AN INVESTIGATION OF STUDENTS’ PERFORMANCE AFTER PEER INSTRUCTION WITH STEPWISE PROBLEM-SOLVING STRATEGIES

abstract

The purpose of this study was to examine the effects of strategic problem solving with peer instruction on college students’ performance in physics. The students enrolled in 2 sections of a physics course were studied; 1 section was the treatment group and the other section was the comparison group. Students in the treatment group received peer instruction with systematic problem-solving strategies whereas students in the comparison group received only peer instruction. Data were collected on physics achievement, problem-solving strategies, homework problems, and students’ opinions about the instruction. Results indicated that the treatment group students’ homework and achievement test performances as well as their visualizing, solving, and checking habits improved relative to the comparison group students, which did not change noticeably. Treatment group students also changed their perspective on solving a problem and found the method helpful to connect the quantitative solution with concepts. These results revealed that the method could be implemented with little effort so as to assess and enhance student performance in science classrooms.

Appendices

APPENDIX A

Sample Problem on the Application of Stepwise Problem-Solving Strategies with Peer Instruction in the Classroom

• A) Identifying Fundamental Principles

  What is (are) the fundamental concept(s) of the problem?

  I. Normal Force II. Pushing Force III. Net Force

  IV. Force of Static Friction V. Force of Kinetic Friction

  • a) Only II

  • b) II and V

  • c) II, III, and IV

  • d) I, II, III, and IV

  • e) I, II, III, and V

• B) Solving

  Qualitative: Which equation(s) do you need to solve the problem?

  I. F _N  = F _g II. F _g  = mg III. F _f  = μ _k mg IV. F _p  = F _f V. F _p  = ma

  • a) Only I

  • b) I and II

  • c) IV and V

  • d) I, II, and V

  • e) I, II, III, and IV

  Quantitative: What is the correct answer to the problem?

  a) 0 b) 5 c) 49 d) 245 e) 294

• C) Checking

  What are the unit and sign of the force?

  • a) kgm/s², negative

  • b) kg, positive

  • c) m/s², negative

  • d) N, positive

  • e) kg/m², negative

  Which one is correct for the solution’s magnitude?

  • a) F _p  > F _N

  • b) 0 < F _p  < F _N

  • c) F _n  = F _p

  • d) F _p  = 0

  • e) F _p  ≤ 0

APPENDIX B

Sample Homework Problem

The sample homework problem was asked of both the TG and CG students; it was given to the CG students without the systematic problem-solving strategy steps.

(Both groups) A 62-kg person on skis is going down a hill sloped at 37 °. The coefficient of kinetic friction between the skis and the snow is 0.15. How fast is the skier going 5.0 s after starting from rest?

(Only for the treatment group)

• A-Identifying fundamental principle: What is (are) the fundamental concept(s) of the problem?

• B-Solving: Which equation (s) do you need to solve the problem? What is the correct answer to the problem?

• C-Checking: What are the unit, sign, and magnitude of the asked variable(s)?

Solution steps

• A-Identifying fundamental principle

After the concepts, known variables, and unknown variable were determined, the problem was visualized with the help of a coordinate system and a motion diagram as shown below.

There is no acceleration in the y-direction (a _y = 0.0 m/s²).

F _gy and F _f are negative because they are in the negative direction; but the directions of the skier’s velocity, acceleration, net force, and F _gx are in the x-direction as defined by the coordinate system. Also, F _N is in the y-direction.

• B-Solving

Qualitative solve for “v _f”

Qualitative solutions were performed with the help of needed equations and formulas. A mathematical model was established for finding desired unknown variable.

y-direction	x-direction	Final velocity “v f”
F net,y = ma y = 0.0N F N − F gy = F net,y F N = F gy = mg cos θ	F net, x = F gx − F f F f = μ k F N ma x = mg sin θ − μ k mg cos θ a x = g(sin θ − μ k cos θ)	v f = v i + a x t v i = 0.0 m/s v f = g(sin θ − μ k cos θ)t

y-direction	x-direction	Final velocity “v f”
F net,y = ma y = 0.0N F N − F gy = F net,y F N = F gy = mg cos θ	F net, x = F gx − F f F f = μ k F N ma x = mg sin θ − μ k mg cos θ a x = g(sin θ − μ k cos θ)	v f = v i + a x t v i = 0.0 m/s v f = g(sin θ − μ k cos θ)t

Quantitative solve for “ v _f”

The desired unknown variable by using the given variables with the help of qualitative section is found in this part. Only the numerical result is needed.

$$ \begin{array}{c}\hfill {v}_{\mathrm{f}}=\left(9.80\;\mathrm{m}/{\mathrm{s}}^2\right)\ \left( \sin {37}^0-0.15 \cos {37}^0\right)\left(5.0\mathrm{s}\right)\hfill \\ {}\hfill {v}_{\mathrm{f}}=24\mathrm{m}/\mathrm{s}\hfill \end{array} $$

• C-Checking

The unit, sign, and magnitude of the variable are controlled in this part.

Unit: Applying dimensional analysis on the unit verifies that v _f is meters per second

Sign: v _f is in the x-direction.

Magnitude: v _f is fast, over 80 km/h, but 37 ° is a steep incline. If the friction between the skis and the snow is not available, v _f is found as 29.4 m/s. In this problem, v _f for the friction is calculated as 24 m/s. This result (v _f,friction < v _{f,no friction})  is reasonable. Also, it could be said that the kinetic friction coefficient between the skis and the snow is not large. The mass of the skier is not necessary for solving.