TEAL at MIT:
Teaching Pedagogy
ISTITUTO "E.Fermi”
Cristina Bonaglia
Dirigente Scolastico
Feb 18, 2015
Dr. Peter Dourmashkin
Physics Department
MIT
padour@mit.edu
MIT Experiment in Active Learning
(Technology Enabled Active
Learning)
TEAL Fourteen Year
Ongoing Experiment at
MIT:
A merger of
presentations, tutorials,
and hands-on laboratory
experience into a
technologically and
collaboratively rich
environment
(Some) Elements of Active Learning
How can you encourage student participation in class?
1. Peer Instruction and Concept Tests
2. Group Problem Solving
3. Discovery Activities
4. Discussion Questions
1. Simulations and Visualizations
Mini-Presentations
Newton’s Second Law
Newton’s Second
Law
Law II: The change of motion is
proportional to the motive force
impressed, and is made in the
direction of the right line in
which that force is impressed.
FDt = Dp
F = dp / dt
F = ma
Newton’s Second Law:
Physics and Mathematics
Newton’s Second Law: equate two sides.
This is a vector equality: equality in magnitude and direction.
Concept
Questions
ConcepTests / Peer Instruction
Model: Eric Mazur’s Peer Instruction based on
ConcepTests using “Clicker” Technology
Methodology:
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Concept Test
Thinking
Individual answer
Feedback: Just in Time Teaching
Peer discussion
Revised group answer
Explanation
Concept Question: Contact Force
Consider a person standing in an
elevator that is accelerating upward.
The magnitude of the upward contact
force on the person is
1.
2.
3.
greater than
equal to
less than
the magnitude of the downward force of gravity on the
person.
Concept Q.: Contact Force Ans.
Answer 1. System: person. The
contact force, N, on the person due to
the interaction between the person’s
feet and the elevator is greater than
the gravitational force, mg, on the
person due to the interaction between
the Earth and the person because the
contact force must also accelerate the
person, as well as oppose the
gravitational force. Thus
N – mg = ma
implies that N > mg. So the magnitude
of the upward contact force is greater
than the magnitude of the downward
gravitational force.
Acceleration diagram
Force diagram
Problem Solving
Students work in small teams at the white boards
solving problems with instructor feedback
Expert Problem Solving
MIT students will solve approximately 10,000 problems in four
years
Students learn to become expert problem solvers through
practice
Develop confidence based on experience
Necessary for innovation and creative thinking
Problem solving requires factual and procedural knowledge,
knowledge of numerous schema, plus skill in overall problem
solving.
Beginner Problem Solvers
Unable to represent physical concepts
Unable to combine multiple ideas
Unable to apply mathematical reasoning
Engage in symbol manipulation
Unable to estimate and make ‘back of the envelope’
calculations
Pólya: How to Solve it!
1. Getting Started – identify assumptions and givens,
identify what information is needed and what should be
neglected
2. Plan the Approach – articulate a strategy that may
involve multiple concepts and problem solving
methodologies,
3. Carry out the plan – use multiple types of
representations, equations, graphs, visualizations
4. Reflection and Modification: Ability to modify plan in a
constructive manner.
5. Review - does the answer make sense? Reflect on the
limits of the model.
Enabling Problem Solving
At the start, teachers
can give some
guidance
During problem solving
teachers circulate and
give guidance
At the end of the
problem solving,
teachers provide
closure
Circular Motion of Satellites About Earth
Circular Motion: Problem Solving
Circular Motion: Force Diagram
System: satellite
GmmE
r̂ : 2
r
v
r̂ : - m
r
2
Newton’s Second Law becomes
2
GmmE
v
=m
2
r
r
Group Problem: Circular Motion
A racing car is rounding a turn. Draw a free-body
diagram showing all the forces you think are acting on
the car. Estimate the magnitude of the acceleration of
the car. What is the direction of the acceleration?
Momentum Principle
20
Momentum Diagrams
1) Identify all objects in system.
2) Choose a reference frame.
3) Draw a momentum diagram for
i) state of the system before interaction
ii) state of system after collision.
Diagram should include
1. Choice of unit vectors
2. Show each mass element. Show the speed and arrow
for direction of velocity of each mass element.
Momentum Principle: Integral Version
Momentum Principle: equate two sides (vector equality)
Group Problem: Heading
Estimate the acceleration
of the head due to the
heading of a soccer ball.
Could this cause brain
injury?
In your groups: discuss
what models and concepts
may apply. What
information is needed and
what can be neglected?
Collision Theory
https://www.youtube.com/watch?feature=player_embedded&x-ytts=1421914688&x-yt-cl=84503534&v=WGTBJHFNywI
24
Collision Theory: Energy
Types of Collisions
Elastic:
Inelastic:
K0sys = K sys
f
1
1
1
1
2
2
2
m1v1,0 + m2 v2,0 + ××× = m1v1, f + m2v2, f 2 + × ××
2
2
2
2
K0sys > K sys
f
Completely Inelastic: Only one body emerges.
Superelastic:
K0sys < K sys
f
Concept Question: Recoil
Suppose you are on a cart, initially at rest on a
frictionless track. You throw balls at a partition that is
rigidly mounted on the cart. After the balls bounce
straight back as shown in the figure, is the cart
1. moving to the right?
2. moving to the left?
3. at rest.
Concept Question: Collision
Answer: 2. Because there
are no horizontal external
forces acting on the system,
the momentum of the cart,
person and balls must be
constant. All the balls
bounce back to the right,
then in order to keep the
momentum constant, the
cart must move forward.
Relative Velocity
Consider two moving objects with velocities in a reference
frame fixed to the ground
v1 = v1,x î and v 2 = v2,x î
The relative velocity of object 1 with respect to object 2 is
defined to be
v rel º v1,2 º v1 - v 2 = (v1,x - v2,x )î
Relative Velocity: Elastic Collision
Energy/Momentum Law
Equation (3):
v1,x,i - v2,x,i = v2,x, f - v1,x, f
Components of
relative velocities:
vrel,x,i º v1,x,i - v2,x,i
Energy/Momentum Law:
vx,rel,i = -vx,rel, f
(3)
vrel,x, f º v1,x, f - v2,x, f
(4)
x-component of initial relative velocity is equal to the
negative of the x-component of the final relative velocity
Demo and Worked Example: Two Ball
Bounce
A football and tennis ball are dropped
from a height h above the ground. The
tennis ball on top has mass M1. The
football on the bottom has mass M2.
Assume that the football collides
elastically with the ground. Then as the
football starts to move upward, it
collides elastically with the tennis ball
that is still moving downwards. How
high will the tennis ball rebound in the
air? Assume that M2 >> M1.
Try this out. Be careful and have fun!
M2>>M1
Hands-on: Two Ball Bounce
Collision of interest is between the falling
tennis ball (labeled 1 in the figure) and
the rising football (2) immediately after
the football reverses direction due to an
elastic collision with the ground. Key
model assumption: change in speed of
football due to collision is negligible.
Two Ball Bounce
Initial velocities: v1,i = -va ĵ with va > 0; v 2,i = +va ĵ
Final velocities: v1, f = vb ĵ with vb > 0; v 2, f = va ĵ
Relative velocities:
v rel ,i = (-va - va ) ĵ= -2va ĵ
v rel , f = (vb - va ) ĵ
Energy/Momentum Law:
Choose positive
y-direction
upwards.
v rel,i = -v rel, f Þ -2va ĵ = -(vb - va ) ĵ (1)
Algebra. Solve Eq. (1) for speed of light ball after collision.
-2va ĵ = -(vb - va ) ĵ Þ vb = 3va
Light ball has three times its original speed.
Two Ball Bounce
Energy law for falling:
2
v
m1ghi = (1/ 2)m1va 2 Þ hi = a
2g
2
v
Energy law for rising: (1/ 2)m1vb2 = m1gh f Þ h f = b
2g
Energy and Momentum Law
Ratio of heights.
hf
(vb )2 (3va )2
=
=
=9
2
2
hi (va )
(va )
Light ball rebounds nine
times as high as it fell.
vb = 3va
Energy Principle
Energy Diagrams
Identify system
Identify interaction
Identify state of system before interaction (initial state) and state
of system after interaction (final state)
Draw energy diagrams for these states including:
1) Choice of zero point P for potential energy for each
interaction in which potential energy difference is well-defined.
2) Draw speed of all objects in system
3) Identify initial and final mechanical energy
mech
i
= K i +U i
mech
f
= K f +U f
E
E
Energy Principle
Energy Principle: Equate two sides
mech
Wnc = E mech
E
f
i
Newton’s Third Law
Newton’s Third Law
To every action there is always opposed an equal
reaction: or, the mutual action of two bodies upon
each other are always equal, and directed to contrary
parts.
F2,1 = -F1,2
Action-reaction pair of forces cannot act on same
body; they act on different bodies.
Concept Question: Interaction Pair
A large truck collides head-on with a small car. During the
collision
1. the truck exerts a greater force on the car than the car
exerts on the truck.
2. the car exerts a greater force on the truck than the truck
exerts on the car.
3. the truck exerts the same force on the car as the car
exerts on the truck.
4. the truck exerts a force on the car but the car does not
exert a force on the truck.
Con. Q. Ans.: Interaction Pair
Answer 3. The car and the truck form an interaction pair.
By Newton’s Third Law the collision forces on each object
have equal magnitude but point in opposite directions.
Hands-on
Discovery
Activity
Group Activity: String Theory
Divide up into pairs.
Each pair pull on the opposite ends of the
provided string.
Group Activity: String Theory
1. Draw three force diagrams on the boards, one for
person A, one for person B, and one for the rope.
2. On each diagram draw all the forces acting on the object
in your diagram.
3. Identify the action-reaction pairs of forces.
Tug of War Force Diagrams
For the rope we have neglected the gravitational force
because we are assuming the rope is very light
Rotational Kinematics
Torque Diagram
1. Identify System
2. Draw free body diagram. For each force F place the arrow
representing the force at the point where the force acts on the rigid
body.
3. Identify rotational coordinate system. Choose rotational angle θ and
draw curved arrow to represent direction in which angle θ is
increasing.
4. Use right hand rule with respect to this direction, to indicate positive
direction for angular acceleration.
5. Select point S about which you are calculating torque and draw it on
the free-body force diagram.
6. For each force, draw the vector
F acts.
rSfrom
the point S to where the force
,F
Torque Law
Torque Law: Equate two sides.
Experiment: Moment of Inertia of
Bicycle Wheel
1. Wrap a string around the rim of a bicycle wheel of
radius R = 20 cm and pass it through the valve hole
and attach it to a spoke.
2. Attach a weight of mass m = 50 g to the end of the
string.
3. Drop the weight from a measured height hf above
the ground and allow it to hit the ground. Measure
the time, tf, it takes to hit the ground.
4. From these measurement we shall determine the
moment of inertia of the wheel about the axle.
Visualizations and Simulations
Visualizations and Simulations
Discovery Activity
Changing magnetic flux induces current
http://public.mitx.mit.edu/gwt-teal/FaradaysLaw2.html
Proposing a
Hypothesis
Propose a qualitative
relationship between
magnetic flux (seen in
top graph) and current
that flows in the ring
(seen in bottom graph).
Demo: Electromagnetic Induction
52
PhET
University of Colorado
PhETs Educational
simulations covering a
diverse topics
designed by the
University of Colorado
available in various
languages
http://phet.colorado.edu/
Discussion
Question:
What types of
active learning
could possibly be
effective in your
classroom?