HCI S3 Physics EOY 2020 P1&P2Answers

1 Hwa Chong Institution Examinations Secondary 3 Integrated Programme CANDIDATE NAME CLASS REGISTER NUMBER PHYSICS Paper 1 and 2 Answers October 2020 -1 mark for any number of sf/dp mistakes -1 mark for any number of unit mistakes (for the final answer) -1 mark for any number of presentation mistakes (no written formulas) Paper 1 1 A 11 B 21 A 2 C 12 A 22 B 3 C 13 D 23 D 4 C 14 D 24 C 5 D 15 C 25 A 6 B 16 A 26 B 7 B 17 B 27 C 8 B 18 A 28 A 9 A 19 C 29 B 10 A 20 B 30 D Paper 2 SECTION A (40 MARKS) Answer all the questions in this section. 1 Fig. 1.1 and 1.2 show the readings on a vernier caliper with the jaws closed and with a test tube clamped between its jaws respectively. Fig. 1.1 Fig. 1.2 0 1 0 10 5 3 4 0 10 5

2 (a) Write down the zero error of the vernier caliper. -0.03 cm [1] [1] (b) Determine the actual diameter of the test tube. Observed reading: 3.20 cm + 0.03 cm = 3.23 cm [1] Actual reading: 3.23 cm – (-0.03 cm) = 3.26 cm [1] [2] (c) Besides zero error, name a systematic error that could occur when conducting this measurement. Parallax error / poor grip of object / object is grip slanted / expansion of scale / vernier scale is calibrated wrongly. [any point is 1 mark] [1] 2 Diamond is useful for industrial applications and jewellery. The cut of a diamond affects how light gets transmitted into the eye of the observer when light passes through diamond. The depth of its pavilion from its crown governs the path light will travel when it enters the diamond from the outside. A diamond that has a good cut is one that allows most light rays to exit from the top of the crown when light is shone from any angle around it. An incident ray of light is allowed to enter the surface of the diamond as shown in Fig. 2. The refractive index of diamond is 2.4 and its critical angle is approximately 25 . Crown Pavilion Fig. 2 Not drawn to scale

3 (a) Calculate the speed of light in diamond given the speed of light in air is 3.0  108 m s-1. v = c / n v = 3.0 x 108 / 2.4 = 1.3 x 108 m/s [1] [1] (b) Given the angle of incidence that light enters the diamond is 45 , find the angle of incidence, x, that the light ray hits surface AB. sin i / sin r = n  r = 17 [1] x = 180 - 17 - 90 - 20 = 53 [1] [2] (c) Complete the path of the light ray in Fig. 2 to show how it exits the top of the diamond. (Note: the diagram is not drawn to scale) [2] (d) The depth of the pavilion plays a pivotal role in how light gets reflected internally and thus how bright a diamond sparkles. With reference to Fig. 2, explain why a diamond with a shallower pavilion will not shine as brightly as one with a deeper pavilion. A shallower pavilion means when light rays hit one of the faces of the pavilion, the angle of incidence on that face will be smaller than the critical angle [1]. Instead of being totally internally reflected and exiting from the crown, more lightrays will get refracted and exit the diamond from the top of the pavilion [1]. [2] 3 Fig. 3 shows a sharp image being formed on a screen placed in front of a thin converging lens. [1] – Total internal reflection at X and Y. [1] – Refracted ray bends away from normal at Z. Subtract 1 mark if any of the three arrows are not drawn. Accept if light ray (after TIR) hits the crown but i = r.

4 (a) Complete Fig. 3 to show how the ray diagram can be used to determine the position of the object. [3] (b) When a permanent marker is used to shade the upper half of the lens, the image on the screen looks different. Identify the difference and suggest a reason for the difference. The image formed will be less bright / dimmer [1] as half / some of the light energy is blocked / absorbed by the marker ink [1]. [Blur not allowed [2] 4 Fig. 4 shows a 400 N weight supported by two strings, inclined at angles 30  and 50  to the vertical respectively. By drawing a scaled diagram, determine the tensions in both strings. State the scale used. Screen Lens Image F F 2F 2F 400 N 30  50  T1 T2 W Fig. 3 Fig. 4 [1] – Light ray from top of object passing through optical centre and top of image. [1] – Light ray from top of object parallel to principal axis, passing through principal focus (after lens) and at top of image. [1] – Object is between F and 2F. Subtract 1 mark if arrows are not drawn for each light ray (before and after the lens)

5 • correct drawing of resultant force with correct arrows (400 N) [1] • correct drawing of T1 and T2 with correct arrows [1] • Appropriate scale used (at least 1.0 cm rep. 50 N) [1] No mark if use = sign. • magnitude of T1 = 310 N (accept 300 to 320 N) [1] • magnitude of T2 = 200 N (accept 190 to 210 N) [1] [5] 5 Fig. 5.1 shows a ball placed at the top of a slope. A block is fixed rigidly to the lower end of the slope. The ball of mass 0.70 kg, initially at rest, is released from the top of the incline and the velocity of the ball rolling down the slope is found to vary with time as shown in the velocity-time graph as shown in Fig. 5.2. (a) Describe the motion of the ball during the following periods: Fig. 5.1 Fig. 5.2

6 (i) 0A Starts from rest and moving down the slope with constant increasing velocity / constant acceleration [1] (ii) AB (Sharp) constant deceleration as the ball hits the block (from A to v = 0) [1] and thereafter, (sharp) constant acceleration from v = 0 to B [1]. or Award [1] if student mentioned there is a change in direction. 0 mark if student wrote deceleration from A to B. [2] (b) Explain why the speed at B is less than the speed at A. Some energy will be lost as heat and sound [1] as the ball hits the block. [1] (c) Calculate the acceleration of the ball down the inclined slope. Acceleration down the incline = (4.0 – 0) / 1.20 [1] = 3.3 ms-2 [1] [2] (d) Sketch a displacement-time graph of the motion of the ball for t = 0 s to t = 2.20 s. The axes are drawn for you below. No penalty if show small horizontal part at 1.20 s [2] s/m t/s 0 1.20 2.40 [1] – Increasing positive gradient for t = 0 to 1.20 s. [1] – Decreasing negative gradient for t = 1.20 to 2.20 s (approx).

7 6 Table 6 shows a table of data of planetary radius and masses. The gravitational field strength, g, depends on the ma ss of the planet (in kg) and the radius of the planet (in m), given by the relationship: g = ( ) 2 11 planetofradius planetofmass1067.6  − (a) Using the above relationship and the information in Table 6, calculate the weight of a 65.0 kg astronaut on earth. g (Earth) = ( ) 26 2511 1038.6 10597.01067.6   − = 9.78 N/kg [1]