JJC H1 PHY P2 Ans

Page 1 of 8 JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT 2016 JC2 Preliminary Examination 8866 H1 Physics Paper 2 Suggested Solutions with Markers’ Comments Qn Suggested solution Remarks 1(a) . ..t 12 01 5 s80 [1] - Sub (b) (. ) (. ) . 2109 8 1 0 1 5 0 1 1 m2h  [1] – Sub [1] – Ans (c) . (. ) (. ) . .. .      1 -1 22 - 1 80 m s 0 9 81 0 15 1 472 m s 80 14 7 2 81 3 m s x y v v v [1] – vy [1] – Ans (d) Both of them have the same vertical acceleration. OR the ball must be targeted at the block. [1] 2(a) As the mass moves upwards at constant speed , the upward force, F, must be equal to the weight of the object, mg. Hence work done on object = Fh = (mg)h By conservation of energy, since Fh is the work done on the object and is equal to the increase in gravitational potential energy, Ep = mgh since its kinetic energy is constant [1] [1] [1] (b)(i) (. ) (. ) (. ) (. ) (. ) (. ) (.) .( . ) . . . 2 By conservation of energy, 10 100 9 81 0 100 9 81 0 50 0 100 1 6 02 0 100 9 81 0 4905 0 128 06 3 m h h h     [1] – Sub [1] – Ans (ii) (. ) (.) (. ) .   -1 By conservation of momentum, 0 100 3 5 0 0 180 19 4 m s v v (. ) (. ) ( ) .   22 By conservation of energy, 110 180 1 94 0 12022 00 7 5 1 m x x [1] - v [1] – Sub [1] – Ans 3(a)(i) . .    3 3 10 1 0 00 1 7 9 56 10 VR I [1] (ii) Into the page/into the plane [1] (b)(i) sin . ( )( . ) .      390 1 12 56 10 0 04 2508 8 2510 N FB I L [1] – Sub [1] – Ans (ii) As the projectile travels along the rail, resistance increases and current I decreases. Therefore force decreases. [1] [1] (iii) To the right. [1]

Page 2 of 8 JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT 2016 JC2 Preliminary Examination 8866 H1 Physics Paper 2 Suggested Solutions with Markers’ Comments Qn Suggested solution Remarks (iv) OR Use stronger voltage/ power supply OR Decrease resistance of track (includes increasing the cross-sectional area of the track). [1] 4(a)(i) The photoelectric current reaches a maximum because it is limited by the rate of emission of photoelectrons which is dependent on the intensity of illumination. Increase in V only increases the acceleration of the photoelectrons but not the rate of emissions of photoelectrons from the metal. [1] [1] (ii) The intensity of radiation is dependent on the rate of incidence of photons. The greater the rate of incidence of photons, the greater the rate of emission of photoelectrons. Since the maximum photoelectric current is dependent on the rate of emission of photoelectrons, increasing the intensity of illumination increases the maximum photoelectric current. Any 2 out of 3 points (b)(i) The current, I = ne where n = number of photoelectrons emitted per second  n = I e = 10 19 4.8 10 1.6 10     = 3.0 × 109 s1 [1] – Ans (ii) Number of photons incident per second N = 2500(3.0 x 109) = 7.5 x 1012 The intensity, i = Nhf A  i = Nhc A = 12 34 8 69 (7.5 10 )(6.63 10 )(3.0 10 ) (24 10 )(410 10 )     = 0.152 W m2 [1] – N [1] - Sub [1] - Ans 5(a)(i) Moon Period T/days mean distance from centre of Jupiter r / 109 m log10 (T/days ) log10 (r/m) Sinope 758 23.7 2.88 10.37 Leda 239 11.1 2.38 10.05 Callisto 16.7 1.88 1.22 9.27 Lo 1.77 0.422 0.25 8.63 Metis 0.295 0.128 -0.53 8.11 1 mark for two correctly filled blanks

Page 3 of 8 JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT 2016 JC2 Preliminary Examination 8866 H1 Physics Paper 2 Suggested Solutions with Markers’ Comments Qn Suggested solution Remarks (ii) [1] - All points plotted correctly. [1] - best fit line (b)(i) 3.0 0.0Gradient of the graph = 1.510.45 8.45   [1] – Sub [1] – Ans (ii) The data support the relation in (a). 23 2 4From rT GM  24Let = which is a constantk GM  T2 = kr3 2lgT = lg k + 3lg r 2lgT = 3 lg r + lg k lg T = 1.5 lg r + ½ lg k ----- (1) Since a straight line graph is obtained and the gradient of the graph is equal to 1.5 which is consistent with the equation (1), thus the data support the relation in (a). [1] - Ans [1] - Expl (c) Given period T = 7.16 days,  lg (7.16) = 0.85. From the graph, lg r = 9.025 Thus the orbital radius of Ganymede = 109.025 = 1.06 x 109 m [1] – Value [1] – Ans 6(a) Rate of change of momentum [1]

Page 4 of 8 JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT 2016 JC2 Preliminary Examination 8866 H1 Physics Paper 2 Suggested Solutions with Markers’ Comments Qn Suggested solution Remarks (b) By definition of force, the body can be instantaneously at zero momentum when its momentum is changing with respect to time. [1] (c)(i) The high rate of change of momentum of her hand will exert a large force to fracture the bricks. [1] (ii) There will be a higher rate of change in momentum and the force exerted on the bricks on impact would be larger. [1] [1] (d)(i) (. ) (.) . 2 2 0 05 0 20 20 W mv mvPF v v tt      [1] – Sub [1] – Ans (ii) (. ) (.) . 2 2 12 1 05 0 202 10 W mv P t   [1] – Sub [1] – Ans (iii) Energy lost due to frictional forces acting on the sand. [1] (e)(i) [-1] for any wrong force (ii) cos cos sin sin 12 12 50 40 (1) 50 40 (2) 125 N oo oo TT W TT W    [2] – Eqn [1] – Ans (iii) Tension in shorter rod will increase as its horizontal component acts in the opposite direction to the wind. [1] [1] (iv) Vertical fall in height = 2.0 – 2.0 cos 50o = 0.71 m By conservation of energy, (. ) (. ) . 2 -1 1 02 22 9 8 1 0 7 1 37 4 m s mv mgh vg h    [1] – h [1] – Ans (v) It will be lower as energy is lost to the drag force due to the surrounding air OR Friction at hinge can lead to energy lost. [1]

Page 5 of 8 JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT 2016 JC2 Preliminary Examination 8866 H1 Physics Paper 2 Suggested Solutions with Markers’ Comments Qn Suggested solution Remarks 7(a) Resistance of a resistor is defined as the ratio of the potential difference across it to the current flowing through it. [1] (b) Volume V = A x l VA l 2 Resistance, 6.0 lllR VAV l     When the length is 3l, 222(3 ) (9 )new resistance 9 9 6.0 54 ll l AAA        [1] – Exp [1] – Sub [1] – Ans (c)(i) Maximum safe current passing through the 1000  resistor, 1000 0.40 1000 P R  I = 0.020 A Maximum safe current passing through the 160  resistor, 160 0.40 160 P R  I = 0.050 A Hence maximum safe current flowing through the circuit without damaging any of the resistor is Imax = 0.020 + 0.020 = 0.040 A Maximum safe potential difference applied between X and Y V = 0.040 x 160 + 0.020 x 1000 = 26.4 V [1] – Value [1] – Sub [1] – Ans (c)(ii) One of the 1000  resistors would be most likely to fail. When the maximum safe potential difference is exceeded, the current flowing in the circuit will be more than the safe current. Thus the current flowing in the 1000  resistor will be more than 0.020 A which will result in exceeding the maximum safe power. [1] [1]

Page 6 of 8 JURONG JUNIOR COLLEGE PHYSICS DEPARTMENT 2016 JC2 Preliminary Examination 8866 H1 Physics Paper 2 Suggested Solutions with Markers’ Comments Qn Suggested solution Remarks (d)(i) Given T BARAeR T B  lnln Temperatures  = 50 0C corresponds T = 50 + 273 =323K and 80 0C correspond to T = 80 + 273 = 353 K From graph, R = 110  at 50 0C and R = 50  at 80 0C respectively. ln110 ln (1) 323 BA --------   ln50 ln (2) 353 BA -------  (1) – (2) gives 35332350ln110ln BB  Solving B  3.0 x 103 K A  1.03 x 10-2  [1] read off values [1] working