation of energ