Physics – 1st, junior
Yossi Quint
[9.14.09]
Metric system = SI = System International = MKS(A)
 Basic units
o meter
m
length, distance
o kilogram kg
mass
o second
s
time
o Ampere
A
electric current
 Unit is capitalized since its named after someone
 Prefixes
o yacto 10^-24 y
o
o
o
o
o
o
o
o
o
zepto
atto
femto
pico
nano
micro
milli
centi
kilo
10^-21 z
10^-18 a
10^-15 f
10^-12 p
10^-9 n
10^-6 mu
10^-3 m
10^-2 c
10^3
k
mega 10^6
M
Giga 10^9
G
Tera 10^12 T
Peta 10^15 P
Exa
10^18 E
Zetta 10^21 Z
Yotta 10^24 Y
 Only need up to 10^12 and as low as 10^-12
Conversions
o
o
o
o
o
o
o

o must do it like – (12 in)(2.54 cm/1 in)
 inches then cancel and left with 30.48 cm
[9.15.09]
Frame of Reference: An environment in which you do an experiment
 DF Position: where an object is
o X(t): position at time t
o Unit: m
 Can be positive, negative or zero



DF Displacement: change in an objects position
o ΔX = Xf – Xi
o Unit: m
 Can be positive, negative or zero
DF speed: how fast an object moves
o Can be positive or negative
DF velocity: how fast and in what direction
o Can be positive, negative or zero
o Avg vel: Vavg = Vav = V (with line over it) = Δx/Δv=
displacement/time
o Unit: m/s
 ex. Travel from NYChicago, 800 km, in 4 hours. Stop
in chi. For 6 h, return to NY in 2h.
 going: Vavg =displacement/time
 800 km/4 h = 200 km/h
 there: 0
 return: 800 km/2 h = 400 km/h
 total: 0
[9.16.09]

continue with velocity
o ex. graphing a car that goes 10km, for 30 min, stops for 1 hr,
continued 40 km in 2 hr,
o It’s a position time graph


On a position time graph, in general the slope equals the velocity
if the graph has a curved line, you can still find the slope between
two points by connecting them
o slope = rise/run =displ./time = avg velocity

on a position time graph, the instantaneous velocity equals the
slope of the tangent line at that point
[9.17.09]
Bike race: 20 km
 Rider 1: velocity = 32.5 km/h
o 37 min
 Rider 2: velocity = 34.0 km/h
o 35.5 min
 Δx = velocity*T
o Often rate

DF Acceleration: a = Δv/Δt
o Acceleration is constant
o Unit: m/s2 (m/s/s)
[9.21.09]
T (s)
V (m/s)
A (m/s/s)
0
1
2
3
0
3
6
9
3
3
3
3
…
t
…
3t
…3…
3
 a=3, since acceleration is constant
ex. a car starts at 25 m/s brakes, slows down at -5 (m/s/s)
T (s)
V (m/s)
A (m/s/s)
0
1
2
3
25
20
15
10
-5
…
t
…
25+t(-5)
…-5…
Final velocity: Vf=vi+at
 Unit: m/s
Average velocity: (Vi+Vf) /2
 Unit: m/s
Displacement = final position
 Same as - Δx =vit +(1/2)at2
 Same as – Xf = xi +vit +(1/2)at2
o Unit m
 Same as = Vf2 = vi2 +2aΔx
o Unit = m2/s2
Ex. a car starts at rest and accelerates at 3 m/s2 for 5 s
 Find final velocity
o Vf=vi+at = 0 + 3*5
o 15 m/s

Find displacement
o Δx =vit +(1/2)at2 = .5*3*5^2
o 37.6 m
Ex. A speeding car is going 30 m/s when it passes a parked police car. The
police car starts accelerating at 2 m/s2 at the instant the speeding car
passes the police car
 What we know
o Speeder – v = 30 m/s
 Δxs = Velocity*Δt
o


o
How
o
o
How
 Δxs = 30t
Police – vi = 0, a = 2 m/s2
 Δxp = vit +(1/2)at2
 Δxp = .5*2*t2 = t2
Both – same time and displacement
long does it take to catch the speeding car?
30t = t2
0 sec and 30 sec
far does it take?
o 30*30= 900 m
[9.22.09]
-went over homework
[9.23.09]
-went over homework
in a velocity-time graph
 slope = acceleration
 displacement = area under the graph
positive velocity + positive acceleration  going faster
negative velocity + positive acceleration  going slower
[9.24.09]
-went over homework
[9.29.09]
-went over homework
acceleration of gravity = -9.8 m/s^2
Free fall: When an object is acting only under the influence of gravity.
 I.e., the only acceleration is -9.8m/s
9/14/2009 4:05:00 PM
[10.12.09]
Vector: A measurable quantity that has both magnitude (how big), and
direction
 Ex. position, displacement, velocity, acceleration, force,
Scalar: A measurable quantity that has only magnitude (and no direction)
 Ex. distance, speed, volume, mass, time, energy
Vectors
 Notated either in bold print, or with an arrow on top (or half an
arrow)
 Graphically is an arrow, with a letter on top of it







Relative

Relative
Finding distance
o When finding the distance graphically between two points,
add the tail of the arrow to the head of another arrow
 Then make a vector connecting the tail of the first to
the head of the last
 Then use Pythagoreans theorem to find the
displacement between starting point and end point
Resultant: The sum of 2 or more vectors
o Δx1+Δx2
Vector addition is commutative
Vector subtraction
o Same thing as addition except what is negated is flipped 180
degrees, so the arrow goes the opposite way on that vector
o And then continue as your normally would
Scalar multiplication
o The direction stays the same, but the magnitude is twice the
size – called dilation
Scalar division
o The direction stays the same, but the magnitude is half the
size – contraction
Magnitude
o Represented by absolute value signs (or a double absolute
value sign) |a|
velocity formula: Vac = Vab + Vbc
Look at notes online for train diagram and problem
velocity in 2-d

A boat on the river is East to west at a velocity of 10 m/s , on a
river that is going south at a velocity of 3 m/s
|Vbg|=10.44 m/s
angle – is tan ϑ = 3 m/s / 10m/s = 17° S of W (197°)


[10.15.09]
Vectors continued


If oblique it makes an angle theta with the axis
Easiest way to deal with it, is to draw two perpendicular lines to both axis from
the end of the component, and put the sums of these to components together
and add them
o Can change everything about a vector, except magnitude and direction

This results in a x and y vector component and add theses to get the original
vectors
[10.19.09]
Given a vector has a magnitude of 8m and is 30 north of east, find the other two
vectors:


Y component is 4m
X component is 6.9m
o Figured out through trig (here it was a 30-60-90 triangle)
A block is on an inclined plane:


Forces:
o W = mg  going straight down
 Weight = mass*gravity
o Fn  normal force, particular to object
Draw an axis which has its origin in the block, and whose x axis is
on the curve, and y axis on the perpendicular force



Now draw components of weight force
Through similar triangles, you find that the components –
o Wy = mgcosϑ and Wx =mgsinϑ
The reason a book sides is Wx =mgsinϑ
If you have two vectors which you want to find their components, you create
a x and y component for each and add the x components with y components

 A + B = (Ax+Ay)+(Bx+By)
Ex. A boat moves southeast at 6 knots. Wind is from the E at .5 knots.
Current is moving SW at 4 knots. What is the total velocity of the boat?
 Draw three vectors
 Add all three y and x components


o X: 4.24-2.84-.5 = .91 kn
o Y: -2.83 - 4.24 = -7.07 km
Then make a triangle with those two sides and add them up (head
to tail)
o The resultant is 7.12 kn
For angle its tan ϑ = 7.07 kn /.94 kn = 82.4° South of East
Acceleration = Fnet / m  fnet =ma
a = gsin ϑ

without friction
[10.21.09]
motion in a vertical plane
Horizontal and vertical motion (without resistanace)
 If object is launched with an initial horizontal velocity (and no
vertical velocity)
o The horizontal velocity is constant, and acceleration is 0
o Vertical acceleration is -9.8 m/s
The vertical acceleration is independent of the
horizontal velocity
o Therefore, if one object was dropped and another thrown,
they would always have the same vertical distance, and reach
the ground at the same



This also shows the vector diagram that can be drawn
The one that is dropped falls linearly
 The one thrown falls in a parabola
 The total velocity vector at any time is tangent to the path
Ex. throw a rock horizontally at 10 m/s from top of a 100 m cliff. Find where
it lands, and its final velocity
 Have to make two lists for two directions
 Horizontal
o Vx = 10 m/s
o ax= 0



o Δx = ?
o t=?
Vertical
o Vi = 0
o ay = -9.8 m/s2
o Δy= -100 m
o t=?
(when given this type of problem, find time first, and do from
vertical)
Δy = Vyit + (1/2)at2
o t = 4.5 s
 Δx = vxt
 Δx = 45 m
 Vf =Vi + at
o Vf = -44.1
[10.26.09]
ex. Throw a penny horizontally off the ESB (380 m high) at a speed of 10
m/s
 horizontal info

o v x =10 m/s
ax = 0
o
Δx = ?
o
t=?
o
vertical info
o vyi =10 m/s
ay = -9.8 m/s^2
o
Δy = -380
o
t=?
o
o Vyf = ?
A) where does it land?
 Δy = vyit+ ½ gt^2
 T = 8.8
 Δx = Vxt = 10*8.8 = 88 m
B) what is its final total velocity, including speed and direction?
 vyf = = vyi + gt = (-9.8)(8.8) = -86.24 m/s
 vf = √(vy2+vx2) = 86.8 m/s
 arctan = 10/86.24 = 6.6° from the vertical
Projectile at an angle
 Vo
o When there is a oblique vector make components of it
o The x = Vx
o The y = Vyi
 Since it changes due to acceleration

[10.28.09]
Projectiles at an angle (continued)
 Usually give: initial angle (ϑ) and initial speed (V0)






o It creates a curve (like the top part of a circle)
Horizontal velocity V0 is constant
The velocity decreases on the way up, and increases on the way
down
o (Insert diagram)
Vyf = 0 at top
Horizontal distance is called range
The vectors on both side of the curve are symmetrical
Ex.
o Given
 V0 = 40 m/s
 ϑ = 30°
o Vx = V0cos ϑ
 = 34. 64 m/s
o Vyi = Vo sin ϑ
o
o
o
o

o
Δx =
 = 20
ax = 0
ay = -9.8 m/s2
Vyf = 0
Vf = Vitat
 0 = 20-9.8t
 t = 2.04 s
 top
t total = 4.08 s
vxt
o = 141 m
ex.

Vx = V0cos ϑ

o = 32.9 m/s
Vyi = V0sin ϑ


o = 26 m/s
hoizontal
o Vx = 32.9
o Ax = 0
o Δx =225
o Δx = vxt
 T = 6.84
Vertical
o Vyi = 26
o ay = -9.8
o T = 6.84
o Δy = Vyit + (1/2)at2
o
= -51.6 m
o  he died since was more then 45 m below surface
[10/30/09]
Ex. throw a ball at 20.0 m/s at an angle of 65 degrees above the horizontal,
and the ball leaves your hand at 1.8 m above the ground. At what height
does the ball strike a vertical wall located 10 m away? Is the ball rising or
falling when it strikes the wall?
 Components
o Vx = 20cos(65) = 8.45 m/s
o Vyi = 20sin(65) = 18.13 m/s





Horizontal
o Vx = 8.45 m/s
o ay = 0 m/s2
o Δx = 10 m
Vertical
o Vyi = 18.13
o Ay = -9.8 m/s2
Finding time
o Δx = Vxt
 10 = 8.45t
 T =1.18
o Δy = Vitt + ½at2
 14.57 m
hits wall
o 14.57 + 1.8 = 16.37 m above ground
check vertical velocity to see if rising or falling
o Vyf = Vyi + at
 18.03 + (-9.8)(1.18) > 0
 since velocity is positive it is rising
Range of projectile
 R = Δx
 Vertical info
o ay = g (-9.8)
o Vyi = V0sinϑ
o Vyf = 0 at top
 This is half of total time





Horizontal info
o ax = 0
o Δx = R?
o Vx = V0cosϑ
Ttop = (V0/g)sinϑ
Ttotal = (V0/g)sinϑ
Range = (V02/g)sin2ϑ
o Only if lands at the same vertical place as launched at
Galileo:
o Angles that have the same excess or defect from 45 digress
have the same range
 Ex. 35 and 55
 Any complementary angles
o Proof: R = (V02/g)sin[2(45 ± α)]
 (V02/g)sin[90 ± 2α]
 = …[sin90cos2α±cos90sin2α]
 = (V02/g)sin[cos2α]

Proof that a projectile follows a parabolic path
o The equation should be y ~ -x2
[11/2/09]
Not on test
Circular motion



Goes around in a anti-clockwise manner, since degrees are getting
bigger
Frequency: the number of revolutions per second
o Unit: 1/s = Hertz (Hz)
o Symbol: f
Period: the time for one revolution
o Unit: seconds


o Symbol: T (upper case)
Formulas
o f = 1/T
o T = 1/f
S= rϑ
9/14/2009 4:05:00 PM
[11/4/09]
DF Forces: A push or pull
 An influence on an object
 A force causes a change in motion i.e., a change in velocity, i.e., an
acceleration
DF Inertia: the resistance of an object to a change in its motion
 Synonym for mass
Newton’s 1st law = law of inertia
 If an object is at rest it will remain at rest, and if an object is
moving with a constant velocity, it will remain moving at a constant
velocity, unless acted on by an external force.
[11/9-11/09]
Missed due to AIPAC
Ax = fx/m
[11/11/09]
continued
…. A = [(m1-m2) / (m1+m2)] * g
… T = 2m1m2g / m1+m2
 tension is in Newton’s
Newton’s 3rd law: For every action (force), there is an equal (in magnitude)
and opposite (in direction) reaction (force)
[11/17-18/09]
An action-reaction pair must act on different objects.
 Ex. two boxes one ways 2 kg, one further away weights 1 kg, a 6 N
force is pushed on it


o Find – acceleration of both boxes
a = 2 m/s2 (6n = 3a)
the two boxes each had a force of 2, that were acting on each other
[11/20/09]
Fnet = ma
 In the homework situations its Fnet = F – W
[11/23/09]
Friction:




All ‘m’ are mu Static friction: when an object is not moving
o fs ≤ MsFn
o unit – N
o
Kinetic friction: object is moving
o fk ≤ MkFn
o unit - N
M = coefficient of friction (roughness coeff)
o Always 0 ≤ mk< ms < 1
o M = 0: no friction
Source of friction:
 Slip-stick model
o When things are at rest, the surface of the two objects get
stuck together at points since both aren’t flat
 There are temporary hydrogen bonds between these two surfaces,
so these bonds have to be broken when starting to move
o But if already moving, harder to form bonds, which is why
kinetic friction is less then static
Ex. a 100 kg block sitting on the floor, with coefficient of static friction 0.5,
and coefficient of kinetic friction 0.4
 A) how much force does it take to initially get the block moving?
 B) Now keep pushing with the same force as found in (a), and what
happens to the block?
 C) how much force do you have to exert to keep it moving at a
constant speed?




W = mg  Fn = 980
We have Ms and Mk
Vi = 0


Fn = w = 980 n
fs max = MsFn
o a) = (.5)(980N )  Push greater then 490 N
b) it will accelerate, since kinetic energy needed is less then static


Fk max = MkFn
o = (.4)(980)  392 N
o Fnet = ma
Push – Fk = ma
 490-392 = 100a
 c) p = 392 N (=Fk)
o b) aa = .98 m/s2
[11/25/09]
ex. an object is sitting on an incline with slowly increasing angle. If the
objects starts to slide at angle ϑ, find the coefficient of static friction


Insert picture





fs =mgsin ϑ
MsFn = mgsin ϑ
Fk = MkFN
Fs ≤ MsFN
Msmgcos ϑ = mgsin ϑ
o  ms = tan ϑ
ex. an object slides down an incline with friction. Find the Acceleration



Insert picture
Y: FN = mgcos ϑ
X: Fn = ma

mgsin ϑ –MkFn = ma

 mgsin ϑ- Mkmcos ϑ = ma  gsin ϑ- Mkcos ϑ = a
o
 a = g(sin ϑ –Mkcos ϑ)
[11/30/09]
terminal velocity – where the air resistance gets to the point that that an object no longer
accelerates or decelerates
9/14/2009 4:05:00 PM
[12/02/09]
Energy
 Work is done on an object by a force when the force acts on the
object, and the object moves through some displacement.
o W=F*Δx
 This only holds if force is in direction of displacement
 If F is parallel to Δx
 Units: N*m = J
o W=F* Δx*cosϑ

Holds if not parallel

If the force is perpendicular to the displacement. Then that force
does no work
[12/07/09]
ex. pulling a wagon at 45° with a force of 12 N, it travels 50m
 w=fΔxcosϑ = 12 N * 50 m * cos 45 = 424.26 J
ex. lifting up versus using an incline to get an object up
 lifting straight up is equal  m*g*h
 pushing up a ramp (w/o friction)  (mgsinϑ)*L (the length of
incline) = mgh

 work done against gravity is path-independent
o gravity is a conservative force
ex. a 2000 cal diet applied to a 75 kg person walking up stairs would allow
him to climb how high per day?
 2000 dietary cal = 2,000,000 real cal (2 mega cal) = 8.37e6 j/day
 w=mgh
o 8.36e6 =75*9.8*h
 h = 11,400 m = 11.4 km
power: the rate at which work is done (the rate at which its energy changes)



P = W/t = FΔx / Δt = Fv
Unit = J/s = Kg*m^2 / s^3 = W (watt)
We know this W is a unit since it can only be after a unit and cannot
stand alone, they have a context.
o Work is a scalar, and weight is a vector
Ex. KW * H = (1000W) (3600 s) = 1000 J/s * 3600 S = 3.6 MJ
 If you pay 20 cents per KW*h then
o A 100 W light bulb for 1000 hr
 .1 KW * 1000 h * 20d/KWh  $20 and $1 for cost
An elevator motor has an output of 12.0 kW and is 25% efficient. How fast
can it lift an elevator (at constant speed) if the elevator has a mass of 500
kg, and the maximum load of people is 900 kg?
 Constant speed : t = mg = 1400*9.8 = 13720 N
 Output (P) = 12,000 W
 P = Fv = mgv
 V = p/mg
o 12,000/ 13,720 = .87 m/s
 .87*.25 = .22
o the energy is lost to the atmosphere, mainly because of heat
due to friction
[12/09/09]
Energy: The ability to do useful work
 Mechanical energy
o 1. Kinetic energy: Energy of motion
 K = ½mv2
 Unit: J
o 2. Potential energy: Energy of position
U = mgh
Unit: J
PE is relative to the zero point
 Can be positive, negative, or zero
 If above 0 pt, then positive, if at, then zero,
if above then negative
Ex. Hit a nail with a hammer that is going 4 m/s when it hits, which drives
the nail in 5 mm. If the mass of the hammer is 600 g, what is the force
exerted by the hammer on the nail?











Initial velocity = 4 m/s
Final velocity = 0 m/s
Δx= .005 m
M = .6 kg
KEi  W on nail
ΔK = W = FΔx
|Δk| = w = FΔx
½(.6)(4^2) = f(.005)
o F = 960 N
Ex. A microwave oven has a power rating of 1000 W. If this energy is all
applied to a 70 kg runner, how much faster can the runner go each second?
 P = 1000 W = 1000 J/s
 E = 1000 J = ΔKE = 1/2 mvf2 – 1/2 mvi2
 1000 = 1/2 70 vf2
o vf= 5.3 m/s
[12/10/09]
Conservation of energy: Energy cannot be created or destroyed; it can only
be transformed from one type to another


If something is about to be dropped it has potential but not
potential
o Right before it hits the ground, the numbers are switched, 0
potential, and the kinetic gains all of that
to find final velocity when knowing the final kinetic energy you:
o K= ½ mv^2
 Use known value of K and M
 Then find the root of V, which can be positive or
negative
 (If going downward its negative)
ex. A 4.0 kg blocks sit at rest at the top of a 2.0 m long 30° incline. The
block slides down, with a coefficient of kinetic friction of 0.20 . Find the final
velocity at the bottom.
 Insert Diagrama
 Ei + Wfr= Ef
 Ki +vi +wfr = Kf+ vf  Ki +vi +wfr = Kf+ vf 
 mgh + fk*Δx = ½mv2f  mgh + Mk*Fn *Δx = ½mv2f
 mgh + Mk*Fn *Δx = ½mv2f 
 mgh + Mkmgcosϑ*Δx =1/2mv2f  can get rid of all m
 Vf = radical (2gh + 2Mkgcosϑ*Δx) = radical (2*9.8*1 –
2*.2*9.8(cos30)*2) = 3.6 m/
o We can compare this to a non fractioned force and this should
be a bit lower
 Vi = kf  mgh = ½ mv2
 …  4.5 m/s
 so looks good
Two types of forces
 Conservative and non conservative
o Conservative force:
 1. A force is conservative if and only if you can define a
potential energy function for that force.
 2. The amount of work done by a conservative force
between 2 points does not depend on the path between
the points.
 3. The work done in any closed path equals zero.
o Conservative forces: ex. gravity, tension, electric, magnetic,
o Non-conservative forces: ex. Normal, friction
Ex. Push a 25 kg box up a 20 m long ramp to a height of 3 m with a
constant force of 120 N. The box starts at rest, and ends up going 2 m/s.
What was the force of friction on the box?
 Ei + Wfr = Wpush = Ef
 Ki + vi + fk Δx + pΔx = Kf + vf
 Fk *20 + 120*20 = ½ * 25 * 22 + 25(9.8)(3)  fk = -80.75 N
[12/11/09]
a force that is perpendicular to the displacement is 0  since cos90 = work
[12/16/09]
Conservation of energy
 Ei + w = ef
 Ki + ui + w = kf + uf
o U = Potential
conservation of energy without potential energy
 W = Kef -Kei
Homework notes

Kinetic energy terms can be dropped, when the original velocity and
final velocity are the same
 if work done is negative then it means that it is losing energy
[12/21/09]
Momentum: How difficult it is to change an object’s motion
 Involves mass and velocity
 P = mv
 Unit: (Kg*m)/s
o It’s a vector
 Ex. a baseball, m = 150g, 20 m/s
o P = mv  (.15*20) = 3 (kg*m)/s
Impulse: Change in momentum
 I = ΔP = Pf – pi = mvf – mvi = mΔv
o Vector
 Unit: (kg*m)/s
 Divide both sides by Δt 
o Fnet = Δp / Δt = mΔv / Δt = ma
 Newtons second law

Impulse momentum theorem
o I = Δp = fΔt
Conservation of momentum
 If there are no outside forces, then the momentum of a system is
the same before and after a collision
 Ex. 70 kg person on ice, throws a .5 kg rock. At what speed must
the person throw the rock in order to have a speed of 1 m/s
afterward.
o Initial momentum for rock and person is 0
 So final momentum is also zero
o Pi = Pf
o PigVig + MirVir = Mfgvfg + mfr+Vfr
o 0 = -70*-1 + .5Vfr
o Vfr = 140 m/s
Collisions (1 dimension)
 1.elastic collision
o a collision in which momentum and kinetic energy is
conserved
 Ki = Kf
 2. Inelastic collision
o kinetic energy is not conserved
 Ki≠ kf
o Totally or perfectly inelastic
 Objects stick together
Ex. totally inelastic

A bullet of mass 10.0 g is fired into and lodges in a bock of wood of
mass 2.00 kg that is initially at rest. The final velocity of the block
with the bullet is .700 m/s. Find the initial velocity of the bullet.
o Pi = pf
[12/22/09]
 continued problem
o MbMvi = MfVf
o .1vb = .201*.7
 vb =140.7 m/s
o Ki = 1/2 mb vb^2 = … 99.4
o Kf = 1/2 mb vb^2 = (1/2)*(2.01)*(.7^2) = .492 J
o Final energy is .492 J/ 99.4 J * 100 % = .5 % of total energy
Ex. two cars. A) 1200 kg, at 10 m/s, east. B) 1000 kg, 15 m/s , north. When
they hit what direction are they at
 Pix = maVai = 12000 (kg*m) /s = pfx
 Piy = mbVbi = 15000 (kg*m) /s = pfx
 Pf = radical(12000^2 + 15000^2) = (19200 kg* m) /s
 Mvf = 2200 vf =19200  vf = 8.7 m/s
 Tan-1 (15000/12000) = 51.3 North of East
Elastic collisions – 1-D
 Kinetic energy is conserved
 Momentum is conserved (always)
Ex. Two identical billiard balls, both mass m
 V1i = initial velocity of ball one
 Conservation of Ke:
o Both balls going at each other have same m and have intial
velocity going against each other
 Ki = kf





½ m v1i2 + ½ m v2i2 = ½ m v1f2 + ½ m v2f2
o all mass cancels out
½ v1f2 - ½v1i2 2 = ½ v2f2 - ½ v2i2
o This can be factored into 
 (V1i + V1f) (V1i - V1f) = (V2i + V2f) (V2i - V2f)
momentum  Pi = Pf
Mv1i + mv2i = mv1f + mv2f  v1i + v2i = v1f + v2f
 v1f - v1i = v2f - v2i

we can then divide the first factored equation by the second
factored equation
o  V1i – v2i = - (vif –v2f)
 this is telling us the velocity as they approach and the
velocity of recession
 V approach = - Vreccesion
[12/23/09]
Vapproach = -Vreccesion
Pi = pf
Ki = kf
ex. 2 balls. A) 1 kg, 3 m/s, east. B) 2kg, 1 m/s, east.
 Elastic 1.dimesnsional
 Pi = Pf
 M1V1i + m2V2i = M1V1f + M2V2f
o 1*3+2*1 = 1*V1f + 2V2f
o equation a: V1f + 2v2f = 5
 Vapproach = -Vreccesion
 V1i – v2i = -(v1f-V2f)
o Eqatuion b: V2f – V1f = 2

Adding the two equation  3v3f = 7
o V2f = 7/3 m/s
o V1f = 1/3 m/s
9/14/2009 4:05:00 PM
[1/05/10]
Fc = centripetal force (seeking the center)
 The force that keen in object in circular motion
 Velocity is tangential
Ac = centripetal acceleration
 Acceleration is radially toward the center
[1/08/09]
If the force is cut off while moving in a circle it will go tangential to the circle
 The distance it goes tangentially that is released equals = vt
 Insert physics picture

R^2
o
o
o
o
+ (VT)^2 = (r+h)^2
H is the ‘fall of the object)
As t  0 (and h^2 goes to 0 a lot quicker so we can drop it)
Then V2T2 = 2rh
H = ½(v2/r)t2 
 ac = v2/r
 Fc = mv2/r
 unit = m/s^2
period of circular motion: The time of one full orbit

If speed is constant:
o V = (2*pi*r) / t
o Ex. speed of earth around the sun:
 Radius 93 million miles = 150 M km = 1.5E11 m
 V = [2pi(1.5e11 m)] / 31446925.9747 s = 29,800 m/s
= 29.8 km/s ≈ 18 mi /s
o ac = 29800^2 / 1.5e11 = .00592 m/s2
[1/12/10]
Amusement park ride that the center drops out, equation needed to not fall INSERT PROBLEM
Angle of banking in a road
 Insert problem
[1/14/09]
tanϑ = v^2 / gr
Motion in a vertical circle

9/14/2009 4:05:00 PM
[10.15.09]
1. Cover sheet
2. Intro
3. Materials
4. Procedure
5. Data
 numerical order
 units in column header
6. Calculations
 (doesn’t have to be typed)
7. Graphs
 Title
 Label axes with variables and units
 Scale: be reasonable, not too many numbers
 Best fit line or curve
o Doesn’t necessarily start in origin
 Don’t label any data points
8. Error analysis
 discussion of what went wrong
o ex. table tilted, if motion friction
 percent error
o % err = [|accepted-experimental| / accepted]* 100 %
9. Question
10. Conclusion
9/14/2009 4:05:00 PM
Displacement
 ΔX = Xf – Xi
Average velocity
 Vavg = Vav = V (with line over it) = Δx/Δt = disp./time = xf-xi/tf-ti
Rate
 Δx = velocity*T
Acceleration
 a = Δv/Δt
Final velocity
 Vf=vi+at
Displacement = final positon
 Δx =vit +(1/2)at2
 Xf = xi +vit +(1/2)at2
 Vf2 = vi2 +2aΔx
o timeless
Relative velocity
 Vac = Vab + Vbc
Range
 Range = (V02/g)sin2ϑ
o Only if lands at the same vertical place as launched at
Circular motion
 f = 1/T
 T = 1/f
Acceleration
 A = [(m1-m2) / (m1+m2)] * g
Forces

Static friction
 fs ≤ MsFn
 no units
Kinetic friction
 fk ≤ MkFn
 no units
Work
 W=F*Δx
o If parallel

W=F* Δx*cosϑ
o If not parallel
Unit: J = Kg*m^2 / s^2

Power
 P = W/t = Fv
 Unit: J/s =watt
Kinetic energy
 K = ½mv2
 Unit: J
Potential energy
 U = mgh
 Unit: J
Conservation of energy
 Ei + w = ef
 Ki + ui + w = kf + uf
Momentum
 P = mv
 Unit: (Kg*m)/s
Centripedal force


ac = v2/r
Fc = mv2/r
o unit = m/s^2
