HCI Sample B Answer Key

HWA CHONG INSTITUTION Sample Paper B Secondary 3 Physics Paper 1 1) D 11) B 21) D 2) A 12) D 22) B 3) A 13) C 23) A 4) D 14) A 24) B 5) C 15) D 25) A 6) C 16) B 26) C 7) A 17) C 27) B 8) A 18) C 28) C 9) C 19) B 29) B 10) D 20) C 30) A

2 Paper 2 SECTION A (40 MARKS) 1) In Oishi Sushi restaurant, after customers had placed their orders via a touch screen menu, shortly after, automated carriers would transport the ordered food from the kitchen to them via high speed tracks. After the customers had removed the food from the carriers, they would press a button located at their booths to send the carriers back to the kitchen. Fig. 1.1 shows two such empty carriers A and B on separate tracks at booths 1 and 2 respectively. Both carriers were at rest and directly below the booth buttons. Image source: http://www.canstockphoto.com/illustration/shinkansen.html Fig. 1.1 Fig. 1.2 represents the motion of carriers A and B over a period of 25 seconds. Fig. 1.2 At t = 0 s, carrier A was at the kitchen whereas carrier B was at booth 2. (a) Describe the resultant force acting on carrier A for its journey from the kitchen to booth 1. [3] t = 0 s to t = 1.5 s: constant positive resultant force [1] t = 1.5 s to t = 7.5 s: zero resultant force [1] t = 7.5 s to t = 9.0 s: constant negative resultant force [1]  to kitchen 1 2 A B high speed tracks buttons Legend: A: B: velocity/m s-1 time/s -0.50 0.50 1.00 0 5.0 10.0 15.0 20.0 -1.00 25.0 30.0

3 (b) Carrier A was transporting a bowl of soup from the kitchen to booth 1. On Fig. 1.3, sketch the soup level if carrier A had (i) moved with constant velocity, [1] (ii) suddenly accelerated from rest. [1] (i) (ii) Fig. 1.3 (c) Both carriers were at rest at booths 1 and 2 during a common time range from t1 s to t2 s. State the values of t1 and t2. [1] From t1 = 9.0 s to t2 = 17.5 s (d) Given carrier B left booth 2 and was directly below carrier A just when it was about to leave booth 1, (i) calculate the distance between the two booth buttons. [2] s = 1/2 [(0.90)(2.5 + 1.0)] [1] = 1.6 m (correct to 2 s.f.) [1] (ii) determine the magnitude and direction of carrier B's initial acceleration. [2] a = (0.90 - 0) / 1.5 = 0.60 m/s2 (correct to 2 s.f.) [1], towards the kitchen/ to the left. [1] or a = - 0.60 m/s2 2) Fig. 2.1 shows a scale diagram of the paths of light rays from an object point P passing through a thin converging lens L of focal point F. Fig. 2.1 Two construction lines: 1 m Two solid rays with arrows: 1 m scale 4.0 cm I F P F L

4 (a) Given the image is 24.0 cm from the lens, on Fig. 2.1, mark out the position of the image and label it I. [1] (b) On Fig. 2.1 , complete the path of the two rays to show how the image I is produced by the lens. Show any construction lines clearly. [2] (c) Explain, in terms of light rays, if image I is a real image or virtual image. [2] Image I is a virtual image. [1] This is because light rays do not meet / converge at the image. [1] [Allow e.c.f.] 3) Fig. 3.1 shows a cycle pump connected to an airtight glass flask and a pressure gauge. Fig. 3.1 (a) State the change in the reading on the pressure gauge when the piston of the cycle pump was pushed in slowly. [1] The reading increases / rises. [1] (b) Using the kinetic model of matter, explain how the pressure change in (a) occurred. [3] When the piston was pushed in, the average separation / distance between air molecules decreases or number of molecules per unit volume increases [1] which causes the frequency of collisions of air molecules with flask's wall to increase [1] Since the force acting on the flask's wall per unit area increases, its pressure increases. [1] (c) Describe another method to bring about the same pressure change in (a) other than pushing in the piston of the cycle pump. [1] Heat up the flask or cycle pump / use a smaller flask glass flask pressure gauge piston cycle pump

5 4) Fig. 4.1 shows a trapezoidal rough wooden block PQRS. A hand exerted a constant force of magnitude f along PQ on a brick of mass 1.2 kg to slide it continuously at constant speed from P to Q. Subsequently, the brick is moved from Q to R and pushed off the edge of side R so that it dropped vertically to the ground as shown. Fig. 4.1 (a) State a pair of action-reaction forces involving the brick as it moves from P to Q. [1] (Normal) force by block on brick and force by brick on block (Normal) force by hand on brick and force by brick on hand (Frictional) force by block on brick and force by brick on block (Gravitational) force by Earth on brick and force by brick on Earth [Any 1] (Type of force need not be explicitly stated.) (b) Calculate (i) the work done against gravity to move the block from P to Q. [2] Work done against gravity = mgh = 1.2 (10)(1.002 - 0.602)0.5 [1] = 1.2 (10)(0.80) = 9.6 J [1] OR Work done against gravity = // component of weight X 1.00 m = 1.2 (10) sin  X 1.00 = 1.2 (10) (0.80/1.00) X 1.00 [1] = 9.6 J [1] P Q R S hand brick wooden block ground 1.00 m 0.40 m 0.10 m 1.00 m  0.25 m Z

6 (ii) the magnitude of constant force f if work done against friction was 2.4 J as it moved from P to Q. [2] Work done by hand = 9.6 J + 2.4 J = 12.0 J [1] = f X 1.00 m f = 12.0 N [1] (c) When the block was at position Z as shown in Fig. 4.1, its velocity was 2.0 m s-1. (i) If the block was falling freely, c alculate the time taken to reach the ground from position Z. [3] Can also solve by conservation of energy: K.E. at Z + G.P.E. at Z = K.E. at ground. v2 = u2 + 2as [u = 2.0 and a = g] = 2.02 + 2(10)(0.80 – 0.10 – 0.25) [1] v = 3.61 m s-1 a = (v – u)/t t = (3.61 – 2.0)/10 [1] = 0.16 s (to 2 s.f.) [1] s = ut + 1/2 a t2 [u = 2.0 and a = g] (0.80 - 0.10 – 0.25) = 2.0 t + 1/2 (10)( t2)[1] By solving quadratic equation, working: [1] t = 0.16 s or t = - 0.56 s (rejected) t = 0.16 s (to 2 s.f.) [1] (ii) Explain, in terms of forces acting on the block, how the time calculated in (c)(i) would differ if air resistance were present as it fell to the ground. [2] If air resistance were present, as air resistance increases with speed, the net force acting on the brick decreases, [1] its magnitude of acceleration decreases, thus the time calculated in (c)(i) would be greater / increase / longer. [1]

7 5) Fig. 5.1 shows the displacement -time graph of a particle of a visible light wave of wavelength . colour wavelength / nm violet 380 – 450 blue 450 – 495 green 495 – 570 yellow 570 – 590 orange 590 – 620 red 620 – 750 (a) (i) Calculate the frequency of this wave. [1] Period = 1.7 X 10-15 s Frequency = 1 / period = 5.88 X 1014 Hz = 5.9 X 1014 Hz (to 2 s.f.) [1] (ii) Given the speed of light in air is 3.0 X 108 m/s, determine , the wavelength of this wave. [2] [E.c.f.: Award 2 m for correct working in (a)(i)] v = f 3.0 X 108 = 5.88 X 1014 () [1]  = 5.1 X 10-7 m (to 2 s.f.) or 510 nm [1] (b) A light ray of wavelength  is incident on a glass prism as shown in Fig. 5.3. Fig. 5.3 displacement time / X 10-15 s 1.0 2.0 3.0 0 Fig. 5.1 Fig. 5.2