NJC 2022 SH1 Promo Paper 2 QnA with Examiner s Comments latest

1 NATIONAL JUNIOR COLLEGE SENIOR HIGH 1 PROMOTIONAL EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER PHYSICS Paper 2 Structured Questions Candidate answers on the Question Paper. No Additional Materials are required. 9749/02 30 September 2020 2 hours READ THE INSTRUCTION FIRST Write your subject class, registration number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answers all questions. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 9 2 / 6 3 / 6 4 / 8 5 / 10 6 / 10 7 / 15 8 / 16 Total (80m) This document consists of xx printed pages and x blank page.

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4 Answer all the questions in the spaces provided. 1 (a) (i) Distinguish between random and systematic error. ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… [3] Suggested Solution: • Systematic errors have same magnitude while random errors have different magnitude. Or Systematic errors are predictable depending on the conditions while random errors are not. • Systematic errors are either consistently larger or smaller than the true value while random errors scatter about a mean value. (refers to sign of the error) • Systematic error can be eliminated by identifying and remo ving the error while random error can only be reduced by averaging. Examiner’s Comments: Many students failed to appreciate that 3 marks are awarded for 3 independent points. Some students tried the give examples to distinguish the two. A handful of students mixed up random and systematic errors (ii) Explain why using a graph a nd drawing line of best fit for the different points will reduce the random error. ……………………………………………………………………………………………… ……………………………………………………………………………………………… [1] Suggested Solution: In drawing the line of best fit, outliers can be eliminated. The line of best fit is drawn to accommodate as much of the data as possible by cutting in between the set of data points. In this way, the data is averaged, with most weighting given to the most similar values. This reduces the effects of random error. Examiner’s Comments: Some students unsuccessfully try to paraphrase the question to answer the question.

5 (b) A ball approaches a ramp with an inclination of 30°. It hits the surface of the ramp at a speed of 5.0 m s −1 at 45° to the normal and bounces off with a speed of 5.0 m s−1 as shown in Fig. 1.1. Fig. 1.1 Determine the change in velocity of the ball. magnitude of change in velocity = ……………………… m s−1 [2] direction of change in velocity is ……………………… ° with respect to the horizontal [1] Suggested Solution: determining the direction of Δv using Δv = vf - vi Since Δv is 90° to the surface of the ramp, Δv is 60° with respect to the horizontal ( )( ) −+= −+= cos9055255Δv 2BCcosACBA 222 222 1sm7.1Δv −=

6 Examiner’s Comments: Most students didn’t show detailed working for either the parts. Many students are able to calculate the change in velocity. Many students got the direction of change in velocity wrong. (c) A physical equation is of the form −= 3 12()r P PQk L where k is a constant, r = (1.55  0.03) mm, P1 = (125  1) kPa, P2 = (100  1) kPa, L = (120  5) m. Determine the fractional uncertainty of Q? fractional uncertainty of Q = ……………………… [2] Suggested Solution: ( ) ( ) ( ) ( ) ( ) ( ) .).1(2.0120 508.055.1 03.033 08.025 2 211 25100125 21 21 21 21 2121 21 21 3 fsL L pp pp r r Q Q pp pp kPapppp kPapp L pprkQ =++    =+− −+= ==− − =+=+=− =−=− −= Examiner’s Comments: Some the students are unable to consider the change in p first, and that that there is a +/- there in the equation. Many students used the max, min, average method. Mostly well executed.

7 [Total: 9] 2 Popcorn is produced in a shallow pan. One piece of popcorn is projected out of the pan from the base of the pan. Fig. 2.1 shows the trajectory of this piece of popcorn. Fig. 2.1 The piece of popcorn reaches a maximum height of 56.0 cm above the base of the pan. You may assume air resistance is negligible. (a) (i) Show that the vertical component uv of the initial velocity of the piece of popcorn is about 3.3 m s-1. [1] Suggested Solution: use 22 2v u as=+ ( ) 220 2 9.81 56 10vu −= +  −   So, 3.3vu = m s-1 OR loss in kinetic energy = gain in gravitational potential energy 21 2 vmu mgh= 22 9.81 56.0 10vu −=    = 3.3 m s-1 Examiner’s Comments: This part is mostly well done.

8 (ii) The piece of popcorn lands at the same level as its starting point in the pan at a horizontal distance of 1.20 m from its starting point. Show that the magnitude of the initial velocity u of the popcorn is about 3.8 m s-1. [3] Suggested Solution: For vertical motion, 21 2s ut at=+ ( ) 210 3.3 9.812tt= +  − t = 0.6728 s Horizontal motion is constant velocity, 1.20 0.6728 hu = = 1.784 m s-1 So, initial velocity u = 221.784 3.3+ = 3.8 m s-1 Examiner’s Comments: A handful of students did not realize that the velocity required is the vector sum of the vector and horizontal components. A number of students cannot appreciate the timing for the different positions like the total time taken is 2 x the time taken for the upwards journey. A few students tried to jump steps and was not awarded the marks.

9 (b) A microphone placed near the pan recorded the sound emitted as the piece of popcorn was produced. Fig. 2.2 shows how the loudness of the popping sound varied with time. Fig. 2.2 Determine the average acceleration of the popcorn as it pops. acceleration = …………………… m s-2 [2] Suggested Solution: Average initial acceleration = 3 3.8 17 10 − = 220 m s-2 Examiner’s Comments: Many students are unable to link the values found in the earlier part to this part. Some students forget to covert time into seconds. [Total: 6]

10 3 (a) When a body is submerged in a fluid, it experiences an upthrust due to the fluid. Explain the origin of upthrust. ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… ……………………………………………………………………………………………………… [2] Suggested Solution: The pressure in the fluid increases with depth. Force by fluid acting upwards on the body is always larger than force acting downwards on the body. (Causing a net upward force on the body) Examiner’s Comments: Answers are not specific. Reference should be made to how pressure varies with height and the force is in the upwards direction! (b) A block of concrete M, of mass 950 kg and density 4750 kg m-3 is held in equilibrium by a vertical cable BE. The block is fully immersed in water of density 1000 kg m-3. Cable BE is attached to a uniform rigid beam AB which is freely hinged to the ground at A and held by another cable CD. The beam makes an angle of 58° with the ground. The angle between cable CD and the beam is 26°. Fig. 3.1 shows the arrangement. A B D