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Long 50slideschapter 5 motion notes [autosaved] Long 50slideschapter 5 motion notes [autosaved] Presentation Transcript

  • Ch. 5 Motion & Forces
    • II. Describing Motion
          • Using Newton’s Laws of Motion
  • A. Motion
    • Problem:
      • Is your desk moving?
    • We need a reference point ...
      • nonmoving point from which motion is measured
  • A. Motion
    • Motion
      • Change in position in relation to a reference point.
    Reference point Motion
  • A. Motion
    • Problem:
    • You are a passenger in a car stopped at a stop sign. Out of the corner of your eye, you notice a tree on the side of the road begin to move forward.
    • You have mistakenly set yourself as the reference point.
  • A. Force
    • Force
      • a push or pull that one body exerts on another
      • What forces are being exerted on the football?
    F kick F grav
  • A. Force
    • Balanced Forces
      • forces acting on an object that are opposite in direction and equal in size
      • no change in velocity
  • A. Force
    • Net Force
      • unbalanced forces that are not opposite and equal
      • velocity changes (object accelerates)
    F friction W F pull F net N N
  • B. Gravity
    • Gravity
      • force of attraction between any two objects in the universe
      • increases as...
        • mass increases
        • distance decreases
  • B. Gravity
    • Who experiences more gravity - the astronaut or the politician?
    • Which exerts more gravity - the Earth or the moon?
    less distance more mass
  • B. Gravity
    • Weight
      • the force of gravity on an object
    MASS always the same (kg) WEIGHT depends on gravity (N) W = mg W : weight (N) m : mass (kg) g : acceleration due to gravity (m/s 2 )
  • B. Gravity
    • Would you weigh more on Earth or Jupiter?
        • greater gravity
        • greater weight
        • greater mass
      • Jupiter because...
  • B. Gravity
    • Accel. due to gravity (g)
      • In the absence of air resistance, all falling objects have the same acceleration!
      • On Earth: g = 9.8 m/s 2
    Animation from “ Multimedia Physics Studios .” elephant feather
  • Newton’s First Law
    • Newton’s First Law of Motion
      • An object at rest will remain at rest and an object in motion will continue moving at a constant velocity unless acted upon by a net force.
    motion constant velocity net force
  • B. Newton’s First Law
    • Newton’s First Law of Motion
      • “ Law of Inertia”
    • Inertia (Greek word meaning Lazyness)
      • tendency of an object to resist any change in its motion
      • increases as mass increases
  • ConcepTest 2
    • You are a passenger in a car and not wearing your seat belt.
    • Without increasing or decreasing its speed, the car makes a sharp left turn, and you find yourself colliding with the right-hand door.
    • Which is the correct analysis of the situation? ...
  • ConcepTest 2
    • 1. Before and after the collision, there is a rightward force pushing you into the door.
    • 2. Starting at the time of collision, the door exerts a leftward force on you.
    • 3. both of the above
    • 4. neither of the above
    2. Starting at the time of collision, the door exerts a leftward force on you.
  • What is a Force?
    • Push or pull
    • Measured in (N) Newton
  • Net forces
    • Net force: all forces exerted on an object
  • Unbalanced and Balanced Forces
    • Unbalanced forces produce a change in motion(acceleration)
    • Balanced Forces produce no change in motion
  • C. Friction
    • Friction
      • force that opposes motion between 2 surfaces
      • depends on the:
        • types of surfaces
        • force between the surfaces
  • C. Friction
    • Friction is greater...
      • between rough surfaces
      • when there’s a greater force between the surfaces (e.g. more weight)
    • Pros and Cons?
  • Types of Friction
    • Sliding Friction
    • Rolling Friction
    • Fluid Friction
    • Static Friction
  • Newton’s Second Law
    • Newton’s Second Law of Motion
      • The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
    • F = ma
  • B. Speed & Velocity
    • Speed
      • rate of motion
      • distance traveled per unit time
    v d t
  • B. Speed & Velocity
    • Instantaneous Speed
      • speed at a given instant
    • Average Speed
  • B. Speed & Velocity
    • Problem:
      • A storm is 10 km away and is moving at a speed of 60 km/h. Should you be worried?
      • It depends on the storm’s direction!
  • B. Speed & Velocity
    • Velocity
      • speed in a given direction
      • can change even when the speed is constant!
  • C. Acceleration
    • Acceleration
      • the rate of change of velocity
      • change in speed or direction
    a : acceleration v f : final velocity v i : initial velocity t : time a v f - v i t
  • D. Calculations
    • Your neighbor skates at a speed of 4 m/s. You can skate 100 m in 20 s. Who skates faster?
    GIVEN: d = 100 m t = 20 s v = ? WORK : v = d ÷ t v = (100 m) ÷ (20 s) v = 5 m/s You skate faster! v d t
  • D. Calculations
    • A roller coaster starts down a hill at 10 m/s. Three seconds later, its speed is 32 m/s. What is the roller coaster’s acceleration?
    GIVEN: v i = 10 m/s t = 3 s v f = 32 m/s a = ? WORK : a = ( v f - v i ) ÷ t a = (32m/s - 10m/s) ÷ (3s) a = 22 m/s ÷ 3 s a = 7.3 m/s 2 a v f - v i t
  • D. Calculations
    • Sound travels 330 m/s. If a lightning bolt strikes the ground 1 km away from you, how long will it take for you to hear it?
    GIVEN: v = 330 m/s d = 1km = 1000m t = ? WORK : t = d ÷ v t = (1000 m) ÷ (330 m/s) t = 3.03 s v d t
  • D. Calculations
    • How long will it take a car traveling 30 m/s to come to a stop if its acceleration is -3 m/s 2 ?
    GIVEN: t = ? v i = 30 m/s v f = 0 m/s a = -3 m/s 2 WORK : t = ( v f - v i ) ÷ a t = (0m/s-30m/s)÷(-3m/s 2 ) t = -30 m/s ÷ -3m/s 2 t = 10 s a v f - v i t
  • D. Calculations
    • What force would be required to accelerate a 40 kg mass by 4 m/s 2 ?
    GIVEN: F = ? m = 40 kg a = 4 m/s 2 WORK : F = ma F = (40 kg)(4 m/s 2 ) F = 160 N m F a
  • D. Calculations
    • A 4.0 kg shotput is thrown with 30 N of force. What is its acceleration?
    GIVEN: m = 4.0 kg F = 30 N a = ? WORK : a = F ÷ m a = (30 N) ÷ (4.0 kg) a = 7.5 m/s 2 m F a
  • D. Calculations
    • Ms. Wills weighs 557 N. What is her mass?
    GIVEN: F(W) = 557 N m = ? a( g ) = 9.8 m/s 2 WORK : m = F ÷ a m = (557 N) ÷ (9.8 m/s 2 ) m = 56.8 kg m F a
  • ConcepTest
    • Is the following statement true or false?
      • An astronaut has less mass on the moon since the moon exerts a weaker gravitational force.
      • False! Mass does not depend on gravity, weight does. The astronaut has less weight on the moon.
  • A. Newton’s Third Law
    • Newton’s Third 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.
  • A. Newton’s Third Law
    • Problem:
      • How can a horse pull a cart if the cart is pulling back on the horse with an equal but opposite force?
      • Aren’t these “balanced forces” resulting in no acceleration?
    NO!!!
  • A. Newton’s Third Law
      • forces are equal and opposite but act on different objects
      • they are not “balanced forces”
      • the movement of the horse depends on the forces acting on the horse
    • Explanation:
  • A. Newton’s Third Law
    • Action-Reaction Pairs
    • The hammer exerts a force on the nail to the right.
    • The nail exerts an equal but opposite force on the hammer to the left.
  • A. Newton’s Third Law
    • Action-Reaction Pairs
    • The rocket exerts a downward force on the exhaust gases.
    • The gases exert an equal but opposite upward force on the rocket.
    F G F R
  • A. Newton’s Third Law
    • Action-Reaction Pairs
    • Both objects accelerate.
    • The amount of acceleration depends on the mass of the object.
    • Small mass  more acceleration
    • Large mass  less acceleration
    F m
  • E. Graphing Motion
    • slope =
    • steeper slope =
    • straight line =
    • flat line =
    faster speed constant speed no motion speed Distance-Time Graph A B
  • E. Graphing Motion
    • 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
    Distance-Time Graph A B
  • E. Graphing Motion
    • Acceleration is indicated by a curve on a Distance-Time graph.
    • Changing slope = changing velocity
    Distance-Time Graph
  • E. Graphing Motion
    • slope =
    • straight line =
    • flat line =
    • acceleration
      • +ve = speeds up
      • -ve = slows down
    constant accel. no accel. (constant velocity) Speed-Time Graph
  • E. Graphing Motion
    • Specify the time period when the object was...
    • slowing down
      • 5 to 10 seconds
    • speeding up
      • 0 to 3 seconds
    • moving at a constant speed
      • 3 to 5 seconds
    • not moving
      • 0 & 10 seconds
    Speed-Time Graph
  • B. Momentum
    • Momentum
      • quantity of motion
    p = mv p : momentum (kg ·m/s) m : mass (kg) v : velocity (m/s) m p v
  • B. Momentum
    • 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: p = ? m = 280 kg v = 3.2 m/s WORK : p = mv p = (280 kg)(3.2 m/s) p = 896 kg·m/s m p v
  • B. Momentum
    • The momentum of a second bumper car is 675 kg·m/s. What is its velocity if its total mass is 300 kg?
    GIVEN: p = 675 kg·m/s m = 300 kg v = ? WORK : v = p ÷ m v = (675 kg·m/s)÷(300 kg) v = 2.25 m/s m p v
  • C. Conservation of Momentum
    • Law of Conservation of Momentum
      • The total momentum in a group of objects doesn’t change unless outside forces act on the objects.
    p before = p after
  • C. Conservation of Momentum
    • A 5-kg cart traveling at 4.2 m/s strikes a stationary 2-kg cart and they connect. Find their speed after the collision.
    BEFORE Cart 1 : m = 5 kg v = 4.2 m/s Cart 2 : m = 2 kg v = 0 m/s AFTER Cart 1 + 2 : m = 7 kg v = ? p = 21 kg·m/s p = 0 p before = 21 kg·m/s p after = 21 kg·m/s v = p ÷ m v = (21 kg·m/s) ÷ (7 kg) v = 3 m/s m p v
  • C. Conservation of Momentum
    • A 50-kg clown is shot out of a 250-kg cannon at a speed of 20 m/s. What is the recoil speed of the cannon?
    BEFORE Clown : m = 50 kg v = 0 m/s Cannon : m = 250 kg v = 0 m/s AFTER Clown : m = 50 kg v = 20 m/s Cannon : m = 250 kg v = ? m/s p = 0 p = 0 p before = 0 p = 1000 kg·m/s p after = 0 p = -1000 kg·m/s
  • C. Conservation of Momentum
    • So…now we can solve for velocity.
    GIVEN: p = -1000 kg·m/s m = 250 kg v = ? WORK : v = p ÷ m v = (-1000 kg·m/s)÷(250 kg) v = - 4 m/s (4 m/s backwards) m p v