Trial Examination 202 VCE Physics Unit 3 Written Examination Suggested Solutions Neap Trial Exams are licensed to be photocopied or placed on the school intranet and used only within the confines of the school purchasing them, for the purpose of examining that school s students only. They may not be otherwise reproduced or distributed. The copyright of Neap Trial Exams remains with Neap. No Neap Trial Exam or any part thereof is to be issued or passed on by any person to any party inclusive of other schools, non-practising teachers, coaching colleges, tutors, parents, students, publishing agencies or websites without the express written consent of Neap. Copyright 202 Neap ABN 49 90 906 643 96 06 Pelham St Carlton VIC 3053 Tel: (03 834 834 Fax: (03 834 8300 TEVPHYU3_SS_202.FM
SECTION A CORE Area of study Motion in one and two dimensions Question a. N = mgcosθ = 800 0 cos( 8 N =.7 0 4 N b. ΣF = ma= T friction mgsinθ Since speed is constant ΣF = 0 0 = T 300 ( 700 0 sin( 8 T = 2.5 0 3 N c. For whole system, ΣF = 0 = F D 850 300 ( 2500 0 sin( 8 F D = 8.9 0 3 N W Fd Power ---- ------ 8.9 0 3 5 = = = ( ----- t t 8 Power =.7 0 4 W Question 2 a. r = 0.8sin( 5 = 0.2 m 2πr ( 2 π 0.2 v = -------- = --------------------------------- T.7 v = 0.77 m s b. Vertically, ΣF = 0 so mg= Tcosθ m 0 = 5.6cos( 5 m = 0.54 kg mv 2 ( 0.54 ( 0.77 2 F c = -------- = -------------------------------------- r 0.2 (Note: give consequential mark if mass calculated is used here F c OR =.5 N F c = T sin5 = 5.6 sin5 =.5 N 2 TEVPHYU3_SS_202.FM Copyright 202 NEAP
c. 2 marks mg T = ------------------ so if m increases, T will increase. sin( 5 mv 2 F c = -------- so if m increases, F r c will increase. d. At the top, when the ball is travelling at its minimum speed T = 0 ΣF = -------- mv2 = mg r v 2 0.54 ------ = ( 0.54 0 0.8 v = 2.8 m s OR At top of flight, v = rg = ( 0.8 0 v = 2.8 m s Question 3 a. Kinetic energy of ball = elastic potential energy stored in spring E k = 0.5mv 2 = 0.5( 0.02 ( 2 2 =.4 J U e = 0.5kx 2 so.4= 0.5k( 0.08 2 K = 450 N m b. Vertically to the highest point: u = 2sin( 35, v = 0, a = 0, x =? v 2 = u 2 + 2ax 0= ( 2sin( 35 2 + ( 2 ( 0 x x = 2.4 m c. To find the speed of ball when it lands: u = 2sin( 35, a = 0 m s 2, x =.2 m, v =? ( v 2 = u 2 + 2ax =( 2sin( 35 2 + ( 2 0.2 v = 4.8 m s or 4.8 m s so on the way down it will be 4.8 m s To find the time of flight: v = u+ at 4.8 = 2sin( 35 + ( 0t so t =.7 seconds Note: the time of flight can also be found by solving a quadratic equation but this method is not required. To find the range, x = ut = 2cos( 35.7 =.5 m Copyright 202 Neap TEVPHYU3_SS_202.FM 3
Question 4 a. P i = P f ( 600 5 + ( m 2 = ( m + 600 3000 2m = m + 600 m = 800 kg b. m v = ΣFt 600 ( 5 = ΣF 0.2 ΣF =.2 0 4 N c. The change in kinetic energy for each car is a fixed quantity, this is equal to the work done. The rigid metal bumper bar will compress less than the rubber bumper bar, hence the car will stop in a shorter distance. Since W = Fx, if W is constant and x is decreased, the Force (F will increase. Note: do not accept answer if it is in terms of momentum change and time rather than energy change and distance. Question 5 a. r 3 GM ---- = ------------ T 2 ( 4π 2 r 3 ( 6.67 0 ( 6.37 0 23 ----------------------------------- = ----------------------------------------------------------------------- ( 24.6 3600 2 4π 2 r = 2.04 0 7 m Altitude = r radius of Mars = ( 2.04 0 7 ( 3.43 0 6 Altitude =.7 0 7 m b. Apparent weight = 0 N The only force acting on the satellite is gravity, so it is in freefall and the apparent weight is zero (there is no normal force which is the apparent weight. c. GM ( 6.67 0 ( 6.37 0 23 g = -------- = ----------------------------------------------------------------------- r 2 ( 3.43 0 6 2 g = 3.6 N kg Note: no marks to be given unless calculations are shown.. 4 TEVPHYU3_SS_202.FM Copyright 202 NEAP
Area of study 2 Electronics and photonics Question a. 4.0Ω Resistors Y and Z are in series and equivalent to 2.0 Ω. The YZ resistor combination is in parallel with resistor X which is 6.0 Ω Using the parallel resistance formula, ----- = ----- + -- = --, R p 2 6 4 which means R p = 4.0 Ω. b..0 A Resistors Y and Z are in series and equivalent to 2.0 Ω and the current through each resistor will be the same as they are in series. The total voltage drop across the YZ combination is 2 V. V 2 I = -- = ----- =.0 A 2 marks R 2 c. 24 W The power dissipated in resistor X is given by P = IV. V 2 I = -- = ----- = 2.0 A R 6 P = IV = 2.0 2 = 24 W. Question 2 a. YES b. Because the LED is in forward bias. c. 22.5 ma From the graph the voltage across the LED is.5 V. Therefore the voltage across the resistor is 6.0.5 = 4.5 V. The current through the LED will be the same as the current through the resistor. V 4.5 I = -- = -------- = 22.5 ma R 200 d. A If the 200 Ω resistor is replaced with a 300 Ω resistor, the current through the LED-resistor series combination will be smaller and therefore the LED will glow less brightly than before. 2 marks Question 3 a. 5.0 kω The value of the resistance of the thermistor when the temperature in the oven is 00 C is read directly from the graph. b. 750 Ω The value of the resistance of the thermistor when the temperature in the oven is 400 C is read directly from the graph (allow a range of 700 Ω 800 Ω. Copyright 202 Neap TEVPHYU3_SS_202.FM 5
c. 6.0 V V OUT R ------------------ T R T + R V = VIN 5 = ----------- 5 + 5 2 V = 6.0 V d. 0.4 V 2 marks V OUT R ------------------ V R T + R V = VIN 5 = ------------------- 5 + 0.75 2 V = 0.4 V 2 marks Note that there will be a small allowable range based on consequentials from Question 3. Question 4 a. C V OUT V IN 3 The gain of the amplifier is given by ----------- = --------- = 00. 2 marks 0.03 Note 30 mv = 0.03 V b. Inverting amplifier. The gradient of the graph is negative. c. The graph of the output voltage versus time graph is marked as follows: for correct voltage values and correct period. for showing inverting nature of amplifier. for showing clipping at the +3 V and 3 V marks. V OUT 3 2 0 20 30 40 t (ms 2 3 3 marks 6 TEVPHYU3_SS_202.FM Copyright 202 NEAP
SECTION B DETAILED STUDIES (2 marks for each correct answer Detailed study Einstein s special relativity Question An inertial reference frame is one which is absolutely stationary or which is moving with constant velocity. Question 2 The Lorentz factor, γ = ----------------- v 2 ---- c 2 γ = --------------------------------- ( 0.95c 2 ------------------- = 3.2 c 2 Question 3 The Michelson Morley Experiment of 887 demonstrated that c is constant everywhere. Question 4 A nuclear power station best relates to Einstein s concept that mass can be converted to energy Question 5 Using the Lorentz equation, γ = ----------------- ---- Therefore solving for v gives v = 0.466 c. An increase of 50% means the new speed is v = 0.6249 c. v 2 c 2.0 = ----------------- v 2 ---- c 2 Substitution into the γ equation gives γ =.28. Question 6 Proper time is best described as the time recorded in the reference frame at rest with respect to the event. Copyright 202 Neap TEVPHYU3_SS_202.FM 7
Question 7 The effect of the proton being accelerated in a new.5 GeV circular particle accelerator is that it will increase its mass (it will also increase its speed but not to the speed of light as in answer C. Question 8 The time dilation factor would be large as the proton travels around the.5 GeV circular particle accelerator because it increases its speed significantly enough in this high energy accelerator and t = -----------------. Question 9 The equivalent energy of an electron of mass 9. 0 3 kg is given by E = mc 2. E = ( 9. 0 3 ( 3.0 0 8 2 = 8.2 0 4 J t 0 v 2 ---- c 2 Question 0 L ---- = -- L 0 γ 20 ----- = -- 50 γ γ = 2.5 Question James Clerk Maxwell s equations demonstrated the existence of electromagnetic waves that can travel at 3.0 0 8 m s. Question 2 A tau meson which has a Lorentz factor of 20 has a speed 0.998749 c. Use the formula γ = ----------------- and solve for v v 2 ---- c 2 8 TEVPHYU3_SS_202.FM Copyright 202 NEAP
Detailed study 2 Materials and their use in structures VCE Physics Unit 3 Trial Examination Suggested Solutions Question The maximum tensile strength is the stress at the point where the material fractures. From the graph, this is equal to.8 0 6 N m. Question 2 The material is ductile as it has a non linear region, indicating that it undergoes plastic deformation. This means that it cannot be brittle. Options C and D are not correct as there is no data which states what defines a tough or strong material. Question 3 Strain = 2.0 0 3 so stress = 8.0 0 5 N m 2 (from graph F Stress = -- so 8.0 0 5 (.0 0 4 = -------------------------- A A A =.3 0 2 m 2 Question 4 Young s modulus = gradient of linear section of graph. ( 6 0 5 Y = ------------------------ = 4.0 0 8 N m 2 ( 4 0 3 Question 5 Material X has a greater area under the graph than material Y so X is tougher. Material Y has a greater maximum tensile strength than material X so Y is stronger. Question 6 Stress =.4 0 6 N m 2 so strain = 2.0 0 3 (from graph l Strain = ---- so 2.0 0 3 l = ---- l 2 l = 4.0 0 3 m = 4.0 mm Question 7 Torque = F.x = ( 5 0 0.5 = 25 N m Copyright 202 Neap TEVPHYU3_SS_202.FM 9
Question 8 Take Στ = 0 around support B. (0.9 20 0 + (.3 5 0 = F A = 36 N F A.8 Question 9 Take Στ = 0 around support B. ( 80 + 3m F books A = ---------------------------------------- so if m increases, will increase..8 books F A Take Στ = 0 around support A. ( 80 + 5m F books B = ------------------------------------- so if m increases, will increase..8 books F B Question 0 Section AB of the shelf will sag under its own weight. This means that the top part of it will be in compression and the bottom region will be in tension. From the diagram, the cable will provide an upwards force on the shelf so it must be in tension. Question Take Στ = 0 around point A. (0 0 0.6 + (5 0.2 = T = 207 N T sin( 40 0.9 Question 2 For the structure to be stable horizontally F = 0. Since cable BC has a component of force to the left, the wall must exert a force component to the right to balance this. For the structure to be stable vertically F = 0. Upwards forces = T sin( 40 = 207 sin( 40 = 33 N Downward forces = ( 0 0 + ( 5 0 = 50 N Since the downward forces are greater than the upward forces, the wall must exert an upwards force component on the shelf. 0 TEVPHYU3_SS_202.FM Copyright 202 NEAP
Detailed study 3 Further electronics Question The frequency of the AC voltage is given by f = -- T = ----- ms 20 = 50 Hz Question 2 The RMS voltage of the power supply is given by V RMS = V PEAK ------ 2 340 = -------- V 2 = 240 V Question 3 V RMS = V PEAK ------ 2 4000 = ----------- V 2 N ----- p N s V IN = ----------- V OUT 4000 = 240 ----------- 2 = :2 Question 4 P = IV 240 = I 240 I =.0 A Question 5 The circuit shown is a half wave rectification circuit. Copyright 202 Neap TEVPHYU3_SS_202.FM
Question 6 A capacitor in parallel with the R LOAD resistor will smooth the output. Question 7 The time constant for the RC circuit is given by τ = RC 2 0 3 = ------------------ 0 4 = 0.2 s Question 8 This circuit models a voltage regulator. Question 9 A Zener diode can be used in an electronic circuit as a voltage regulator by using the Zener diode in reverse bias. This is because in breakdown mode it has a constant voltage across it. Question 0 The value of the resistive load can be found using V = IR 9.0 R = --------- = 60.0 Ω 0.5 Question One time constant represents a 63% decay in the voltage or a voltage of 7.4 V which is closest to a time value of 0.02 s as determined from the graph. Question 2 Heat adversely affects the performance output of the transistor. 2 TEVPHYU3_SS_202.FM Copyright 202 NEAP