Physics (2007) Sample assessment instrument and indicative responses Supervised assessment 3 December 2010 Purposes of assessment 1 The purposes of assessment are to: promote, assist and improve student learning inform programs of teaching and learning provide information for those people — students, parents, teachers — who need to know about the progress and achievements of individual students to help them achieve to the best of their abilities provide information for the issuing of certificates of achievement provide information to those people who need to know how well groups of students are achieving (school authorities, the State Minister for Education and Training and the Arts, the Federal Minister for Education). It is common practice to label assessment as being formative, diagnostic or summative, according to the major purpose of the assessment. The major purpose of formative assessment is to help students attain higher levels of performance. The major purpose of diagnostic assessment is to determine the nature of students’ learning, and then provide the appropriate feedback or intervention. The major purpose of summative assessment is to indicate the achievement status or standards achieved by students at a particular point in their schooling. It is geared towards reporting and certification. Syllabus requirements Teachers should ensure that assessment instruments are consistent with the requirements, techniques and conditions of the Physics syllabus and the implementation year 2007. Assessment instruments 2 High-quality assessment instruments: 3 have construct validity (the instruments actually assess what they were designed to assess) have face validity (they appear to assess what you believe they are intended to assess) give students clear and definite instructions are written in language suited to the reading capabilities of the students for whom the instruments are intended are clearly presented through appropriate choice of layout, cues, visual design, format and choice of words are used under clear, definite and specified conditions that are appropriate for all the students whose achievements are being assessed have clear criteria for making judgments about achievements (these criteria are shared with students before they are assessed) are used under conditions that allow optimal participation for all are inclusive of students’ diverse backgrounds allow students to demonstrate the breadth and depth of their achievements only involve the reproduction of gender, socioeconomic, ethnic or other cultural factors if careful consideration has determined that such reproduction is necessary. 1 QSA 2008, P–12 Assessment Policy, p. 2. 2 Assessment instruments are the actual tools used by schools and the QSA to gather information about student achievement, for example, recorded observation of a game of volleyball, write-up of a field trip to the local water catchment and storage area, a test of number facts, the Senior External Examination in Chinese, the 2006 QCS Test, the 2008 Year 4 English comparable assessment task. 3 2 QSA 2008, P–12 Assessment Policy, pp. 2–3. | Physics (2007) Sample assessment instrument and indicative responses Supervised assessment Physics (2007) Sample assessment instrument and indicative responses Supervised assessment Compiled by the Queensland Studies Authority October 2010 The QSA acknowledges the contribution of the State Review Panel Chair for Physics in the preparation of this document. About this assessment instrument The purpose of this document is to inform assessment practices of teachers in schools. For this reason, the assessment instrument is not presented in a way that would allow its immediate application in a school context. In particular, the assessment technique is presented in isolation from other information relevant to the implementation of the assessment. For further information about those aspects of the assessment not explained in this document, please refer to the assessment section of the syllabus. This sample provides opportunities for students to demonstrate: reproduction and interpretation of complex and challenging concepts and principles comparison and explanation of complex processes linkage and application of algorithms and concepts in complex and challenging situations systematic analysis of secondary data to identify patterns and trends analysis and evaluation of complex scientific interrelationships exploration of scenarios and possible outcomes with justification of conclusions discriminating selection, use and presentation of scientific data. To complete this assessment, students would have access to generic formulae, and a table of physical constants would be allowed. This sample assessment instrument is intended to be a guide to help teachers plan and develop assessment instruments for individual school settings. Queensland Studies Authority December 2010 | 3 Assessment instrument The student work presented in this sample is in response to an assessment task which involves students applying and using relevant knowledge and skills to create a response to a problem or issue. Question 1 The force of a tennis racket hitting a ball increases the velocity from rest to 5 m s-1 in a distance of 6 metres. If the ball has a mass of 60 g calculate the force acting on the ball. Question 2 An object is dropped from the top of a cliff 125 m high. Using g = 9.8 m s-2, find the time it takes for the object to reach the bottom of the cliff. Question 3 A tennis racket of mass 1.5 kg falls off the shelf in the garage, 2 m above the floor, onto the floor. a. Calculate the kinetic energy gained by the tennis racket in falling to the floor. b. Calculate the gravitational potential energy of the tennis racket half way to the floor. c. At what velocity will the tennis racket hit the floor? Question 4 A squash ball of mass 0.4 kg hits the back wall in a squash court at an angle of 300 to the normal, at a speed of 25 m s-1, and rebounds at the same angle and speed. Calculate the change in momentum of the squash ball. N 300 300 Question 5 A cricket ball of mass 156 g is held at a height of 1.00 m above the ground. a. Calculate the gravitational potential energy of the cricket ball at this height. b. Determine how high above the ground a squash ball of mass 24 g must be positioned in order for it to have the same gravitational potential energy as the cricket ball. 4 | Physics (2007) Sample assessment instrument and indicative responses Supervised assessment Question 6 In cricket, the speed of the ball and height it bounces off the grass pitch are affected by various factors (e.g. type of grass, density of the soil, dryness of the pitch). In order to determine how fast a cricket ball is likely to travel as it bounces off the pitch, a “drop test” was devised. A cricket ball is dropped from a height of 3.00 m and the height of the bounce is measured. The height of the bounce is used to determine the “speed” of the pitch, as shown in the table below. Mass of a cricket ball = 156 g Radius of cricket ball = 3.6 cm Pitch speed Height of bounce Very fast 76 cm and greater Fast 64 cm to 76 cm Moderately fast 51 cm to 64 cm Easy paced 38 cm to 51 cm Slow Less than 38 cm a. If a cricket ball has a coefficient of restitution of 0.48 when dropped from 3.00 m onto a pitch, what is the “pitch speed”? b. Describe the energy transformations that occur during the “drop test” and hence explain how the bounce height of the cricket ball can be used to determine how fast the cricket ball is likely to travel as it bounces off the pitch. Question 7 A spin bowler intentionally causes the ball to spin in order to make it more difficult for the batsman to hit the ball. One of the results of making the ball spin is that the ball “dips” as it travels through the air, as shown in the diagram below. Using your knowledge of relevant Physics principles, justify the ball’s path. Question 8 Brett Lee was able to bowl a cricket ball at speeds around 150.0 km h-1. If a bat is in contact with the ball for 1.30 ms, what force is needed from the bat to hit the ball straight back at a speed of 70.0 km h-1? Queensland Studies Authority December2010 | 5 Question 9 Ricky Ponting hit a ball with a speed of 23 m s-1 at an angle of 680 above the horizontal. The ball goes straight towards Michael Clarke who was standing 67 m away. When the ball was hit, it was at a height of 1.0 m above the ground. Calculate the range of the ball and hence determine if Michael Clarke is likely to catch the ball before it bounces, given that he can run at a speed of 6.5 m s-1. (Note: Ignore air resistance.) Question 10 A golf ball is hit at 40 m s-1 at an initial direction of 35° to the horizontal. Find a. the maximum height of the ball b. the velocity of the ball at maximum height c. the horizontal range (assuming a level golf course) and d. the direction of motion after three seconds. Question 11 The following table shows the most recent Olympic records for the men’s track events of 100 m, 200 m and 10 000 m. (Note: 26:17.563 means 26 minutes and 17.563 seconds) Athlete Event (in m) Time (in mins and s) Usain Bolt 100 9.58 Usain Bolt 200 19.19 Kenenisa Bekele 10000 26:17.53 a. In which race, the 100 m or the 200 m, did Usain Bolt have the fastest average speed? Justify your answer. b. What was the average speed for Kenenisa Bekele in the 10,000 m race? Question continues next page. 6 | Physics (2007) Sample assessment instrument and indicative responses Supervised assessment Question 11 continued During the 10,000 m race, the following times were recorded at 1000 m intervals. 1000 m 2000 m 3000 m 4000 m 5000 m 6000 m 7000 m 8000 m 9000 m 2:46.24 5:34.24 8:19.55 11:04.75 13:40.45 16.18.57 18.57.63 21:37.80 24:26.31 c. "The fifth kilometre was the fastest yet, as 13:40.45 meant they were down to 2:35 pace". Justify this statement by showing the calculations. What final time would Kenenisa Bekele have been aiming for at this pace? If he could have run that time, would the time be better than the previous record of 26:22.75 set by Haile Gebrselassie in 1998? d. Kenenisa Bekele covered the last 400 m in 57.4 seconds. If he had run at this pace for the whole race what would have been his final time? Justify your answer. The following table shows the 10 m split times for two 100 m races for Usain Bolt. Portion of the race Usain Bolt Usain Bolt 9.69 s race (in s) 9.58 s race (in s) Beijing Olympics Berlin World Championships Reaction time 0.165 0.146 0–10 1.85 1.89 10–20 1.02 0.99 20–30 0.91 0.90 30–40 0.87 0.86 40–50 0.85 0.83 50–60 0.82 0.82 60–70 0.82 0.81 70–80 0.82 0.82 80–90 0.83 0.83 90–100 0.90 0.83 e. What factors have contributed to Bolt having a faster time of 9.58 s for the second race? Question continues next page. Queensland Studies Authority December2010 | 7 Question 11 continued The following table shows the 100 m Men’s record times under the 10 minute time since 1968. Electronic timing began in 1977. Maximum wind factor for a time to be recorded is 2.0. Time f. Year Runner Wind factor ms-1 9.58, 9.69, 9.72 2008 Usain Bolt 0.9, 0.0, 1.7 9.74, 9.77 2008, 2005 Asafa Powell 9.78 2002 Tim Montgomery 2.0 9.79 1996 Maurice Greene 0.1 9.84 1996 Donovan Bailey 0.7 9.85 1994 Leroy Burrell 1.2 9.86 1991 Carl Lewis 1.0 9.90 1991 Leroy Burrell 1.9 9.92, 9.93 1988, 1988 Carl Lewis 9.83 1987 Ben Johnson 1.0 9.93 1983 Calvin Smith 1.4 9.95 1968 Jim Hines 0.3 1.7, 1.6 1.1, 1.0, 1.1 It has been stated that other runners have run as fast a speed as Usain Bolt but do not have as good an overall time for the 100 m. You are to respond to this statement by comparing and contrasting the race times for the 100 m in both tables. You will need to draw graphs to support your answer. g. Do you think that the time set by Usain (9.58 s) could be broken in the future? Predict the fastest time possible for the 100 m in the future. Refer to your graphs to answer this question. 8 | Physics (2007) Sample assessment instrument and indicative responses Supervised assessment Instrument-specific criteria and standards Schools draw instrument-specific criteria and standards from the syllabus dimensions and exit standards. Schools will make judgments about the match of qualities of student responses with the standards descriptors that are specific to the particular assessment instrument. While all syllabus exit descriptors might not be assessed in a single assessment instrument, across the course of study, opportunities to demonstrate all the syllabus dimensions and standards descriptors must be provided. The assessment instrument presented in this document provides opportunities for the demonstration of the following criteria: Knowledge and conceptual understanding Investigative processes Evaluating and concluding. This document provides information about how the qualities of student work match the relevant instrument-specific criteria and standards at standards A and C. The standard A and C descriptors are presented below. The complete set of instrument-specific criteria and standards is on page 24. Key: Qualifier Characteristic of general objective Standard A Knowledge and conceptual understanding Investigative processes Evaluating and concluding Standard C reproduction and interpretation of complex and challenging concepts and principles in relation to motion comparison and explanation of complex processes in relation to motion linking and application of algorithms and concepts to find solutions in complex and challenging scenarios in relation to motion reproduction of concepts and principles in relation to motion explanation of simple processes in relation to motion application of algorithms to find solutions in a simple scenario in relation to motion systematic analysis of secondary data to identify patterns and trends in relation to motion analysis of secondary data to identify obvious patterns and trends in relation to motion analysis and evaluation of complex scientific interrelationships between the research and the scenarios in relation to motion exploration of scenarios and possible outcomes with justification of conclusions/ recommendations in relation to motion discriminating selection, use and presentation of scientific data and ideas to make meaning accessible to intended audiences through innovative graphs description of scientific interrelationships in the research and scenarios in relation to motion description of scenarios and possible outcomes with statements of conclusion/ recommendation in relation to motion selection, use and presentation of scientific data and ideas to make meaning accessible in graphs Queensland Studies Authority December2010 | 9 Sample indicative responses: Standard A Note: Exerpts from the exam questions are set bold. Comments on the student response within the student response column are set italic. Standard descriptors Student response A Question 1 u = 0, v = 5, s = 6, a = 2.08 F=ma = 0.06 x 2.08 Application of algorithms to find solutions to simple situations = 0.12 N Question 2 s = ut + ½ at2 125m = 0 + ½ x 9.8 x t2 125 = 4.9 t2 25.5 = t2 t = 5.05 s Question 3 a. KE = ∆ PE = mgh = 1.5 x 9.8 x2 Linking and application of algorithms to find solutions to complex situations = 29.4 J b. PE = ½ mgh = 14.7J c. KE = ½ m v2 29.4 = ½ x 1.5 x v2 29.4 = 0.75 x v2 29.4/0.75 = v2 therefore, v = 4.7 m s-1 Interpretation of complex concepts Question 4 v = 25 x 0.866 = 21.65 m s-1 Application of algorithms to find solutions to simple situations ∆p=m∆v = 0.4 x 21.65 = 8.66 kgm s-1 Note: is positive. is negative. h = horizontal, v = vertical. Queensland Studies Authority December 2010 | 10 Standard A (continued) Standard descriptors Student response A Question 5 Linking and application of algorithms and concepts to find solutions in a complex situation a. mc = 156 g = 0.156 g, hc = 1.00 m GPEc = m g hc = 0.156 x 9.8 x 1.00 = 1.53 J b. ms = 24.0 g = 0.0240 kg GPEs = ms g hs hs = GPEs / ms g = 1.53/0.024x9.8 = 6.50 m Question 6 a. e = 0.48 Application of algorithms to find solutions to simple situations h1 = 3.00 m e = √ h 2 / h1 therefore h2 = e2 x h1 = (0.48)2 x 3.00 = 0.69 m = 69 cm Therefore, the pitch would be fast (refer to the table) Reproduction of concepts and principles b. Initially the ball has only GPE (gravitational potential energy) at the starting height. As the ball falls it gains KE (kinetic energy) until the instant of impact when GPE has all been converted to KE. As the ball bounces it “loses” energy as heat and sound. When it starts moving back up the KE it possesses is now less than before. As the KE is converted back to GPE, the ball will not bounce as high. As such the height reached by the ball related to GPE can be related to the speed with which the ball leaves the pitch. This is assuming that energy “loses” as the ball moves through the air are negligible. GPE = KE mgh = ½ mv2 gh = ½ v2 Linking and application of algorithms and concepts to find solutions in a complex situation Therefore the speed of the ball can be calculated, e.g. using the figures from part (a) 9.81 x 0.6912 = ½ v2 v2 = 6.78/0.5 v = 3.68 m s-1 Therefore the ball would have been travelling at approx 3.68 m s-1 as it bounced off the pitch. Queensland Studies Authority December 2010 | 11 Standard A (continued) Standard descriptors Reproduction and interpretation of complex and challenging concepts and principles Exploration of a scenario with justification of a conclusion Student response A Question 7 Bernoulli’s Principle states that faster moving air creates a region of lower pressure. As a result a ball with a difference in pressure between two sides will experience a force towards the area of lower pressure. As the ball rotates the boundary layer of air moves with the ball. Above the ball, air pressure is higher as the relative air speed is slower due to the boundary layer opposing the direction of air flow. Below the ball air pressure is lower due to an increased air speed as the boundary layer of air flow are in the same direction. Therefore the ball will experience a downwards force in addition to its weight and will dip as shown. Question 8 v = 150.0 km h-1 = 41.7m s-1 Linking and application of algorithms and concepts to find solutions in a complex situation m = 156 g = 0.156 kg pi = m vi = 0.156 x 41.7 = 6.50 kg m s-1 pf = m vf = 0.156 x (-70/3.6) = - 3.03 kg m s-1 ∆ p = pf - p i Analysis and evaluation of complex scientific interrelationships = - 3.03 – 6.5 = -9.53 kg m s-1 ∆ p = I = Ft therefore, F = ∆ p/ t = - 9.53 / (1.3 x10-3) = -7.33 x 103 N ie 7.33 x103 N in the direction of the final velocity Students need to recognise that they have to calculate pf and pi before ∆ p and hence F. B standard students may forget to convert km h-1 to m s-1 or forget to convert ms to s. C standard studenst may try to calculate F by using F = ma and use “v” as “a”. 12 | Physics (2007) Sample assessment instrument and indicative responses Supervised assessment Standard A (continued) Standard descriptors Student response A Question 9 Note: is positive. is negative. h = horizontal, v = vertical. Linking of complex and challenging concepts Step 1. uh = 23 cos 680 = 8.6 m s-1 uv = -23 sin 680 = -21 m s-1 sv = 1.0 m Step 2. Time to hit the ground sv = uv t + ½ a t2 1.0 = -21 t + ½ 9.8 t2 0 = 4.9 t2 -21t -1.0 Using the quadratic equation t = 4.3 s Linking and application of algorithms and concepts to find solutions in a complex and challenging situation Step 3. Horizontal distance travelled s h = uh t = 8.6 x 4.3 37 m Exploration of a scenario with justification of a conclusion Step 4. In the same time, assuming Michael Clarke reacts immediately, he could run smc = u mc t = 6.5 x 4.3 = 28 m If he was initially 67 m away he would end up 67-28 = 39 m away from where Ricky Ponting hit the ball. This is 2 m further than where the ball will land and so it is unlikely he will catch the ball. Queensland Studies Authority December 2010 | 13 Standard A (continued) Standard descriptors Student response A Question 10 a. Initial vertical velocity = u sin θ = 40 x sin 350 = 22.94 m s-1 Initial horizontal velocity = u cos θ = 40 x cos 350 = 32.77 m s-1 At maximum height, vertical velocity is zero. Consider vertical motion uy= 22.94 m s-1 Linking of concepts in a complex and challenging situation a = g = -9.8 m s-2 vy = 0 vy2 = uy +2as 2 0 = (22.94)2 + (2x-9.8xs) 19.6s = (22.94)2 s = 26.8 m Therefore, the maximum height is 26.8 m b. Linking and application of algorithms and concepts to find solutions in a complex and challenging situation The velocity of the ball at maximum height = 32.77 m s-1 The horizontal velocity is unchanged c. To find the horizontal range the time of flight must be calculated. The time of flight is the time taken for the ball to reach its final vertical displacement (sy = 0) Need to find t when sy = 0, uy= 22.94 m s-1, a = = -9.8 m s-2 s = ut + ½ a t2 0 = 22.94 t – (½ 9.8 t2 ) 0 = t (22.94 – 4.9t) t = 4.68 seconds Horizontal range = horizontal velocity x time of flight = 32.77 x 4.68 =153 m 14 | Physics (2007) Sample assessment instrument and indicative responses Supervised assessment Standard A (continued) Standard descriptors Student response A d. The direction of motion is the direction is the direction of the vertical vector. Linking and application of algorithms and concepts to find solutions in a complex and challenging situation After three seconds, horizontal velocity = 32.77 m s-1 Vertical velocity after 3 s is given by t = 3 s, vy = ?, uy = 22.94 m s-1, a = -9.8 m s-2 vy = uy + at = 22.94 +(-9.8)3 = -6.4 m s-1 To find θ tan θ = 6.46/32.77 θ = tan-1 6.46/32.77 θ = 11.20 Therefore, after three seconds, the direction of motion is 11.20 down from the horizontal. Question 11 Application of algorithms to find solutions in simple situations a. For the 100 m race Speed = distance /time = 100/9.58 = 10.438 m s -1 For the 200 m race Speed = distance /time Application of algorithms to find solutions in simple situations = 200/19.19 = 10.422 m s -1 Fastest average speed was for the 200 m race b. What was the average speed for Kenenisa Bekele in the race? Speed = distance /time = 10000/26:17.563 Systematic analysis of secondary data in a complex situation Comparison and explanation of a complex process = 6.338 m s -1 c. The time at the 4000 m mark was 11:04.75 or 664.75 seconds and the time at the 5000 m was 13:40.45 or 820.45 seconds. Therefore the time for the fifth kilometre was 155.7 seconds or 2:35.7 minutes and seconds. Hence the term 2.35 pace. Speed = distance /time = 1000/155.7 = 6.423 m s -1 At this pace the race time would be expected to be approximately 13:40.35 + ( 5x 2:35) = 26:25.45. This would not be better than the previous record. Queensland Studies Authority December 2010 | 15 Standard A (continued) Standard descriptors Analysis of secondary data to identify a trend Student response A d. Speed = distance /time = 400/57.4 = 6.968 m s -1 Therefore time for the 10 000 m race would have been 10000 x 6.968 Exploration of a scenario with discussion of possible outcomes with discussion of conclusions = 1435.13 seconds = 23:55.13 s e. The reaction time of 0.146 was less for the second race. This may have contributed to the fast starting speed. The wind factor was greater ie 0.9 for the 9.58 s race against 0.0 for the 9.69 s race. The consistent times of less than 0.85 for longer periods of the race and the time of 0.81 for the 60-70 m section. f. Students will need to calculate the speed for the intervals of the race. Portion of the race Systematic analysis of secondary data to identify patterns and trends Reaction 9.58 s race (in s) Set in 2009 Time to get to this point Velocity for this split time (ms-1) 0.146 time Discriminating selection, use and presentation of data to make meaning accessible to the intended audience 16 | Physics (2007) 0-10 1.89 1.89 5.29 10-20 0.99 2.88 10.10 20-30 0.90 3.78 11.11 30-40 0.86 4.64 11.63 40-50 0.83 5.47 12.05 50-60 0.82 6.29 12.20 60-70 0.81 7.10 12.35 70-80 0.82 7.92 12.20 80-90 0.83 8.75 12.05 90-100 0.83 9.58 12.05 Sample assessment instrument and indicative responses Supervised assessment Standard A (continued) Standard descriptors Student response A f. (continued) Discriminating selection, use and presentation of data to make meaning accessible to the intended audience Students may produce different styles of graphs. The above graph shows the speed that Usain Bolt was travelling at during the race. The graph shows that Usain Bolt ran consistently above the 12 m s-1 from the 40m mark. Other runners have reached that speed during races but have not been able to sustain this speed for as long as Bolt during a race. Responses might mention reaction times. Usain Bolt substantially lowered his response time in the 9.58 race. This may have contributed to the fast increase in speed during the first 10 m. Maximum velocity was during then 60 to 70 m split. Exploration of a scenario with possible outcomes and justification of conclusions s = 10m, t = 0.81 Speed = distance /time = 10/0.81 = 12.35 m s -1 The average velocity was the same from 40 to 50 m and from 80 to 100 m. Queensland Studies Authority December 2010 | 17 Standard A (continued) Standard descriptors Student response A g. Students may produce different graphs for this answer, depending on what data has been chosen to graph. The graph should show a downward trend although the slope may be different depending on the data graphed. Discriminating selection, use and presentation of data to make meaning accessible to the intended audience The scale for the horizontal axis may be in years since 1968 or by runner. Exploration of a scenario and possible outcomes with a justified conclusion If the downward trend in the times for the 100m continues the limit of 0 seconds would eventually be reached. This would be impossible. The use of hand timers pre 1977 also makes the slope of 7.7 milliseconds fall per year seem greater than it really is. The greatest decrease in time has occurred during 2008 when Usain Bolt’s times went from 9.72 to 9.58. Predictions using logistic fit are that the limit is 9.48 in 500 years. There are many factors which students might mention eg wind factor, type of track surface, starting blocks, running spikes, coaching methods. 18 | Physics (2007) Sample assessment instrument and indicative responses Supervised assessment Sample indicative responses: Standard C Note: Exerpts from the exam questions are set bold. Comments on the student response within the student response column are set italic. Standard descriptors Student response C Question 1 u = 0, v = 5, s = 6, a = 2.08 F=ma = 0.06 x 2.08 Application of algorithms to find solutions to simple situations = 0.12 N Question 2 s = ut + ½ at2 125m = 0 + ½ x 9.8 x t2 125 = 4.9 t2 25.5 = t2 t = 5.05 s Question 3 Application of algorithms to find solutions to simple situations a. KE = ∆ PE = mgh = 1.5 x 9.8 x2 = 29.4 J Application of algorithms to find solutions to simple situations b. PE = ½ mgh = 14.7 J c. A C response might get part way through finding a solution or forget to square a figure or to take the square root of another. Question 4 Interpretation of complex concepts v = 25 x 0.866 = 21.65 m s-1 No change in momentum calculated Application of algorithms and concepts to find solutions in a simple situation Question 5 a. mc = 156 g = 0.156 g, hc = 1.00 m GPEc = m g hc = 0.156 x 9.8 x 1.00 = 1.53 J b. No attempt at this part of the question. Sample assessment instrument and indicative responses Supervised assessment | 19 Standard C (continued) Standard descriptors Student response C Question 6 a. e = 0.48 h1 = 3.00 m Application of algorithms to find solutions to simple situations e = √ h 2 / h1 therefore h2 = e2 x h1 = (0.48)2 x 3.00 = 0.69 m = 69 cm Therefore, the pitch would be fast (refer to the table). b. During the drop test some energy transferred. The energy is transferred from gravitational potential energy to kinectic energy. Reproduction of concepts and principles Description of a scenario with statements of a conclusion Question 7 The bowler caused the ball to spin forwards, therefore according to Bernoulli’s principle the air flowing over the top of the ball is moving much quicker than the air following on the under side of the ball therefore causing it to dip and shortening the path of the ball. Without spin the air flowing on either side of the ball is equal causing it to only dip as it loses speed. Question 8 v = 150.0 km h-1 = 41.7 m s-1 m = 156 g = 0.156 kg Linking and application of algorithms and concepts to find solutions in a complex situation pi = m vi = 0.156 x 41.7 = 6.50 kg m s-1 pf = m vf = 0.156 x (-70/3.6) = - 3.03 kg ms-1 No attempt at this part of the question. 20 | Physics (2007) Sample assessment instrument and indicative responses Supervised assessment Standard C (continued) Standard descriptors Student response C Question 9 Analysis of a complex scientific interrelationships v = 23 m s-1 θ = 680 h=1m Assume Michael Clarke catches the ball at the same height as it was hit from ie 1.0 m vh = 23 cos 680 assume a = -9.8 m s-2 when ball reaches max height, v = 0 m s-1 v = u + at 0 = 23 + -9.8 t -23 = -9.8 t t = 2.35 s Total time = 2 t = 4.69 s No attempt to complete the question. Question 10 a. Initial vertical velocity = u sin θ = 40 x sin 350 = 22.94 m s-1 Linking of complex and challenging concepts Initial horizontal velocity = u cos θ = 40 x cos 350 = 32.77 m s-1 At maximum height, vertical velocity is zero. Consider vertical motion uy= 22.94 m s-1 a = g = -9.8 m s-2 vy = 0 vy2 = uy +2as 2 0 = (22.94)2 + (2x-9.8xs) 19.6s = (22.94)2 s = 26.8 m Therefore, the maximum height is 26.8 m No attempt at other parts of the question. Queensland Studies Authority December 2010 | 21 Standard C (continued) Standard descriptors Student response C Question 11 a. For the 100 m race Application of algorithms and concepts to find solutions in a simple situation Speed = distance /time = 100/9.58 = 10.438 m s -1 For the 200 m race Speed = distance /time = 200/19.19 = 10.422 m s -1 Fastest average speed was for the 200 m race b. What was the average speed for Kenenisa Bekele in the race? Speed = distance /time = 10000/26:17.563 = 6.338 m s -1 Application of concepts in a simple situation c. The time at the 4000 m mark was 11:04.75 or 664.75 seconds and the time at the 5000 m was 13:40.45 or 820.45 seconds. Therefore the time for the fifth kilometre was 155.7 seconds or 2:35.7 minutes and seconds. Hence the term 2.35 pace. Speed = distance /time = 1000/155.7 Explanation of a simple process = 6.423 m s -1 A C response may not give an answer to this part of the question. d. Speed = distance /time = 400/57.4 = 6.968 m s -1 A C response may not give an answer to this part of the question. e. A C response may not give as many factors for this part of the question. The reaction time of 0.146 was less for the second race. This may have contributed to the fast starting speed. The wind factor was greater, i.e. 0.9 for the 9.58 s race against 0.0 for the 9.69 s race. The consistent times of less than 0.85 for longer periods of the race and the time of 0.81 for the 60 to70 m section. Application of concepts in a simple situation 22 | Physics (2007) Sample assessment instrument and indicative responses Supervised assessment Standard C (continued) Standard descriptors Student response C f. A C response may not calculate the speed for the intervals of the race as would an A response and so not produce a graph such as the one below. Application of algorithms and concepts to find solutions in a simple situation Students may produce different styles of graphs. The above graph shows the speed that Usain Bolt was travelling at during the race. The graph shows that Usain Bolt ran consistently above the 12 m s-1 from the 40m mark. Other runners have reached that speed during races but have not been able to sustain this speed for as long as Bolt during a race. This may be calculated. Maximum velocity was during then 60 to 70 m split. s = 10 m, t = 0.81 Speed = distance /time Application of algorithms to find solutions in simple situations = 10/0.81 = 12.35 m s -1 The average velocity was the same from 40 to 50 m and from 80 to 100 m. g. Students may produce different graphs for this answer, depending on what data has been chosen to graph. A C response may produce a graph, which should show a downward trend, although the slope may be different depending on the data graphed. The scale for the horizontal axis may be in years since 1968 or by runner. A comment may be made about the predicted time in the future but possibly won’t be justified by a graph. Queensland Studies Authority December 2010 | 23 Instrument-specific criteria and standards Key: Qualifier Characteristic of general objective Standard A Knowledge and conceptual understanding reproduction and interpretation of complex and challenging concepts and principles in relation to motion comparison and explanation of complex processes in relation to motion Standard B reproduction and interpretation of complex or challenging concepts and principles in relation to motion comparison and explanation of processes in relation to motion Standard C Standard D Standard E reproduction of concepts and principles in relation to motion reproduction of simple ideas and concepts in relation to motion reproduction of isolated facts in relation to motion explanation of simple processes in relation to motion description of simple processes in relation to motion recognition of isolated simple phenomena in relation to motion linking and application of algorithms and concepts to find solutions in complex and challenging scenarios in relation to motion linking and application of algorithms and concepts to find solutions in complex or challenging scenarios in relation to motion. application of algorithms to find solutions in a simple scenario in relation to motion. application of algorithms in relation to motion application of simple given algorithms in relation to motion Investigative processes systematic analysis of secondary data to identify patterns and trends in relation to motion analysis of secondary data to identify patterns and trends in relation to motion analysis of secondary data to identify obvious patterns and trends in relation to motion identification of obvious patterns in relation to motion recording of data in relation to motion Evaluating and concluding analysis and evaluation of complex scientific interrelationships in the scenarios in relation to motion analysis of complex scientific interrelationships in the scenarios in relation to motion explanation of scenarios and possible outcomes with discussion of conclusions/ recommendations in relation to motion identification of simple scientific interrelationships in the scenario in relation to motion exploration of scenarios and possible outcomes with justification of conclusions/ recommendations in relation to motion description of scientific interrelationships in the scenarios in relation to motion description of scenarios and possible outcomes with statements of conclusion/ recommendation in relation to motion identification of scenarios or possible outcomes in relation to motion identification of obvious scientific interrelationships in the scenario in relation to motion statements about outcomes in relation to motion presentation of scientific data or ideas. discriminating selection, use and presentation of scientific data and ideas to make meaning accessible to intended audiences through innovative graphs. selection, use and presentation of scientific data and ideas to make meaning accessible to intended audiences in graphs. selection, use and presentation of scientific data and ideas to make meaning accessible in graphs. presentation of scientific data or ideas in graphs. Queensland Studies Authority December 2010 | 24

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