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Vectors and Linear Motion
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Vector Quantities: Have a magnitude And direction ex: meters, velocity, acceleration Scalar Quantities: Have only a magnitude ex: temperature, speed, time
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Distance A length between two points Scalar quantity Unaffected by direction
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Displacement Distance and direction between 2 positions Vector quantity Ex: 5m West
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Acceleration
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Motion of acceleration
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Dimensional Analysis
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Step 1: Multiply by 1
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Dimensional Analysis Step 2: Cancel units
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Dimensional Analysis Step 3: Repeat
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Dimensional Analysis Step 3: Repeat
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Dimensional Analysis Step 4: Solve
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Unit 2: Acceleration and Freefall
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Gravity Pulls on all objects with the same acceleration (neglecting air resistance) Earth’s gravitational field decreases as you travel above Earth’s surface g = -9.8m/s 2
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An object is dropped from an airplane and falls freely for 20s before hitting the ground. A)How far did the object fall? B)What was it’s velocity after 20s?
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Objects Thrown Straight Up For objects launched upward, the time to reach the top of their flight is half the total flight time V y @ top is 0m/s a=-9.8m/s 2 always v i = -v f
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A physics student throws a tennis ball into the air. It takes 6s for the tennis ball to land. A) How high did the tennis ball go? B) What was it’s initial velocity?
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Velocity and Acceleration Graphs
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Velocity and Distance Graphs Distance Velocity
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Projectile Motion
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Independence of Dimensions Vertical and Horizontal motions are treated separately We calculate one at a time Use the X-Y chart to show your givens
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The X-Y Chart XY a vivi vfvf t d
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Horizontal Values v i = v f in the x direction a is always 0m/s 2 in the X direction time is the same as the vertical time
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Vertical Values a is always -9.81 m/s 2 in the Y direction Find time or distance using: d=v i t+(1/2)at 2
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A ball is rolling 5m/s and rolls off of a 10m ledge. How far from the ledge does the ball land? 5 m/s 10 m
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A ball is rolling 5m/s and rolls off of a 10m ledge. How far from the ledge does the ball land? 5 m/s 10 m
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A ball is rolling 5m/s and rolls off of a 10m ledge. How far from the ledge does the ball land? 5 m/s 10 m Velocity Vectors Velocity vectors
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A ball is rolling 5m/s and rolls off of a 10m ledge. How far from the ledge does the ball land? 5 m/s 10 m Acceleration Vectors Acceleration vectors (Notice they don’t change)
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A ball is rolling 5m/s and rolls off of a 10m ledge. How far from the ledge does the ball land? 5 m/s 10 m XY a vfvf vivi t d
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A ball is rolling 5m/s and rolls off of a 10m ledge. How far from the ledge does the ball land? 5 m/s 10 m XY a vfvf vivi 5 t d10
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A ball is rolling 5m/s and rolls off of a 10m ledge. How far from the ledge does the ball land? 5 m/s 10 m XY a0-9.81 vfvf 5 vivi 50 t d?10
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A ball is rolling 5m/s and rolls off of a 10m ledge. How far from the ledge does the ball land? 5 m/s 10 m XY a0-9.81 vfvf 5 vivi 50 t1.43 d?10
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A ball is rolling 5m/s and rolls off of a 10m ledge. How far from the ledge does the ball land? 5 m/s 10 m XY a0-9.81 vfvf 5 vivi 50 t1.43 d7.1510
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Try this one on your own A careless driver left a car in neutral near the edge of a 15m high cliff. The car rolled at a velocity of 3m/s as it rolled off the cliff. A) How long did it take the car to reach the bottom? B) How far from the cliff did it land?
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XY a vfvf vivi 3 t d15 Givens:
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XY a0-9.8 vfvf 3 vivi 30 t d15 We also know:
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XY a0-9.8 vfvf 3 vivi 30 t1.75 d5.2515
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Objects Launched at An Angle Break the velocity down into x and y components 30° 10m/s 10sin(30°) 10cos(30°)
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θ Opposite Adjacent
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Objects Launched at An Angle Break the velocity down into x and y components 30° 10m/s 10sin(30°) 10cos(30°)
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Objects Launched at An Angle Break the velocity down into x and y components 30° 10m/s v y = 5m/s V x = 8.66m/s
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Start filling in your X-Y chart Solve for one dimension at a time
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XY a vfvf vivi t d
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XY a vfvf vivi 8.665 t d
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XY a0-9.81 vfvf 8.66-5 vivi 8.665 t d
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XY a0-9.81 vfvf 8.66-5 vivi 8.665 t1.01 d8.5710.05
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What happens as θ increases? 10m/s 10°
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What happens as θ increases? 10m/s 10° 10m/s sin(10°) 10m/s cos(10°)
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What happens as θ increases? 10m/s 10° 1.74 m/s 9.84 m/s
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What happens as θ increases? 10m/s 55° 10m/s sin(55°) 10m/s cos(55°)
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What happens as θ increases? 10m/s 55° 8.19 m/s 5.74 m/s
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Range Horizontal Distance a projectile travels before landing For a given velocity, 45° produces the greatest range Range
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Air Resistance No Air Resistance
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Air Resistance With Air Resistanc e No Air Resistance
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Forces And Newton’s Laws
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A Force is a push or a pull on an object Units- Newtons (N) Forces only exist as a result of an interaction
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Contact Force- Results from 2 objects touching each other Field Force- A push or a pull between 2 bodies that aren’t touching. What field force have we discussed so far in class? Can you think of any others?
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FNFN FgFg For an object resting on a level surface F N = -F g
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Concurrent Forces Two or more forces acting on the same object at the same time are considered to be concurrent with each other. They are sometimes said to be acting concurrently
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When you will see the term “Concurrent”: The magnitude of the resultant of two concurrent forces is a minimum when the angle between them is: A) 0°B) 45°C) 90°D) 180°
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When you will see the term “Concurrent”: The magnitude of the resultant of two concurrent forces is a minimum when the angle between them is: A) 0°B) 45°C) 90°D) 180°
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Resultant Forces are vector quantities. They can be expressed and added using arrows just like distance and velocity vectors. We call this sum the resultant 2N+ 2N = 4N
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Net Force F Net Sum or “Net” of all forces acting on an object F Net = F 1 + F 2 + F 3 … Not in reference table
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Equilibrium A state where F Net is 0N A force that causes equilibrium is called an equilibrant An equilibrant is equal in magnitude to, but opposite in direction of a resultant (or F Net )
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In other words: What force would we need to cancel out or neutralize the forces we have now?
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Components We can find the components of forces the same way we found components of velocities and distances F FyFy FxFx F x = F cos(θ) F y = F sin(θ) θ
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Free Body Diagrams (FBD’s) FBD’s are used to show the forces acting on an object. Start with a square and draw all forces originating from the center of the square pointing outward **Ignore MC questions that show it differently**
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What does the FBD look like for an object that weighs 1 N and is being pulled on to the right with a force of 1N, but being held still by a force of friction of 1N?
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2 N 1 N 2 N 1 N Find F Net :
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2 N 1 N 2 N 1 N Step 1: Find F y and F x :
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1 N F Net Step 2: Add your vertical and horizontal vectors
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1 N 1.4N Step 3: Solve
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1 N 1.4N
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Newton’s First Law “An object at rest will stay at rest and an object in motion will stay in motion unless acted on by an unbalanced force.” Inertia- The resistance of an object to a change in motion Inertia is directly proportional to mass
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Which Object Has More Inertia?
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AB 20 m/s Right10 m/s Right
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Newton’s Second Law The acceleration of an object is directly proportional to the net force acting on it and indirectly proportional to it’s mass. Think of it this way: Bigger forces cause bigger accelerations Bigger masses cause smaller accelerations
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Newton’s Second Law
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Mass vs Weight Mass- Amount of matter in an object – Units- kg – Unchanged by gravitational field Weight- Force of pull on an object caused by gravity – Units- N – Depends on gravitational field strength
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Weight (F g ) Always directed towards Earth (down) Force of Gravity F g = mg g=F g /m Units: m/s 2 = N / Kg g = 9.8m/s 2
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Newton’s Third Law “For every action there is an equal and opposite reaction”
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If a person weighing 700N is standing in an elevator that’s accelerating up, what force does the elevator apply on the person?
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FgFg F Net = a = + 2 m/s 2 m = 70kg FN=FN=
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F f = µF N Solve this equation for µ and predict what you think the units for µ will be µ = F f / F N µ is unit-less
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Inclined planes We will call this force F II FNFN FgFg F II
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Inclined planes F II is a component of the object’s weight along with F ˔ FNFN FgFg F II F˔F˔
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Calculating F II and F ˔ Sin(θ) = F II / F g F II = F g sin(θ) Cos(θ) = F ˔ / F g F ˔ = F g cos(θ)
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What do F II and F ˔ mean? F ˔ has the same magnitude as the normal force FNFN F˔F˔ θ
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What do F II and F ˔ mean? F II is the component of the weight that is parallel to the plane F II θ
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F II = -F f when F Net = 0N F II θ FfFf
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