CLASS+NOTES

Projectile Motion Help
PLEASE POST DAILY NOTES HERE. ORGANIZE THEM AS FOLLOWS:

Unit: Topic: Text Reference: Idea: Definition: Key Examples: Additional Information/Interesting Links/etc.

SEPTEMBER 13 - Maria
Unit 1 Topic: Introduction Text Reference: None. See handouts Idea: An Introduction to SPH 4U - Syllabus - What is Group Work all about - Inquiry cube activity - TED Talk - [|Beauty and Truth in Physics]

SEPTEMBER 14 - Patrick
Unit: 1 Topic: Introduction? Text Reference: Pgs 5-9

My apologies If the writing is unreadable. = = =

SEPTEMBER 15 - Shagufta
Appendix E: Helpful Mathematical Equations and Techniques Text Reference: pg 765-768 Idea: - Measurements - Significant Figures - Rounding - Quadratic Formula - Graphs - Solving Equations - Exponents

Definitions:

Measurements: record to the smallest interval PLUS one extra digit. Ex: using a metre stick: smallest interval is the mm = 3 signigicant digits + 1

Significant figures: They are the number of digits that are known with CERTAINTY. Digits include:

1) All non zero digits. Ex: 1234 m = 4 s.d

2) All embedded zeroes. Ex: 1204 m = 4 s.d

3) All trailing zeroes after decimal. Ex: 1.23400 = 6 s.d

4) All trailing zeroes known to be measured. Ex: 12 000 = 5 s.d if specified. Otherwise assume 2 s.d Always assume trailing zeroes in whole numbers are NOT significant unless you know something about the measuring device.

5) Preceding zeroes in numbers less than one are not significant. Ex: 0.00123 = 3 s.d

6) Counted values are infinitely significant.

Scientific notation includes ALL significant digits Ex: 1.2000 x 10^4m if all zeroes are significant or 1.2 x 10^4 if zeroes are not significant

When adding and subtracting : the least number of decimal places is used Ex: 1.2m +1.22m = 3.642m --> 3.6 m --> LEAST number of decimal places used

When multiplying and dividing : the least number of s.d is used Ex: 1.2m x 1.333m = 1.5996m^2 --> 1.6m^2 --> LEAST number of s.d is two, therefore answer uses two s.d as well. Rounding: Ex: 1.236 --> 3 s.d --> 1.24 WHY: the 6 is greater than 5, therefore you round up Ex: 1.234 --> 3 s.d --> 1.23 WHY: the 4 is less than 5, therefore the 3 stays the same.

If the digit to determine rounding is 5, round to EVEN. Ex: 1.235 --> 3 s.d --> 1.24 WHY: the 3 is odd, therefore you round up Ex: 1.225 --> 3 s.d --> 1.24 WHY: the 2 is even, therefore it stays the same.

Graphs: General line equation: y = mx + b where m is the slope and b is the y- intercept. Slope = rise/run

Refer to page 767 in the text for a graph and calculations.

Solving Equations: Solve equations using either of two methods:

1) When you have two equations and two unknowns, combing the two equations through a common variable. You will then have one equation and one variable.

2) When you have the appropriate equation, but the term you need is not isolated, rearrange the equation. Remember: move terms separated by +/- first. When a term moves across the equal sign, its sign changes. Then separate desired term by multiplying and dividing.

Exponents: - simplify multiplication using whole numbers, decimals, fractions, unknowns

Ex: 10 * 10 * 10 * 10 = 10000 = 10^4 = 1.0 x 10^4 Ex: A * A * A * A = A^4

Interesting Link: Having difficulties with sig figs? This song just might help. . . []

How to Solve Fermi Questions:

SEPTEMBER 17 - Shagufta
Unit: Forces and Motions Dynamics Topic: 1.8 : A Graphical Analysis of Linear Motion Text Reference: pg 24- 32 Idea: - 3 main graphs - Slopes and Areas of graphs and their significance

Definition:

__Position-time graphs:__ The slope is the VELOCITY of the object.

Uniform motion is represented by a STRAIGHT line graph.

If the slope is 0, the object is at REST.

If the magnitude of the slope increases, there is POSITIVE ACCELERATION. If not, there is NEGATIVE ACCELERATION.

Reference: pg 25 in the text for a summary of p-t graph analysis.

__Velocity-time graphs:__ If it is a straight line velocity-time graph:

The slope is the constant acceleration of the object.

If it is a curve:

The slope of its tangent at any given point is the INSTANTANEOUS ACCELERATION of the object.

Area under v-t graph: DISPLACEMENT of object.

Reference: pg 31 in the text for a summary of v-t graph analysis.

__Acceleration-time graphs:__ Area under an a-t graph is the CHANGE IN VELOCITY of the object.

__Remember:__

When graphing always indicate the point on the graph. The point should be recorded with one extra digit. (Ex: 1.0, 3.2)

Refer to: Review: Graphing Kinematics __Homework:__ Graph Analysis Assignment - due Wednesday

__Additional Information:__ For more info about any of the above graphs check this site out: []

=September 20- Wade= Unit 1 Topic: Algebraic Description of Uniform Linear Acceleration Text Reference: Pg 10-19 Idea: Using the main 5 equations to solve word problems. The equations are as follows:


 * Equation || Variable Eliminated ||
 *  v 2 =v 1 +a∆t || ∆d ||
 *  ∆d=1/2(v 2 +v 1 )∆t || a ||
 *  ∆d=v 1 ∆t+1/2a∆t  2 ||  v 2  ||
 * <span style="display: block; font-family: Helvetica; font-size: 14px; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal; margin: 0px;"> ∆d=v <span style="font-family: Helvetica; font-size-adjust: none; font-size: 9px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">2 ∆t-1/2a∆t <span style="font-family: Helvetica; font-size-adjust: none; font-size: 9px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;"> 2 || <span style="display: block; font-family: Helvetica; font-size: 14px; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal; margin: 0px;"> v <span style="font-family: Helvetica; font-size-adjust: none; font-size: 9px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">1  ||
 * <span style="display: block; font-family: Helvetica; font-size: 14px; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal; margin: 0px;"> v <span style="font-family: Helvetica; font-size-adjust: none; font-size: 9px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">22 =v <span style="font-family: Helvetica; font-size-adjust: none; font-size: 9px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">1 2 +2a∆d || <span style="display: block; font-family: Helvetica; font-size: 14px; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal; margin: 0px;">∆t ||

The rules for solving Uniform Acceleration problems are: 1. Identify a reference point, generally the initial position so d=0

2. Identify a reference direction, generally the direction the object is initially travelling.

3. Assign +/- sign to all values given.

4. Solve problem and then interpret the solution, often includes making a therefore statement.

Refer to ALGEBRAIC DESCRIPTION OF UNIFOM LINEAR ACCELERATION handout for sample problems or in the text on pg 13-18.

Homework: Practice Problems 1-4 (handout).

Topic 2: Bodies in free fall/ Acceleration due to gravity Text reference: Pg 19-23

Idea : Galileo Galilei's experiments proved that a body's rate of fall is not proportional to it's mass. Examples: Coin and feather dropped in vacuum hit the ground at the same time, a feather and a hammer were dropped on the moon and hit the ground at the same time etc.

Acceleration due to gravity is 9.81 m/s [down] (Use 10m/s [down] in this class)

Examples in text on pg 20-23

Additional information: http://www.youtube.com/watch?v=WyBYVQzvGdI

=September 21- Sangitha=

Unit 1 Topic: Introduction to Cooperative Group Problem Solving Text Reference: See handout

Idea: Develop a strategy to tackle the problem through 5 steps: pictorial/verbal, schematic/algebraic, algebraic manipulation, numerical calculation and double checks.

Definition: 1. Describe the Problem: Use the given information to sketch the problem including all relevant and required information. Include statements that describe what needs to be found and which physics concepts that may be useful in solving the problem. You will know this step is complete if you do not need to refer to the original question ever again.

2. Define the Physics: Simplify the drawing into a mathematical drawing while keeping all relevant information. Write down the standard equations to show the relation of the mentioned physical quantities. You should not longer need to refer to step 1.

3. Plan the Solution: Use the identified equations to create an expression to solve for the unknown. You may then need to create additional expressions for any sub-problems that may be embedded in the original problem. Always rearrange for the unknown and you should have created a logical chain of equations.

4. Execute the Plan: Plug in all known values into the equations in step 3 and solve for the unknown variables.

5. Evaluate the Answer: Check over your solution to make sure that the value is reasonable and provide a simple justification for your solution.

//Refer to Handout "Introduction to Cooperative Group Problem Solving" and the accompanying in class problem.//

Additional Information/Interesting Links/etc. View this link for additional problem solving tips and other physics related information: [|Physics Problem Solving]

=__September 22nd, 2010__= Clinton D'Silva

Group Activity: Solving Two-Body Uniform Acceleration Problems - The Washer and String experiment!!!!!! (refer to handout)

Use the five step method: 1. DESCRIBE THE PROBLEM 2. DEFINE THE PHYSICS 3. PLAN THE SOLUTION 4. EXECUTE THE PLAN 5. EVALUATE THE ANSWER

Refer to Sangitha's explanation (previous note) and follow the steps to solve problems in a systematic fashion.

Homework Practice problems #5-10

Need some Physics Review?? []

Physics Kinematics Practice []

REMINDER Vector Analysis has been moved to Monday
 * KINEMATICS TEST next week* study study study!!!

=September 24- Shagufta=

Unit 1 Topic: Uniform Acceleration

Idea: Working with your physics group on 4 uniform acceleration problems which were handed in at the end of the period.

Definition: Problems were solved in a systematic manner, using the steps described in the previous notes. Refer to handout: SPH4U: Algebraic Description of Uniform Linear Acceleration for equations and sample problems if needed.

Homework: Handout: Uniform Acceleration Worksheet: Questions 5 and 6

Additional Information: Kinematics test is on Wednesday! Don't forget to study :)

Remember to choose and sign up for a physicist for the rant by Wednesday as well!

Explore acceleration: []

=September 27-Wade=

Topic: Vector Analysis
Idea: Addition and subtraction of vectors including RATs and non-RATs.

Text Reference: Pg 64-67 and appendix F Pg 768

When drawing vectors length indicates the magnitude and the arrow points to the direction of the vector, so if you have 50m/s [E] it would point to the right and if you have 25m/s [S] it would be half as long and point down.

When writing vectors follow the following form o <span style="font-family: Helvetica; font-size-adjust: none; font-size: 12px; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; letter-spacing: 0px; line-height: normal;">V L O representing the object moving, so if it was a person use p or a car c etc, L represents what it is moving in reference to, resultant vectors should always be in reference to land.

When adding vectors add TIP TO TAIL, when subtracting reverse the second vector and then add tip to tail.

<span style="display: block; font-family: Helvetica; font-size: 12px; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal; margin: 0px;">

Here Alpha represents the angle north of east for the first vector (a), beta represents the angle north of east for the second vector (b) and theta represents the angle north of east for the resultant vector (l).

For Right-angle triangles or RATs use Pythagorean Theorem to find the resultant vector and then use trigonometry (SOHCAHTOA) to find it's angle. For non right-angle triangles or Non-RATs use the Sin or Cosine laws. If you cannot remember them you can find them on pg 768.

Homework: Handout: //PROBLEMS ON VECTORS AND RELATIVE VELOCITY// # 1ab, 2ab and 4

Additional Information: http://www.physicsclassroom.com/class/vectors/u3l1b.cfm

TEST ON KINEMATICS WEDNESDAY. GET STUDYING!

=<span style="display: block; font-size: 1.4em; margin: 0px; padding-bottom: 0px; padding-left: 0px; padding-right: 0px; padding-top: 5px;">September 29- Liam = Test on uniform acceleration today! (no new material) Hope you all acceled!

October 1 - Gwynne Finlay
Unit: 1

Topic: Vector Components

Text Reference: Notes on board and handouts.

Idea: Calculating the vector components by the resultant vector and finding the resultant vector by calculating and adding all the different vector components in a vector diagram.

Definition: Components: two vectors that add to the resultant vector, when they are at a right angle to each other and the directions (N, S, E and W).

Key Examples: --If you have one resultant vector, make a triangle with component vectors and use trigonometry to solve for the two component vectors. Make sure that you draw the component vectors parallel and at right angles to the x and y-axis (N, S, E and W).

EG. To solve for the component vectors. V=40m/s [E 40 N] --First: draw given vector on a grid --Second: draw component vectors going across x-axis and up, parallel to the y-axis meeting at the tip of the previous resultant vector. --Third: calculate the two component vectors by using the trig functions i.e. --Vx=Vcos(angle) --Vy=Vsin(angle)

EG. To solve for the resultant vectors by adding all the component vectors. --First: draw all given vectors and all component vectors --Second: calculate all component vectors and add them together --Third: draw final diagram showing dR and the two totals of the component vectors (dRx and dRy) and solve for dR by trig functions.
 * E/W Components || N/S Components ||
 * d1x=5cos53=3km [E] || d1y=5sin53=4km [N] ||
 * d2x=13cos22=12km [E] || d2y=13sin22=5km [N] ||
 * d3x=10sin37=6km [W] || d3y=10cos37=-8km [N] ||
 * dRx=9km [E] || dRy=17km[N] ||

Homework: Group work handout <span style="color: #000000; display: block; margin-bottom: 10pt; margin-left: 0cm; margin-right: 0cm; margin-top: 0cm; padding: 0px;">Oct 6th 2010 <span style="color: #000000; display: block; margin-bottom: 10pt; margin-left: 0cm; margin-right: 0cm; margin-top: 0cm; padding: 0px;"> <span style="color: #000000; display: block; margin-bottom: 10pt; margin-left: 0cm; margin-right: 0cm; margin-top: 0cm; padding: 0px;">Homework: Questions on Handout! <span style="color: #000000; display: block; margin-bottom: 10pt; margin-left: 0cm; margin-right: 0cm; margin-top: 0cm; padding: 0px;">[]

Friday, October 8 - Drew
We didn't do much in terms of a lesson; we got into groups of three and measured the mass of Maria's car using Newton's Laws of Motion.

At the end of class, Maria showed us the megawoosh video, where the guy in a neoprene suit slides down a ramp and is launched into a kiddie pool (caution: it's fake). Check the link out to see how it was done.

HAPPY THANKSGIVING!

Monday, October 9 - Scott
//Unit:// 1 //Topic:// Projectile Motion //Idea:// Finding Velocity

Velocity of a projectile at any point is always going to be the net velocity of both horizontal and vertical velocity components.

__Finding velocity at a certain time__ Always find components of initial velocity using the projection angle and the initial velocity. v//h// = v(cos θ) v//v// = v(sin θ) Remember horizontal velocity stays constant throughout the whole projection.

Find the velocity of the projectile when time = t v//v1// = initial velocity = v(sin θ) v//v2// = ? a//g// = -10 m/s2 t = t v//v2// = v//v1// + a//g//t This formula will always be used to find vertical velocity at a certain time.

__Finding initial velocity with total time, horizontal distance and projection angle given__ Known: t = given value d//h// = given value θ = given value

Unknown: v = ? v//h// //= ?// v//v// //= ?//

We know: v//h// = v(cos θ) v//v// = v(sin θ)

So all that is needed is either v//h// or v//v// to find v

Since we know horizontal distance travelled and the time travelled, we can find the horizontal component for velocity. v//h// = d//h///t The horizontal component remains the same throughout the entire projection so we can use v//h// = v (cos θ) to find the initial velocity.

__Finding initial velocity with vertical distance, horizontal distance and projection angle given__ Known: d//v// = given value d//h// = given value θ = given value v//h// = v(cos θ) v//v// = v(sin θ) total time travelled horizontally = total time travelled vertically (t = t)

Unknown: v = ? t = ? Since we have two unknowns, we have to get rid of the unknown time to find the initial velocity. To do that, we equate time in terms of horizontal projection and time in terms of vertical projection. This requires to formulas:

Horizontal: t = d//h// / vh  and since we know vh = v(cos θ)… t = d//h// / v(cos θ)

Vertical: //*Note*// d//v// is relative to where your reference point is, so if you label up as a positive direction and the projectile lands below where you started, dh will be negative. d//v// = v//v// t + ½at^2 and we know v//v// = v(sin θ) … d//v// = v(sin θ) t + ½at^2

And because t in the horizontal equation equals t in this equation, we end up with this: d//v// = [v(sin θ)] [ d//h// / v(cos θ) ] + ½a [ d//h// / v(cos θ) ]^2 after that substitution mess, you can just find the initial velocity by isolating v as all other variables are given in the question.

Wednesday, October 13 - Clinton
RANGE -the max range of a projectile comes when the angle is 45 degrees assuming there is no wind resistance -complementary angles also have the same range
 * the range equation can only be used OVER LEVER GROUND Dy=0

Range =(V^2 sin2(angle)) / gravity* Gravity is a positive value here *

To see the derivation check your notes Or you can check Wikipedia but I don't know if maria would like this. []

Unit: Forces Topic: Drawing FBD's and Newtons Three Laws of Motion Text Reference: 32-47

Idea: Gurty The Cow

When drawing FBD's set the object to a shape, then add vectors to the object to show what forces are acting upon it. Gravity is always perpendicular to the ground, while normal force is always perpendicular to the surface of contact Vectors can also so angles

Newtons 3 Laws of Motion

Aristotle - dealt with forces acting upon objects in order to make it move Galileo - was the first to state the Law of Inertia If you role a ball down a ramp, it will finish at the same height going up a ramp no matter what angle (only the total distance traveled changes) If the ball is only set down a ramp and onto a flat surface, in theory it will role for eternity

Newton (1687)

1st Law of Motion - LAW OF INERTIA Every body perseveres in it's state of rest or of uniform motion in a right line unless it is compelled to change it's state by forces impressed upon it. *This is showed through the paper being pulled from under a bottle* Only works if a sudden immediate force acts upon it

2nd Law of Motion - Fnet = ma (linear equation which is also the sum of all the forces in one direction) Acceleration is directly proportional to the net force Acceleration is inversely proportional to the mass

* Acceleration takes it direction directly from the net force*

3rd law of Motion - Action Reaction Law For every force there is a same force in magnitude but in the opposite direction. Eg watter forces our a beer can through the hole causing it to spin!!!

Additional Information/Interesting Links/etc.

Newton's Three Laws of Motion

[] []

Drawing Free Body Diagrams [] []

Friday, October 15 - Drew

Newton's First Law - Sample Problem (1-D)
Find T1 and T2: __Method One - Components (YAY!)__

//Perpendicular:// ΣFy = ma = 0 T1y + T2y - Fg = 0 T1y + T2y = Fg T1sin30 <span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">° + T2sin45° = 200

// Parallel: // // ΣFx = ma = 0 // T2x - T1x = 0 T2x = T1x T1cos30<span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">° = T2cos45°

// Solve: // T1 = (<span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">T2cos45°) / ( cos30<span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">°)

<span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">Substitute that into the perpendicular equation: T1sin30 <span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">° + T2sin45° = 200 ==> [(T2cos45°)(sin30°)] / ( cos30<span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">°) + T2sin45° = 200 (0.408)T2 + T2sin45<span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">° = 200 T2 = 179.36 N [R 45°U]

Solve for T1: T1sin30 <span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">° + T2sin45° = 200 ==> T1  sin30 <span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">° + 179.36 sin45<span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">° = 200 T1 = 146.35 N [L3 0°U]

__Method Two - Vector Addition__ Then use the sin law to solve:

T1/sin45<span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">° = T2/sin60° = 200/sin75°

<span style="font-family: arial,helvetica,sans-serif;">T1/sin45 <span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">° = 200/sin75° T1 = 146.41 N [L3 0°U]

<span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">T2/sin60° = 200/sin75° T2 = 179.32 N [R45°U]

Newton's Second Law - Sample Problem (1-D)
Find the cow's acceleration:

// Parallel: // // ΣFx = ma // // Fa1x + Fa2x - Ff = ma // // 20cos30 //<span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">° +10cos20° - 5 = 5a <span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">a = 4.34 m/s 2

The cow's acceleration is 4.34 <span style="font-family: arial,sans-serif; font-size: small; line-height: 15px;">m/s 2 [R]

Homework: The syllabus says to read pages 85-88 and answer questions 1-4 on page 89. Also remember there is a TEST ON MONDAY; review the vector and projectile motion notes, practice test, and textbook questions.

Links: [|Forces in 1-D Simulation] Vector Addition Slideshow Projectile Motion Help

Monday, October 18 - Scott
Vectors and Projectile Motion test. I hope your resultant test mark was the sum of hard studying and paying attention in class.

Wednesday, October 20 - Patrick Newtons 3rd Law or the Action Reaction Law " For every action, there is an equal and opposite reaction"

Friday, October 22, 2010: __Newton's Third Law and Pulleys__ By: Eva Klimova

OSCSS physics students, I was unable to post the wiki update directly on this page, so I attached it! Sorry for the inconvenience, but attached is today's lesson, extra infromation about the lesson and also the homework!

Unit: Forces (1) Topic: Newton's Third Law and Pulleys Text Reference:Page 93-98 and 130-132 Idea, Key Examples and Additional Information/Interesting Links/etc. : In the attached files



October 25th 2010: Fixed and Free Pulleys By: Leila Meema-Coleman Unit: Forces Topic: Pulleys Idea: By combing pulleys you can make the the workload easier

A Free pulley is a pulley where it is not attached to an fixed point and can move freely on the rope In a free pulley you only have to pull with half the force: Fg=2T A fixed pulley is when the pulley is attached to a fixed point. It feels easy to lift the weight because you are pulling with gravity. Fg=T

A combined system is where free and fixed pulleys are used to reduce the force required to lift the weight. It is easier to lift the weight because you pull with half the force and you are pulling with gravity. Fg=1/2T Additional Links and Interesting Information: Very simplified pulley: [] If you need help with calculations: [] OCTOBER 26th 2010: Circular Motion By: Samantha McDermott We had the pleasure of welcoming Paul into our class today ! Sorry for the poor illustrations, paint isnt very user friendly when it comes to drawing things!

OCTOBER 27th 2010: Circular Motion By: Clinton D'Silva

Unit:Forces Topic:Circular Motion Text Reference: Idea:The concept of Circular motion and the relation to its acceleration

Key Examples: USE SAMS EXAMPLE ABOVE

Fnet=ma Fnet=ma(c) where a(c) is the circular acceleration Fn - Fg =ma(c)

You know that the magnitude of the acceleration does not change just the direction so to figure out "a" use the following formulas

a(c) = (V^2)/r
you know that circumference/period is V so sub in equation

a(c) = (4(3.14)r) / (T^2)
you also learned that T = total time/ number of cycles and f = number of cycles / Time soo.......

a(c)= 4(3.14)^2rf^2
You can also say that

Fn = [(mV^2) / (r)] + mg
Additional Information/Interesting Links/etc.

Dynamics of Circular Motion []

Circular Motion and Gravity []

FOR THE REST OF THE PERIOD WE DID OUR PHYSICS RANT WHICH IS DUE IN 2 WEEKS!!!
November 2: Review Day :) Marshell Kurniawan

Unit:Forces Topic:Review Text Reference: Idea:Reviewing Contact forces via past assignments and Reviewing and asking questions for the Forces Test for November 3, 2010

(I'll be back for more)

= November 3 - Drew = We had our forces unit test!

Some circular motion questions were assigned that are to be completed BY FRIDAY! They are the ones on the circular motion handout we received.

On the first side of the circular motion handout, complete:
 * 1) 3 (practice 1), #1 & 3 (practice 2), #55

On the second side of the circular motion handout, complete:
 * 4, 7, and 9

SparkNotes has a great section on their site with an introduction to the concept of circular motion with some practice problems, check it out: [|SparkNotes Uniform Circular Motion]

I hope the test went well!

November 8 - Sangitha
__Unit:__ 3 __Topic:__ Newton's Universal Gravitation __Text Reference:__ Read pages 48-51 for further clarification

__Idea:__ As we had previously learned, Newton had discovered that between all objects there is a force of attraction which he called gravity. This force of attraction (gravity) is present between any two or more objects regardless of their size, shape, material etc.However, Newton discovered that the strength of this force of attraction could vary depending on two important factors: distance and mass.

Newton summarized his findings in an algebraic expression in his //law of universal gravitation//: Fg = GMm/r^2

Fg is the gravitational force on M (the larger mass) and m (the smaller mass) and r is the distance between the two masses from their centers. G is the universal gravitational constant that is applicable everywhere in the universe.

G was first measured by Henry Cavendish who measured the twist in a thin wire due to the force of attraction between two lead spheres to discover the universal gravitational constant. G= 6.67 x 10^-11 N(m^2/kg^2)

This equation, developed by Newton, is applicable anywhere in the universe. It can be used to understand the force of gravity between any two objects to further understand why objects move or behave in a certain manner (i.e. why a pen falls from a person's hand who is standing on earth).

__Definition:__ Law of Universal Gravitation: It the force of attraction between to objects in the universe and states that the force of gravity is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centers.

Key Examples: In class we tested this force of attraction/gravity between some of the students. To give you an example of the application of newton's law of universal gravitation below is an example of the force of gravity between two students: Clinton and Marshell.

We know.. Fg= GMm/r^2 G= 6.67 x 10^-11 N(m^2/kg^2)

What is Fmarshellclinton=?

First we must do some measurements.. If.. Fmarshellclinton= GMm/r^2 M= Marshell's mass M= 85 kg m= Clinton's mass m= 75 kg and the distance between their centers r= 4.2 m

Fmarshellclinton= ((6.67 x 10^-11)(85)(75))/(4.2^2) Fmarshellclinton= 2.4 x 10^-8 N

Therefore Marshell and Clinton have a force of gravity of 2.4 x 10^-8 N between them. The reason why Marshell and Clinton don't want to smooch together is because there force of attraction is relatively insignificant compared to the force of attraction they experience with the earth.

This equation can even be used with multiple objects by creating a vector diagram and adding up the vectors of each force of gravity to find the resultant force of gravity.

__Additional Information/Interesting Links/etc.:__

This is a detailed video of Newton's law. media type="youtube" key="TEhLOuiziSA?fs=1" height="385" width="640"

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=<span style="font-size: 1.4em; margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; padding-bottom: 0px; padding-left: 0px; padding-right: 0px; padding-top: 5px;">**__November 9 - Liam Barkley__** =

<span style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; padding-bottom: 0px; padding-left: 0px; padding-right: 0px; padding-top: 0px;">Today we worked on our gravitational and electrostatic forces problem set in groups for the whole period. We watched a video of the feather and wrench drop on the moon(where there is no atmosphere and thus no air resistance) to show demonstrate that Galileo was correct in assuming that gravitational fields cause all objects, regardless of their mass to accelerate at the same rate. <span style="-webkit-text-decorations-in-effect: none; margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; padding-bottom: 0px; padding-left: 0px; padding-right: 0px; padding-top: 0px;">media type="youtube" key="5C5_dOEyAfk?fs=1" height="385" width="480"

<span style="-webkit-text-decorations-in-effect: none; margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; padding-bottom: 0px; padding-left: 0px; padding-right: 0px; padding-top: 0px;">We also watched a video that showed a demonstration of laminar flow. <span style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; padding-bottom: 0px; padding-left: 0px; padding-right: 0px; padding-top: 0px;">__Laminar Flow__: The smooth flow of fluids in layers or streams which do not collide or cross each other's paths. It generally occurs at slow velocities. <span style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; padding-bottom: 0px; padding-left: 0px; padding-right: 0px; padding-top: 0px;">__Turbulent Flow:__ The flow of fluids at high velocities, and exactly the opposite of laminar flow in which velocities and pressures vary seemingly randomly throughout fluids. <span style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px; padding-bottom: 0px; padding-left: 0px; padding-right: 0px; padding-top: 0px;">media type="youtube" key="HKori-K-zJA?fs=1" height="307" width="512"

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November 12



=__November 15:__=

Review of Unit for Test: Refer to powerpoint Universtal Law of Gravitation: Fg=GM1M2/r^2 Electrostatic Force: Fe= kq1q2/r^2 Kepler's Laws: 1. Planets move around in elliptical paths around the sun. The sun is in one of the focal points of the ellipse 2. The radius vector drawn from the sun to a planet seeps out equal areas in equal time intervals (Law of equal areas) 3. The square of the orbital period, T, of any planet is proportional to the cube of the semimajor axis of the elliptical orbit, a. Which basically means that T^2 is proportional to radius^2. T^2/r^3 = a constant for any system Problem Set Answers:
 * 1) 6.0 x 10^-9 N
 * 2) It would have to be 4 times further. right? (2.64 x 10^6 m)
 * 3) a) it's 400kg everywhere. b) on earth it's 3920N and on the other planet it's 800 N
 * 4) 3.46 x 10^8m away from Earth
 * 5) 1.068 x 10^9m is the radius
 * 6) It's averagely 1.61 x 10^11m away
 * 7) Estimated time is 225.3 days
 * 8) The radius is 1.5 x 10^-8 m
 * 9) The net force is 8.77N away from the other charges
 * 10) C=3/2Q - 3Q/l r= ((l^2-2l)/2)^1/2 (r = .366 x l)
 * 11) The net force is 1.91 x 10^5N [right 45 up] (1.78 x 10^5 N)

=__November 16__= __Unit: 3__ __Topic: Paraphrasing a Given article relating to Universal Gravitation (Answer questions)__ __Text Reference: Click Below__



=November 17th, 2010 - Zachary Moore=

Text Reference: 189-210


=November 19th, 2010 - Zachary Moore=

Today was the physics problem set, the homework we have is on the sheet Maria gave to us after the assignment. The questions due are #1, 3, 4, 5, 6, 9

http://dev.physicslab.org/Document.aspx?doctype=3&filename=Momentum_Momentum.xml

Here is a good website with many good momentum problems and the answers and solutions are hidden so that you don't have to worry about seeing the answer while working on the problem.

=November 22th, 2010 - Chris Chung=

How to solve Perfectly Elastic Collisions:

=?m/s= || V2* ?m/s ||
 * M1 = 4kg || M2 = 2.0kg || M1 = 4kg || M1 = 2.0kg ||
 * V1 = 8m/s || V2 = 0m/s || V1*

__ P = P* __ M1V1 + M2V2 = M1V1* + M2V2* 4x8 + 2x0 = 4V1* + 2V2* 16 = 2V1* + V2* V2* = 16 - 2V1*

__ Conservation of Ek x 2 __

M1V12 + M2V22 = M1V1*2 + M2V2*2 4x64 + 2 x 0 = 4V1*2 + 2V2*2 256 = 4V1*2 + 2V2*2 128 = 2V1*2 + V2*2

__ Solve __ 128 = 2V1*2 + ( 16 - 2V1*) 2 128 = 2V1*2 + 4V1*2 + 64V1* + 256 0 = 6V1*2 + 64V1* + 128 V1* = 2.6 and 8 <- initial so don’t count it

V1* = 2.6

==

November 23rd - Sangitha Mensingh
__Unit:__ Conservation of Energy __Topic:__ Collisions and Momentum in 2-D __Text Reference:__ pg. 203-210 and pg. 260-266

__Idea:__ Objects involved in a non head-on collision will move in a direction at an angle according to their initial direction, mass and initial velocity.

__Key Examples:__ See document below.

__Additional Information/Interesting Links/etc.:__ This site gives a good refresher on momentum in one dimension, key terms and momentum in two dimensions. It also includes some practice questions and helpful animations. Link: [|Conservation of Momentum]

A couple words on momentum in general by Bill Nye: media type="youtube" key="y2Gb4NIv0Xg?fs=1" height="385" width="480"

==

November 24th - Wade Walker
Today we received a problem work sheet. Do it.

December 1st Travis Guy
Collisions Lab with the Air table

December 8th - Scott Mastromatteo
Egg drop lab. I know what Maria's having for dinner tonight.

=__**January 7th 2010**__ **by**: Joanna Decc=

-Universal Wave Equation -Snell's Law -Double slit interference
 * Unit:** Optics?
 * Topic:** Light as a wave
 * Text Reference:** None.See handouts below.
 * Ideas:**

In class we just added the notes.
 * What we did in class:** The following 2 documents attached are taken from the Ripple Tank Investigation Sheet (which was given to us on January 5th 2010).




 * What we got for homework:**

[] (Look under Lesson 3: Mathematics of 2 Point Source Interference)
 * Helpful Website:**

=**January 11th 2010 by: Joanna Decc**=


 * No Class! HUBBLE INSTEAD (Y).**