CJC.H2.PRELIM.2023.P1.Suggested Solutions

[Turn over PHYSICS 9749/1 Paper 1: Multiple Choice Questions September 2023 1 hour Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST Write your name and tutorial group on this cover page. Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write and shade your name, NRIC / FIN number and HT group on the Answer Sheet (OMR sheet), unless this has been done for you. There are thirty questions on this paper. Answer all questions. For each question, there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet (OMR sheet). Read the instructions on the Answer Sheet carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. Q 1 Q 2 Q 3 Q 4 Q 5 Q 6 Q 7 Q 8 Q 9 Q 1 0 Q 1 1 Q 1 2 Q 1 3 Q 1 4 Q 1 5 Q 1 6 Q 1 7 Q 1 8 Q 1 9 Q 2 0 Q 2 1 Q 2 2 Q 2 3 Q 2 4 Q 2 5 Q 2 6 Q 2 7 Q 2 8 Q 2 9 Q 3 0 D C C D A B A B A B D C A A B C D B C B A A B B A D B D A D Suggested Solutions This document consists of 21 printed pages and 1 blank page. NAME CLASS 2T Catholic Junior College JC2 Preliminary Examinations Higher 2

2 DATA speed of light in free space c = 3.00 x 108 m s-1 permeability of free space µ0 = 4π x 10-7 H m-1 permittivity of free space ε0 = 8.85 x 10-12 F m-1 (1/(36π)) x 10-9 F m-1 elementary charge e = 1.60 x 10-19 C the Planck constant h = 6.63 x 10-34 J s unified atomic mass constant u = 1.66 x 10-27 kg rest mass of electron me = 9.11 x 10-31 kg rest mass of proton mP = 1.67 x 10-27 kg molar gas constant R = 8.31 J K-1 mol-1 the Avogadro constant NA = 6.02 x 1023 mol-1 the Boltzmann constant k = 1.38 x 10-23 mol-1 gravitational constant G = 6.67 x 10-11 N m2 kg-2 acceleration of free fall g = 9.81 m s-2

3 [Turn over FORMULAE uniformly accelerated motion s = u t + ½ a t2 v2 = u2 + 2as work done on / by a gas W = p ∆V hydrostatic pressure p = ρgh gravitational potential φ = - Gm r temperature T / K = T / ˚C + 273.15 pressure of an ideal gas p = 1 3 Nm V 〈c2〉 mean translational kinetic energy of an ideal gas molecule E = 3 2 kT displacement of particle in s.h.m. x = x0 sin ωt velocity of particle in s.h.m. v = v0 cos ωt = 22 0 x x− ±ω electric current I = Anvq resistors in series R = R1 + R2 + ... resistors in parallel 1/R = 1/R1 + 1/R2 + ... electric potential V = Q 4πεor alternating current / voltage x = x0 sin ωt magnetic flux density due to a long straight wire B = μoI 2πd magnetic flux density due to a flat circular coil B = μoNI 2r magnetic flux density due to a long solenoid B = μonI radioactive decay x = x0 exp(-λt) decay constant λ = 1 2 ln2 t

4 1 A man of mass 75.2 kg uses a set of weighing scales to measure his mass three times. He obtains the following readings. mass / kg reading 1 80.2 reading 2 80.1 reading 2 80.2 Which statement describes the precision and accuracy of the weighing scales? A not precise to ± 0.1 kg and accurate to ± 0.1 kg B not precise to ± 0.1 kg and not accurate to ± 0.1 kg C precise to ± 0.1 kg and accurate to ± 0.1 kg D precise to ± 0.1 kg and not accurate to ± 0.1 kg Answer: D Since the weighing scale’s reading is bout 5 kg away from the man’s true mass, the scale reading is not accurate. However, repeated readings are close to one another with 0.1 kg deviation. Hence the readings are precise. 2 An object is launched with a speed of 50.9 m s-1 at an angle θ above the horizontal, as shown. The ground is level. The object lands on the ground 100 m from its initial launch position. What is the value of the angle θ? You may need to make use of the double angle formula: 22sin( ) sin cosθ θθ= A 2.8° B 5.5° C 11° D 79° Solution: C In the horizontal direction: 50.9 t cos θ = 100 In the vertical direction (taking upwards as positive): 0 = 50.9 t sin θ – (0.5)(9.81)t 2 = 50.9 sin θ – ½ (9.81)t 50.9 sin θ = (0.5)(9.81)t 50.9sin (0.5)(9.81)t θ= 50.9 m s-1 100 m

5 [Turn over Substituting into the previous equation: 2250.9 sin cos 50.9 (0.5)(sin 2 )100 (0.5)(9.81) (0.5)(9.81) 2 22 11 θθ θ θ θ = = = =   3 Two spheres with different masses are initially moving towards each other with the same speed. The two spheres collide elastically. Which of the following statements is true? A The two spheres can be at rest at the same time because they were initially moving in opposite directions. B The two spheres can be at rest at the same time because momentum is always conserved in an elastic collision. C The two spheres cannot be at rest at the same time at any point in time because the total momentum of the system is not zero. D The two spheres cannot be at rest at the same time at any point in time because energy is always conserved in an elastic collision. Solution: C Since the two spheres do not have the same momentum (since their momentum are unequal in magnitudes), the total momentum of the system is not zero, and hence they cannot be at rest at the same time. 4 A rower is sitting on a boat in the middle of a calm lake. Which of the following forces forms a Newton’s third law pair with the upthrust acting on the boat? A The weight of the boat alone. B The weight of the rower alone. C The combined weight of the boat and the rower. D The force of the boat on the lake’s water to displace some of the lake’s water. Solution: D The action-reaction pair to the upthrust (which is exerted by the water on the boat) is the downwards force exerted by the boat on the water when it displaced some of the lake’s water.

6 5 A fully submerged buoy is tethered by a rope to a riverbed. The current of the river exerts a constant horizontal force of F on the buoy causing the rope to make an angle of 26.6° to the vertical, as shown. The buoy is in equilibrium. The buoy has a volume of 0.500 m3 and a density of 390 kg m-3. The river water has a density of 1000 kg m-3. What is the force F exerted by the river current on the buoy F? A 1500 N B 3830 N C 6000 N D 9810 N Solution: A The vertical forces on the buoy are the vertical component of tension, upwards upthrust and the downwards weight. The net force in the vertical direction is: () (0.500)(9.81)(1000 390) 3000N ρ ρρ ρρ = − = − = − = − = − = y w wb wb F UW V g mg V gV g Vg The tangent of the angle that the rope makes with the vertical can be expressed as the ratio of the magnitude of the resultant horizontal force acting on the buoy to the resultant vertical force: N tan 26.6 0.5 3000 1500 = = = x y x x F F F F

7 [Turn over 6 An object moves along a frictionless track from A to C, through B, which is the lowest point of the motion. Which graph shows how the kinetic energy of the object varies with the vertical distance from A? A B C D Solution: B The gravitational potential energy will decrease linearly with the vertical distance from A since the change in GPE = mgΔh The kinetic energy will increase at the same linear rate, and therefore, the answer is B.

8 7 A rubber strip is stretched by a force. The graph of the applied force against its length is shown. What is the work done in stretching the strip to the first point where it no longer obeys Hooke’s Law? A 0.70 J B 6.3 J C 700 J D 6300 J Solution: A The limit of proportionality is when the strip no longer obeys Hooke’s Law. This occurs at 40mm of length