Heather Fisher & Alex MacDonaldWith Marina Milner-Bolotin

Proudly supported by TLEF

Thursday, February 28, 2013SyMETRI Meeting

Math & Science Teaching and Learning through Technology

http://scienceres-edcp-educ.sites.olt.ubc.ca/

Question TitleQuestion TitleBlocks and a Pulley

m2

m1

Question Title

Two blocks are connected via a pulley. The blocks are initially at rest as

block m1 is attached to a wall. If string A breaks, what will the accelerations

of the blocks be? (Assume friction is very small and strings don’t stretch)

Question TitleBlocks and a Pulley II

m2

m1A B 1 2

1 2

1 2

1 2

A. 0; 0B. ; C. 0; D. ; 0E. None of the above

a aa g a ga a ga g a

Why are the assumptions above important?

Comments

Answer: EJustification: None of the above answers is correct. Consider two blocks as one system: one can see that the system has a mass of (m1+m2), while the net force pulling the system down is m1g. Therefore, applying Newton’s second law, one can see that the acceleration of the system must be less than g:

Some people think that the acceleration will be g. They forget that the system consists of two blocks (not just m1) and the only pulling force is m1g. Thus the system is NOT in a free fall. Compare this questions to the previous one to see the difference.

CommentsSolution

2 2

1 2 1 2

m g ma g gm m m m

Objectives

Assist teacher-candidates through four platforms:

Within their B. Ed Methods courses

(as learners)

Developing personal teaching style

Providing resources for teacher-candidates

(as teachers)

Creating a stronger connection to the UBC Community

Activity

Logical reasoning

If m and p are positive integers and (m + p) x m is even, which of the following must be true?

(A) If m  is odd, then p  is odd.(B) If m  is odd, then p  is even.(C) If m  is even, then p  is even.(D) If m  is even, then p  is odd.(E) m  must be even.

Question Title

A. 24 m2

B. 76 m2

C. 100 m2

D. 124 m2

E. Not enough information

What is the area of the figure below?

Question TitleArea

10 m

6 m

4 m10 m

Comments

Answer: BJustification: The easiest way to find the area is to imagine a 10 m by 10 m square and subtracting a 4 m by 6 m rectangle.

CommentsSolution

A = 10 m x 10 m – 4 m x 6 m = 76 m2

Alternatively, the shape’s area can be found by dividing it into 2 rectangles and adding the areas together.

10 m

6 m 4 m

4 m

6 m

10 m

Logical reasoning

In the figure below, a square is inscribed in a circle with diameter d. What is the sum of the areas of the shaded regions, in terms of d ?

2

2

2

2

2

1( ) 4 2

1( ) 4 4

1( ) 2 4

( ) 2

( ) 1

a d

b d

c d

d d

e d

1. Which questions generated the most productive discussion? Why?

2. How do you think clickers supported/hindered problem solving? How did clickers affect class dynamics?

3. Were there multiple ways to solve the problems? How were multiple interpretations dealt with?

4. What level of understanding did each question address (ex. conceptual understanding, application, recall, etc.)? How did students engage with each type of question?

5. What did the teacher do to support the students during the activity? What could have been done differently? Were all students supported?

Discussion

Next Steps

ResourcesSMARTboardsPrimary Grades

OutreachFaculty and community

awareness

• Beatty, I., Gerace, W., Leonard, W., & Defresne, R. (2006). Designing Effective Questions for Classroom Response System Teaching. American Journal of Physics, 74(1), 31–39.

• CWSEI Clicker Resource Guide: An Instructors Guide to the Effective Use of Persnal Response Systems (Clickers) in Teaching. (2009, June 1).

• Lasry, Nathaniel. (2008). Clickers or Flashcards: Is There Really a Difference? The Physics Teacher, 46(May), 242-244.

• Milner-Bolotin, Marina. (2004). Tips for Using a Peer Response System in the Large Introductory Physics Classroom. The Physics Teacher, 42(8), 47-48.

• Mishra, P., & Koehler, M. J. (2007). Technological pedagogical content knowledge (TPCK): Confronting the wicked problems of teaching with technology. In Society for Information Technology & Teacher Education International Conference (Vol. 2007, pp. 2214–2226). Retrieved from http://www.editlib.org/p/24919/

Resources

Proudly supported by the

Teaching & Learning Engagement Fund

Appendix

Logical reasoning

If & is defined for all positive numbers a and b by

Then 10 & 2 =

& aba ba b

5( ) 35( ) 2

( ) 520( ) 3

( ) 20

a

b

c

d

e

Question TitleQuestion TitleRatios in a Bag of Candy

The ratio of red to green to blue candies in a bag of candy is 2 : 3 : 4. Jeremy eats 1 red, 1 green, and 1 blue candy. Does the ratio of red to green to blue candies change?

A. Yes

B. No

C. Not sure

–

Comments

Answer: AJustification: The ratio of red to green to blue candies will change. After removing 1 of each color:

2 – 1 = 1 red candy

3 – 1 = 2 green candies

4 – 1 = 3 blue candies

The ratio 2 : 3 : 4 is not the same as 1 : 2 : 3. In most cases, only multiplying or dividing the terms in a ratio will result in an equivalent ratio.

CommentsSolution

The Unit Circle

A circle with radius 1 is drawn with its center through the origin of a coordinate plane. Consider an arbitrary point P on the circle. What are the coordinates of P in terms of the angle θ?

)sin,(cosE.

)cos,(sinD.

)sin,(cosC.)cos,(sinB.

),(A.

11

11

P

P

PPP

x

y

θ

1

P(x1,y1)

Press for hint

P(x1,y1)

1

θ

x1

y1

x

y

Solution

Answer: CJustification: Draw a right triangle by connecting the origin to point P, and drawing a perpendicular line from P to the x-axis. This triangle has side lengths x1, y1, and hypotenuse 1.

P(x1,y1)

1

θ

x1

y1

x

y

)sin(1

)sin(

)cos(1

)cos(

11

11

yy

xx

Therefore, the point P has the coordinates (cos θ, sin θ).

The trigonometric ratios sine and cosine for this triangle are: