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2006 VCAA Physics Exam 1 Solutions © Copyright 2006 itute.com Do not photocopy Free download and print from www.itute.com SECTION A – Core Area of study 1 – Motion in one and two dimensions Q1 70 N Q9 Relative to Mary, the ball has a horizontal velocity component to the rear of the train. E. Q10 Conservation of momentum: 6000× + 5.00 + M × 0 = (6000 + M )× + 0.098 , ∴ M = 3.0 × 10 5 kg. Q11 Shuttle: impulse = change in momentum, r r Fav × 20 = 6000× + 0.098 − 6000× + 5.00, ∴F av = − 1471 . 190 N Average force on shuttle = 1.5 × 10 3 N. F (driving force on the rear tyre by the road) Constant speed, no acceleration, ∴ Fnet = 0, + F + − 70+ − 190 = 0, ∴ F = 260 N, ∴driving force on the road by the rear tyre is 260 N. Q2 70 N T Constant speed, no acceleration, ∴ Fnet = 0, ∴T = 70 N. + T + − 70 = 0, Q12 Total work done on Sam = heat energy generated due to friction + increase in Sam’s gravitational potential energy. 22720 = 300 × L + 13720 , ∴ L = 30 m. Q13 Increase in gravitational potential energy = mgh , ∴13720 = 70 × 10 × h, ∴ h = 19.6 m. Q14 Change in elastic potential energy = change in gravitational 1 potential energy, i.e. kx 2 = mgh, 2 1 2 ∴ k 8 = 70 × 10 × 18, hence k = 3.9 × 10 2 Nm-1. 2 Q15 a = g ∝ Q3 When the cyclist stops pedalling, driving force = 0. Work done by opposing forces = ∆E K , (70 + 190)x = 1 (90 + 40) × 6.0 2 , x = 9.0 . 2 Distance travelled during slowing = 9.0 m. Q16 Since a = g in orbit, ∴ + − 112 v2 = 60 × = 9.7 × 10 2 N 7.5 r 4π 2 r GM = 2 , T2 r 4π 2 r 3 = 1.6 × 10 5 s. GM ∴T = Q4 Friction force F = weight force W = mg = 60 × 10 = 600 N Q5 Fnet = ma, C = m M 1 1 , ∴ a = 10 × × 2 = 4 ms-2. 2 1 10 ( 2 ) r Area of study 2 – Electronics and photonics Q1 Point 1 + Q6 Fnet = 22+ 0.5 × 10= 17 N, resultant force = 17 N. + Fnet 17 + = = 34 ms-1 (constant upward acceleration), m 0.5 u = 0, t = 1.5 s, s = ? 1 1 s = ut + at 2 = 0 + × 34 × 1.5 2 = 38.25 . Height = 38 m. 2 2 2 Q7 a = 3 Voltage (V) 20 1 × 20 = 4 1+ 4 20 − 8 mA × 1 kΩ = 12 Q2 ∆vOUT = −200 × ∆v IN = −200 × 50 mV = −10 V, ∆v R + 10 ∴ ∆v R = +10 V. ∴ ∆iC = = = +0.01 A = +10 mAp-p. RC 1000 C Q8 Horizontal momentum = horizontal impulse = 22 × 1.5 = 33 kgms-1. p 33 = 66 ms-1. After 1.5 s, horizontal velocity = = m 0 .5 Vertical component: u = 0, a = −10, t = 1.5 , v = u + at = −15 . ∴speed = 15 2 + 66 2 = 68 ms-1 15 θ = tan −1 = 13 o . 66 Copyright itute.com 2006 C Q3 v IN is too large and clipping occurs. No change in period. 8 vOUT (V) 0 θ 4 time (ms) − 12 1 Q4 A decoupler separates AC and DC components of signals in a transistor amplifier. For a correctly biased transistor amplifier only AC signals are fed to the input. To ensure this a capacitor (CIN) acts as a decoupler to filter out the DC component from a signal. Another capacitor (COUT) ensures only amplified AC signals at the output. Capacitor CE filters out the AC component across RE to maintain a constant biased DC voltage at the input. Q5 Voltage across the 100-ohm resistor = 5.0 − 0.7 = 4.3 V, V 4.3 ∴I = = = 0.043 A = 43 mA. R 100 Q9 L ′ = L 1 − ∴1 − v2 L′ v2 , ∴ = 1 − 2 = 99% = 0.99 , 2 L c c v2 = 0.9801 , ∴ v = 0.0199 × c = 4.23 × 10 7 ms-1. c2 Q10 As the electron is accelerated to greater and greater speeds close to c, its mass becomes larger and larger approaching infinity. B Q11 mc 2 = mo c 2 + E K . C Q6 Read from graph, 100Ω. Q7 At T = 10 o C, RTHERM = 400 Ω, VTHERM = VOUT = 4.0 V, ∴ V R = 12 − 4.0 = 8.0 V. Since R RTHERM = VR VTHERM , ∴R = 8.0 × 400 = 800 Ω. 4.0 Detailed study 2 – Investigating materials and their use in structures Q1 B Q2 Polyethylene – it absorbs more energy before breaking (larger area under σ-ε graph). Q8 Modulation is a particular form of coding. Information signal (sound signal) is used to modulate the light signal, which is then transmitted through the fibre-optic cable. Demodulation is the reverse process of modulation in which information carried by the modulated light signal is recovered and decoded. Q3 When placed under a tensile stress of 20 MPa, ∆L ε= ≈ 2% = 0.02 (from graph) ∴ ∆L = 0.02 × 5.0 = 0.10 m. L Q9 At P, LED; at Q, LDR. Q4 Young’s modulus = SECTION B – Detailed studies Detailed study 1 – Einstein’s special relativity Q5 σ = Q1 The experimenters did not detect any difference in the speeds of two light beams (one parallel and one perpendicular to the direction of the rotation of the Earth). One of the possible explanations for the null result was that ether does not exist. Q2 An inertial frame of reference is one in which Newton’s first law (the law of inertia) is valid. Q3 L is the proper length because it was measured by Kris who was at rest in her rocket ship. 60 MPa = 3000 MPa. 0.02 F mg = , A A σA 62 × 10 6 2.0 × 10 −4 ∴m = = = 1.2 × 10 3 kg. 10 g ( )( ) Q6 Work = area under σ-ε graph × volume of material 1 = × 62 × 10 6 × 0.021 × 2.0 × 2.0 × 10 − 4 ≈ 260 J. 2 ( ) Q7 Steel rods are strong in tension and used to reinforce concrete under tensile stress. Q8 C Q4 L ′ = L 1 − 2 v , ∴L = c2 L′ v2 1− 2 c . B Q9 D Q10 FB 8.0 m Q5 Lee: 340 − 30 = 310 ms-1; Sung: 340 + 30 = 370 ms-1. Q6 Light propagates through empty space with a definite velocity c = 3.0 × 10 8 ms-1 independent of the velocity of the source and/or observer. A Q7 Dilated time = 2.2µs c ) 1 − ( 0.995 c 2 = 22 µs 1.0 m B FA 5000 N Net torque about point B = 0, + FA × 8.0+ − 5000 × 1.0 + FB × 0 = 0 , ∴ FA = 625 N. 2 0.995c Q8 Contracted height = 2627 1 − = 262 m c Copyright itute.com 2006 Q11 D 2 Detailed study 3 – Further electronics V 2 12 2 = = 0.60 W R 240 Q1 P = N S VS 18 , ∴ NS = = × 4800 = 360 240 N P VP Q2 Q3 V peak = 18 2 ≈ 25.5 V, period = 1 1 = = 0.02 s = 20 ms. f 50 B Q4 None of the sketches shown. Voltage between points 2a and 2b is the same as that between points 3a and 3b. See Q5. Q5 The capacitor smooths the voltage. D Q6 The voltage regulator keeps the output voltage at 12 V. 360 × 140 = 10.5 VRMS = 14.8 Vpeak. 4800 After rectification this peak voltage falls below the voltage required by the voltage regulator to operate in the voltageregulating region, causing the voltage output of the power supply to fall below 12 V and possibly the appearance of ripples. Q7 Output of transformer = Q8 A Q9 Output current increases; output voltage remains the same. Q10 Time constant = RC = 2000 × 10000 × 10 −6 = 20 s Q11 Voltage(V) 10 8 6 4 2 Time(s) 0 0 10 20 30 40 50 60 Q12 After 10 s the capacitor discharges by 63% of 10 V, i.e. 6.3 V. Voltmeter reading = 10 − 6.3 = 3.7 V. Please inform [email protected] re conceptual, mathematical and/or typing errors Copyright itute.com 2006 3