Problem solving: Physics modeling-based interactive engagement

A very straightforward check of validity of the codes from the rubric was used to analyze data is to see if the codes can be consistently applied by other people. This is called intra-rater reliability (Patton, 2002). For this study, the students’ solutions to question 1, 2, and 3 were coded. Later, a graduate student who was in the physics department coded the solutions from three problem-solving sections of six students separately. To establish inter-rater reliability, the Spearman’s correlation r_s for each dimension was calculated. Spearman’s r_s correlation instead of Pearson’s correlation was used because the sample size (N=6) was small and there were lots of ties.

The first problem-solving protocol: The Spearman’s correlation r_s for general approach (GP) is 0.98; r_s for specific application of physics (SAP) is 0.95; r_s for logical progression (LP) is 0.83. Since the first question is conceptual, appropriate mathematics (AM) is not included.

The second problem-solving protocol: The Spearman’s correlation r_s for general approach (GP) is 0.97; r_s for specific application of physics (SAP) is 1; r_s for logical progression (LP) is 0.89;r_s for appropriate mathematics is 0.99.

The third problem-solving protocol: The Spearman’s correlation r_s for general approach (GP) is 1 which is perfect correlation; r_s for specific application of physics (SAP) is 0.73, r_s for logical progression (LP) is 0.82, r_s for appropriate mathematics (AM) is 0.78. Therefore using the codes is reliable and consistent.

The following table shows the scores of students for questions 1, 2, and 3 by using Foster’s rubric. Table 2 shows the general approach (GA) scores, specific application of physics (SAP) scores, logical progression (LP) scores, and appropriate mathematics (AM) scores respectively. The numbers in the Table 2 show average scores in the first, second, and third interviews. The average scores were calculated by using the rubric in Appendix D. For each code, the average of 6 students’ scores was calculated based on the rubric. For example, each student got 7 points based on the rubric in Appendix D. Six students were involved in think-aloud problem solving. So, the average score= 6x7/6=7. For each dimension, the scores were calculated in the same way.

Table 2. Students' scores for questions 1, 2, and 3.

SCORES
Dimensions	Q1	Q2	Q3
GA	4.83	6.60	4.67
SAP	6.16	7.50	5.92
LP	5	7	5.66

The only significant differences using Wilcoxon signed ranked tests are those between Q1 and Q2 for GA, SA, and LP.

Appropriate mathematics (AM) scores were obtained from only the second interview since only the second question includes mathematical calculations. The total score is 7.5. The averages of scores of six students are 6, 7, and 7.5. It seems that students used the appropriate math.

In addition to showing how students progress in Table 2, the data obtained from students’ interviews were reported to show how students progressed throughout the course. The transcripts contain the following shorthand notation: [ ] represents comments about the interview added after the fact, {…} indicates that unimportant words were omitted from the transcript, and inaudible words or sentences were not included. Names used are pseudonyms.

Due to space constraints, only one student’s physics problem solving think-aloud protocols was considered in detail in this paper. From the six students, Clark was randomly chosen by one person who did not have any input in the study and was given information in detail concerning his performance on the physics problem solving.

Clark

1^st interview: Clark solves the problem correctly. He drew and identified some forces except for the gravitational forces in a correct free-body diagram for two cars. He solved the first part of the problem by means of Newton’s third law instead of the momentum principle containing Newton’s third law. Later and for the second part, he used the momentum principle and got Newton’s third law from the principle. The excerpt below is taken from the interview.

C: Alright … Uh, diagram..car one..car two..toward the other car. [Sighs] It’s got no motion since it is at rest. It’s zero. So each car during the collision. Showing all forces. Car one..car two.. [sighs]. The force exerted..on car two by car one is equal and opposite to the force on car one by car two. That’s Newton’s Third Law. [Pause] [mumbles something] [pause] [sighs]. Actually-

Clark stopped talking while he was solving the problem. I asked him what he was doing. After that, he started to talk again. He used the momentum principle to solve the second part of the question. And he got the correct answer. He gave explanations step by step to make everything clear to be understood.

C: Alright. Um, just drew, uh… So I just drew a line at the top that shows the direction of the force- forces acting on the cars. ‘Cause I want it to be clear that they are- if they were reading it they’d understand that there’s a forces acting this way on this- on this car and forces on this one acting on this car. Um, the car during the collision... [pause] that would be F. Equal magnitude but opposite direction. The forces, uh, rank the magnitude of the horizontal forces. Give your reasoning. Uh, so the momentum principle of p equals f net dt. [Sighs] And..[makes noises] be that one car... Say car two and delta p equals f net times dt. Uh, time for the collision for both of them is equal. So we have...the equations form... [mumbles]. Relate the change in momentum in net force acting on the car. [Sighs] equal to f net..[mumble] change in momentum of one car over the net force on one- on that same car is equal to the change in momentum over the net force, um, of the other car.

C: The delta t is the same for the collision so the force is acting for the same period of time on both cars. D- during the collision. So... [mumbles] This is just the magnitude and this would be direction. Which...[pause] …The original car is moving in the positive x direction. So the forced applied on car one would be negative. P sub one and p sub two..delta p sub one will be opposite of delta p sub two. So on delta p it can be- since- since they are not relativistic model and the mass- non relativistic model and the masses are not changing you can take mass as a delta p. You have m times negative delta p. Then equals m sub two times p. And since the masses of both cars are equal... v- I mean delta v not delta p on that. And because the mass of the cars are then same get this down to the change in velocity. For car two is equal and opposite in the change of velocity of car one. So..magnitude. Horizontal forces and give you reasoning. I just went through that already, ok. Um, the magnitudes. Magnitudes of all the horizontal forces are equal.

At this point he also used Newton’s third law as if he wanted to makes sure he was doing it right.

C: … Um, the magnitude of the forces are equal and in opposite direc- but in opposite directions. By Newton’s Third Law it said that the forces would be equal in that case.

For the second part of the question, Clark uses the linear momentum principle.

C: …it’s because theFord Escort, um, has a change in momentum which by the momentum principle says that there is a force acting over a certain period of time. And there’s- the only..other object in this system that it would interact with would be the moving van. So the moving van would s- is applying the force..against the Ford Escort. I mean which the Ford Escort is experiencing. Uh, (b). Does the Ford Escort exert a force on the moving van? Uh, yes it does. Same thing as the last case. There’s a change in momentum of the moving van. And other objects being acted- only other object being considered in this diagram would be the Ford Escort. And the force is a, uh, is proof of an interaction. So any other way it could interact with the Ford Escort is through physical contact. Which it would be during the collision. And (c). If the answers to (a) and (b) are yes which force is larger? Explain your answers to (a), (b), and (c). Alright here’s the fun part. Which force is larger? So now delta v and delta p equals f net delta t that would be our Ford escort here. And delta p of the moving van f knot and then delta t. So, same way the collision happens over the same period of time. So delta p of the Ford Escort over f net Ford Escort equals delta t. And same thing for the moving van, delta t. Delta p of van over f net. Moving van equals delta t. And since those two equations are equal to each other you can pull them together. And delta p over f net escort. Equals delta p moving van over f net. Moving van. Alright. Um... [pause] [mumbles] The force- I forgot which one I’m thinking of. So... [mumbles] [sighs] The net force- the force that the...I think that that’s right. Newton’s third law- by Newton’s Third Law ... [sighs] the force f e. Escort- the Ford Escort on the moving van equal- would be equal and opposite to the force of the van on the Ford Escort.

C: Ok, delta- delta p of the Ford Escort... net Escort equals p over net- net van. Ok. Since this is a non-relativistic you can move the mass of each. So it will be mass of Ford Escort..delta v Ford Escort. M v. Mass of Escort equals mass of moving van delta v. F from that. Moving van would be..initially moving at the same speed. So..um...(mumble) Um..where I’m going with this but... V is..moving van that. Let me give one equals one which proves the equation is equal. ‘Cause the delta- delta p f e is equal to the delta p of the moving van. But the forces are not equal because by the equals f equals m a...this takes place because of the Ford Escort. Acceler- acceleration is..it being experience is going to be equal and opposite to. I know this problem as I’m sure most people do. So by Newton’s Third Law force is equal and opposite to each other. And-

Clark is really confused about Newton’s third law and in what situations it can be applied. He started to think he cannot apply Newton’s third law because of the different masses of cars. Then he does correct himself soon after.

C: … So Newton’s Third Law doesn’t really apply because, um, actually yeah it does apply. But the amount of de- the amount of deceleration that the Ford Escort and the moving van experiences is equal, but the masses are different. And since the mass of the moving van is a lot larger than the Ford Escort the amount of force applied to the moving van is greater than the amount of the force applied to the Ford Escort. But…[mumbles] By that, um, the force of the moving van would be a lot larger because they would- the basic assumption right here is that acceleration- is their velocity would drop down to zero in collision. They would- they wouldn’t just keep going through each other. Or they wouldn’t keep going a certain direction at a certain speed. But that the accelerations would drop down to zero and because the truck was more massive than the Ford Escort the amount of force that would be required would be larger than the Ford Escort. No no that’s not right… The amount of force required on the Ford Escort is equal to the amount of force required on the van.

In last part, Clark tried to explain Newton’s third law in his own words. Although it is not very clear, it can be understood that this explanation says Newton’s third law.

C: [pause] [sighs] one f equals f two and one two. Much that pushes back. The answer now is, um, the basic equation is used is f one on two is equal and opposite to f two on one.

Clark solved the problem correctly, but he was always in a dilemma. He was therefore exhibiting some profound conceptual difficulties with Newton’s Law because he thought he could just use Newton’s third law when the masses are equal. He has the p-prim which states that since the truck is more massive, it requires larger force. He could not make links between the pieces pf knowledge. He cannot make some connection between the momentum principle and Newton’s third law correctly. Finally, after a long process he got the correct answer by struggling back and forth between Newton’s third law and the momentum principle. The idea of momentum principle concept and Newton’s third law concept is not clear in his mind.

2^nd Interview: Clark solved the problem concerning the energy principle in a short amount of time. As in his first problem solving attempt, he was aware of what principle he needed to apply. Clark answered correctly without hesitating by applying the integrated knowledge regarding choosing a system which makes the problem easy to solve, making approximations, and then applying the energy principle. He was really good at explaining each step that he followed while he was solving the problem. For example, he explained why he chose a system which is reasonable for solving the problem. The researcher did not have to talk or prompt him by asking any questions. The excerpt below is taken from the interview.

C: … Umm, objects in the system would be block, spring, and earth. Umm, well since the block is falling there’s interaction between the block and the earth, and it would just be like, to contain that system. And the spring is also involved even though, because the spring is resting on the earth, and at one point the block interacts with the spring. So once you start using, once you start going in to the computations and everything, the change in energy equations will be very simple because there are no outside interactions with the system I’ve chosen…

At this point, he started to apply the energy principle and explain. Also, he mentioned that the speed is not relativistic, which is one of approximations. He wrote formulas correctly and did calculations. He did not make any calculation errors.

C: … You must do the analysis in terms of the system chosen in part A. All right. So…first there’s no outside work done on the system. So delta E on the system is zero. So… that means that E total initial and E total final are equal, since there’s no change. What you have to worry about, with the E total, is the potential gravitational energy of the block, kinetic energy of the block, energy of the spring. There is the energy of the block, but that is ignored since we’re not dealing with relativistic speeds, other than change. Umm, so, plug everything in…(calculation mumbling). Potential energy of spring initial is zero since it’s at its relaxed length, there is no energy stored in each spring. (…More calculation mumbling…). So it ends up being that gravitational potential energy initial plus kinetic energy initial equals gravitational potential energy final plus kinetic energy final plus spring potential energy final. At this point, just plug in all the earlier equations. (…Calculation mumbling and writing…). So at this point you plug everything in.

C: So…I take what is on the right side of the equation, which is 8.82 + 10 joules plus 1.5 times velocity squared final, add like terms together, and the on the left side, which is initial energy…so I just take the out energy now on the right side, and subtract it from the energy on the left side, so its 29.52 minus 18.82 which equals 10.7 joules equals 1.5 times velocity final squared. 10.7 divided by 1.5, which equals 7.13 repeating. Take that and square root of it. And the answer would equal 2.67, basically one sig fig, so it would be about 3 m/s at that point in time, and because it’s in the downward direction it would be, the velocity would equal the speed.

In the last part, Clark talked about the approximations which he made. He made all approximations correctly.

C: Umm, I assumed there was no air resistance, because I am given no information on that. I assumed there were no outside interactions with the system that might have been causing work to be done, to change the energy. I assumed the spring was mass less because it would change the, how much force is on it throughout the spring. Umm, there’s no energy lost to sound or any of that fun stuff. Umm, let me think. Earth is stationary.

There is no more to say about Clark’s performance because he was very good at solving this problem. He used the energy principle correctly, made correct approximations, and chose the correct system. It was not possible to tell if Clark changed his problem solving skills because he also did very well in the first interview concerning the linear momentum principle.

3^rd interview: Clark solved the problem correctly. Moreover, he used vector and matrices notions to solve it. He knew the concepts well and integrated them with the mathematics very well. He chose the matrices to do the cross product. Actually, it was a long way to complete the problem, but a more sophisticated way to solve it. In other words, the way in which he used shows how to do cross product. On the other hand, it is not required to solve the problem.

C: … Uh, object and disk system so, uh, energy... angular momentum, and momentum are conserved. So the immediately after the collision was the angular momentum of the combined-

C: Alright. Immediately after the collision what is the angular momentum of the disk plus mass m. This should be the same of the sys- same angular momentum of the system before, um, they collided. So... sighs.. Angular momentum initial equals angular momentum final so m initial. Uh...[some noise]... mass m. That’s angular momentum disk. Angular momentum ball. So that would be the disk is stationary so that means there is no spin and there is no translational motion which the center of the disk is taken to be the point at which we are taking angular momentum from. So that momentum initial is zero for the disk. And angular momentum ball- or object- we’ll call it ball right now- um, [mumble] Anyway initial. So and that’s all translational because there’s no, uh, there’s no speed on it- the ball. I mean there’s no spin on the ball because there is no rotational. But there is one to the- there’s- there’s angular momentum relative to the point relative to the center of the disk. So, and so just before the collision which would be..when is was at distance R from the center- from that point. So... uh that equals R- uh distance from that point crossed with..momentum of the object. And ok, I’m going to assume that the object is a point mass so there is no radius. So the distance at which right before it collides is the radius of the object. So [sighs].. Um, theta R... that would be v o it can have any direction. So the momentum of the ball- the momentum of the points would be the mass times the x component times the y component and the z component of the velocity vector. Put the m in it’s m x , m v y, m v z is the momentum. So it would be m v x, v y, v z. For R. Distance from the- the direction I forgot how to do that one. Um...it’s theta- trying to remember how to so the vector of the angle…

Clark now finds initial momentum using matrix. The following excerpt shows how he used matrix and found the initial momentum.

C: Yeah. Um, times x component of the radius zero plus one phi dx prime. Um, the other two parts.. Um... [mumbles] Ok… So the y component is, um, according to this would be R cosine theta. Yeah. And the x component would be R sine theta. Ok. [Laughs] [Mumbles] So we get zero for this matrix. This is the fun part. R sine theta R cosine theta, zero, zero. Going to be.. first one he’s out since there’s a zero R. Next one is another zero. Last one would be r sine theta time m v- uh, v knot…

C: Z component of the angular momentum. So, um, immediately after the collision the uh, angular momentum would be R sine theta time m v knot.

At this point he answered the second part of the question which asks to estimate the angular velocity of the disk. Since it is not required to know the rotational inertia for different objects, I told him the rotational inertia of a disk. On the other hand, he was supposed to know how to find the rotational inertia of combined systems. He did not know how to do that. After some assistance, he got the right answer.

C: In the z direction. Uh, two. Immediately after the collision estimate the angular velocity of the disk. Um, let’s see. Angular velocity of the disk… Alright so angular velocity can be shown like this: angular velocity final. So angular of the disk would be.. Um, this would be the initial…

I: For the disk is one half, uh, m R squared.

C: One half m R squared is the moment of inertia for the disk. Um, so that means that is an initial for that last one. So it would be fi- angular momentum final would be angular momentum ball plus angular momentum disk. Its uh, angular momentum disk…

C: … With angular momentum there would be no translational because it would be all rotating at a point. So, um, momentum of the disk would equal moment of inertia times angular velocity…

I: Yeah, m R squared.

C: Be m R squared?

I: Umhm, m R squared.

C: Alright. (Sighs) Alright, so angular momentum ball equals m R squared…

C: Ok, ok. Um.. So that total momentum for the disk is one half m R squared equals angular velocity. And then angular velocity is the same moving at the same rate, so the ball and disk [mumbles]... Plug everything back in. So R sine theta time m v y initial. So that’s angular momentum final. Equals, um, m, uh, times R squared, angular velocity of system plus one half m disk R squared.

Finally, he found the angular velocity of the disk. Also, he answered the last part of the question correctly.

C: The kinetic energy of combined system should be less because it is an inelastic collision. Some internal energy will be lost. The lost energy can be transferred to sound, vibration, or thermal energy.

Clark’s performance for this interview was very good even though he had one difficulty concerning the rotational inertia for combined system. He did not have any conceptual difficulties and he made connection between energy principle, momentum principle, or approximations correctly. He did a good job in all three interview questions. He grasped the ideas of making approximations and using principles which are the linear momentum, energy, and angular momentum principles. Also, he could make links between the different concepts easily.