2024 CJC H1.Phy.PRELIM P1 ans

PHYSICS 8867/1 Paper 1 Multiple Choice 12 September 2024 1 hour Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name and tutorial group on this cover page. Write and/or 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). 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. MARK SCHEME This document consists of 19 printed pages and 1 blank page. [Turn over NAME CLASS 2T Catholic Junior College JC2 Preliminary Examination Higher 1

2 Data speed of light in free space, c = 3.00  108 m s1 elementary charge, e = 1.60  1019 C unified atomic mass constant, u = 1.66  1027 kg rest mass of electron, me = 9.11  1031 kg rest mass of proton, mp = 1.67  1027 kg the Avogadro constant, NA = 6.02  1023 mol1 gravitational constant, G = 6.67  1011 N m2 kg2 acceleration of free fall, g = 9.81 m s2 Formulae uniformly accelerated motion, s = ut + 1 2 at2 v2 = u2 + 2as resistors in series, R = R1 + R2 + . . . resistors in parallel, 1/R = 1/R1 + 1/R2 + . . .

3 [Turn over 1 Which estimate is not realistic? A The power of a hair dryer is 150 W. B The kinetic energy of a running man is 2000 J. C The weight of a can of soft drink is 4 N. D The density of ice is 900 kg m─3. L2 Answer: A Option A: Hair drye r uses a voltage of 240 V . I f the power is 150 W, then the current is power 150 0.625 Avoltage 240  , which is a low current. The heating coils will not heat up properly. Option B: Assume the average mass of the running man as 70 kg and running at a speed of 8 m s–1. K.E. of man =     21 70 8 2240 J2   . Option C: The volume of a canned drink is approximately 330 ml ~ 400 g in mass. Weight of a can of soft drink = mg = 0.400  9.81 = 3.9 N ~ 4 N. Option D: Ice is less dense than water, which has a density of 1000 kg m–3. 2 Prefixes are often used to represent powers of 10 when writing the units of quantities. For example, 6.0 microampere, that is, 6.0  10-6 A, is written as 6.0 μA, where μ is the prefix symbol for 10-6 A. Which prefix is not one of the standard symbols? A g B p C m D T L1 Answer: A Option B: p is a prefix that represents pico. Option C: m is a prefix that represents milli. Option D: T is a prefix that represents tera. 3 To find the resistivity of a semi-conductor, a student makes the following measurements of a cylindrical rod of a material. length = (25 ± 1) mm diameter = (5.0 ± 0.1) mm resistance = (68 ± 1) Ω He calculates the resistivity to be (5.34  10-2) Ω m. How should the uncertainty be included in the student’s statement of the resistivity of the semi-conductor? A (5.34 ± 0.07)  10-2 Ω m

4 B (5.34 ± 0.05)  10-2 Ω m C (5.3 ± 0.4)  10-2 Ω m D (5.3 ± 0.5)  10-2 Ω m L2 Answer: D Let R be the resistance, A the area, L the length and d the diameter of the semi-conductor. Resistivity 2 22 4             dR RA Rd L L L 2             R d L R d L 1 0.1 1268 5.0 25        = 0.09471   20.09471 5.34 10     0.005 Ω m 20.5 10 Ω m    25.3 0.5 10 Ωm     4 A ball is released from rest at time zero. After 1.0 s, it hits a horizontal surface and rebounds, reaching the top of its first bounce after 1.5 s. What is the displacement of the ball from its original position after 1.5 s? A 1.25 m B 3.75 m C 5.00 m D 6.25 m L2 Answer: B Downward displacement of ball from the drop point = area under the graph from t = 0 s to t = 1.0 s 1 1.0 10 5.00 m2    velocity / ms1 time / s −5 0 0.5 1.0 1.5 5 10 0

5 [Turn over Upward displacement of ball from the surface when it reaches the top of its first bounce after 1.5 s = area under the graph from t = 1.0 s to t = 1.5 s 1 0.5 5 1.25 m2    Displacement of the ball from its original position after 1.5 s = 5.00 m – 1.25 m = 3.75 m 5 A car travelling at speed u comes to a complete halt after a distance s when the driver applies a uniform braking force. A second car, travelling at speed 2u, comes to a halt when the driver applies a uniform braking force with twice the magnitude to that of the first car. You may ignore the reaction time of the driver. What is the distance, in terms of s for the second car to come to a halt? A 0.5s B 2s C 3s D 4s L2 Answer: B Let v be the final velocity and a be the acceleration of the car. For the first car, 2 2 2 v u as   20 2  u a s 2 2 us a For the second car,     2 20 2 2 2  u a s     2 2 2 2 2 2  u us a a = 2s 6 A projectile is fired at an angle of 40.0 above the horizontal, and leaves the gun with a speed of 150 m s-1. The projectile strikes a wall, which is 500 m away, at a vertical height of h. Neglecting the effects of air resistance, what is the vertical height h? A 96.4 m B 327 m C 420 m D 512 m L2 Answer: B wall 500 m 150 m s-1 40.0 h

6 Let sx be the horizontal displacement, ux the initial horizontal velocity of the projectile and t be the time taken for the projectile to strike the wall. sx = uxt   0500 150 cos 40.0 t   0 500 4.3513576 s 150cos 40.0  t Let sy be the vertical displacement, ax the vertical acceleration of the projectile and t be the time taken for the projectile to strike the wall. Taking upwards as positive,       22 0 y y y 1 1 150sin40.0 4.3513576 9.81 4.35135762 2    s u t a t = 326.68 m sy = h = 327 m (3 s.f.) 7 Which of the following pairs of forces is not an example of Newton’s third law of motion? A the force exerted on a man’s feet by the floor and the weight of the man standing on the floor B the forces of repulsion experienced by each of two parallel wires carrying currents in opposite directions C the forces of attraction between an electron and a proton in a hydrogen atom D the force of repulsion between an atom in the surface of a table and an atom in the surface of a book resting on the table L1 Answer: A By Newton’s third law of motion, the opposite reaction force to the force exerted on a man’s feet by the floor is the force exerted on the floor by the man’s feet. 8 What happens to the apparent weight of an object falling freely in an elevator? A It becomes zero. B It decreases. C It increases. D It remains the same. L1 Answer: A When an object is inside a freely falling elevator, it is in a state of free fall along with the elevator. Both the elevator and the object inside the elevator are accelerating downwards at the same rate. As a result, the object experiences weightlessness, and its apparent weight becomes zero. This is because there is no normal force acting on the object to provide a sensation of weight.

7 [Turn over 9 A football of mass 0.42 kg is travelling towards a player at 3.0 m s-1. The player kicks the football with an impulse of 6.3 N s, returning it in the direction of approach. What is the new speed of the football? A 3.3