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) gravitational force provides centripetal force ( )( ) ( ) 2 2 32 22 11 24 2 77 3 7 3 3 3 E 6 67 10 6 0 10 60 60 10 m 10 m 10 35 2 2 24 4 23 4 2298 640 9 10 m 0 GMm mrr GM r GM GM . r T .. . rR . hr ω ω ω π π − = =  =  = = ×× ×× ×× −× = ×  = = = − M1 C1 (r value) A0 3(b)(i) = +   =         = = + = +− +− +        = −− − total K P 2 2 2 11 22 1 2 2 2 E r EE GMm mv GMmmv rrr GMm GMm GMm GMmr rr G r Mm r r M1 (KE) A1 (Summation) 3(b)(ii) ( )( ) ( ) ( ) − − − ×× = × = −× 11 24 2 9 2 6 67 10 6 0 10 1000 2 4 2298 10 4 73 10 J GMm r .. . . A1 3(b)(iii) w.d. against drag so total energy decreases GPE decrease so satellite lowers in height KE increase so linear speed increase    satellite spirals towards Earth with increasing speed Note: see topical compilation (gravitational field) page 13 to understand how the change in radius affects the various energies B1 M1 A1 4(a) magnetic force on particle is always normal to direction of motion so path is arc of a circle in magnetic field after leaving field, Newton’s first law, no net force on particle so path is straight line after leaving field B1 B1

5 ©EJC 2020 8867/J2H1JCT/2020 Qns Answer Marks 4(b)(i) momentum/speed decreasing so the radius of circular motion ( Bq mv r/= ) is becoming smaller B1 B1 4(b)(ii)1. Spirals are in opposite directions so oppositely charged B1 4(b)(ii)2. equal initial radius so equal initial speeds B1 4(c) F = BlLsinθ F = 0.4 x 6 x 0.14sin60° F = 0.29 N C1 A1 5(a) region of space in which a force acts on a stationary charge B1 5(b) B1 arrow directions B1 normal to surfaces B1 symmetry above and below sphere 5(c)(i) as electron moves right towards -50V plate, kinetic energy decreases as electric potential energy increases at closest distance of approach from -50 V plate, kinetic energy is minimum while electric potential energy is maximum as electron moves left towards + 50 V plate, kinetic energy increases as electric potential energy decreases B1 B1 B1 metal sphere + + + + +

6 ©EJC 2020 9749/J2H2JCT/2020 5(c)(ii) E field strength is uniform between charged parallel plates so acceleration a is constant ( ) e horizontal horizontal ee 19 15 2 31 3 1 6 10 8 7816 10 50 50 9 11 m s10 0 102 F qE m a a mm . q qVE x . .. − − −− ∆= ∆  ×= = × = = = −− ××  at closest distance, horizontal 0v = ( ) ( ) ( ) ( )( ) ( ) 222 horizontal fr centre 2 33 fr centre horizontal 26 3 19 31 3 4 2 sin45 2 sin451 0 10 1 0 10 2 5 6 10 sin 45 1 50 50 9 11 0 10 1 6 102 10 1 20 0 1 m 110 u as v a vx. . a . . . vx x .. . −− − − −− − = + = °+ °= ×× ×° ×  ×  ××  −= − = − − = × − OR By conserving energy, consider change in potential from centre of plates, loss in KE = gain in EPE ( ) ( ) ( ) ( ) ( )( ) ( ) 2 closest 2 closest 231 6 19 2 1 sin2 sin 9 11 10 5 6 10 sin 45 44 162 64 10 mv V mvV .. . q q . θ θ − − ∆ ∆=  ×× °= = = ×   E field strength is uniform between charged parallel plates: ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) closest 3 fr centre 3 closest fr centre 2 3 231 6 3 19 4 10 10 10 sin 9 11 1 50 50 20 20 100 20 2 100 20 100 0 5 6 10 sin 45 10 1 6 10 8 928 m 2 10 VV xx V x mv .. . E . . . . q . θ − − − − − − − ∆∆ = =∆× ∆× =  ×=   ×× ° × = ×  = −= × − ( ) 3 fr centre 4 10 10 m 10 11 x. x . − − = − = × × M0 (kinematics) C1 (accel value) C1 (substitute into 22 2 0u as= + ) A1 (need units) M0 (PCE) C1 ( V∆ value) C1 (x from centre) A1 (need units)

7 ©EJC 2020 8867/J2H1JCT/2020 Qns Answer Marks 5(c)(iii) The change in EPE is due to changing KE affected by acceleration horizontally. B1 (parabola) B1 (y labels) B1 (x = 3.57) B1 (x = 4.39) 6(a)(i) E OR droplet is in equilibrium so vector sum of downward acting weight and upward electric force is zero V W F QE VWQ Q dd V +− = = =      −∆     B1 B1 0 Ep / eV d / mm 44.5 -50 3.57 4.39

8 ©EJC 2020 9749/J2H2JCT/2020 Qns Answer Marks 6(a)(ii) upthrust due to air negligible compared to gravitational force/weight because density of air negligible to density of oil at same temperature    B1 6(b) ΔV T W / 10-14 N Q / 10-19 C N 770 11.2 2.9 1.66 1 230 10.0 3.4 6.53 4 1030 9.4 3.7 1.59 1 470 7.6 5.2 4.89 3 820 6.9 5.9 3.18 2 395 6.2 7.0 7.83 5 B1 (both W’s) B1 (both Q’s) B2 (each N) 6(c) ( ) ( ) 19 19 19 1 66 6 53 1 59 4 89 3 18 7 83 10 1413 25 1 61 10 C accept 1 605 10 C Q N ...... . e . − −− +++++ ×= +++++ = ×× = ∑ ∑ M1 (substitution) A1 6(d)(i) ( ) 14 14 4 8 10 lg 4 8 10 8 13 319 W. .. T, − − = × ×= − = (do not accept -13.3 be