What is gravity? • Where does it come from? • What kinds of things have gravity? • I have two masses. One is a basketball. The other is a 5lb bag of potatoes. If I hold them at the same height and drop them at the same time, which one hits the ground first? • Do they have a constant velocity? • Do they have the same gravitational force acting on them?
Equations!!! • Equations in IB look different. • Fill in the following chart based on the given equations.
Vocabulary • Distance – a scalar quantity of position changed • Displacement - a vector quantity of a change in the position of an object. • Speed – is the rate of change of distance with respect to time. (scalar) • Velocity – is the rate of change of displacement with respect to time. (vector quantity) • Acceleration – is the rate of change of velocity with respect to time. (vector quantity)
Distance and Displacement • A mass initially at 0m moves 10m to the right and then 2m to the left. What is the final displacement? What is the final distance traveled? • A mass initially at 0m, first moves 5m to the right and then 12m to the left. What is the total distance covered by the mass and what is the change in displacement?
Distance and Displacement • Distance = 12m, Displacement = 8m • Distance = 17m, Displacement = -7m
Speed and Velocity • Speed is nornally given in m/s or km/h. • OR ms-1 or kmh-1 • What the difference? • Speed is a scalar, velocity is a vector.
Speed and Velocity • A car of length 4.2m travelling in a straight line takes 0.56s to go past a mark on the road. What is the speed of the car? • A car starts out from the 100km mark in a straight line and moves a distance of 20km towards the right, and then returns to its starting position 1h later. What is the average speed and the average velocity for this trip?
V = 7.5 m/s • Avg speed = 40km/h, avg velocity = 0km/h • Starting point is irrelevant.
Acceleration • We mostly look at constant acceleration situations. • In this case the instantaneous acceleration and average acceleration are the same thing. • Acceleration due to gravity is 9.8m/s2 or 10m/s2 or 9.81m/s2. • Don’t forget about the ball thrown upward. • Don’t forget about the variable table!
Acceleration • An object starting with an initial velocity of 2 m/s undergoes constant acceleration. After 5s its velocity is found to be 12m/s. What is the acceleration? • A ball is thrown downward from a 70m tower with an initial velocity of 3m/s. How fast would it be going after 2 seconds? What would it’s position be after 2seconds?
Acceleration • 2m/s2 • 22.6m/s, 44.4m high
Everything • A mass has an initial velocity of 10m/s. It moves with acceleration -2m/s2. When will it have a zero velocity? • What is the displacement after 10s of a mass whose initial velocity is 2m/s and moves with acceleration of 4m/s2? • A car has an initial velocity of 5m/s. When its displacement increases by 20m, its velocity becomes 7m/s. What is the acceleration? • A body has initial velocity of 4m/s and a velocity of 12m/s after 6s. What is the displacement?
t = 5s • s = 220m • a = .6m/s2 • s = 48m
Test yourself • Two balls start out moving to the right with constant velocities of 5m/s and 4m/s. The slow ball stars first and the other 4s later. How far from the starting position are they when they meet? • A mass is thrown upwards with an initial velocity of 30m/s. A second mass is dropped from directly above, a height of 60m from the first mass, 0.5s later. When do the masses meet and how high is the point where they meet?
t=20s so s = 80m • t=2.35s so s is 42.9m
The slope of d-t graph gives the value of the v-t graph • The slope of v-t graph gives the value of the a-t graph • Area under a-t graph gives the change in velocity • Area under v-t graph gives the change in displacement.
Bell RingerSept 4/5 • Consider the following graph
Bell RingerSept 4/5 • What is the initial displacement? • What is the velocity for the first 10s? Second 5s? • When is the object at the origin? • What is Δs? What is the total distance traveled? • What is the avg velocity?
Initial displacement is -10m • 2m/s, -2m/s • 5s and 15s • Δs = 10m, total distance = 30m • Avg speed = 2m/s, avg velocity = .66m/s
Interpreting graphs • A mass starts out from zero with velocity 10m/s and continues moving at this velocity for 5s. The velocity is then abruptly reversed to -5m/s and the object moves at this velocity for 10s. For this event find: • The graph • The change in displacement • The total distance travelled • The average speed • The average velocity
See previous slide • 10 x 5m = 50m, -5 x 10 = -50m, Δs = 0 The object moved toward the right, stopped and returned to its starting position. • 50m to the right, 50m to the left, 100m total • Avg velocity =0 • Avg speed= 100m/15s = 6.7m/s
Use the graph and your understanding of slope calculations to determine the acceleration of the rocket during the listed time intervals. • t = 0 - 1 s, t = 1 - 4 s, t = 4 - 12 s
a = 40 m s-2 • b = 20 m s-2 • c = - 20 m s-2
Describe the motion depicted by the following velocity-time graphs. In your descriptions, make reference to the direction of motion (+ or - direction), the velocity and acceleration and any changes in speed (speeding up or slowing down) during the various time intervals (e.g., intervals A, B, and C).
The object moves in the + direction at a constant speed - zero acceleration (interval A). The object then continues in the + direction while slowing down with a negative acceleration (interval B). Finally, the object moves at a constant speed in the + direction, slower than before (interval C). • The object moves in the + direction while slowing down; this involves a negative acceleration (interval A). It then remains at rest (interval B). The object then moves in the - direction while speeding up; this also involves a negative acceleration (interval C). • The object moves in the + direction with a constant velocity and zero acceleration (interval A). The object then slows down while moving in the + direction (i.e., it has a negative acceleration) until it finally reaches a 0 velocity (stops) (interval B). Then the object moves in the - direction while speeding up; this corresponds to a - acceleration (interval C)
Consider the velocity-time graph below. Determine the acceleration (i.e., slope) of the object as portrayed by the graph.
The acceleration (i.e., slope) is 4 m/s/s. If you think the slope is 5 m/s/s, then you're making a common mistake. You are picking one point (probably 5 s, 25 m/s) and dividing y/x. Instead you must pick two points (as in the discussed in this part of the lesson) and divide the change in y by the change in x.
Determine the displacement of the object during the time interval denoted by the shaded area.
a = 90m • b = 45m • c = 40m
A car (A) moves to the left with speed 40km/h (with respect to the road). Another car (B) moves to the right with speed 60km/h(also with respect to the road). Find the relative velocity of B with respect to A.
Types of forces • Gravitational – between objects as a result of their masses. (also called weight) • Normal reaction – between two surfaces that are touching • Frictional- force that opposes the relative motion of two surfaces. (Includes Air resistance/drag) • Applied – force of an outside object pushing or pulling • Tension- when a string/spring is stretched, it is equal and opposite the force acting on it. • Compression – opposite of tension (applies to solid objects) • Upthrust – the upward force that acts on objects when it is submerged in a fluid. Causes objects to float. • Lift – caused when air flows over an aircraft’s wing. Causes an upward force.
Free Body Diagram • Five Steps. • Is there gravitational force? (Fg) • Is it sitting on a surface? (Fn) • Is there some thing pushing or pulling? (Fapp) • Is there friction? (Ff) • Is it accelerating? a = ? http://www.youtube.com/watch?v=BuPfDI7TyL0
Forces are vectors • Magnitude and direction • Many times the direction is the x or y • This means they can be resolved into components. • This also means they can be added and subtracted. http://www.youtube.com/watch?v=IrY-FlJ0c7Y
Practice/Application • A person is trying to lift a heavy concrete block without success. The upward force exerted on the block by the person is P, the contact force on the block by the floor is C, the weight of the block is W. • Which one of the following is true about the magnitudes of the forces on the block while the person is trying to lift it? • P + C =W • P + C < W • P+ C > W • P = C = W
Practice/Application • Consider the vector with a magnitude of 7N acting at an angle of 140° to the horizontal. What will the horizontal and vertical components of this vector be? • An object O is acted upon by three forces as shown in the diagram. What is the magnitude of the resultant force acting on O?
Answers • Since the person is not succeeding to lift the box, the weight of the box must be equal to the sum of the upward force and the floor contact force. Answer: A • y= 5.4 x = -4.5 • The 6N and 3N forces are acting against each other. This results in a 3N force to the left. This resulting 3N force left and the 4N upward force can be resolved using Pythagoras. Answer = 5N
Practice/Application • A block of wood of mass 4kg rests on a slope, inclined at an angle of 25° to the horizontal as shown. Calculate: • The weight of the block • The normal reaction force of the plane acting on the block. • The resultant accelerating force down the slope • The acceleration of the block down the slope
Answers • There are several • Fg = mg => Fg = 40N • FN = Fgcosθ => FN = 36.6N • FR = Fg sinθ => FR = 16.9N • Fg = mg => a = 4.2m/s2 • d
Equilibrium • Occurs when the net force on an object is zero. • An object can move and still be at equilibrium. • Neutral equilibrium(translational) – an object is at equilibrium, it is then moved and then it is still at equilibrium. A displacement results in another equilibrium position. The net force acting on an object is zero.