EJC Physics 2021 J2 H1 MYE P2 MS

EUNOIA JUNIOR COLLEGE JC2 MID YEAR EXAMINATION 2021 General Certificate of Education Advanced Level Higher 1 PHYSICS MARK SCHEME Section A 8867/02 July 2021 Question Solution Marks 1 (a) Rate of change of velocity (with time) B1 (b) Constant acceleration OR Motion in a straight line B1 (c) (i) =xxs ut ( )=25 50cos20 o t = 0.53ts C1 A1 (ii) = +y yyv u at = + = -1 50sin20 (9.81)(0.53) 22.3 m s o y y v v 22.3tan 50cos20 y o x v vθ = = 25 oθ = C1 C1 A1 (iii) = + 212yy ys ut at = + 21(50sin20 )(0.53) (9.81)(0.53)2 o ys =ys 10.4 m = + 22 xysss = + 2225 10.4s = 27 ms C1 C1 A1

(iv) Accept positive or negative velocities. B1 B1 velocity time 0 A B

Question Solution Marks 2 (a) Total momentum of a system remains constant provided no external resultant force acts on the system. B1 (b) (i) ∆𝑃𝑃 = 𝑃𝑃𝑓𝑓 − 𝑃𝑃𝑖𝑖 = 25 × 103 − 50 × 103 = −25 × 103 kg m s−1 A1 (ii) Initial momentum of Y is 30 × 103 kg m s-1 Momentum of Y increases by 25 × 103 kg m s-1 B1 B1 (iii) 𝑢𝑢𝑋𝑋 = 25 m s−1, 𝑢𝑢𝑌𝑌 = 10 m s−1, 𝑣𝑣𝑋𝑋 = 12.5 m s−1 , 𝑣𝑣𝑌𝑌 = 18.3 m s−1 total kinetic energy before collision = 775 kJ total kinetic energy after collision = 660 kJ Since the total kinetic energy before collision (775 kJ) is not equal to the total kinetic energy after collision (660 kJ), the collision in not elastic. OR relative speed of approach = 15 m s−1 relative speed of separation = 5.83 m s−1 Since relative speed of approach (15 m s−1) is not equal to relative speed of separation (5.83 m s−1), the collision is not elastic. M1 A1 (M1) (A1) (iv) ( ) 22511 (1240) 22 2 3055608 1022 KE mv . .= = = × J C1 momentum / 103 kg m s−1 time / s 0 0.4 20 40 60 0.0 0.8 1.2 1.6 X X

2K K E mv Emv ∆ ∆∆= + 5 20 0823055608 10 1240 22 2 KE . .. ∆ = +× ∆EK = 0.3 × 105 J EK = (310 ± 30) kJ C1 A1 Examiner’s Comments

Question Solution Marks 3 (a) F = ma Driving force – mgsin9.0o = ma Driving force – (2500)(9.81)sin9.0o = (2500)(2.0) Driving force = 8837 N Power = Fv Power = (8837)(8.5) Power = 7.5 × 104 W C1 C1 A1 (b) loss in KE = gain in GPE + work done against friction 2211 ( )( )22mu mv mgh friction s −= + ( ) 2211(2500)(12) (2500)(8.0) (2500)(9.81) sin9.0 (500)( )22 ss −= + s = 23 m C1 A1 Examiner’s Comments

Question Solution Marks 4 (a) gravitational force provides centripetal force GMm / r2 = mrω2 ω = 2π/T Showing algebra clearly to reach answer B1 B1 (b) Any two of the followings: appears to remain above the same point on the Earth has a period of 24 hours. equatorial orbit/orbits (directly) above the equator from west to east B1 B1 (c) (24 × 3600)2 = 4 π r3 ÷ (6.67 × 10-11 × 6.0 × 1024)