SPH 3UO LESSON PLANS

Text: Physics 11 (Addison-Wesley)

Review the language and skills of physics
(observations, measuring, scientific notation, significant figures, prefixes, conversions
 rounding off, accuracy, error)

position (d): location relative to a reference point
                    -it is a vector
    eg. school-----7 km-----house
position of your house: d = 7 km [E] of school (includes measurement, unit, direction and reference point)

displacement (/\d): change in position
                            -it is a vector but does not need a reference point
/\d = d_f - d_i

Drawing a vector:    tail -------------> tip (length of vector shows magnitude)

eg. Draw 7 km [W]
        1) choose scale
        2) draw a line to scale in the right direction
        3) draw an arrow and label
                    1 cm = 1 km             W<------>E

                                            <------
                                        /\d

Adding Collinear Vectors (in the same line)
    -use algebra                Collinear: ----> + ----->
                                                      ----> + <----
eg.     /\d₁ = 500 km [N]
         /\d₂ = 200 km [S]
    find /\d₁ + /\d₂
     1) use negatives to put them in the same direction
    2) add the numbers to get the answers, keep the directions
    3) change to a positive number

 /\d₁ + /\d₂= 500 km [N] + 200 km [S]
1) = 500 km [N] - 200 km [S]    or    = -500 km [N] + 200 km [S]
2) = 300 km [N]                              = -300 km [S]
3) = 300 km [N]                              = 300 km [N]

Adding Non-collinear Vectors
    -use analytical techniques

eg.     /\d₁ = 5 km [S]
         /\d₂ = 2 km [W]
     find /\d₁ + /\d₂
    1) draw vectors to scale from tip to tail
    2) connect tail to the tip
    3) measure length and angle
    4) label
                    ________/\d₂
                    l            /                1 cm = 1 km
          /\d₁     l           /                W<------->E
                    l        /
                    l O^o /         /\d₁ + /\d₂   = 5.4 km O^o E of N
                    l    /
                    l  /
                    l/

Velocity and Uniform Motion

Velocity: change in position per unit time

If the velocity is constant then it is called "uniform motion"

If the velocity is changing, then:

1) average velocity = change in position/elapsed time
             v_av = /\d//\t
 
2) instantaneous velocity is the velocity at one point in time
 
most of the time -> average velocity is not equal to instantaneous velocity

-you run 10 km in 2 hours
v_av  = ?

eg. a cyclist goes 20 km [W] in 1.5 h and then 15 km [S] in 2 h. Find the average velocity.

                    v_av = /\d//\t

a) find  /\d            5 km = 1 cm

                                      ____________20 km
                                      l               O  /
                           15 km l                 /
                                      l             /
                                      l         /      25 km [37^o S of E]
                                      l      /
                                      l   /
                                      l/

b)  /\t = 1.5 h + 2 h
         = 3.5 h

v_av = /\d//\t
      = 25 km [37^o S of E] / 3.5 h
      = 7.1 km [37^o S of E]

Acceleration = a
-acceleration: change in velocity / time
-acceleration is a vector
-units for acceleration are m/s² or km/h/s

Constant Acceleration
-velocity changes by equal amounts every second

eg. object is in free fall a = 10 m/s² [down]
      object is at rest     a =   0 m/s²

Changing Acceleration
-most common
eg. car going from 0-60 km/h [W] in 6 s. Find the a:

                 a = 10 km/h/s [W]  is the average acceleration

What was the a at 1 s?
Probably not 10 km/h/s [W].
Probably higher at the start, eg. 15 km/h/s at 1 s, which is called the instantaneous acceleration

Velocity-Time Graphs
Velocity is on Y axis, Time is on X axis

Graph 1.6 m/s [Down]  in 4 s, and 2.0 m/s [Down] in 5 s as a straight line
Slope = rise/run
         = /\v/t
         =acceleration

Slope of a v vs. t graph gives acceleration
-a straight line on this graph means acceleration is constant

eg. slope = rise/run
              = 20 m/s[Down]/5 s
              = 4 m/s² [Down]

-the area under this curve is the displacement

eg. displacement after 5 s = area of triangle
                                      = 1/2 bh
                                      = 1/2 (5s)(20m/s[Down])
                                      = 50 m [Down]

eg. displacement after 4 s = area of triangle
                                      = 1/2 bh
                                      = 1/2 (4 s)(16 m/s[Down])
                                      = 32 m [Down]

Relative Velocity
eg. a pedestrian and a passenger observe the driver travelling at different speeds

Motion depends on the observer or motion is relative

eg. observer O watches a train move 25 m/s [S] (train B), and another train move 45 m/s [N] (train A)

-to someone on train A, observer O is moving at 45 m/s [S] and train B is moving at 70 m/s [S]
-to someone on train B, observer O is moving at 25 m/s [N] and train A is moving at 70 m/s [N]

eg. Car Y goes 85 km/h [W] and Car X goes 100 km/h [E], and both cars start together

Forces (why things move)

force:    -a push or pull on an object
            -results in a change in shape or motion of the object
            -is a vector (magnitude and direction)
            -is measured in Newtons
            (describe spring scale)

Four Fundamental Forces
-gravity (weakest)
-strong nuclear force
-weak nuclear force
-electro-magnetic force

Free-Body Diagrams
-rarely is there only one force acting on an object (usually at least two)
1) Force of Gravity
2) Normal Force

eg. Free body diagram of box           /\    6 N    F_N
                                                      l
                                                   BOX
                                                      l
                                                     V    5 N    F_g

1) draw the object isolated from its surroundings
2) start from the center of mass [center of object]
3) draw arrow in the direction of the forces
4) length of lines represent the magnitudes of the force
5) only draw forces ON the object

Net Force is the total of the forces acting ON an object
F_net =  6 N [up] + 6 N [down]
       =  0 N

Car that is coasting         /\  F_N
                                                   l
                            F_f <------  CAR
                                                   l
                                      V  F_g

F_net = F_N [up] + F_g [down] + F_f [back]
F_net = F_f [back] if the other forces are equal

Adding collinear forces is easy, and needed often in physics

eg. two bumper cars hit a third car. What is the F_net on the bumper car?
                       1000 N [N] /\
                                        l
                                    CAR ---> 600 N [E]

        method 1:   F_net = 1166 N [O^o = 59^o] found with scale diagram, ruler and protractor

                                           /l\
                          F_net            /  l     1000 N
                                                    /_>l
                                                  600 N

        method 2:                   / l
                                        /Bl
                             c        /     l    a
                                            /A_ Cl
                                       _b

SINE LAW               sin A/a = sin B/b = sin C/c

COSINE LAW          c² = a² + b² - 2ab cos C

                                                 /l
                      F_net                      /  l     1000 N
                                                        /O_l
                                                      600 N

 Use cosine law to find c:  
                         c² =     a²      +   b²    - 2ab cos C
                     F_net² =  (1000)² + (600)² - 2(1000)(600) cos 90^o
                     F_net² =  (1000)² + (600)²
                             =  1 360 000
                     F_net  =   1166.2 N

Use sine law to find O:
                            sin O/1000 = sin 90^o/1166.2

Gravity: -the amount of material in an object
             -it is the same anywhere in the universe
             -measured in kg

Weight: -the force that a celestial body exerts on an object.
            -it is measured in N

Newton's Law of Universal Gravitation
F_g = Gm₁m₂//\d²
        where F_g = force of gravity (N)
                 m₁ = mass of one object (kg)
                 m₂ = mass of other object (kg)
                 /\d = distance between centers of objects (m)
                   G = 6.67 X 10^-4 Nm²/kg²

eg. find the force of gravity on you if you were at sea level:
                r_E   = 6.38 X 10⁶ m (radius of earth)
                m_E = 5.98 X 10²⁴ kg(mass of earth)
                your mass = 78 kg

eg. Find the force of gravity from you on the Earth

F_g = _____N [towards the center of the earth]