2013_IJC_H2_Physics_P3 Solutions

2013 H2 Physics Prelim 2 Exam – Paper 3 Solutions 1 (a) (i) The reaction force must have a horizontal component because the horizontal component of the tension is unbalanced (or needs to be balanced by another horizontal force). It must also have a vertical force to produce an anti-clockwise moment in order to balance the clockwise moment caused by the tension when moment is taken about his C.G.. OR since all forces must pass through a common point, the reaction force from the wall on the person is at an angle to the horizontal. Hence there are horizontal and vertical components. [B1] [B1] (ii) The reaction force must point about northwest, towards the intersection of the weight and tension. [B1] (b) Equilibrium of horizontal forces: (FR)X = 610 sin20° (208.6 N leftward) Equilibrium of vertical forces: (FR)X + 610 cos20° = 590 (16.79 N upward) FR = squareroot [(FR)X 2 + (FR)Y 2] = 209.3 N Angle with vertical = tan-1 [(FR)X / (FR)Y] = 85.4° [M1] [M1] [A1] [A1] 2 (a) (i) ( ) Hz (ii) √ m = 0.0788 kg (b) (i) Amplitude of the vertical oscillations will also increase. [A1] Driving force is larger due to the larger amplitude of wave, OR more energy transfer. [M1] (ii) wavelength of wave increases and since the speed remains the same, by frequency of wave will decrease. [M1] (wavelength has increased, driving frequency will decrease, for the same speed. Since driving frequency is not equal to natural frequency, amplitude will decrease. [A1] A.k.a. no more resonance) (iii) Damping force increases [M1]  amplitude decreases [A1] 3 (a) The graph in Fig. 3.1 curves towards the current axis Hence the ratio of V to I will decreases with increasing voltage OR the line drawn from origin to a point on the curve becomes gentler as voltage increases. Hence the ratio of V to I, which is the inverse of the gradient of this line, decreases. [B1] [B1] [B1] [B1] (b) The semiconductor has a band structure with a small band gap between the fully-filled valence band and the totally-unfilled conduction band at zero kelvin. With an increase in temperature, some electrons from the valence band can [B1]

easily gain enough thermal energy to promote to the conduction band via the small gap. The electrons in the conduction band and holes in valence band behave as mobile charge carriers, increasing the number of mobile positive and negative charge carriers. [B1] [B1] (c) (i) Current through X = 90 – 47 = 43 mA From Fig. 3.1, p.d. across X = 6.2 V [C1] [A1] (ii) p.d. across R and X = p.d. across 180 Ω resistor = I x R = 47 x 10-3 x 180 = 8.46 V p.d. across R = p.d. across R and X – p.d. across X = 8.46 – 6.20 = 2.26 V R = V / I = 2.26 / 0.043 = 52.6 . [C1] [A1] (iii) Since the voltage