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Motion Chapter 11 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 Observing Motion • Motion- change in position in relation to a reference point. Measuring Motion: Distance • Distance- how far an object moves on a path • Displacement- how far between starting and ending points on a path Measuring Motion: Speed • Speed – rate of motion – distance traveled per unit time speed = distance time Measuring Motion: Speed (cont’d) • Instantaneous Speed – speed at a given instant • Average Speed= total distance total time Measuring Motion: Velocity • Velocity – speed in a given direction – can change even when the speed is constant! Calculating Velocity: Distance/Speed/Time Triangle Graphing Speed/Velocity • X axis- usually independent variable (time) • Y axis- usually dependent variable (distance) • Slope of straight line= vertical change horizontal change Graphing (cont’d) • slope = velocity • steeper slope = faster velocity • straight line = constant velocity • flat line = 0 velocity (no motion) Calculating Slope Distance (m) 1. Choose two points on graph to calculate slope. 2. Calculate the vertical and horizontal change. 3. Divide the vertical change by the horizontal change. Distance Vs Time 16 14 . 12 10 8 . 6 4 2 0 1 2 3 Time (s) 4 5 Calculating Slope(cont’d) 16 14 . 12 Distance (m) 1. point 1: d= 6m t= 1s point 2: d= 12m t= 4s 2. vert ∆- 12m-6m= 6m horiz ∆- 4s-1s= 3s Distance Vs Time 10 Vertical change 8 . 6 Horizontal Change 4 3. slope= 6m = 2m/s 3s 2 0 1 2 3 Time (s) 4 5 Practice Problem A B • Who started out faster? – A (steeper slope) • Who had a constant speed? –A • Describe B from 10-20 min. – B stopped moving • Find their average speeds. – A = (2400m) ÷ (30min) A = 80 m/min – B = (1200m) ÷ (30min) B = 40 m/min Measuring Motion: Acceleration • 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 Measuring Motion: Acceleration • Positive acceleration -“speeding up” • Negative acceleration -“slowing down” Calculating Acceleration • a= vf – vi t • a= ∆ v t • • • • a- acceleration vf- final velocity vi- initial velocity t- time Calculating Acceleration (cont’d) 1. List given, then unknown values. 2. Write equation for acceleration. 3. Insert known values into equation and solve. t Vf-Vi a t 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 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 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 + 11.5m/s t= 5.0s vf= 16m/s Graphing Acceleration Distance/Time 400 On Distance-Time graph: • Acceleration is shown as a curved line Distance (m) 300 200 100 0 0 5 10 Time (s) 15 20 Graphing Acceleration Speed/Time 3 Speed (m/s) 2 1 0 0 2 4 6 Time (s) 8 10 On a Speed-Time graph: • Slope of straight line= acceleration • Positive slopespeeding up • Negative slope- slowing down • Flat line- constant velocity (no acceleration) 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 Force Force- push or pull that one body exerts on another Fundamental Forces • 4 types: 1. gravity 2. electromagnetic 3. weak nuclear 4. strong nuclear • Vary in strength • Act through contact or at a distance 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 Force Pair Examples 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) Friction Facts • Necessary for all motion • Rougher surfaces create greater friction • Greater mass creates greater friction • Lubricants reduce friction Newton’s First and Second Laws Chapter 12 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. 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) 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 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 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=mxa F a= 0.657m/s2 F = 175 kg x 0.657m/s2 F F= ? m a F = 11.49 N 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: F= 1.4 kg/m/s2 F a= 9.8 m/s2 m= ? m a Solve: m=F a m = 1.4 N 9.8 m/s2 m = .14 kg 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) w m g free-fall acceleration Gravity • Force of attraction between 2 objects in the universe • Increases as: - mass increases - distance decreases • Affects all matter Gravity Quiz 1 Who experiences more gravity - the astronaut or the politician? More distance Less distance Gravity Quiz 2 Which exerts more gravity, your hand or your pencil ? Gravity Quiz 3 • Would you weigh more on Earth or Jupiter? (Hint: Which planet has the greater mass?) 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 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 Air Resistance (cont’d) • Terminal velocity= maximum velocity reached by a falling object • Reached when… F gravity = F air resistance (no net force) 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 Projectile Motion (cont’d) • Horizontal Velocity – depends on inertia – remains constant • Vertical Velocity – depends on gravity – accelerates downward at 9.8 m/s2 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 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. 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 Newton’s Third Law Problem: How can a horse pull a cart if the cart is pulling with equal force back on the horse? Newton’s Third Law Answer: 1. Forces are equal and opposite but are acting on different objects 2. Forces are not balanced 3. The movement of the horse depend on the forces working on the horse. Momentum Momentum• quantity of motion • = mass x velocity • units: kg x m/s p m v 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 m = 280 kg v = 3.2 m/s P=? Remember p m vv v Solve p = mv p =(280kg)(3.2m/s) P = 896 kg m/s 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 p = 675 kg·m/s p m = 300kg m v v=? Solve v=p÷m v =(675kg·m/s)÷(300kg) v = 2.25m/s 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). Jet Car Challenge CHALLENGE: Construct a car that will travel a least 3 meters using only the following materials: • scissors • 2 skewers • tape • 1 balloon • 4 plastic lids • 1 cardboard base • 2 plastic straws Use your knowledge of Newton’s Laws to make this thing go!