Physics
THE AGE MIDYEAR VCE EXAM GUIDE 2012
Physics MELANIE LANE The mid-year Physics exam will look a little different this year, with minor but significant changes to the format of the paper. The numbering of questions in Section A Core-Areas of study - ‘Motion’ and ‘Electronics and Photonics’ - will continue to be grouped according to the information to which they relate, but they will now be presented in multiple parts. For example, rather than the preamble “The following information relates to Questions 1-3’, the questions will be presented as Question 1a, Question 1b, Question 1c. Hence the total number of questions will be reduced, but the overall length of the paper and the style of questions being asked will remain the same. Motion will have eight questions for a total of 40 marks, while Electronics and photonics will have five questions for 26 marks. For more details please check the cover of the June examination published on the VCAA website at: http://www.vcaa.vic.edu.au/vcaa/vce/exams/examcovers/2012_June_covers/2012physics1cover.pdf The detailed study section will still be examined by 12 multiple choice questions. Even though June 12 seems just around the corner there is still LOADS of time to prepare. No excuses! The saying ‘Practice makes perfect’ could never ring more true than when preparing for the Physics paper. Go to the VCAA website and make sure you attempt all past papers and don’t forget to read the chief assessor’s report! Here are some general tips to ensure you are fully rewarded for your written answers: Make your answers clear and concise. Don’t ramble! Rambling answers, which try to touch on many, often irrelevant, points can end up being contradictory and do not score well. Don’t simply copy material from your A4 sheets. It’s a good idea to re-read your final answer and check that you have actually answered the question asked. Don’t hesitate to include clear, concise and labelled diagrams as part of an explanation. Show all working. Motion in one and two dimensions: For students sitting the 2011 paper in June the circular motion questions proved difficult. Questions 4-6 involved a car cornering on a banked road at the ‘design speed’. Question 4 was done well and students were able to use the formula to find the net force, 3375 N. Question 5 proved more difficult: the bike and the rider had two forces acting on them as seen in figure 2b: the weight force and the normal reaction force .). Students were asked to draw an (i.e. one only) arrow to show the direction of the non-zero net force responsible for circular motion. A horizontal arrow to the left was required. Students were also required to explain how this non-zero net force came about. Many could not. In this situation, the normal reaction force has a horizontal
component which provided the centripetal force. The vertical component of the normal reaction force was balanced out by the weight force of the bike and rider (mg). Many students knew the net force was centre- seeking and to the left, but where unsure where the centre was, and many did not draw a horizontal arrow. Many students were able to use either = or to get 48.4 0, the answer required. (Note: because of the new format these three questions would now be, say, Q 4 a, b, c , because they all relate to the same situation.) Question 17-20 were also challenging. An understanding between the independent variable ‘displacement’ ( ), and the dependent variable (‘energy’ in Joules), was needed. It was important to decide if it was a linear or squared relationship, and whether the type of energy being investigated was increasing or decreasing in magnitude as it moved upward from 80 cm to 60 cm. Graph D best showed the variation of the kinetic energy of the system plotted against the length of the stretched spring. was maximum at zero displacement (i.e. 70 cm) when the spring is at its fastest, and is minimum at maximum displacements (i.e. 60 cm and 80 cm). Graph C best represented the variation of the total energy of the spring/mass system plotted against the length of the stretched string, as the total energy remained constant throughout the experiment. It is considered a closed system. Graph B best showed the variation of the gravitational potential energy of the spring/mass system (measured from the lowest point with zero energy) plotted against the length of the stretched string. Note that = and hence varied directly to (or in this case and therefore the relationship was linear. Graph F best shows the variation of the spring (strain) potential energy vs the length of the string. As = , then is directly proportional to and therefore the graph is parabolic. However, it was important to recognise that the spring was always under strain while extended between 60 cm and 80 cm, because the spring had been stretched from its original length of 40 cm by the hanging mass. Hence the strain energy will be minimum at 60 cm but not zero. Hence, the most incorrect response chosen was often E. This question was not done well by students sitting the 2011 paper. Electronics and photonics: In this area of study many students sitting the June 2011 paper found questions 1-4 challenging. Question 4 proved to be one of the most difficult of the paper. Question 1 was a ‘show that….’ question. The answer was given and students needed to show working to find that given answer. (This avoided students getting stuck on the first question and not being able to answer the next three! Show every step and set out your work very clearly to ensure full marks are awarded. Resistors RC and RD were in series so their resistance should have be added algebraically, i.e. 4 . Both these resistors were in parallel with RB and hence had an equivalent resistance of 1.3 . (Note: watch when using the reciprocal formula for effective resistance of resistor in parallel, it can be problematic for some!) When this equivalent resistance was added to RA which is 2 . This resulted in answer 3.3 “as required”. Most students easily found IT = 3.03 A using Ohms law and the RTfrom Q 1, but they were unable to calculate the current through RB. Many students simply divided the current by two. However, the branch containing RB had a lower resistance than the branch with RC and RD, and hence more current flowed through RB . Specifically, the current divide in a 2:1 ratio. Question 3 was better done with students using Ohm’s Law,
where I = 3.03 A and R = 2.0 to get V = 6.0 V across RA. However, Question 4 proved difficult and generally students sitting the paper scored low. Students needed to know there was 1.0 A through this branch and hence 1.0 A was flowing through R D. Also, if there was 6.0 V across RA, there would be 4 V across both branches, and hence 4 V shared across R C and RD. Since they had the same resistance, they would have dissipated the same voltage, hence 2 V. Using two of the following V = 2.0 V, I = 1.0 A or R = 2 for resistor D, and the appropriate power equation, the power dissipated in resistor D was 2.0 W. A basic understanding and application of Kirchoff’s voltage and current laws in series, parallel and combination circuits is vital! Make sure you know all your power equations. Detailed Studies: Einstein’s Special Relativity: In the past students answering questions regarding this detailed study often had difficulty with the substitution and transposition of the time dilation and length contraction formulae. Usually students have been able to identify the appropriate formula, but they have difficulty in determining which is the proper frame of reference. Investigating materials and their use in structures: Every year questions involving translational and rotational effects of forces (torque) in this area of study proved difficult. Students attempting this section must make sure they set their answers out correctly and clearly identify the points on which they are taking torques. Finally, don’t forget to include the gravitational field strength “g” when converting from mass into weight force and to multiply W by r when finding the torque. Further Electronics: Students from previous years have often incorrectly substituted and transposed the transformers turn/voltage ratio formula NP/NS = VP/VS. Therefore, the reciprocal of the correct answer was often given. Recognising the differences between linear components, where Ohm’s law can be applied, and non-linear components, where the relationship is acquired from a graph (I-V characteristics), still proves to be an area of concern in both this detailed study and area of study 2 in the core. Finally: Attempt only one of the Detailed Study sections. Take one scientific calculator into the examination (Graphics calculators are not allowed). Ensure your scientific calculator is in degree mode. Melanie Lane is head of science at Haileybury.
Physics MELANIE LANE The mid-year Physics exam will look a little different this year, with minor but significant changes to the format of the paper. The numbering of questions in Section A Core-Areas of study - ‘Motion’ and ‘Electronics and Photonics’ - will continue to be grouped according to the information to which they relate, but they will now be presented in multiple parts. For example, rather than the preamble “The following information relates to Questions 1-3’, the questions will be presented as Question 1a, Question 1b, Question 1c. Hence the total number of questions will be reduced, but the overall length of the paper and the style of questions being asked will remain the same. Motion will have eight questions for a total of 40 marks, while Electronics and photonics will have five questions for 26 marks. For more details please check the cover of the June examination published on the VCAA website at: http://www.vcaa.vic.edu.au/vcaa/vce/exams/examcovers/2012_June_covers/2012physics1cover.pdf The detailed study section will still be examined by 12 multiple choice questions. Even though June 12 seems just around the corner there is still LOADS of time to prepare. No excuses! The saying ‘Practice makes perfect’ could never ring more true than when preparing for the Physics paper. Go to the VCAA website and make sure you attempt all past papers and don’t forget to read the chief assessor’s report! Here are some general tips to ensure you are fully rewarded for your written answers: Make your answers clear and concise. Don’t ramble! Rambling answers, which try to touch on many, often irrelevant, points can end up being contradictory and do not score well. Don’t simply copy material from your A4 sheets. It’s a good idea to re-read your final answer and check that you have actually answered the question asked. Don’t hesitate to include clear, concise and labelled diagrams as part of an explanation. Show all working. Motion in one and two dimensions: For students sitting the 2011 paper in June the circular motion questions proved difficult. Questions 4-6 involved a car cornering on a banked road at the ‘design speed’. Question 4 was done well and students were able to use the formula to find the net force, 3375 N. Question 5 proved more difficult: the bike and the rider had two forces acting on them as seen in figure 2b: the weight force and the normal reaction force .). Students were asked to draw an (i.e. one only) arrow to show the direction of the non-zero net force responsible for circular motion. A horizontal arrow to the left was required. Students were also required to explain how this non-zero net force came about. Many could not. In this situation, the normal reaction force has a horizontal
component which provided the centripetal force. The vertical component of the normal reaction force was balanced out by the weight force of the bike and rider (mg). Many students knew the net force was centre- seeking and to the left, but where unsure where the centre was, and many did not draw a horizontal arrow. Many students were able to use either = or to get 48.4 0, the answer required. (Note: because of the new format these three questions would now be, say, Q 4 a, b, c , because they all relate to the same situation.) Question 17-20 were also challenging. An understanding between the independent variable ‘displacement’ ( ), and the dependent variable (‘energy’ in Joules), was needed. It was important to decide if it was a linear or squared relationship, and whether the type of energy being investigated was increasing or decreasing in magnitude as it moved upward from 80 cm to 60 cm. Graph D best showed the variation of the kinetic energy of the system plotted against the length of the stretched spring. was maximum at zero displacement (i.e. 70 cm) when the spring is at its fastest, and is minimum at maximum displacements (i.e. 60 cm and 80 cm). Graph C best represented the variation of the total energy of the spring/mass system plotted against the length of the stretched string, as the total energy remained constant throughout the experiment. It is considered a closed system. Graph B best showed the variation of the gravitational potential energy of the spring/mass system (measured from the lowest point with zero energy) plotted against the length of the stretched string. Note that = and hence varied directly to (or in this case and therefore the relationship was linear. Graph F best shows the variation of the spring (strain) potential energy vs the length of the string. As = , then is directly proportional to and therefore the graph is parabolic. However, it was important to recognise that the spring was always under strain while extended between 60 cm and 80 cm, because the spring had been stretched from its original length of 40 cm by the hanging mass. Hence the strain energy will be minimum at 60 cm but not zero. Hence, the most incorrect response chosen was often E. This question was not done well by students sitting the 2011 paper. Electronics and photonics: In this area of study many students sitting the June 2011 paper found questions 1-4 challenging. Question 4 proved to be one of the most difficult of the paper. Question 1 was a ‘show that….’ question. The answer was given and students needed to show working to find that given answer. (This avoided students getting stuck on the first question and not being able to answer the next three! Show every step and set out your work very clearly to ensure full marks are awarded. Resistors RC and RD were in series so their resistance should have be added algebraically, i.e. 4 . Both these resistors were in parallel with RB and hence had an equivalent resistance of 1.3 . (Note: watch when using the reciprocal formula for effective resistance of resistor in parallel, it can be problematic for some!) When this equivalent resistance was added to RA which is 2 . This resulted in answer 3.3 “as required”. Most students easily found IT = 3.03 A using Ohms law and the RTfrom Q 1, but they were unable to calculate the current through RB. Many students simply divided the current by two. However, the branch containing RB had a lower resistance than the branch with RC and RD, and hence more current flowed through RB . Specifically, the current divide in a 2:1 ratio. Question 3 was better done with students using Ohm’s Law,
where I = 3.03 A and R = 2.0 to get V = 6.0 V across RA. However, Question 4 proved difficult and generally students sitting the paper scored low. Students needed to know there was 1.0 A through this branch and hence 1.0 A was flowing through R D. Also, if there was 6.0 V across RA, there would be 4 V across both branches, and hence 4 V shared across R C and RD. Since they had the same resistance, they would have dissipated the same voltage, hence 2 V. Using two of the following V = 2.0 V, I = 1.0 A or R = 2 for resistor D, and the appropriate power equation, the power dissipated in resistor D was 2.0 W. A basic understanding and application of Kirchoff’s voltage and current laws in series, parallel and combination circuits is vital! Make sure you know all your power equations. Detailed Studies: Einstein’s Special Relativity: In the past students answering questions regarding this detailed study often had difficulty with the substitution and transposition of the time dilation and length contraction formulae. Usually students have been able to identify the appropriate formula, but they have difficulty in determining which is the proper frame of reference. Investigating materials and their use in structures: Every year questions involving translational and rotational effects of forces (torque) in this area of study proved difficult. Students attempting this section must make sure they set their answers out correctly and clearly identify the points on which they are taking torques. Finally, don’t forget to include the gravitational field strength “g” when converting from mass into weight force and to multiply W by r when finding the torque. Further Electronics: Students from previous years have often incorrectly substituted and transposed the transformers turn/voltage ratio formula NP/NS = VP/VS. Therefore, the reciprocal of the correct answer was often given. Recognising the differences between linear components, where Ohm’s law can be applied, and non-linear components, where the relationship is acquired from a graph (I-V characteristics), still proves to be an area of concern in both this detailed study and area of study 2 in the core. Finally: Attempt only one of the Detailed Study sections. Take one scientific calculator into the examination (Graphics calculators are not allowed). Ensure your scientific calculator is in degree mode. Melanie Lane is head of science at Haileybury.
