EJC 2023 J1 H2 Promo P1 MS

©EJC 2023 9749/J1H2Promo/2023 EUNOIA JUNIOR COLLEGE JC1 Promotional Examination 2023 General Certificate of Education Advanced Level Higher 2 PHYSICS MARK SCHEME 9749 Sep/Oct 2023 Paper 1 Multiple Choice Question Key Question Key Question Key 1 C 6 A 11 B 2 C 7 C 12 C 3 A 8 B 13 A 4 D 9 C 14 B 5 C 10 D 15 D 16 B 21 B 26 B 17 C 22 D 27 B 18 B 23 A 28 D 19 B 24 B 29 C 20 A 25 D 30 C 1 Typical length = 1 m = 100 cm Typical breadth = 3 cm Typical height = 0.5 cm Hence, estimate volume = 100 × 3 × 0.5 = 150 cm3 2 Option A: Incorrect based on definition of accuracy and precision. Option B: Checking for zero error prevents systematic error. Option C: 5 5 20 20 5 20 5 5 20 20 5 5 vs 20 20 vs = ⇒∆ = ∆ = ⇒∆ = ∆ ∆∆ ∆∆∆∆ = = = = t Tt T t Tt T tt ttTT Tt t T t t 5 20 5 20 Note: 1. The uncertainty for a stopwatch timing is fixed, whether it is timing 5 or 20 osc. Hence 2. Since fractional uncertainty of is smaller w hen the number of osc is larger. ∆= ∆ = ∆ < T t t t. t t, Option D: 1 123 4 123 4 C e ompare 1 vs 4 measurements, take uncertainty of the instrument as Case 1 (1 measurement): Case 2 4 Hence taking more m (4 measure s me ea u n r ts): 4 ∆ ∆= ∆ ∆ +∆ +∆ +∆⇒ + ∆= = ∆ ++=ave ave d d ,d ,d ,d dddd d d d d d d dddd ments to find average has no impact on uncertainty. 3 Using v2 = u2 + 2as 702 = 402 + 2(a)(300) a = 5.5 m s-2

2 ©EJC 2023 9749/J1H2Promo/2023 4 Taking downward as positive: s = ut + ½ at2 5 = 0 + ½ (9.81) t2 t = 1.00964 Consider the horizontal: sx = vx × t = 0.8 × 1.0 = 0.8 m 5 The sandbag is travelling upwards with a velocity of 4.0 m s-1 when it is first released. Hence the displacement of the sandbag will increase in the upwards direction before it eventually starts falling (since acceleration of the sandbag is 9.81 m s -2 downwards). Since s is defined as the displacement of the sandbag from the point of release, s = 0 at the point of release. 6 Method 1: Each mass is a different system ( ) 11 22 12 12 12 12 (1) (2) (1) (2): ( ) ( ) ( ) mg T ma T mg ma m mg m ma m mga mm −= −= + −= + −= + Method 2: Both masses belong to the same system Mass of system= (m1 + m2) Net force on system = (m1g – m2g) Acceleration of the two objects, ( )1 2 12 12 12 = −= + −= + netF ma mg mg m m a mg mga mm 7 Since total momentum before the collision is 0, total momentum after the inelastic collision should also be 0. Total kinetic energy before the collision is 27 J. Since collision is inelastic, total kinetic energy after collision should be less than 27 J. Students who thought KE = 0 assumed incorrectly that it was a perfectly inelastic collision. 8 At equilibrium, let U be the amount of upthrust required to balance the weight W of the veh