EJC 2022 J1 H2 PROMO MS (P1 &P2)

©EJC 2022 9749/J1H2MYE/2022 EUNOIA JUNIOR COLLEGE JC1 Promo Examination 2022 9749 PHYSICS MARK SCHEME Paper 1 – Multiple Choice Questions 1 Answer: D (Left hand side): v has units of m s-1. (Right hand side): -2 -1 -2 22 -3 -1 -2 2 -2 -3 -1 units of N kg m sunits of = = = = kg m sunits of m m units of = kg m units of kg m s= = m sunits of kg m units of = m sunits of Fp A d p d p d Units on LHS = Units on RHS Therefore, γ has no units. 2 Answer: A 2 2 2 2 2 2 -1 =- = +(- ) = + - 2( )( )cos150° = 50 +80 - 2(50)(80)cos150° =126 km h 80 126=sin sin150° θ =18.5° LC L c Lc LC L C L C LC LC v v v vv v v v v v v v θ Bearing = 360°-18.5°-60° = 281.5° 𝜃 60° -Vc = 80 km h -1 150° VLC VL = 50 km h-1 N

2 ©EJC 2022 9749/J1H2MYE/2022 3 Answer: A Student A Student B Student C Student D Accuracy 0.85 0.85 0.8625 0.85 Precision (range) 0.17 0.06 0.17 0.04 4 Answer: C Minimum distance occurs when both cars have the same speed. ( ) ( ) ( ) ( )     1Car X : × 3.0+5.0 ×20 + 55×5 = 355 m2 1Car Y : 35×2.0 + × 35+55 ×3.0 = 205 m2 Minimum distance = (450+ 205)- 355 = 300 m 5 Answer: B Can be obtained by looking at the gradient of the s-t graph. 6 Answer: C 2 2 1 2 1 2 s ut at s at =+ = 2s at 2 2 2 2 31 66 3 m m m E EE m E m s a t s at t t tt = = = 7 Answer: A Action and reaction forces should act on opposite bodies. For Option A, the correct statement should look like “the gravitational forces of attraction between satellite and Earth” 8 Answer: B By Newton’s Second Law, rate of change of momentum is the resultant force acting on the body. In this case, that would be 4N.

3 ©EJC 2022 9749/J1H2MYE/2022 9 Answer: C Fnet = m/t vperpendicular = m/t (2v sin30) = 2.5 × (2 × 5.0sin30) = 13 N Option A: Forgot factor of 2 Option B: Forgot factor of 2 and considered cosine instead of sine Option C: Correct Option D: Considered consine instead of sine 10 Answer: B For Option A, on both arms at level where the two liquids intersect, pressures are the same. Since pressure decrease (by hg) as we progress upwards, pressure at Pt Q would be diff from pressure at Pt P since decrease in pressure is hg (same h but diff ). For Option B, Pt R and Pt S are at the same pressure. Consider the pressure at bottom - most of tube, the same decrease in liquid pressure by hg (same h,) For Option C, since the surface level of both arms experience the same pressure (atmospheric) and both liquids are of different densities, the surface levels cannot be changed. For Option D, since the surface level of both arms experience the same pressure (atmospheric) and both liquids are of different densities, the surface levels cannot be aligned. 11 Answer: C Given that there are 3 forces acting on the beam to bring about equilibrium, the lines of action of these