CJC 2024.H2.Phy.PRELIM.P1 - Solns

PHYSICS 9749/01 Paper 1 Multiple Choice Questions 12 September 2024 1 hour Additional Materials: Multiple Choice Answer Sheet READ THESE INSTRUCTIONS FIRST Write your name and class in the spaces at the top of this 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. MARK SCHEME This document consists of 22 printed pages and zero blank page. [Turn over 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 0 = 4 x 10-7 H m-1 permittivity of free space 0 = 8.85 x 10-12 F m-1 (1/(36)) 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 = gh gravitational potential  = - 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 t velocity of particle in s.h.m. v = v0 cos t = 22 0 xx − 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 t 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(-t) decay constant λ = 1 2 ln2 t

4 1 A car is travelling west with a speed of 15 m s-1. A drone moving south flies over the car with a speed of 20 m s-1. At this instant, which arrow represents the velocity of the drone relative to the car? A B C D L2 Answer: A Vdrone rel car = Vdrone – Vcar = Vdrone + (-Vcar) Answers B and D are in the wrong direction. Answer A is closer to 36.9° east-of-south compared to Answer C. 2 A basketball player throws a ball with an initial velocity of 6.5 m s -1 at an angle of 50° to the horizontal. The ball is 2.3 m above the ground when released and passes through the basket on its way down. car 15 m s-1 - Vcar Vdrone Vdrone rel car 36.9° Vcar = 15 m s-1 Vdrone = 20 m s -1 2.3 m 3.0 m 6.5 m s-1 50° basket ground

5 [Turn over What is the time taken for the ball to reach the top of the basket which is 3.0 m above the ground? A 0.17 s B 0.36 s C 0.54 s D 0.85 s L2 Answer: D Take upwards as positive direction. ( ) ( ) 2 2 2 2 1 2 13.0 2.3 6.5sin50 ( 9.81) 2 0.7 4.97929 4.905 4.905 4.97929 0.7 0 0.16858 ( ) 0.84657 0.85 (2 s.f.) y y ys u t a t tt tt tt t s reject or s s =+ − =  + − =− − + = = = Accept the answer where the ball reaches displacement of 0.7 m the 2nd time. 3 An object of mass 20 kg moves along a straight line on a smooth horizontal surface. A force F acts on the object in its direction of motion. A graph of F against time t is shown below. If the velocity of the object at t = 4.0 s is 4.0 m s-1, what is its velocity at t = 6.0 s? A 3.0 m s-1 B 3.3 m s-1 C 4.7 m s-1 D 5.0 m s-1 L2 Answer: C Area underneath the F – t graph gives the change in momentum of the object. Between 4.0 s to 6.0 s, area = 1 2 (6 − 4) ( 2 3 (20)) = 13.333 𝑁𝑠 Let positive direction be the direction of the object’s motion. Since F acts in the direction of motion, it acts in the positive direction, and the change in momentum is also of positive direction, i.e. ∆p = +13.333 Ns. Final momentum - Initial momentum = +13.333 𝑚(𝑣𝑓 − 𝑣𝑖) = +13.333 (20) (𝑣𝑓 − (+4.0)) = +13.333 𝑣𝑓 = +𝟒. 𝟕 𝒎 𝒔−𝟏 F / N 20 0 3.0 4.0 6.0 t / s 0

6 4 Water is ejected at a speed of 0.5 m s -1 onto a wall from the nozzle of a hose with a diameter of 0.01 m. The density of water is 1000 kg m-3. If the water does not rebound, what is the force exerted by the water on the wall? A 5.0 x 10-3 N B 2.0 x 10-2 N C 2.5 x 10-2 N D 7.9 x 10-2 N L2 Answer: B Force on water by wall = Momentum change per unit time for the water = (mass per unit time) x (change in velocity of water) = (density x volume per unit time) x (change in velocity of water) = (density x cross-sectional area x speed) x (final velocity – initial velocity of water) = (1000)(𝜋(0.005)2)(0.5) x (0 - 0.5) = - 0.0196 N = - 2.0 x 10-2 N (2 s.f.) By Newton’s third law, Force on water by wall = - Force on wall by water 5 A particle moving with kinetic energy K undergoes a head-on perfectly inelastic collision with an identical particle that is initially at rest. What is the total kinetic energy of both particles, in terms of K, after the collision? A 0.25 K B 0.5 K C K D 2K L2 Answer: B Let m be the mass of each particle and u be the initial speed of the first particle. K = ½ m u2 In a perfectly inelastic collision, the two particles stick together and move with a common velocity V after the collision. By principle of conservation of momentum, mu + 0 = (m + m) V mu = (2m) V V = ½ u Thus total KE of both particles after the collision = ½ (2m) V2 = m (½ u)2 = ½ (½ m u2) = ½ K = 0.5 K

7 [Turn over 6 One end of a spring is fixed to a support. A mass is attached to the other end of the spring as illustrated below. The variation of mass M with length L is shown in the graph below. What is the energy stored in the spring when it is extended to a length of 35 cm? A 0.00750 J B 0.0315 J C 0.132 J D 0.309 J L2 Answer: C Energy stored in the spring = Work done by the mass on the spring = (Area bounded by the graph and horizontal axis) x (acceleration due to gravity) = 1 2 ( 35−20 100 )( 180 1000)(9.81) = 0.132 J 7 A uniform cube of volume 0.729 m3 is floating in water. The density of water is 1000 kg m-3. A load of 400 N is then placed onto the cube. The cube remains afloat. What is the change in the depth of the cube submerged in the water after the load is added? A 0.0503 m B 0.0559 m C 0.494 m D 0.900 m L2 Answer: A Let the change in depth be d, Cross-sectional area of the cube = ( ) 23 0.729 = 0.81 m2 By the principle of flotation, Weight of added load = Additional weight of the water displaced due to the load’s weight 400 = (0.81d) (1000) (9.81) d = 0.0503 m L L / cm M / g M mass

8 8 A right-angle rule hangs at rest from a peg P as shown below. The rule is uniform i n density and cross-sectional area. One arm is of length L while the other arm is of length 2L. What is the angle θ at which the rule will hang in equilibrium