υ = (b) area under graph = 0.80±0.05 (1) (kgms 1 )

Transcription

1 1. (a) (i) points plotted correctly (deduct one for each incorrect) sensible scales chosen line of best fit (ii) change in momentum [or impulse] (accept 0.8) max 4 (b) area under graph = 0.80±0.05 (kgms 1 ) υ = mυ = 1.6 ms 1 m alternative: state average force = 0.10(N) leading to correct derivation of 1.6ms 1 (c) (i) mυ = 0 [or statement] υ = 0.40ms 1 (ii) kinetic energy = 0.16 J (iii) initial kinetic energy = 0.64 (J) kinetic energy lost so inelastic 5 [11]. (a) (i) Q = C (ii) E = J (b) (i) V = 50 V (ii) (E 1 = 1 QV = J) E =.5 10 J (iii) current flows (when capacitors connected together) (energy lost due to) heat in wires 4 [6] James Allen's Girls' School 1

2 3. (a) (uniformly) curved path continuous with linear paths at entry and exit points arrow marked F towards top left-hand corner (b) into (the plane of) the diagram (not accept downwards ) 1 (c) F(= BQυ) = = N (d) B must be in opposite direction 1 (much) smaller magnitude 000 [7] 4. (a) (i) force per unit positive charge [force on a unit charge only] vector (ii) + + N overall correct symmetrical shape outward directions of lines spacing of lines on appropriate diagram neutral point, N, shown midway between charges max 6 James Allen's Girls' School

3 (b) (i) E AP Q = 4πε 0r = 150 V m = 1 4π (0.1) 3 10 (ii) E PB = 1 4π (0.16) 9 = 1050Vm E (iii) θ P 1050 allow e.c.f. from wrong numbers in (i) and (ii) E = Vm 1 θ = tan = 50.0 to line PB and in correct direction max (c) (i) potential due to A is positive, potential due to B is negative at X sum of potentials is zero (ii) = 0 4πε 0 ( x) 4πε 0 (0.0 x) gives AX (= x) = 0.080m (only from satisfactory use of potentials) 4 [16] 5. (a) attractive force between two particles (or point masses) proportional to product of masses and inversely proportional to square of separation [or distance] GMm (b) (for mass, m, at Earth s surface) mg = R rearrangement gives result James Allen's Girls' School 3

4 (c) M gr moon 6 = = 1.6 ( ) G = kg M moon = 4 M (= 0.013) 1.3% 3 earth [7] 6. (a) (i) ε r = 6.3 ε 0 = (Fm 1 ) εa C = d = (F) Q = CV = C (ii) ρ = (Ωm) ρ l R = = Ω 6 A (b) I = Q t = = A (c) (as potential difference between plates increases) electric field strength inside dielectric increases limit when breakdown occurs rapid discharge of capacitor 3 [11] 7. (a) (i) Q = 0.4(3)C (ii) E = 19 J (iii) I = 14A 3 (b) E = 1 C (90 80 ) [or E 80 = 15(J)] leading to 4.0 J [5] 8. (a) (i) equation showing momentum before = momentum after correct use of sign James Allen's Girls' School 4

5 (ii) no external forces (on any system of colliding bodies) 3 (b) (i) (by conservation of momentum m 1 υ 1 + m υ = 0) 0.5. = ( )0.50υ υ = ( )1.1(0)ms l (ii) = total k.e. = = 0.91J 4 (c) (i) mass of air per second = ρaυ correct justification, incl ref to time (ii) momentum per second (= Mυ = υ Aρ) = υ Aρ (iii) force = rate of change of momentum (hence given result) upward force on helicopter equals (from Newton third law) downward force on air 5 (d) υ Aρ = mg (for 50% support) υ = gives υ = 7.ms 1 (or 7.3, g taken as 10) if not 50% of weight, max 1/3 provided all correct otherwise (gives 10.) 3 [15] 9. (i) kinetic energy = mgh = 0.37 J (ii) υ = E =. ms 1 m (iii) F c =.9 N [or 3.0 N if g = 10 used] James Allen's Girls' School 5

6 (iv) T = F c + W = 4.4 N [6] 10. (a) (i) remains constant since connected to constant p.d. (ii) decreases because C d 1 (iii) decreases because Q = CV and C has decreased (iv) decreases because E = 1 CV and C has decreased 4 ε (b) (i) C = 0 A = d (= F) 1 E (= CV 1 ) = ( ) = 1890J 8 3 (ii) I Q Q = = use of Q = CV use of I = t t = 157 A 6 [10] 11. (a) (i) V V g = 19 = ( ) gives V = 190 J kg 1 x 10 (ii) W(= m V) = = 1710J [or mgh = = 1710J] (iii) on mountain, required energy would be less because gravitational field strength is less max 3 (b) 1 1 GMm g (or F or correct use of F = ) r r r 19 g = = 4.75(Nkg 1 ) [5] µ 0 1. (i) B = lni = 7 4π = T James Allen's Girls' School 6

7 (ii) Φ(= BA) = = Wb [3] 13. A 14. D 15. B 16. (a) (i) equation showing momentum before = momentum after correct use of sign (ii) no external forces (on any system of colliding bodies) 3 (b) (i) (by conservation of momentum m 1 υ 1 + m υ = 0) 0.5. = ( )0.50υ υ = ( )1.1(0)ms 1 (ii) allow e.c.f from (i) min. stored energy = total k.e. = = 0.91J 4 (c) (i) mass of air per second = ρaυ correct justification, incl ref to time (ii) momentum per second (= Mυ = υ Aρ) = υ Aρ James Allen's Girls' School 7

8 (iii) force = rate of change of momentum (hence given result) upward force on helicopter equals (from Newton third law) downward force on air 5 (d) υ mg Aρ = (for 50% support) υ = gives υ = 7.ms 1 (or 7.3, g taken as 10) if not 50% of weight, max 1/3 provided all correct otherwise (gives 10.) 3 [15] 17. (a) (i) free: system displaced and left to oscillate (ii) forced: oscillation due to (external) periodic driving force [or oscillation at the frequency of another vibrating system] (b) (i) k = = Nm 1 (ii) T = π m 9000 = π 4 k g giving 0.78 s 3 s 16 (c) (i) t = = = 0.80 s υ 0 (ii) time period of free oscillations, resonance i.e. large amplitude oscillations 3 [8] 18. (i) 1000 km hr 1 = ms flux cut per second = B area swept out per second or = 0.5Wb 3 James Allen's Girls' School 8

9 (ii) induced e.m.f. equals flux cut per second [or equation and symbols defined] E = 0.5V (iii) direction of p.d. reversed [6] 19. (a) Q E V capacitance [or charge per volt or Q/V] 3 V (b) (i) Q = CV (= )=4.1C 1 1 (ii) E = QV = =1J [5] 0. (i) 3000 f = = 50 (Hz) 60 ω (= πf) = 314 (rad s 1 ) (ii) α = (rω )= = ms (iii) (inwards) towards axis of rotation 5 [5] ð1. (a) vibrations are forced when periodic force is applied frequency determined by frequency of driving force resonance when frequency of applied force = natural frequency when vibrations of large amplitude produced [or maximum energy transferred at resonance] max 3 (b) (i) damping when force opposes motion [or damping removes energy] James Allen's Girls' School 9

10 (ii) damping reduces sharpness of resonance [or reduces amplitude at resonant frequency] [5]. (a) N kg 1 electric field strength gravitational constant N C 1 or V m 1 N m kg mass kg charge C distance (from distance mass to point) (from m charge to m point) 4 (b) (i) none 1 1 both F E and F G r (hence both reduced to 4 [ affected equally] (ii) charge on B must be doubled 3 [7] 3. (a) arrow towards left 1 (b) I E.0 = = = 5 (A) R F(= BIl) = = 0.04 N [3] James Allen's Girls' School 10

11 I F 4. (a) S f (i) F is straight line through 0 (ii) S is straight line through 0 with smaller (half) gradient (iii) I = V = ( 1 πfc) X c V = πfcv gradient gradient is πcv and C = 5 πv (b) X C = V I r.m.s. r.m.s. = Ω [7] 5. (a) (i) k = T = π 0.5 = 0.70 s time/s (ii) 0.70s a correct (= 0mm) x = ±0 mm at t = 0 T correct (= 0.70 s) 5 James Allen's Girls' School 11

12 (b) (i) vibrates at 0.5 Hz with low amplitude (ii) vibrates with high amplitude at natural frequency resonates max 3 [8] 6. (a) (i) parallel (near centre), perpendicular to and touching plates arrows away from positive plate E V 1500 = = d 0.00 = V m 1 [or N C 1 ] V (ii) straight line from origin 5 d (b) (i) F(=Ee) = = (N) 4 α F = =.5 10 m = 0.45 (m s ) 3 t = s = (= 0.0s) a 0.45 ball towards positive plate (ii) on contact, acquires same charge as plate hence repelled away [or attracted to other plate] at other plate, charge change again, process repeats max 6 (c) no (resultant) force charge must be moving for magnetic force [or weight balanced by tension] [13] 7. (a) (i) 31 MJ kg 1 James Allen's Girls' School 1

13 (ii) increase in potential energy = m V = 100 (6 1) 10 6 = J 4 V (b) (i) g = x (ii) g is the gradient of the graph = =.44 N kg 1 6 (iii) g 1 R expect g to be and R is doubled =.45 N kg 1 [alternative (iii) 1 g and R is halved R expect g to be.44 4 = 9.76 N kg 1 ] 5 [9] 8. (a) output voltage proportional to current in circuit (*) resistance of resistor is independent of frequency (*) reactance of capacitor falls as frequency rises (*) at low frequencies, little current so output voltage small (*) at high frequencies, max current flows so output voltage high (*) limited only by resistance (*) (*) 3 (b) V R = I out r.m.s. = = 10 kω 1 James Allen's Girls' School 13

14 (c) V =.5 V from graph I =.5 = 0.5 ma V 5.0 Z = = = 0 kω 4 I r.m.s Z = R + X 4 1 = ( 10 ) + = π 900 C (gives C = F) max 3 Vout V (d) P = P so = R R 5 = 3.54 V which corresponds to a frequency of 1.6kHz (± 0. khz) max [9] 9. B 30. C 31. C 3. (a) (i) parallel (near centre), perpendicular to and touching plates arrows away from positive plate E V 1500 = = d 0.00 = V m 1 [or N C 1 ] James Allen's Girls' School 14

15 V (ii) straight line from origin 5 d (b) (i) F(= Ee) = = (N) 4 a F = =.5 10 m = 0.45(m s ) 3 t = s = (= 0.0 s) a 0.45 ball towards positive plate (ii) on contact, acquires same charge as plate hence repelled away [or attracted to other plate] at other plate, charge change again, process repeats not SHM, valid explanation 8 (c) no (resultant) force charge must be moving for magnetic force [or weight balanced by tension] [15] 33. (a) (i) interaction between current and B-field gives force on wire equal and opposite force on magnet (down) (ii) force on wire must be up current right to left by left hand rule (iii) (force = BIl = mg = change in mass 9.8) B = B = T [50.3 mt] max 6 James Allen's Girls' School 15

16 balance reading (b) I straight line intercept, upward slope [8] 34. (a) momentum before collision = momentum after collision provided no external force acts (b) (i) p = mυ =.(0) kg ms 1 (Ns) (ii) total mass after collision = 0.40 kg 0.40 υ =.0 gives υ = 5.(0) ms 1 (allow e.c.f. from (i)) 4 (c) (i) kinetic energy = 1 mv = (= 00 J) (ii) kinetic energy = (= 5.0 J) (iii) Q = 00 5 = 195 (J) = mc θ θ = = 78 K (allow e.c.f. for incorrect Q) 5 (d) kinetic energy lost (= potential energy gained) = mgh h = m [13] James Allen's Girls' School 16

17 35. (a) kinetic energy is not conserved 1 (b) (i) (p = mv gives) p = =. N s (.16 N s) (ii) p = 0.1 ( 15) = 1.8 N s (iii) p =. ( 1.8) = 4.0 N s (3.96 N s) (allow e.c.f. from(i) and(ii)) (iv) ( mv) (F = t gives) F = = 8 N (8.3 N) (allow e.c.f from(iii)) (v) (E k = ½mv gives) E k = (18 15 ) = 5.9 J 6 [7] 36. (a) 360 N 1 (b) (i) (E p = mgh gives) E p = = J (ii) T cos 0 = 360(N) T = 380 N (allow e.c.f from(a)) 3 (c) (potential energy) changes centre of mass/gravity moves upwards QWC [6] 37. D 38. C 39. B 40. A 41. D James Allen's Girls' School 17

18 4. C 43. B 44. B 45. A (a) use of mg = ke gives k = = 56 N m 1 [or kg s ] (b) (i) 8 (N m 1 ) (unit to be given in either (a) or (b)) (allow C.E. from (a)) (ii) m 0.50 (use of T = π gives) T =π = 0.84 (s) k 8 (allow C.E. for value of k from (b)(i)) 60 number of oscillations per minute = = (allow C.E. from (b)(ii)) 3 [5] 47. (a) graph to show: straight line from origin end point at 4.5 (V), 9.0 (µf) James Allen's Girls' School 18

19 (b) (i) W = V Q explained energy stored or total work done in charging = area under graph or charge average voltage energy stored = work done (= ½QV) (ii) Q = = 3.0 (µc) E (=½ QV) = ½ = J [or E = (½CV = ½ = J] 5 [7] 48. (a) (i) (force) to the right (ii) electrons accelerate or speed increases (b) (i) sketch to show path curving upwards in the field (must not become vertical) (ii) horizontal component of velocity is unchanged vertical or upwards acceleration (or force) parabolic path described (or named) max 3 QWC [5] 1.5 = J [or E = (½CV = ½ = J] 5 [7] 49. (a) (i) length of card [or distance travelled by trolley A] time at which first light gate is obscured [or time taken to travel the distance] (ii) time at which second light gate is obscured [or distance travelled after collision and time taken] 3 James Allen's Girls' School 19

20 (b) momentum = mass velocity mass of each trolley (check whether) p initial = p final max (c) incline the ramps until component of weight balances friction [or identify where the friction occurs sensible method of reducing ] [7] 50. B 51. B 5. D 53. B 54. D 55. A 56. C James Allen's Girls' School 0

21 57. C 58. D 59. A 60. (a) forced vibrations or resonance 1 (b) reference to natural frequency (or frequencies) of structure driving force is at same frequency as natural frequency of structure resonance large amplitude vibrations produced or large energy transfer to structure could cause damage to structure [or bridge to fail] max 4 (c) stiffen the structure (by reinforcement) install dampers or shock absorbers [or other acceptable measure e.g. redesign to change natural frequency or increase mass of bridge or restrict number of pedestrians] [7] 61. (a) Q = CV (= ) = C or 8 µc (b) E = ½CV = ½ = J [or E = ½QV = ½ = J ] 3 James Allen's Girls' School 1

22 (c) time constant is time taken for V to fall to V o e V must fall to. V time constant = 3 ms [or draw tangent at t = 0 intercept of tangent on t axis is time constant accept value ms ] [or V = V 0 exp( t/rc) or Q = Q 0 exp( t/rc) correct substitution time constant = 3 ms ] 3 (d) time constant = RC R = = 6800 Ω (allow C.E. for value of time constant from (c)) [10] 6. (a) θ = 90 (or 70 or π or 3π ) 1 (b) Φ = BA cosθ = cos 30 = Wb (c) Φ max = (Wb) (= ) flux linkage = = 1.1(4) 10 3 (Wb turns) [5] 63. (a) 1 83 Bi α Tl either (for both atomic mass numbers, 4 and 08) and (for both atomic numbers, and 81) 08 [or for 81 Tl and incorrect α] (b) (i) E k = (½mv ) = (J) substitution for m = (kg) v = / (= m s 1 ) James Allen's Girls' School

23 (ii) correct use of conservation of momentum m Tl v recoil = m α v substitution of m Tl = 08u (allow C.E. for mass = 08) 7 v recoil = = m s 1 08 (allow C.E. for value of v) 6 [8] 64. (a) (i) uud (ii) u d (b) (i) mv = Bev [or mv r = ] r Be m = mv r = = 19 Be = 0.98 m (ii) pion path more curved than proton path (iii) path more curved [or radius (of path) smaller] for both paths 7 [9] 65. (a) gravity or force acts towards centre force acts at right angles to velocity or direction of motion [or velocity is tangential] no movement in direction of force no work done so no change of kinetic energy so no change in speed 3 (b) (i) B = ( ) ½ = 59 µt James Allen's Girls' School 3

24 17 (ii) tanθ = 56 θ = 17 (± 1 ) (iii) rod sweeps out or cuts (magnetic) flux [or rod cuts field] 4 [7] 66. (a) kinetic energy changes to potential energy potential energy calculated by measuring h equate kinetic energy to potential energy to find speed [or use h to find s use g sinθ for a use v = u + as ] [or use h to find s time to travel s and calculate v av v = v av ] 3 (b) (i) p(= mv) = 0.5(0) 0.4(0) = 0.(0) N s(or kg m s 1 ) (ii) (use of m p v p = m t v t gives) 0.00(0) v = 0.(0) v = 100 m s 1 4 (c) (i) kinetic energy is not conserved (ii) initial kinetic energy = = 10 (J) final kinetic energy = = (J) hence change in kinetic energy (allow C.E. for value of v from (b)) 4 [11] 67. (a) displacement is a vector ball travels in opposite directions max 1 (b) velocity is rate of change of displacement average speed is rate of change of distance velocity is a vector [or speed is a scalar) velocity changes direction any two James Allen's Girls' School 4

25 ( ) (c) (i) a = 0.10 = ( )140.m s 1 (allow C.E. for incorrect values of v) (ii) F = 0.45 ( ) 140 = ( ) 63N (allow C.E for value of a) (iii) away from wall at right angles to wall [or back to girl ] [or opposite to direction of velocity at impact ] 5 [8] 68. B 69. A 70. B 71. C 7. A 73. B James Allen's Girls' School 5

26 74. A 75. B 76. (a) (use of T = π g l gives) T = π = 1.8 s (b) mgh = ½ mv 3 v = ( ) (= 0.63 m s -1 ) πa v max = πfa = T A = (= 0.18m) π [or by Pythagoras A = 800 gives A = ( ) ( = 180 mm) (or equivalent solution by trigonometry ) πa v max = πfa or = T π 0.18 = (= 0.63 m s -1 ) (c) tension given by F, where F mg = F = mv l = 0.6 N [8] Q 77. (a) (i) E (= 4πε r ) = π ( ) = Vm -1 (or (NC -1 ) James Allen's Girls' School 6

27 GM (ii) V(= ) = ( ) r = ( ) sign and J kg -1 5 (b) arrow pointing to the right 1 [6] 78. (a) (i) acceleration (ii) both represent acceleration of free fall [or same acceleration] (iii) height/distance ball is dropped from above the ground [or displacement] (iv) moving in the opposite direction (v) kinetic energy is lost in the collision [or inelastic collision] 5 (b) (i) v = v = 4.9 m s 1 (4.85 m s 1 ) (ii) u = u = 3.8 m s 1 (3.84 m s 1 ) (iii) change in momentum = = 1.3 kg m s 1 (1.5 kg m s 1 ) (allow C.E. from (b) (i) and (b)(ii)) (iv) F = = 13 N (allow C.E. from (b)(iii)) 8 [13] 79. A 80. B James Allen's Girls' School 7

28 81. A 8. B 83. A 84. D 85. C 86. C 87. D 88. A 89. (a) period = 4 hours or equals period of Earth s rotation remains in fixed position relative to surface of Earth equatorial orbit same angular speed as Earth or equatorial surface max James Allen's Girls' School 8

29 (b) (i) GMm = mω r r T = π ω 1/ 3 1/ (4 3600) r = GMT = 4π 4 π (gives r = km) (ii) V = GM 1 1 R r = = (J kg 1 ) E p = m V (= ) = J (allow C.E. for value of V) [alternatives: calculation of GM ( ) or GM ( ) R r or calculation of GMm ( ) or GMm ( ) R r calculation of both potential energy values subtraction of values or use of m V with correct answer 6 [8] 90. (a) units: F - newton (N), B - tesla (T) or weber metre (Wb m ), I - ampere (A), l - metre (m) condition: I must be perpendicular to B (b) (i) mass of bar, m = ( ) 8900 l (= 5.56l) weight of bar (= mg) = 54.6l mg = BIl or weight = magnetic force 54.6l = B 65 l gives B = T (ii) arrow in correct direction (at right angles to I, in plane of bar) 5 [7] James Allen's Girls' School 9

30 91. (a) (i) r = 0.01 (m) (use of v = πfr gives) v = π = 3.8 m s 1 (3.77 m s 1 ) (ii) correct use of a = v or a = r = m s [or correct use of α = ω r] (allow C.E. for value of v from (i) 5 (b) panel resonates (because) motor frequency = natural frequency of panel QWC [7] 9. (a) mg = T cos 6 F = T sin 6 hence F = mg tan 6 [or correct use of triangle: for sides correct, for 6, for tan 6 = F/mg tan 6 = F 3 mg or F x = mg h, tan θ = x h (b) (i) (use of E = V gives) E = 400 = V m 1 d (ii) (use of Q = F mg tan 6 4 gives) Q = = tan 6 E E = C (allow C.E. for value of E from (i)) 3 [6] 93. C 94. A James Allen's Girls' School 30

31 95. D 96. A 97. C 98. C 99. C 100. A 101. C 10. (a) quantity SI unit (gravitational potential) J kg 1 or N m kg 1 scalar (electric field strength) N C 1 or V m 1 vector (magnetic flux density T or Wb m or N A 1 m 1 vector 6 entries correct 4 or 5 entries correct or 3 entries correct 3 James Allen's Girls' School 31

32 (b) (i) mg = EQ mg 9 E = = (V m 1 ) 1 Q (ii) positive 3 [6] 103. (a) deflects one way then the other way (b) (i) acceleration is less than g [or reduced] suitable argument (e.g. correct use of Lenz s law) (ii) acceleration is less than g [or reduced] suitable argument (e.g. correct use of Lenz s law) 4 (c) magnet now falls at acceleration g emf induced but no current no energy lost from circuit [or no opposing force on magnet, or no force from magnetic field or no magnetic field produced] 3 QWC [9] 104. (a) Q (= CV = ) = 970 (µc) E (= ½QV) = ½ _ = J [or E (= ½CV ) = ½ = J ] (b) time constant (= RC) = = 155 s 1 t / RC (c) Q( Q e ) = 0 = 970 e 60/155 = 00 (µc) (allow C.E. for time constant from (b)) Q V = 00 = = 6.11V C 330 (allow C.E. for Q) [or V = V 0 e t/rc = 9.0 e 60/155 = 6.11 V ] 3 [6] James Allen's Girls' School 3

33 105. (a) increased impact time same loss of momentum force = change of momentum/impact time force is reduced [alternative for nd and 3rd : reduced deceleration of body force = mass acceleration ] [or area of contact increased force on driver spread out over larger area pressure or force/unit area on driver reduced ] [or air bag absorbs E k of driver over a greater distance E force = k distance force is reduced ] 4 QWC (b) (use of v = u (0) 18 + as gives) a = v u = s.5 a = 65 m s ( 64.8 m s ) (hence deceleration = 65 m s ) [6] 106. (a) kinetic energy not conserved [or velocity of approach is equal to velocity of separation] 1 (b) (i) (use of p = mv gives) p = =.7 kg m s 1 (ii) ( mv) (use of F = gives) F =.7 t = 180 N [or a = v u = 60 t = 4000 (ms 1 ) F = (ma) = = 180 N] 4 James Allen's Girls' School 33

34 (c) (i) 180 N (allow C.E. for value of F from (b) (ii) in opposite direction (to motion of the club) (ii) body A (or club) exerts a force on body B (or ball) (hence) body B (or ball) exerts an equal force on body A (or club) correct statement of Newton s third law max 4 QWC 1 [9] 107. C 108. D 109. D 110. B 111. A 11. B 113. B 114. A 115. B James Allen's Girls' School 34

35 116. C 117. (a) (i) out of plane of diagram (ii) circular path in a horizontal plane [or out of the plane of the diagram] BQv = mv r 5 5 radius of path, r mv = BQ = 0.91(4) m max 5 (b) (i) radius decreased halved [or radius is halved ] (ii) radius increased doubled [or radius is doubled ] max 3 [8] 118. (a) work = force distance moved in direction of force (in circular motion) force is perpendicular to displacement no movement in direction of force (hence no work) [or speed of body remains constant (although velocity changes) kinetic energy is constant potential energy is constant ] [or gravitational force acts towards the Earth Moon remains at constant distance/radius from Earth since radius is unchanged, gravitational force does no work or E p of Moon is constant ] 3 QWC 1 (b) (i) any suitable example of circular motion (ii) any SHM example at maximum displacement [or any other suitable example, e.g. car starts from rest] [5] James Allen's Girls' School 35

36 119. (a) (i) h (= ct) (= ) =.0(4) 10 7 m (ii) g = ( ) GM r r (= ) = (m) (allow C.E. for value of h from (i) for first two marks, but not 3rd) g = ( ) 4 (= 0.56 N kg 1 ) 4 (b) (i) g = v r v = [0.56 ( )] ½ = m s 1 ( m s 1 ) (allow C.E. for value of r from a(ii) [or v = GM = r 11 v = = m s 1 ] 4 1/ (ii) 7 T = πr = π v = 4.3(5) 10 4 s (1. hours) (use of v = gives T = s = 1.0 hours) (allow C.E. for value of v from (I) [alternative for (b): 3 (ii) T = 4π r GM = 4π 7 ( ) T = 4.3(6) 10 4 s = ( (s ) (i) v πr = π T = 3.8(6) 10 3 m s 1 ] 7 (allow C.E. for value of r from (a)(ii) and value of T) 5 [9] James Allen's Girls' School 36

37 10. B 11. C 1. D 13. D 14. D 15. C 16. (a) acceleration is proportional to displacement acceleration is in opposite direction to displacement, or towards a fixed point, or towards the centre of oscillation 5 (b) (i) f = = 1.1 Hz (or s 1 ) (1.09 Hz) 3 (ii) (use of a = (πf) A gives) a = (π 1.09) = 3.6 m s (3.56 m s ) (use of f = 1.1 Hz gives a = 3.63 m s ) (allow C.E. for incorrect value of f from (i)) (iii) (use of x = A cos(πft) gives) x = cos(π ) = ( ) m (43 mm) (use of f = 1.1 Hz gives x = ( )4.0(7) 10 m (41 mm)) direction: above equilibrium position or upwards 6 James Allen's Girls' School 37

38 (c) (i) graph to show: correct shape, i.e. cos curve correct phase i.e. (cos) (ii) graph to show: two cycles per oscillation correct shape (even if phase is wrong) correct starting point (i.e. full amplitude) max (a) work done/energy change (against the field) per unit mass when moved from infinity to the point GM E GM (b) V E = and M VM = RE RM M E V M = G = V E 81 RE 81 = ( 63) =.9 MJ kg 1 (.88 MJ kg 1 ) 3 (c) 0 Surface of Earth Surface of Moon 63 limiting values ( 63, V M ) on correctly curving line rises to value close to but below zero falls to Moon from point much closer to M than E max 3 [8] James Allen's Girls' School 38

39 18. (a) Φ ( = BA) = π ( ) = Wb ( Wb) (b) (i) N Φ ( = NBA 0) = = 0.59 (Wb turns) (0.589 (Wb turns)) (if Φ = , then (Wb turns)) (allow C.E. for value of Φ from (a)) Φ (ii) induced emf ( = N ) = t 0. 1 = 4.9 V (4.91 V) (allow C.E. for value of Wb turns from (ii) 4 [6] 19. (a) (i) change of momentum (= ) = 14(.1) kg m s 1 ( mv) 14.1 (ii) (use of F = gives) F = 3 t = 1.5(3) 10 3 N (allow C.E. for value of (mv) from (i) 3 (b) (i) deceleration = = m s ( m s ) (ii) (use of a = v r gives) (iii) 4 centripetal acceleration = = m s 0.6 ( m s ) before impact: radial pull on knee joint due to centripetal acceleration of boot during impact: radial pull reduced 4 [7] 130. (a) (i) (change in momentum of A) = kg m s 1 (or N s) (ii) (change in momentum of B) = kg m s 1 4 James Allen's Girls' School 39

40 (b) initial vel/m s 1 final vel/m s 1 initial k.e./j final k.e./j truck A truck B (c) not elastic because kinetic energy not conserved kinetic energy is greater before the collision (or less after) [or justified by correct calculation] 3 [11] 131. B 13. A 133. C 134. D 135. D 136. A 137. A 138. B James Allen's Girls' School 40

41 139. (a) reference to resonance air set into vibration at frequency of loudspeaker resonance when driving frequency = natural frequency of air column more than one mode of vibration stationary wave (in air column) (or reference to nodes and antinodes) maximum amplitude vibration (or max energy transfer) at resonance [alternative answer to (a): first two marks as above, remaining four marks for wave reflected from surface (of water) interference/superposition (between transmitted and reflected waves) maximum intensity when path difference is nλ maxima (or minima) observed when l changes by λ/ ] Max 4 QWC 1 (b) (i) λ = (= 355 mm) λ = 710 mm [if 4 λ = 168, giving λ = 670 mm, (1 max) (67 mm)] (ii) c( = fλ) = = 341 m s 1 (allow C.E. for incorrect λ from (i)) [allow = 30 m s 1 (1max) (3 m s 1 )] 4 [8] 140. (a) (i) straight line through origin (ii) 1 capacitance (iii) energy (stored by capacitor) (or work done (in charging capacitor)) 3 (b) (i) RC = (= 38.1 s) V(= V 0 e t/rc ) = 1 e 6/38.1 = 6.1 V (6.06 V) [or equivalent using Q = Q 0 e t/rc and Q = CV] James Allen's Girls' School 41

42 (ii) (RC) = (= 19.0 s) V (= 6.06 e 14/19 ) =.9(0) V (use of V = 6.1 V gives V =.9() V) (iii) pd/v time/s 7 [10] 141. (a) attractive force between point masses proportional to (product of) the masses 3 inversely proportional to square of separation/distance apart (b) mω GMm mv R = ( ) or = R R π 4π GM (use of T = gives) = 3 3 ω T R G and M are constants, hence T R 3 (c) (i) (use of T R 3 gives) T m = 87(.5) days 365 ( ) 3 T m = ( ) 3 (ii) 1 ( ratio = 11 ) = 3 R N 1 11 (gives R N = m) = 30(.1) 4 [10] James Allen's Girls' School 4

43 14. C 143. C 144. D 145. B 146. B 147. A 148. C 149. A 150. B James Allen's Girls' School 43

44 151. (a) (i) mg = ke k = = 61(.3) N m m 0.69 T = π = π (= s) k 61.3 (ii) 1 1 f = = (= 1.50 Hz) 4 T (b) (i) forced vibrations (at 0. Hz) amplitude less than resonance ( 30 mm) (almost) in phase with driver (ii) resonance [or oscillates at 1.5 Hz] amplitude very large (> 30 mm) oscillations may appear violent phase difference is 90º (iii) forced vibrations (at 10 Hz) small amplitude out of phase with driver [or phase lag of (almost) π on driver] Max 6 [10] 15. (a) E V (or E = 1 / CV ) pd after 5 s = 6 V (b) (i) use of Q = Q 0 e t/rc or V = V 0 e t/rc 5 (e.g. 6 = 1e 5/RC 1 5 ) gives e RC = and = 1n 6 RC (RC = 36(.1) s) [alternatives for (i): V = 1 e 5/36 gives V = 6.0 V (5.99 V) or time for pd to halve is 0.69RC RC = 5 = 36(.) s ] (ii) R = = 5.3(0) 10 4 Ω 4 [6] James Allen's Girls' School 44

45 153. (a) orbits (westwards) over Equator maintains a fixed position relative to surface of Earth period is 4 hrs (1 day) or same as for Earth s rotation Max 3 offers uninterrupted communication between transmitter and receiver steerable dish not necessary (b) (i) Mm G = mw + ( R + h) π use of w = T ( R h) GM 4π = T (ii) gives ( ) 3 R + h, hence result (iii) limiting case is orbit at zero height i.e. h = 0 6 ( ) 3 3 4π R 4π T = = 11 4 GM T = 5090 s (= 85 min) 6 (c) speed increases loses potential energy but gains kinetic energy [or because ν 1 GMm mv from = ] r r r [or because satellite must travel faster to stop it falling inwards when gravitational force increases] [11] V (a) (i) = = 3 d E ( = Vm 1 ) James Allen's Girls' School 45

46 (ii) 3 l t = = = s 7 v (iii) ma y = Ee ay = (= m s ) acceleration is upwards [or towards + plate] 5 (b) vy (= a y t) = (= m s 1 ) 7 v = ( ) + (3. 10 ) = m s 1 7 at tan = 6 3. above the horizontal 3 [8] 155. (a) momentum kinetic energy (b) (i) 450ms 1 in the opposite direction p = = Ns 4 (c) force is exerted on molecule by wall to change its momentum molecule must exert an equal but opposite force on wall in accordance with Newton's second or third law 4 [10] 156. B James Allen's Girls' School 46

47 157. C 158. B 159. C 160. D 161. D 16. B 163. C 164. D 165. (a) f 1 g = π l Oscillations must be of small amplitude James Allen's Girls' School 47

48 (b) (i) f = = 0.53(8)( s 1 ) [or T = = 1.8(6) (s)] 5 g 9.81 l = = 4π [or f 4π l = 0.85(9)m (allow C.E. for values of or T) T g l = 4π ] 4π (ii) a max {= ( )(πf) A} = (π 0.538) (= ms ) (allow C.E. for value of from (i)) F max (= ma max ) = = N ( N) 51 [or F max (= mg sin θ max ) where sin θ max = 859 = = N ] 6 [8] 166. (a) vibrates or oscillates or moves in shm vibration/oscillation is vertical/perpendicular to wave propagation direction frequency (=c/λ) = 3.0(Hz) (or same as P) amplitude = 90 (mm) (or same as P) Q has a phase lag on P (or vice versa) phase difference of 0.4 π = π (rad) or 10º James Allen's Girls' School 48

49 (b) use of f =3.0(Hz) v max (= πfa) = n = 1.7(0)ms 1 3 [8] 167. (a) (i) force per unit charge acting on a positive charge (ii) vector 3 (b) (i) 9 9 Q1Q F = = πε 0r 4π (80 10 ) =4.5(0) 10 5 N (ii) (use of Q V = gives) 4πε 0 x = 3 4πε 0 x 4πε 0 (80 10 x 4 8 or = x 80 x x = 6.7mm 4 (c) correct directions for E 4 and E s E 8 approx twice as long as E 4 correct direction of resultant R shown 3 E 8 P R E 4 [10] James Allen's Girls' School 49

50 168. (a) greater flux (linkage) or more flux lines (at same distance) [or stronger magnet produces flux lines closer together] greater rate of change of flux (linkage) [or more flux lines cut per unit time] emf rate of change of flux (linkage) φ [or using = N, where Φ = A B, v and t are the same t B is larger since magnet is stronger N and A are constant, is larger ] 3 (b) (i) area swept out, A = lv t Φ (= B A) = Blv t φ Blv t = ( N) = gives result t t 3 (c) (i) w(=πf) = π l6 = 101 rads 1 (ii) v(=rw) = = 3.(3)ms (allow C.E. for value of w from (i)) (iii) (= Blv) = = 5.7(9) 10 3 V 5 (allow C.E. for values of v from (ii)) (solutions using = Bfnr to give 5.7(6) 10 3 V acceptable) [11] 169. D 170. B James Allen's Girls' School 50

51 171. C 17. A 173. A 174. B 175. B 176. C 177. A 178. B 179. C 180. D James Allen's Girls' School 51

52 181. B 18. A 183. D 184. B 185. D 186. A 187. C 188. D 189. C 190. D James Allen's Girls' School 5

53 191. C 19. B 193. C 194. (a) kinetic energy is not conserved (or velocity of approach equals velocity of separation) 1 (b) (i) (use of p = mv gives) p = =.7kg m s 1 (ii) (use of F = [or a = v u t ( mv) t 60 = gives) F = = 180 N 3 = 400 (m s 1 ) F = ma = = 180N 4 [5] 195. (a) (i) mg = ke k = = 61(.3) N m 1 (ii) T = m = π = π k (= s) 1 1 f = = (= 1.5(0) Hz) 4 T James Allen's Girls' School 53

54 (b) The marking scheme for this part of the question includes an overall assessment for the Quality of Written Communication (QWC). There are no discrete marks for the assessment of QWC but the candidates QWC in this answer will be one of the criteria used to assign a level and award the marks for this part of the question. Level Good 3 Modest Limited 1 Descriptor an answer will be expected to meet most of the criteria in the level descriptor answer supported by appropriate range of relevant points good use of information or ideas about physics, going beyond those given in the question argument well structured with minimal repetition or irrelevant points accurate and clear expression of ideas with only minor errors of spelling, punctuation and grammar answer partially supported by relevant points good use of information or ideas about physics given in the question but limited beyond this the argument shows some attempt at structure the ideas are expressed with reasonable clarity but with a few errors of spelling, punctuation and grammar valid points but not clearly linked to an argument structure limited use of information or ideas about physics unstructured errors in spelling, punctuation and grammar or lack of fluency Mark range 0 incorrect, inappropriate or no response James Allen's Girls' School 54

55 examples of the sort of information or idea that might be used to support an argument forced vibrations (at 0. Hz) amplitude fairly large ( 30 mm) in phase with driver resonance (at 1.5 Hz) amplitude very large (> 30 mm) oscillations may appear violent phase difference at 90º forced vibrations (at 10 Hz) small amplitude out of phase with driver or phase lag of π on driver [10] 196. (a) period is 4 hours (or equal to period of Earth s rotation) remains in fixed position relative to surface of Earth equatorial orbit same angular speed as Earth (or equatorial surface) max (b) (i) GMm = mω r r π T = ω (4 1/ 3 GMT 3600) r = = 4 4 π π (gives r = km) 1 1 (ii) ΔV = GM R r = = (J kg 1 ) ΔE P = mδv (= ) = J (allow ecf for value of ΔV) 6 James Allen's Girls' School 55

56 (c) (i) signal would be too weak at large distance (or large aerial needed to detect/transmit signal, or any other acceptable reason) the signal spreads out more the further it travels (ii) for road pricing would reduce congestion stolen vehicles can be tracked and recovered uninsured/unlicensed vehicles can be apprehended against road pricing would increase cost of motoring possibility of state surveillance/invasion of privacy any valid points (must be for both for or against) (a) T cos 6º = mg T sin 6º = F hence F = mg tan 6º [or by use of triangle: sides correct 6º correct tan 6º = F/mg ] 3 V 400 (b) (use of E = gives) E = d = V m 1 4 F mg tan tan 6 (use of Q = gives) Q = E 4 E (allow ecf for value of E from (i)) = C 3 [6] 198. (a) (i) E (= ½ CV = ) = 0.90J (ii) W (= QV = CV = ) = 1.8 J (b) (i) (V = V 0 e t/rc ) gives 30 = 100 e t/rc t = ( RC ln (30/100) = s) = s James Allen's Girls' School 56

57 (ii) image would be less sharp (or blurred) because the discharge would last longer and the image would be photographed as it is moving image would be brighter because the capacitor stores more energy and therefore produces more light 4 [6] 199. (a) greater flux (linkage) or more flux lines (at same distance) [or stronger magnet produces flux lines closer together] greater rate of change of flux (linkage) [or more flux lines cut per unit time] induced emf = [or =] rate of change of flux (linkage) B [or using = NA ΔB is larger since magnet is stronger t N, A and Δt are the same at the same speed is larger ] 3 (b) area swept out ΔA = lvδt ΔΦ (= BΔA) = BlvΔt Φ Blv t = ( N) = gives result 3 t t (c) (i) ω (= πf) = π 16 = 101 rad s 1 (ii) v (= rω) = = 3.(3) m s 1 (allow ecf for value of ω from (i)) (iii) (= Blv) = = 5.7(9) 10 3 V (allow ecf for value of v from (ii)) [or accept solutions using = Bfπr to give 5.7(9) 10 3 V] 5 [11] James Allen's Girls' School 57