NJC 2021 SH1 Promo P2 V3 (Solutions) w markers comments

[Turn over 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 29 September 2021 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 / 6 2 / 7 3 / 7 4 / 7 5 / 13 6 / 10 7 / 10 8 / 20 Total (80m) This document contains 23 printed pages and 01 blank page.

2

3 [Turn over

4 1 A small wooden block is held stationary on a rough slope at a distance of 2.0 m from the bottom of the slope. At time t = 0, it is projected up the slope with an initial speed of 5.0 m s-1 as shown in Fig. 1.1. The deceleration of the block is 6.0 m s-2. (a) Show that the distance travelled by the block, relative to its initial position at t = 0, before it comes to rest momentarily at the top of the slope is 2.1 m. [1] Comments: Most students got this part correct. (b) Determine the time taken for the block to travel to the top of the slope. time = ...................... s [1] Comments: Most students got this part correct. (c) After reaching the top of the slope, the block starts to slide down with an acceleration of 3.8 m s-2. Determine (i) the time taken for the block to travel from the top of the slope to the bottom of the slope. t = ........................... s [1] Comments: Most students got this part correct. Those who got it wrong, mostly substitute s = 2.1 m instead of 4.1 m as shown in the solution above. Other methods to solve are accepted. Fig. 1.1 𝑣2 = 𝑢2 + 2𝑎𝑠 0 = 5.02 + 2(−6.0)𝑠 𝑠 = 25 12 = 2.08 ≈ 2.1 m (shown) 𝑣 = 𝑢 + 𝑎𝑡  0 = 5.0 − 6.0𝑡 𝑡 = 5.0 6.0 = 0.83 s 𝑠 = 𝑢𝑡 + 1 2 𝑎𝑡2 4.1 = 1 2 (3.8)𝑡2 𝑡 = 1.47 s

5 [Turn over (ii) the speed of the block on reaching the bottom of the slope. speed = ................. m s-1 [1] Comments: Most students got this part correct. There were error-carried-forward marks awarded from (c)(i). (d) On Fig. 1.2, sketch the variation with time, of the displacement of the block (relative to the start point) for the entire motion starting from t = 0 to the instant the block reaches the bo