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