VJC 2022 Prelim P3 Guide

1 2022 VJC Prelim H2 P3 Suggested solution 1(a) At maximum height, Loss in KE of mass = gain in GPE of mass + gain in EPE in cord [CONCEPT] 2211 022mv mgh kx− = + 3 2 3 211(150 10 )(5.7 ) 0 (150 10 )(9.81)(1.12) (45)22 x−− − =  + 0.19x= m (b) Length of unstretched cord = 1.12 – 0.19 = 0.93 m (c) (G should end lower so that G + E = initial K at 1.12 m) (d) Since the tension acting on the ball is always towards the center and perpendicular to the displacement of the ball [APPLICATION], there is no work done by the tension. [CONCLUSION] [CONCEPT: WD = force x displacement in the direction of the force] 2(a) When the sound wave hits the wall, it will be reflected in the opposite direction and interfere with the original wave. [APPLICATION] Now we have 2 waves of the same amplitude, frequency and wavelength moving in opposite directions interfering with each other [APPLICATION], giving rise to a stationary wave. [CONCLUSION] [CONCEPT: conditions for formation of stationary wave] Energy/J displacement/m K G E 0.93 1.12

2 (b) (c) From above diagram,  = 10 cm Frequency f = v  = 360 0.10 = 3600 Hz d(i) After hitting the wall, the wave will continue to spread out as before, but in the opposite direction. From diagram, reflected wave travels 25 cm. r1 = 5 cm, amplitude A1 = 3.0 x 10-5 m r2 = 25 cm, A2 = ? Intensity I  A2  2 1 r [CONCEPT]  A  1 r 21 12 Ar Ar = A2 = A1 1 2 r r = 3.0 x 10-5 x 5 25 = 6.0 x 10-6 m Source Wall 15 cm  = 10 cm Source X 5 cm 10 cm

3 (ii) Since X was originally a node, it is a place of destructive interference.  resultant amplitude = (3.0 – 0.6) = 2.4 x 10-5 m 3(a) The electric field strength at a point is defined as the electric force exerted per unit positive charge placed at that point. (b)(i) (Before x = 10 cm, the electric field points to the right. After 10 cm, it points to the left.) Since the electric field changes direction between A and B, the charges have the same sign. OR At a point between A and B, the electric fields due to each charge cancel out to give a resultant field strength of zero. So the charges have the same sign. (ii) Any 1 point: • Charge A is not the only charge present. • The electric field strength is also influenced by charge B. • The electric field strength is due to two/both charges. • The electric field strength is the resultant of the two fields due to charges A and B. (iii) From the graph, E = 1.8  103 N C-1 where x = 6.0 cm. Electric force on proton, FE = qE = (1.60  10−19)(1.8  103) = 2.88  10−16 N Acceleration of proton, 16 27 2.88 10 1.67 10 EFa m − − ==  = 1.72  1011 m s−2 (iv) At x = 10 cm, field strength due to A = field strength due to B (in magnitude) ABEE= [CONCEPT] 22 0044 AB AB QQ rr = 2 2 22 (10.0 cm) 4(5.0 cm) AA B