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.

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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 bottom of the slope. Take displacement upslope to be positive. [2] Comments: There is acceleration, clearly the graph cannot be straight lines. There were graphs showing “teleportation” as well. Many students also didn’t add the duration up to get about t = 2.3 s. 𝑣 = 𝑢 + 𝑎𝑡 = 0 + 3.8(1.47) = 5.6 m s-1 0 Fig. 1.2 Displacement / m Time / s 2.0 − 2.0 Correct shape B[1] t = 0.83 s OR 2.30 s labelled B[1] 0.83 2.30

6 2 (a) Derive, from the definitions of pressure and density, the equation for pressure due to a column of liquid is given by =p h g [3] Definition: Pressure is given by force per unit area, where the force is acting at right angles to the area. p = 𝐹 𝐴 Definition: Density is the mass per unit volume of the substance.  = m/V Consider a column of liquid of cross-sectional area A and depth h from the surface, in a container. The pressure, p, at the base would be given by the definition of pressure, p = 𝐹 𝐴 where F is the force acting on the base due to the weight of the column of the liquid. (Since the block of liquid is stationary) That is, F = mg But from the definition of density, mliquid = liquid Vliquid, thus p = 𝜌𝑙𝑖𝑞𝑢𝑖𝑑 𝑉𝑙𝑖𝑞𝑢𝑖𝑑 𝑔 𝐴 = 𝜌𝑙𝑖𝑞𝑢𝑖𝑑 (ℎ 𝐴) 𝑔 𝐴 = liquid h g So, the pressure due to a column of liquid is given by 𝑝 = ℎ𝜌𝑔 h: depth from the surface ρ : density of the liquid g: gravitational field strength A h

7 [Turn over (b) Fig. 2.1 shows a simplified catapult used to hurl projectiles a long way. Fig. 2.1 The counterweight is a wooden box full of stones to one end of the uniform beam. The projectile, usually a large rock, is in a sling hanging vertically from other end of the beam. The weight of the sling is negligible. The beam is held horizontal by a rope attached of the frame. The stones and the wooden box in the counterweight have a total mass of 1000 kg, the mass of the beam is 500 kg and the projectile weighs 250 N. Calculate the tension in the rope. Explain your working clearly. tension = ………………………………… N [4] Taking moments about the pivot, B1 Clockwise moments = 1000 x 9.81 x 1.5 = 14715 B1 Anti-Clockwise moments = 250 x 4 + Tsin50(4) + 1.25(500)(9.81) 1000 x 9.81 x 1.5 = 250 x 4 + Tsin50(4) + 1.25(500)(9.81) B1 - 250 x 4+1.25(500)(9.81) T = 2475 N = 2480 N A1

8 3 (a) Two ma sses, 2.5 kg and 4.5 kg , are connected by an inextensible cord that runs over a frictionless and massless pulley as shown in Fig. 5.1. Both masses are released from rest and they moved a distance of 0.50 m. (a) Determine the final speed of the system after travelling 0.50 m. speed = ..................... m s-1 [3] Comments: Some students uses 2.5 kg for the mass for KE. Many also made mistake for the Gain/Loss in GPE. In the above method, the “Loss” in energy is all placed in the LHS, while all the “Gain” in energy is placed on the RHS. Students should stay focus on one metho d and stick to it. There were also students solving this using kinematics correctly. (b) The power of the engine of a sports car of mass 1500 kg is 200 kW in normal driving mode. It can cruise at speed of 90 km h-1. (i) Show that the force that is delivered by the engine to the car is 8000 N. [2] Comments: By now, students should not have issue recognizing the speed were given in km h-1. There were also a handful of students who did not remember the formula. 30 2.5 kg 4.5 kg Fig. 5.1 Loss in GPE by 4.5 kg = Gain in GPE by 2.5 kg + Total Gain in KE [C1 - GPE] [C1 – KE] 4.5 × 9.81 × 0.50 = 2.5 × 9.81 × 0.50 sin30 + ½ (4.5 +2.5) v2 v = 2.13 m s−1 [A1] 90 km h-1 = 25 m s-1 [B1] P = F v 200 000 = F (25) [M1] F = 8000 N [A0]

9 [Turn over (ii) When the car suddenly switches to turbocharging mode, the power doubles. Assuming the resistive force remains the same at that instant, determine the acceleration. acceleration = .................... m s-2 [2] Comments: Many students didn’t calculate the resultant, thus only got half of the marks. 4 A circus performer is riding his motorcycle with uniform speed such that its period is 20.0 s in a horizontal circle of radius 83 m on the inner surface of a cylindrical wall, as shown in Fig. 4.1. The orientation of his motorcycle is shown in Fig. 4.2. The two forces acting on the motorcycle -man system are the reaction force R acting at an angle  with the vertical and the weight W. (a) R is the resultant of two forces. They are perpendicular to each othe