2013 JJC H2 Physics P2 solutions

Page 1 of 10   JURONG JUNIOR COLLEGE JC2 Preliminary Examination 2013 H2 Physics Paper 2 solutions Qn Suggested solution Remarks 1(a) ll lTT g gg T 22 2 2 442( 1 )      (2) 0.1 0.0052 2 0.015820.6 0.910 gl T gl T      =1.6 % or 2 % (only accept 2 sig. fig. or 1 sig. fig.) Comments: Two common mistakes observed. First, g was not made the subject (1) and hence (2) was wrong. Second, the final answer was left as “1.58%”. A significant number of candidates mixed up absolute error and fractional/percentage error. Some candidates could have scored at least “[1] sub” but full substitutions of values were not presented. [1] sub [1] ans (b)    22 22 4 0.2064 9.821 0.91 lg T   m s-2   0.0158 9.821 0.2g  m s-2 g = (9.8  0.2) m s-2 Comments: Some candidates did not compute the value of g and erroneously assumed the value of 9.81 m s-1. A significant number did not present g to 1 sf. [1] g value 3-4 s.f. [1] g value 1 s.f. [1] g value 1 d.p. 2(a) Taking moments about the elbow, Clockwise moment = anti-clockwise moment T cos 20° x 3 = (60 x 9) + (20 x 34) T = 433 N Comments: Many students failed to indicate the position at which moment is taken. [1] statement [1] sub

Page 2 of 10   JURONG JUNIOR COLLEGE JC2 Preliminary Examination 2013 H2 Physics Paper 2 solutions Qn Suggested solution Remarks (b) Tx = 433 sin 20 = 148.1 N  Fx = 0 Tx = Rx = 148.1 N  Fy = 0 Ty = Ry + 60 + 20 Ry = 433 cos 20° - 60 – 20 = 326.9 N Resultant force R acting at elbow (pivot) = 22 xyRR = 359 N Angle = tan-1( 1 326.9)t a n ( ) 148.1 y x R R  = 65.6° below the forearm Comments: A significant number of students failed to consider the concept of translation equilibrium ( Summation of horizontal forces / Summation of vertical forces = 0 ). Additionally, students do not consider the direction of the horizontal force resulting in the wrong angle direction even though the magnitude of the angel is correct. [1] value of R x [1] value of R y [1] value of R [1] correct direction 3(a)(i) F = eE = e V d   = (1.6 x 10 -19) -2 80 0.50x10   = 2.6 x 10-15 N Comments: Quite a number of students used F =qV. [1] sub [1] ans (ii) a = F m   = -15 -31 2.6 x 10 9.11 x 10    = 2.8 x 1015 m s-2 [1] ans (iii) Using v = u + at  v = at = (2.8 x 10 15)( 6.5 x 10-10) = 1.8 x 106 m s-1 Comments: Many students substituted the horizontal velocity for the initial vertical velocity. [1] sub [1] ans

Page 3 of 10   JURONG JUNIOR COLLEGE JC2 Preliminary Examination 2013 H2 Physics Paper 2 solutions Qn Suggested solution Remarks (b) Let y = deflection of electron beam on screen x = distance of plates to screen = 15 cm Geometrically : y x vy x u  6 7 18 x 1 0 15 3 1 x 1 0 y. . y = 0.87 cm [1] sub [1] ans (c) [1] correct position with label (ii) When Vr.m.s. = 80 V, maximum deflection = ( 2 )(y) = 1.2 cm Comments: Many students drew sinusoidal waves. [1] value of maximum deflection [1] correct sketch with label c.r.o. screen x = 15 cm electrons - + y ux vy 1 cm 1 cm undeflected beam A 1 cm 1 cm undeflected beam S

Page 4 of 10   JURONG JUNIOR COLLEGE JC2 Preliminary Examination 2013 H2 Physics Paper 2 solutions Qn Suggested solution Remarks 4(a) The tesla is the magnetic flux density of a magnetic field in which the force per unit length per unit current acting on a straight conductor placed perpendicular to the magnetic field is one newton per metre per ampere. Comments: Very few provided answers that emphasise the significance of the ratio aspect i.e. force per unit length per unit current. Reference to “straight conductor placed perpendicular to the magnetic field” must be made. All units used (newton, metre and ampere) must be defined clearly and not written in symbols. NOTE: The following version is accepted for only this Prelims. The tesla is the magnetic flux density of a magnetic field in which the force per unit length acting on a straight conductor placed perpendicular to the magnetic field is one newton per metre when the current flowing through the conductor is one ampere. [1] [1] just for this Prelims; being the 2009 TYS solution provided to students (b)(i) Comments: Very few realised that (1) the position of the neutral point has shifted towards the left; (2) the spacing of field lines on the left and right side of X or Y are different. [1] direction of magnetic field around both wires [1] spacing of magnetic field lines (ii)1. oxMagnetic flux densityat Y = ( )2π μ I r [1] 2. ox 12 yMagnetic force = = ( ) 2π μ IBI L I L r Magnetic forceper unitlength ( )2π ox y I Ir  [1] for final expression

Page 5 of 10   JURONG JUNIOR COLLEGE JC2 Preliminary Examination 2013 H2 Physics Paper 2 solutions Qn Suggested solution Remarks (iii) -7(4π×10 )(100)Magnetic forceper unitlength ( ) (200)2π 2π(5) ox y I Ir   - 4 - 18. 00 ×10 Nm The magnitude of the force is small and hence the movement of the wires will be small compared to separation of wires. Comments: Many candidates did not realise they were expected to complete some calculations as part of their explanation even though values were provided in the question. [1] ans [1] explanation 5(a)(i) Vacuum gap between STM probe and surface. Comments: Many students wrote “distance between…”, but the distance is a measurement of the barrier, rather than the barrier itself. [1] (ii) The tunnelling current between the STM probe and the surface is sensitive to small variations of the width of the energy barrier, or vacuum gap. By moving the probe along the surface, the varying currents at each point is mapped out as varying depth/height of each point on the surface. By combining these points, an atomic-scale image of the surface is obtained. OR : By using a feedback mechanism, the probe is moved up or down to maintain constant current. The vertical motion is then plotted to obtain an atomic-scale image of the surface. [1] [1] (b) Valence band is the highest occupied energy band whereas conduction band is the lowest unoccupied energy band. Comments: Many students missed out the word “energy”. [1] [1] (c)(i) In the p-type region, the majority charge carriers are holes and in the n-type region, the majority charge carriers are electrons. Diffusion of the mobile charge carriers occurs. Holes at the p-type region diffuse across the junction to the n-type region and electrons at the n-type region diffuse across the junction to the p-type region. As holes and electrons diffuse across in opposite directions, most of them meet and recombine near the junction. This resulted in the junction being depleted of mobile charge carriers hence forming the depletion region. [1] [1] [1] [1]

Page 6 of 10   JURONG JUNIOR COLLEGE JC2 Preliminary Examination 2013 H2 Physics Paper 2 solutions Qn Suggested solution Remarks (c)(ii) Comments: A number of students did not draw a proper circuit for the battery. [1] 6(a) The half-life of a radioactive nuclide is the average time taken for the activity to fall to half (its original value). Points to note: First there is the problem of the meaning of words such as isotopes, nuclide and nucleus. Here nucleus or its plural, nuclei, is usually required. Then there is the problem of what actually does halve in one half-life. The answers ‘amount’ or ‘mass’ or ‘quantity’ are not acceptable. Probably the simplest solution is to state that it is the activity which halves and include in the definition should be the comment that half-life is the average time for the activity to halve. Comments: Many students failed to include “average”. [1] (b)(i) The decrease in output