Goals: SWBAT...
1. solve 1-D free-fall kinematics problems using the UAM equations.
2. state whether vx and vy are constant or accelerating for a projectile
Warm-Up (4min)
Read the notes in #031A, then do G U E with the warm-up (we'll do S S together)
A baseball is popped straight up in the air and has a hang-time of 6.25s. Determine the max height (ymax).
Hint: the time to max height is 1/2 the total hang time
CLASSWORK
1. #031A: Hangtime and Max Height
Notes:
Ymax is the maximum height, and is achieved 1/2 way through the flight of the projectile
Ymax occurs at 1/2 the total flight time (often called "hang-time")
At Ymax, the y velocity is zero. You can use this as your initial velocity, and then treat the problem like a free-fall problem
HW Review: LINK
Practice #1: A football is kicked and reaches ymax at 2.0s. How high is ymax?
Practice #2: With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.
(you can check your answer/get hints here: LINK)
2. #031B: Notes: Preparation - "Ball in a Cup" Lab
- What is a projectile? (write your best educated guess)
- Notes: https://www.flippingphysics.com/projectile-motion.html
-EdPuzzle: https://edpuzzle.com/media/57bc6dea3715279a24e9fdf6
3. Review: What is constant for a projectile? vx or vy? ax or ay?
4. Prep for Monday (I'll be at a MLG training)
- preview the assignment (LINK)
- preview the setup (demo table)
- groups (write down your groupmates names)
At Home Learning (HW)
1. #031C: Monday you will work on your pre-lab as we build towards our ramp/cup/ball bearing challenge later this week.
Take Notes: LINK 1
Watch the video, and answer the EdPuzzle Questions: LINK 2
NEED HELP WITH FREE-FALL? Check out Step-By-Step Science here: https://www.youtube.com/watch?v=62BymeTQ_SM&index=2&list=PLcJ1NYursFxhEH7nL60_Q8SFSpY_jXIre
#Goals: SWBAT...
1. Use FBD's to find Net Force
2. Solve for forces using mass, net force, and acceleration
3. Define Newton's 3rd law
4. List interaction force pairs (Action-Reaction pairs)
Warm-Up (4min)
I'm still sick, so I won't be talking loud today. Keep your voices low.
Read the scenario below, then answer the questions at the end
"Lee Mealone is sledding with his friends when he becomes disgruntled
by one of his friend's comments. He exerts a rightward force of 9.13 N
on his 4.68-kg sled to accelerate it across the snow. If the acceleration
of the sled is 0.815 m/s/s, then what is the coefficient of friction between
the sled and the snow?"
List the forces present in this example
Draw the FBD
Which equations would you use in your process to find the coefficient
of friction (µ)
CLASSWORK
1. #031A: Practice with Solving FBD's
our last two FBD's
A. find the coefficient of friction for the warm-up problem
B. In a Physics lab, Ernesto and Amanda apply a 34.5 N rightward force to a
4.52-kg cart to accelerate it across a horizontal surface at a rate of 1.28 m/s/s.
Determine the friction force acting upon the cart.
2. #031B: Newton's 3rd
Definition (notes): For every force applied by an object, there's an equal and opposite force on that object.
More Notes: link
Edpuzzle: EDpuzzle
3. #031C: Interaction Force Pairs
Read the following, and list all the interaction pairs you find:
A variety of interaction force pairs are evident in nature. Consider the propulsion of a fish through the water. A fish uses its fins to push water backwards. But a push on the water will only serve to accelerate the water. Since forces result from mutual interactions, the water must also be pushing the fish forwards, propelling the fish through the water. The size of the force on the water equals the size of the force on the fish; the direction of the force on the water (backwards) is opposite the direction of the force on the fish (forwards). Interaction force pairs make it possible for fish to swim.
Consider the flying motion of birds. A bird flies by use of its wings. The wings of a bird push air downwards. Since forces result from mutual interactions, the air must also be pushing the bird upwards. The size of the force on the air equals the size of the force on the bird; the direction of the force on the air (downwards) is opposite the direction of the force on the bird (upwards). For every force, there is an equal (in size) and opposite (in direction) force. Interaction force pairs make it possible for birds to fly.
Consider the motion of a car on the way to school. A car is equipped with wheels that spin. As the wheels spin, they grip the road and push the road backwards. Since forces result from mutual interactions, the road must also be pushing the wheels forward. The size of the force on the road equals the size of the force on the wheels (or car); the direction of the force on the road (backwards) is opposite the direction of the force on the wheels (forwards). For every force, there is an equal (in size) and opposite (in direction) force. Interaction force pairs make it possible for cars to move along a roadway surface.
#031D: More Interaction Pairs
Solve the Following:
1. While driving down the road, a firefly strikes the windshield of a bus and makes a quite obvious mess in front of the face of the
driver. This is a clear case of Newton's third law of motion. The firefly hit the bus and the bus hits the firefly. Which of the two forces is greater: the force on the firefly or the force on the bus?
2. For years, space travel was believed to be impossible because there was nothing that rockets could push off of in space in order to provide the propulsion necessary to accelerate. This inability of a rocket to provide propulsion is because ...
a. ... space is void of air so the rockets have nothing to push off of.
b. ... gravity is absent in space.
c. ... space is void of air and so there is no air resistance in space.
d. ... nonsense! Rockets do accelerate in space and have been able to do so for a long time.
a. greater than the acceleration of the bullet.
b. smaller than the acceleration of the bullet.
c. the same size as the acceleration of the bullet.
4. In the top picture (below), Kent Budgett is pulling upon a rope that is attached to a wall. In the bottom picture, Kent is pulling upon a rope that is attached to an elephant. In each case, the force scale reads 500 Newton. Kent is pulling ...
a. with more force when the rope is attached to the wall.
b. with more force when the rope is attached to the elephant.
c. the same force in each case.
At Home Learning (HW)
Complete #031C and #031D
#031E: Watch the following, take notes, and answer questions. A Common Misconception about Newton's Third Law Force Pairs (or Action-Reaction Pairs) - EDpuzzle
What is the misconception?
NGSS Standard (this is what we're learning with this unit)
Analyze data to support the claim that Newton’s second law of motion describes the mathematical relationship amongthe net force on a macroscopic object, its mass, and its acceleration. [Clarification Statement: Examples of data could include tables or graphs of position or velocity as a function of time for objects subject to a net unbalanced force, such as a falling object, an object sliding down a ramp, or a moving object being pulled by a constant force.] [Assessment Boundary: Assessment is limited to one-dimensional motion and to macroscopic objects moving at non-relativistic speeds.]