EJC Physics 2020 J2 H1 JCT P2 MS

©EJC 2020 9749/J2H2JCT/2020 EUNOIA JUNIOR COLLEGE JC2 June Common Test 2020 General Certificate of Education Advanced Level Higher 1 PHYSICS MARK SCHEME 8867 June/July 2020 Paper 1 Longer Structured Questions Qns Answer Marks 1(a) Consider vertical motion, taking downwards as positive ( )( ) ( ) 22 vertical 11 2 9 81 8 2 2 40 m s 39 618 m 0 s v us v as . . a −− = + = = = direction: 30.6° below the horizontal to the left OR 69.4° from the normal/vertical to the left speed: ( )( ) 22 2 horizontal vertical 22 horizontal vertical 2 1 67 2 9 81 80 77 8 m s vv vv v . . v − = = = + = + + (accept method by PCE) C1 (vertical velocity value) A1 (direction) A1 1(b) both start from parachutist horizontally, Q same shape but less range: A1 vhorizontal = 67 m s-1 vvertical = 39.618 m s-1 θ P Q

2 ©EJC 2020 9749/J2H2JCT/2020 Qns Answer Marks 1(c)(i) air resistive force is equal and opposite to weight of parachutist [accept if free body diagram is clear, labelled and correct] [0 if velocity appears as a vector on free body diagram] A1 (magnitude and direction 1(c)(ii) loss in GPE is w.d. against force due to air resistance converted into heat KE remains same    B1 1(c)(iii) ( ) ( ) ( )( ) ( ) 2final initial avg final initial avg By N2L, 0 or 28 m s 82 0 7 0 25 2296 57 N 4 accept 2300 N dpF dt pppFa tt t mv v t F . − = −∆− ≈= = =∆∆ ∆  −= ∆ −= = C1 (terminal momentum) A1 2(a) resultant force in any direction is zero resultant moment about any point is zero B1 B1 2(b)(i) (by Principle of Moments) take moment about hinge sum of clockwise moments = sum of anticlockwise moments    ( ) ( )( ) ( )( ) A A 15 700 sin 40 700 sin 40 15 5 5 150 N (shown) T T ° ° = = = M1 M1 A0

3 ©EJC 2020 8867/J2H1JCT/2020 Qns Answer Marks 2(b)(ii) system is in equilibrium so resultant force in any direction is zero horizontally: ( ) ( ) ( ) [ ] A A 700 sin 40 700 cos 40 700 cos 40 N to the ri 150 300 ght x x TF TF = = = °= + °− °− vertically: ( ) ( ) [ ] 700 cos 40 700 cos 40 253 2000 N vertically upwards6 y yF WF°+ = °+= = 222 22 1 2550 N tan tan 83 3 to the horizontal xy xy y x y x FF F F F F F . F F F θ θ − = = +  +  =  =   =    = °  C1 (300N, right) or (2536N, up) A1 (2550 N) A1 (83.3°) 2(c) Let T be new tension along half the wire 2 sin 1 5 10 2sin 1 10 191 5 N T. .T °= = ° = C1 (eqn) A1 θ Fy Fx F

4 ©EJC 2020 9749/J2H2JCT/2020 Qns Answer Marks 3(a)(i) same direction as Earth’s rotation about its own axis so satellite begins launch with some speed in correct direction satellite already has some kinetic energy so less fuel needed to raise the gravitational potential energy M1 A1 (energy) 3(a)(ii) gra