2 Standards Students will: SPS8. Determine the relationship between force, mass and motionSPS8a Calculate velocity and accelerationSPS8c Relate falling objects to gravitational forceSPS8d Explain difference in mass and weight
3 Observing MotionMotion- change in position in relation to a reference point.
4 Measuring Motion: Distance Distance- how far an object moves on a pathDisplacement- how far between starting and ending points on a path
5 Measuring Motion: Speed rate of motiondistance traveled per unit timespeed = distancetime
6 Measuring Motion: Speed (cont’d) Instantaneous Speedspeed at a given instantAverage Speed-= total distancetotal time
7 Measuring Motion: Velocity speed in a given directioncan change even when the speed is constant!
11 Calculating Slope Choose two points on graph to calculate slope. Distance Vs Time16Choose two points on graph to calculate slope.Calculate the vertical and horizontal change.Divide the vertical change by the horizontal change.14.1210Distance (m)8.64212345Time (s)
12 Calculating Slope(cont’d) Distance Vs Time161. 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 3s14.1210VerticalchangeDistance (m)8.6HorizontalChange4212345Time (s)
13 Practice Problem A B Who started out faster? A (steeper slope) Who had a constant speed?ADescribe B from min.B stopped movingFind their average speeds.A = (2400m) ÷ (30min) A = 80 m/minB = (1200m) ÷ (30min) B = 40 m/minAB
14 Measuring Motion: Acceleration the rate of change of velocitychange in speed or directionCentripetal accelerationcircular motion (even if speed is constant, direction is always changing)ex. Moon accelerates around Earth
16 Calculating Acceleration a= vf – vita= ∆ va- accelerationvf- final velocityvi- initial velocityt- time
17 Calculating Acceleration (cont’d) List given, then unknown values.Write equation for acceleration.Insert known values into equation and solve.Vf-Via tt
18 Practice Problem 1A 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
19 Practice Problem 2Joseph’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
20 Practice Problem 3A 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
21 Graphing Acceleration Distance/TimeOn Distance-Timegraph:Acceleration is shown as a curved line
23 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 ororF= ma
24 ForceForce- push or pull that one body exerts on another
25 Fundamental Forces 4 types: 1. gravity 2. electromagnetic 3. weak nuclear4. strong nuclearVary in strengthAct through contact or at a distance
26 Forces Force Pairs: forces moving in opposite directions Balanced forces: do notmove; push equally on each otherUnbalanced forces:acceleration (movement) in the direction of larger force
28 Friction Friction: force that opposes motion between 2 surfaces Static Friction: nonmoving surfacesKinetic Friction: moving surfaces- sliding or rolling(sliding friction is greater than rolling friction)
29 Friction Facts Necessary for all motion Rougher surfaces create greater frictionGreater mass creates greater frictionLubricants reduce friction
31 Newton’s Laws of Motion 1st Law of Motion: Law of InertiaAn 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.
32 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 forcelarge mass can be accelerated by large force)
33 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 ororF= ma
34 Newton’s Second Law For equal forces, large masses accelerate less Force is measured in newtons (N)1N= 1kg x 1m/s2Fma
35 Practice Problem 1Zoo 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 = NFF
36 Practice Problem 2A baseball accelerates downward at 9.8 m/s2. Ifgravity is the only force acting on the baseball and is1.4N, what is the baseball’s mass?Given: Remember: Solve:F= 1.4 kg/m/s m = Fa= 9.8 m/s am= ? m = 1.4 N9.8 m/s2m = .14 kgFma
37 Weight and Mass Weight- measure of gravity on an object Not equal to mass (constant everywhere)Measured in Newtonsweight = mass x free-fall acceleration (9.8m/s2)free-fallm g accelerationw
38 Gravity Force of attraction between 2 objects in the universe Increases as:- mass increases- distance decreasesAffects all matter
39 Gravity Quiz 1Who experiences more gravity - the astronaut or the politician?More distanceLess distance
40 Gravity Quiz 2Which exerts more gravity, your hand or your pencil ?
41 Gravity Quiz 3 Would you weigh more on Earth or Jupiter? (Hint: Which planet has the greater mass?)
42 Free Fall Acceleration Occurs when Earth’s gravity is only force acting on an objectIn absence of air resistance, all objects accelerate at same rateg = 9.8 m/s2g = 9.8 m/s2 g = 9.8 m/s2 g = 9.8 m/s2 g = 9.8 m/s2 g = 9.8 m/s2
43 Air ResistanceForce of air on a moving object which opposes its motionAka fluid friction or dragDepends on objects:- speed- surface area- shape- density
44 Air Resistance (cont’d) Terminal velocity=maximum velocity reached by a falling objectReached when…F gravity = F air resistance(no net force)
45 Projectile Motion Projectile- Any object thrown in air Only acted on by gravityFollows parabolic path- trajectoryHas horizontal andvertical velocitiesV oyV ox
46 Projectile Motion (cont’d) Horizontal Velocitydepends on inertiaremains constantVertical Velocitydepends on gravityaccelerates downward at 9.8 m/s2
47 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 truckB) Behind the truckC) In the truckAnswer: C because horizontal and vertical velocities areindependent of each other
48 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.
49 Newton’s Third Law Forces always occur in pairs Forces in a pair do not act on the same objectEqual forces don’t always have equal effects
50 Newton’s Third LawProblem: How can a horse pull a cart if the cart is pulling with equal force back on the horse?
51 Newton’s Third Law Answer: Forces are equal and opposite but are acting on different objectsForces are not balancedThe movement of the horse depend on the forces working on the horse.
52 Momentum Momentum- quantity of motion = mass x velocity units: kg x m/spmv
53 Practice Problem 1Find 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/svvpmv
54 Practice Problem 2The 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/spmv
55 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).
56 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:scissorstape4 plastic lids2 plastic strawsUse your knowledge of Newton’s Laws to make this thing go!