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Publicado: 13 de julio de 2011
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How do you approach a physics problem?
This article has been downloaded from IOPscience. Please scroll down to see the full text article. 1999 Phys. Educ. 34 16 (http://iopscience.iop.org/0031-9120/34/1/014) View the table of contents for this issue, or go to the journal homepage for more
Download details:IP Address: 148.228.125.38 The article was downloaded on 07/03/2011 at 21:11
Please note that terms and conditions apply.
How do you approach a physics problem?
Bob Kibble
Moray House Faculty of Education, Edinburgh University, UK
A group of PGCE trainee physics teachers was set a straightforward problem to solve. Their answers revealed six quite different approaches to a solution.One of the regular features on the PGCE Physics noticeboard is the ‘problem for the week’. I pin up a different problem each Monday and invite solutions by Thursday. Chocolate bar prizes are offered at an awards ceremony the following Tuesday. I decided to develop this feature with the aim of drawing my students’ attention to one use for a noticeboard and to experience the motivational effect ofan element of mild challenge/reward. The resulting discussions both in class and around the lunch table have helped us all to focus our attention on aspects of problems— their structure, language and modes of solution. Such was the response to this initiative that when I decided after eight weeks to stop the questions (I had run out of ideas) the students put up their own questions. Two of theproblems were stimulated by the Minds on Physics (MOP) curriculum group of the University of Massachusetts. Dufresne et al
16
Phys. Educ. 34(1) January 1999
(1997) suggest that, when analysing a problem, we interpret it in different ways by associating it with different pieces of knowledge. These associations, be they mathematical, pictorial or conceptual, determine the way we construct solutionsand how much we learn from the process of solving the problem. This article will concentrate on one particular problem, taken from the MOP work, which I set to my PGCE group via the noticeboard in October 1998.
The problem
Janet and her brother John decided to race along to the end of their street and back again. The street was 80 m long. Janet ran at 2.5 m s−1 whilst John’s speed was 1.5 ms−1 .
Where were they when they passed each other?
TEACHING PHYSICS
The solution
I received correct solutions from eight students. Four students had offered more than one route to the solution. One student solved the problem in four different ways. What was interesting to me was the variety of ways in which these students had approached what was the same problem—but of course seen throughdifferent eyes. Altogether there were six different methods of solution, each of which I’ll try to illustrate here.
Using a table
Distance from start Time (s) 0 10 20 30 40 50 60 John 0 15 30 45 60 75 70 Janet 0 25 50 75 60 35 10
They cross at 60 m. Using algebra: 1 They meet after John has travelled X metres. Using a graph
So the time taken for them reach this point is X s 1.5 80 + (80− X) for Janet: = s 2.5 but this time is the same for both, so for John: = 80 + (80 − X) X = 1.5 2.5 X = 60 m.
Relative speed On the outward journey their relative speed is 2.5 − 1.5 = 1 m s−1 . So when Janet reaches the far end in T seconds, John is T metres behind. On Janet’s return their relative speed is 1.5 + 2.5 = 4 m s−1 . The further time before their meeting is T /4 seconds, so the totaltime for the journey before meeting is T + T /4 = 5T /4 seconds. Now considering Janet’s motion, T = 80/2.5, so the time before meeting is 5T /4 =
5 4
Using algebra: 2 They meet after Janet has travelled Y metres from the end of the street.
Therefore John has travelled (80 − Y ) m and Janet has travelled (80 + Y ) m. They run for the same time, so 80 − Y 80 + Y = 1.5 2.5 Y = 20 m so...
Search
Collections
Journals
About
Contact us
My IOPscience
How do you approach a physics problem?
This article has been downloaded from IOPscience. Please scroll down to see the full text article. 1999 Phys. Educ. 34 16 (http://iopscience.iop.org/0031-9120/34/1/014) View the table of contents for this issue, or go to the journal homepage for more
Download details:IP Address: 148.228.125.38 The article was downloaded on 07/03/2011 at 21:11
Please note that terms and conditions apply.
How do you approach a physics problem?
Bob Kibble
Moray House Faculty of Education, Edinburgh University, UK
A group of PGCE trainee physics teachers was set a straightforward problem to solve. Their answers revealed six quite different approaches to a solution.One of the regular features on the PGCE Physics noticeboard is the ‘problem for the week’. I pin up a different problem each Monday and invite solutions by Thursday. Chocolate bar prizes are offered at an awards ceremony the following Tuesday. I decided to develop this feature with the aim of drawing my students’ attention to one use for a noticeboard and to experience the motivational effect ofan element of mild challenge/reward. The resulting discussions both in class and around the lunch table have helped us all to focus our attention on aspects of problems— their structure, language and modes of solution. Such was the response to this initiative that when I decided after eight weeks to stop the questions (I had run out of ideas) the students put up their own questions. Two of theproblems were stimulated by the Minds on Physics (MOP) curriculum group of the University of Massachusetts. Dufresne et al
16
Phys. Educ. 34(1) January 1999
(1997) suggest that, when analysing a problem, we interpret it in different ways by associating it with different pieces of knowledge. These associations, be they mathematical, pictorial or conceptual, determine the way we construct solutionsand how much we learn from the process of solving the problem. This article will concentrate on one particular problem, taken from the MOP work, which I set to my PGCE group via the noticeboard in October 1998.
The problem
Janet and her brother John decided to race along to the end of their street and back again. The street was 80 m long. Janet ran at 2.5 m s−1 whilst John’s speed was 1.5 ms−1 .
Where were they when they passed each other?
TEACHING PHYSICS
The solution
I received correct solutions from eight students. Four students had offered more than one route to the solution. One student solved the problem in four different ways. What was interesting to me was the variety of ways in which these students had approached what was the same problem—but of course seen throughdifferent eyes. Altogether there were six different methods of solution, each of which I’ll try to illustrate here.
Using a table
Distance from start Time (s) 0 10 20 30 40 50 60 John 0 15 30 45 60 75 70 Janet 0 25 50 75 60 35 10
They cross at 60 m. Using algebra: 1 They meet after John has travelled X metres. Using a graph
So the time taken for them reach this point is X s 1.5 80 + (80− X) for Janet: = s 2.5 but this time is the same for both, so for John: = 80 + (80 − X) X = 1.5 2.5 X = 60 m.
Relative speed On the outward journey their relative speed is 2.5 − 1.5 = 1 m s−1 . So when Janet reaches the far end in T seconds, John is T metres behind. On Janet’s return their relative speed is 1.5 + 2.5 = 4 m s−1 . The further time before their meeting is T /4 seconds, so the totaltime for the journey before meeting is T + T /4 = 5T /4 seconds. Now considering Janet’s motion, T = 80/2.5, so the time before meeting is 5T /4 =
5 4
Using algebra: 2 They meet after Janet has travelled Y metres from the end of the street.
Therefore John has travelled (80 − Y ) m and Janet has travelled (80 + Y ) m. They run for the same time, so 80 − Y 80 + Y = 1.5 2.5 Y = 20 m so...
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