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Motion Chapter 11

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**Standards Students will:**

SPS8. Determine the relationship between force, mass and motion SPS8a Calculate velocity and acceleration SPS8c Relate falling objects to gravitational force SPS8d Explain difference in mass and weight

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Observing Motion Motion- change in position in relation to a reference point.

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**Measuring Motion: Distance**

Distance- how far an object moves on a path Displacement- how far between starting and ending points on a path

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**Measuring Motion: Speed**

rate of motion distance traveled per unit time speed = distance time

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**Measuring Motion: Speed (cont’d)**

Instantaneous Speed speed at a given instant Average Speed- = total distance total time

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**Measuring Motion: Velocity**

speed in a given direction can change even when the speed is constant!

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**Calculating Velocity: Distance/Speed/Time Triangle**

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**Graphing Speed/Velocity**

X axis- usually independent variable (time) Y axis- usually dependent variable (distance) Slope of straight line= vertical change horizontal change

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**Graphing (cont’d) slope = velocity steeper slope = faster velocity**

straight line = constant velocity flat line = 0 velocity (no motion)

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**Calculating Slope Choose two points on graph to calculate slope.**

Distance Vs Time 16 Choose two points on graph to calculate slope. Calculate the vertical and horizontal change. Divide the vertical change by the horizontal change. 14 . 12 10 Distance (m) 8 . 6 4 2 1 2 3 4 5 Time (s)

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**Calculating Slope(cont’d)**

Distance Vs Time 16 1. point 1: d= 6m t= 1s point 2: d= 12m t= 4s 2. vert ∆- 12m-6m= 6m horiz ∆- 4s-1s= 3s 3. slope= 6m = 2m/s 3s 14 . 12 10 Vertical change Distance (m) 8 . 6 Horizontal Change 4 2 1 2 3 4 5 Time (s)

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**Practice Problem A B Who started out faster? A (steeper slope)**

Who had a constant speed? A Describe B from min. B stopped moving Find their average speeds. A = (2400m) ÷ (30min) A = 80 m/min B = (1200m) ÷ (30min) B = 40 m/min A B

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**Measuring Motion: Acceleration**

the rate of change of velocity change in speed or direction Centripetal acceleration circular motion (even if speed is constant, direction is always changing) ex. Moon accelerates around Earth

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**Measuring Motion: Acceleration**

Positive acceleration -“speeding up” Negative acceleration -“slowing down”

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**Calculating Acceleration**

a= vf – vi t a= ∆ v a- acceleration vf- final velocity vi- initial velocity t- time

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**Calculating Acceleration (cont’d)**

List given, then unknown values. Write equation for acceleration. Insert known values into equation and solve. Vf-Vi a t t

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Practice Problem 1 A flowerpot falls off a second-story windowsill. The flowerpot starts from rest and hits Mr. Mertz 1.5s later with a velocity of 14.7m/s. Find the average acceleration of the flowerpot. Given: Remember: Solve: t= 1.5s a= vf – vi a= 14.7m/s-0m/s Vi= 0m/s t 1.5s Vf= 14.7m/s a= 14.7m/s a= ? 1.5s a= 9.8m/s2

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Practice Problem 2 Joseph’s car accelerates at an average rate of 2.6m/s2. How long will it take his car to speed up from 24.6m/s to 26.8m/s2? Given: Remember: Solve: a= 2.6m/s2 t= (vf-vi) ÷ a vf= 26.8m/s2 vf-vi t= 26.8m/s2-24.6m/s2 Vi= 24.6m/s2 a t 2.6m/s2 t= ? t= 2.2m/s 2.6m/s2 t= 0.85s

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Practice Problem 3 A cyclist travels at a constant velocity of 4.5m/s westward and then speeds up with a steady acceleration of 2.3m/s2. Calculate the cyclist’s speed after accelerating for 5.0s. Given: Remember: Solve: vi= 4.5m/s vf= vi + a x t vf= ? vf-vi vf= 4.5m/s + (2.3m/s2 x 5.0s) a= 2.3m/s2 a t vf= 4.5m/s m/s t= 5.0s vf= 16m/s

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**Graphing Acceleration**

Distance/Time On Distance-Time graph: Acceleration is shown as a curved line

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**Graphing Acceleration**

Speed/Time On a Speed-Time graph: Slope of straight line= acceleration Positive slope- speeding up Negative slope- slowing down Flat line- constant velocity (no acceleration)

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**Newton’s Laws of Motion**

2nd Law of Motion: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Force = mass x acceleration or or F= ma

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Force Force- push or pull that one body exerts on another

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**Fundamental Forces 4 types: 1. gravity 2. electromagnetic**

3. weak nuclear 4. strong nuclear Vary in strength Act through contact or at a distance

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**Forces Force Pairs: forces moving in opposite directions**

Balanced forces: do not move; push equally on each other Unbalanced forces: acceleration (movement) in the direction of larger force

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Force Pair Examples

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**Friction Friction: force that opposes motion between 2 surfaces**

Static Friction: nonmoving surfaces Kinetic Friction: moving surfaces- sliding or rolling (sliding friction is greater than rolling friction)

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**Friction Facts Necessary for all motion**

Rougher surfaces create greater friction Greater mass creates greater friction Lubricants reduce friction

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**Newton’s First and Second Laws**

Chapter 12

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**Newton’s Laws of Motion**

1st Law of Motion: Law of Inertia An object at rest will remain at rest; An object in motion will continue moving at a constant velocity unless acted upon by a net force.

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**Inertia Objects move only when net force is applied.**

Objects maintain state of motion. Inertia is related to mass. (small mass can be accelerated by small force large mass can be accelerated by large force)

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**Newton’s Laws of Motion**

2nd Law of Motion: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Force = mass x acceleration or or F= ma

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**Newton’s Second Law For equal forces, large masses accelerate less**

Force is measured in newtons (N) 1N= 1kg x 1m/s2 F m a

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Practice Problem 1 Zoo keepers lift a stretcher that holds a sedate lion. The lion’s mass is 175 kg and the upward acceleration of the lion and stretcher is 0.657m/s2. What force is needed to produce acceleration of the lion and stretcher? Given: Remember: Solve: m= 175 kg F = m x a a= 0.657m/s2 F = 175 kg x 0.657m/s2 F= ? m a F = N F F

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Practice Problem 2 A baseball accelerates downward at 9.8 m/s2. If gravity is the only force acting on the baseball and is 1.4N, what is the baseball’s mass? Given: Remember: Solve: F= 1.4 kg/m/s m = F a= 9.8 m/s a m= ? m = 1.4 N 9.8 m/s2 m = .14 kg F m a

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**Weight and Mass Weight- measure of gravity on an object**

Not equal to mass (constant everywhere) Measured in Newtons weight = mass x free-fall acceleration (9.8m/s2) free-fall m g acceleration w

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**Gravity Force of attraction between 2 objects in the universe**

Increases as: - mass increases - distance decreases Affects all matter

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Gravity Quiz 1 Who experiences more gravity - the astronaut or the politician? More distance Less distance

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Gravity Quiz 2 Which exerts more gravity, your hand or your pencil ?

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**Gravity Quiz 3 Would you weigh more on Earth or Jupiter?**

(Hint: Which planet has the greater mass?)

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**Free Fall Acceleration**

Occurs when Earth’s gravity is only force acting on an object In absence of air resistance, all objects accelerate at same rate g = 9.8 m/s2 g = 9.8 m/s2 g = 9.8 m/s2 g = 9.8 m/s2 g = 9.8 m/s2 g = 9.8 m/s2

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Air Resistance Force of air on a moving object which opposes its motion Aka fluid friction or drag Depends on objects: - speed - surface area - shape - density

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**Air Resistance (cont’d)**

Terminal velocity= maximum velocity reached by a falling object Reached when… F gravity = F air resistance (no net force)

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**Projectile Motion Projectile- Any object thrown in air**

Only acted on by gravity Follows parabolic path- trajectory Has horizontal and vertical velocities V oy V ox

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**Projectile Motion (cont’d)**

Horizontal Velocity depends on inertia remains constant Vertical Velocity depends on gravity accelerates downward at 9.8 m/s2

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**Projectile Motion Quiz**

A moving truck launches a ball vertically (relative to the truck). If the truck maintains a constant horizontal velocity after the launch, where will the ball land (ignore air resistance)? A) In front of the truck B) Behind the truck C) In the truck Answer: C because horizontal and vertical velocities are independent of each other

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**Newton’s Laws of Motion**

3rd Law of Motion: When one object exerts a force on a second object, the second object exerts an equal but opposite force on the first.

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**Newton’s Third Law Forces always occur in pairs**

Forces in a pair do not act on the same object Equal forces don’t always have equal effects

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Newton’s Third Law Problem: How can a horse pull a cart if the cart is pulling with equal force back on the horse?

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**Newton’s Third Law Answer:**

Forces are equal and opposite but are acting on different objects Forces are not balanced The movement of the horse depend on the forces working on the horse.

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**Momentum Momentum- quantity of motion = mass x velocity**

units: kg x m/s p m v

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Practice Problem 1 Find the momentum of a bumper car if it has a total mass of 280 kg and a velocity of 3.2 m/s. Given Remember Solve m = 280 kg p = mv v = 3.2 m/s p =(280kg)(3.2m/s) P = ? P = 896 kg m/s vv p m v

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Practice Problem 2 The momentum of a second bumper car is 675 kg·m/s. What is its velocity if its total mass is 300 kg? Given Remember Solve p = 675 kg·m/s v = p ÷ m m = 300kg v =(675kg·m/s)÷(300kg) v = ? v = 2.25m/s p m v

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**Conservation of Momentum**

Law of Conservation of Momentum: Total momentum of two or more objects before a collision is the same as it was before the collision (momentum is conserved).

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**Jet Car Challenge 2 skewers 1 balloon 1 cardboard base CHALLENGE:**

Construct a car that will travel a least 3 meters using only the following materials: scissors tape 4 plastic lids 2 plastic straws Use your knowledge of Newton’s Laws to make this thing go!

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