Momentum Lab

Today, we did a Momentum lab. The lab is from Chapter 9 (green text book). It was "The Explosion" lab. We had to calculate how far and fast the two carts went until they hit the bricks/books etc.

We had no time to finish it, due to the Pep Rally. I guess we will continue tomorrow with the lab.

Have a good evening, :D

September 29, 2010

Hello.

Today in class we finished up the video about Momentum and handed in 5-7 facts about it. Then we went over a few notes on deriving the impulse-momentum equation from Newton's 2nd law.



To go over it, the notes were;

There were also questions in the textbook;
pg287 question 6

pg308-309, questions 50-51


Check Your Answers.
On pg287: Find the impulse:
6. a) 125 N∙s [S]
b) 1875 N∙s [W]
c) 45 N∙s [W]

Pg208-309: Find the Impulse.
50. a) 1750 N∙s
b) 6∙10^-1 N∙s
c) -225 N∙s

51. 22.5 N∙s



*Sorry for the late post.

Dynamics Test

Dynamics test

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Momentum - September 28, 2010

Hey Grade 12's!
Today we started the class off by doing the "Dynamics" test corrections on the smart board which Ms. Kozoriz will post sometime today I believe. After that we began our new unit "Momentum" by watching a film about it, we were told to copy down 5-7 facts that we personally found interesting to us but since we didn't finish the whole film we are going to continue watching the rest of it tomorrow.
Enjoy the rest of your evening! :)

Dynamics Review

Hi Guys.
Last Friday in class we did some reviews in dynamics. We did the case questions 1,4,5.

We also did the question from the book in page 196 from 1 to 3.

1. A copy of Physics: Concepts and Connections leaves the printing press and slides down a 4.0 m long ramp into the arms of an eager physics student. If the ramp is inclined at an angle of 25 degrees to the horizontal ans has a coefficient of kinetic friction of 0.10, how long will it take the 2.0 kg textbook to reach the student?

2. The plastic case from your least-favorite CD recording is flung up a frictionless ramp, inclined at an angle of 20 degrees to the horizontal. If the case leaves your hand at a speed of 4.0 m/s, how long will it take before the case comes to rest?

3. A skateboarder slides down a frictionless ramp inclined at an angle of 30 degrees to the horizontal. He then slides across a frictionless horizontal floor and begins to slide up a second inclined at an angle of 25 degrees to the horizontal. The skateboarder starts at a distance of 10 m from the bottom of the first incline. hoe far up the second incline will he go if the coefficient of kinetic friction on the second incline is 0.10?

Lastly, we did a worksheet on transparency.

Hi Everyone!

Today we went over the "Dynamics Equlibrium Problems" worksheets. The answer is kinda blurred sorry guys i wrote it on pencil.

After we went over the worksheets, we did the 2 questions on pages 13 (#2 and 4) Dynamics booklet (the rest of the answer is on the booklet on page 22) the answer of those 2 questions are:

Lastly, we went over The Atwood's Machine on page 11 and she gave us a yellow booklet about that lessons and ask to answer some question on that.

If you don't have the yellow sheets , see Ms. Kozoriz tomorrow.

Our test will be on Monday on this unit.

September 22

Hey
So today in class we did a lab showing the forces working on a block moving down an inclined plane. We needed to find the FORCE OF GRAVITY (Fw), FORCE OF PARALLEL (Fgx), FORCE OF PERPENDICULAR (Fgy), the ANGLE, and COEFFICIENT OF FRICTION (µ).

The Objective of the lab was to find the value of the coefficient of sliding function using an inclined plane



To learn the steps involved read page 19 in the Dynamics Unit

Questions

1.) Calculate Fgx, Fgy, and µ

Fgx = (sin24)(o.75N) [24 being the angle at which the block moved, and .75 being the force of the block]

Fgx = 0.31 N

Fgy = cos24(o.75N)

Fgy = 0.69N

The formula for finding µ is: µ = Fgx/Fgy

µ = 0.31N/0.69 N

µ = o.45

2.) Show that µ = tanθ
tan24 = 0.45



Next we read pages 172-179, and 194- 198, and doing questions 1-4 on pg 96 of the textbook

Questions

1.) A book leaves the printing press and slides down a 4.0m long ramp into the arms of a student. If the ramp is inclined at an angle of 25 degrees to the horizontal and has a coefficient of kinetic friction of 0.10, how long will it take the 2.0kg textbook to reach the student?

To be honest I don't know how to get the answer, so here's a basic diagram

For better information please as Ms. K.

Tuesday, September 21, 2010

Hi,

Today we went straight into a worksheet that Ms K walked us through called the Transparency worksheet. As we worked through it, we learned how to find and use forces on an incline plane. For the sheet, see Ms K!

After that, we went back into our Dynamics Unit booklet and did the example question on page 7.

Problem

A 475 N trunk is resting on a plane inclined 40.0° above the horizontal. Find the components of weight force parallel and perpendicular to the plant.

To solve inclined plane problem, you could start by drawing a diagram and or a free-body diagram

(image not drawn to scale)

Firstly, we know Fg = 475N and angle FgFgy = 40°

To find Fgx, we multiply sin40° by 475N

Fgx = sin40°(475N) = 305N

To find Fgy, we multiply cos40° by 475N

Fgy = cos40°(475N) = 364N

And the answer is Fgx = 305N, Fgy = 364N


After Ms K went over this example question, she assigned practice problems in the booklet which are on page 18. The answers to these questions are in the back. She also assigned a worksheet called Equilibrium Problems. If you don't have it, see her as soon as possible tomorrow. Both of these assignments are to be done for tomorrow.

Monday, September 20, 2010

Grade 12 Physics: Dynamics Unit

Hi everyone.

Today in class we went over the 'Free-Body Diagrams' on pages 12 and 13 of the Dynamics booklet. Unfortunately I don't have a scanner. So if anyone has some images of those diagrams, please stick them here:



After that we went over the 'Static Equilibrium Problems' on page 16 and 17. The answers to those questions are:

1. Fg = 42N

2. FT = 980N

3. FT = 346N

4. FT = 281N

5. FT = 57N


On page 15 you'll find the 'Net Force' questions which we also did in class. To do these questions you must find the net force of the forces acting on a box.

Ex: 3N <---O----> 4N
Fnet = 3N, left + 4N, right
Fnet = 1N right

Hint: use the Pythagorean Theorem (c² = a² + b²) for triangular vectors.

Answers:
1. 0N
2. 1N, right
3. 5N, 53°
4. 5N, 217°
5. 5N, 53°
6. 5N, 217°
7. 10N, 45°
8. 10N 45°
9. 0N
10. 7N, 270°

Lastly we were assigned some problems on page 5 as homework.

Friday, September 17th 2010

Grade 12 Mechanics Unit: Acceleration Test (Answers):

1) Step 1: Square both sides of the equation
V2² = V1² + 2V1at + a²t²
Step 2: Factor out 2a from the last two terms
V2² = V1² + 2a(V1t + ½at²)
Step 3: Substitute a term in the equation given in step 2
V2² = V1² + 2ad
Step 4: The equation is _____
V2² = V1² + 2ad

2) V1 = 40.0km/h ÷ 3.6 = 11.11m/s
a = -2.3m/s²
t = 2.7s
d = ?
V2 = ?

2a) d = V1t + ½at²
d = (11.11)(2.7) + ½(-2.3)(2.7)²
d = 21.6m

2b) V2 = V1 + at
V2 = 11.11 + (-2.3)(2.7)
V2 = 4.89m/s

3) V1 = 0m/s
a = 0.300m/s²
d = 25.0m
V2 = ?

V2² = V1² + 2ad
V2² = 0² + 2(0.300)(25.0)
V2² = 15
V2 = 3.87m/s

4) motorist = 80.0km/h(W)
police car = 95.0 km/h(W)

4a) motorist - police car
80.0 - 95.0 = 15km/h(E)

4b) police car - motorist
95.0 - 80.0 = 15km/h(W)

5) airplane's airspeed = 2.0 x 10²km/h(E)
approching destination = 15°(N of E)

5a) tan15° = 200 ÷ wind speed
wind speed = 200 ÷ tan15°
wind speed = 53.6km/h

5b) c² = a² + b²
speed approaching destination = (2.0 x 10²)² + (53.6)²
speed approaching destination = 207km/h

6) V1 = 25m/s
t = 1.45s
a = -8.5m/s²
V2 = 0m/s
d = ?

reacting to child and applying the brakes:
d = Vt
d = (25)(0.45)
d = 11.25m
after applying the brakes:
V2² = V1² + 2ad
d = V2² - V1² ÷ 2a
d = (0)² - (25)² ÷ 2(-8.5)
d = 36.76m
total distance = 11.25m + 36.76m = 48.0m

_____________________________________________________

Equilibrium: When an object has zero acceleration, meaning that the object is at rest or moving at constant velocity
Free Body Diagram: A drawing that represents the object and the forces acting on it

Steps to Drawing a Free Body Diagram:

1) Focus on the object you will be studying. If there is more than one object in the drawing, do each seperately.
2) Begin drawing the free body diagram by looking at only the desired object. The drawing will show where forces are applied to the object, so draw those forces on your free body diagram in exactly the same direction thats shown on the drawing.
3) Draw x and y axes on your free body diagram so you will be able to analyse the diagram. Try to draw your axes so that as many forces as possible are close or directly on the axes. It will help when calculating.
4)When the situation is equilibrium, make sure that the x and y components of the forces will be equal to zero when added.
5) Finally, solve for the unknown quantities.

_____________________________________________________

In class we received the 'Grade 12 Physics: Dynamics Unit' booklet, some of us recieved this booklet after finishing the Acceleration Test on Thursday. We read pg.1, completed pg.14, and completed #1 on pg.16. For homework we were assigned pg.12 & 13 (Free Body Diagrams) and #2-5 on pg.16 & 17 (Static Equilibrium Problems).

Test Day

Test on Acceleration. How did you do?

Relative Motion Problems

okay, so today we didn't really learn something new. we just went over the questions on page 10 of the first booklet we were given: Acceleration & Relative Velocity

I don't know about you guys but questions 4-7 gave me a hard time solving them, until Ms. Kozoriz gave me some really important tips on how to solve them.

so here are some steps of how you can solve them.

1) draw your x-axis and y-axis

2) draw the first information that you are give on the graph

3) draw the second information that you are given, but draw it from the end tip of the first information. i will show you what i mean by this

4) now you use this equations to solve for y and x.

y = sin(angle) * given distance or velocity

x = cos (angle) * given didstance or velocity

5) now you make sure that you have the total x or y

6) use the x and y to find the resultant velocity if you are asked to find it.

7) after the above steps you now have enough information to solve for tan (angle)

okay let us put the above steps into actions

question: A dog's owner walks 20 meters due west and then turns and walks 45 degrees south of west a distance of 150. where is he with the respect to where he started?

y = sin45 (150) = -106

x = cos45 (150) = -106

the total value for x is -126

this is because we had -20m which we were given in the question, plus the -106 that we just solved for.

total value for y is -106

* okay now you might be thinking why this values are negative, and that is becuase if you look on the graph the side that they are drawn on is negative for both y and x axis

now to find the resultant distance you use pythegorean theorem.

r = 165

now the last thing is find the angle.

tan (angle) = 106/126

tan (angle) = 40s degrees south of west

so now this is what you do when you get a question like that. just solve it step by step.

Cosine Law

Cosine law example

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Scribe List

This is The Scribe List. Every possible scribe in our class is listed here. This list will be updated every day. If you see someone's name crossed off on this list then you CANNOT choose them as the scribe for the next class.

jad YASSIN(2) marinhel Alex mark e

Justin Aruni(3) sakhone(3)

shahani katrina(3) chemutai(3)

This post can be quickly accessed from the [Links] list over there on the right hand sidebar. Check here before you choose a scribe for tomorrow's class when it is your turn to do so.

Monday, September 13, 2010

First we began our class with handing in the assignment from Friday, titled "Grade 12 Physics: Acceleration Assignment" (which if you haven't, reminder that you should!) and copied and checked answers 8 - 11 on the practice worksheet titled "Grade 12 Physics: Acceleration II."

The answers are as followed :
(Note: Thanks Amy, and I hope you like the colours added.)

8b. = 121m
8c. = 391m
8d. = 946m
9. = 15m
10. = 1350m
11a. = 15.4s
11b. = 427m
11c. = 200km/h

For full, see a friend or Ms.K.

Once finished, we went back to the first booklet we received titled, "Grade 12 Physics Acceleration & Relative Velocity" and went over page 6 - 9, doing the questions along the way. Don't worry if you weren't there, all the answers are stapled along with the booklet on page 11.

Page 6 extending to 7 is the lesson about relative velocity.

The concept of relative velocity can be a bit confusing, but with a little more understanding it can be easy.

It's the relationship between 3 different velocities, that can be different depending on the question. In the example included in the booklet, there are the velocities of a passenger on a train, the train itself, and someone standing on the ground watching the train.

To the passenger, who is walking to the front of the train, their velocity is +2.0 m/s.
Note: The + sign indicates direction, so if the passenger was walking towards the rear, the velocity would be -2.0 m/s.

The train's velocity, to the person on the ground, is +9.0 m/s.

We can calculate the velocity that the person the ground can see, or any of the three as long as we have any 2 velocities.

The formula that can be used is:

Vpg = Vpt + Vtg

Vpg = (2.0 m/s) + (9.0 m/s) = 11 m/s

p = passenger, t = train, g = ground.

Each velocity has letters in the subscript that are in relation to the objects in the question. Look at the bottom of page 6 for more details.

At the end of page 7 there is a small chart which can be filled out to find the answer.

So on the next page, 8 , there are two diagrams with a matching sheet of questions. The questions are pretty straight forward, but it's never bad to take a second check on the answer key.

For homework, we were assigned two things:

1) Page 10 of the booklet, for practice. (Answers included in the key)

2) A relative velocity sheet with 7 questions which you can get from Ms.K, if you don't already have it, which we will correct in class.

Well that concludes our day, and I will come back to fill in those answers, sorry everyone! Reminder, TEST ON THURSDAY! So study, study!

-Sakhone P

FRIDAY, SEPTEMBER 10, 2010

For the start of the class, we went over one of the worksheets that was given out the yesterday that didn't have the answers.

[Please see Amy's post for the original sheet].

We worked it out and put all the answers on the white boards.

I've recorded down the answer for those who missed the class or missed an answer and would like to check it over:

We then received another worksheet, but this time instead of working it at home, we worked it together as a class with Mrs. Kozoriz on the SmartBoard.

The sheet that was given out:

Answers to the following worksheet:

Here, we use one of the formulas of finding velocity.

D.B. stands for Donovan Bailey and M.J. stands for Michael Johnson.

Mrs. Kozoriz also asked us how come Michael Johnson ran the 200-m faster than the 400-m. Well, because he had to do both runs! Total of 600-m and of course he would be tired and slow down !

Here, we used the same formula as we used for question one, but we rearranged it so that we could find the distance instead of the velocity.

To change the units of km/h to m/s, we have to divide the number in units of km/h with 3.6 .

** We have to make sure to change the units or else our answer will be wrong ! **

Here we also have to find the distance, but this time we use a different formula to find the distance. This formula uses both the velocity instead of one to find the distance.

Here, we now have acceleration given to us to help us find the final velocity.

Here, we use another formula to find the length of time.

It's also the same formula that we used to find the answer to the previous question, but again, we rearranged it to help find the time instead of the velocity.

What's different here, is that we have a negative acceleration.

Although it's negative, we don't do anything different, so don't let the negative sign throw you off!

For this question, it consists of 2 steps.

First we find the distance using a different formula.

Make sure you use a formula that doesn't use the final velocity to find the distance because we have to find that also.

Here, we assume that the final velocity equals to zero because it slows down to a stop.

Since the time was not given to us in the question, we have to use this formula to help us find the distance instead.

Although question 8 has four parts, we only went over how to do the first one, because once you know how to do the first one, the rest are worked out the same and should be easy to solve.

The worksheet [Gr.12 Physics Acceleration Assignment] that was given out on Wednesday, September 8th had answers given except for question 5. a) & b).

Amy had posted the worksheet and the answers but she didn't have the answers for number 5 because it was given in today's class.

So I'm just reposting the worksheet with the answers for those who might not have seen on Amy's previous post followed by the answers for question 5:

HOMEWORK :

Since we stopped after question 8) a. in class with Mrs. Kozoriz, we're left of to finishing the rest of question 8 to question 11.

She also gave us another handout to work on which is due on Monday, September 20! :

That's all that we did today !

Hope you enjoyed this post. Good luck and see you on Monday ! (:

- Kim

Amy's Scribe Post (Thursday, September 9, 2010)

Hello, my fellow students. I am gladly to present you our class' very first scribe post on the blog today. To start things off, the first thing we did in class was have Ms. Kozoriz review some simple formulas that we learned from grade 11 physics.

These equations can be found from finding the slope from a distance-time graph (for velocity) or a velocity-time graph (for acceleration).

The second thing that Ms. Kozoriz explained in class was how to derive equations from our handout given to us yesterday. These equations can be found on the formula sheet given as the first four under the section "Mechanics."

These equations can be rearranged to find the thing we need by simply rearranging the constants (using either multiplication/division) to find what we want.

Ms. Kozoriz also pointed out that velocity 1 equals to zero.

The next thing we learned was the Derivation of Acceleration Formulae. This can be explained through the diagram below:

As you can see, we can obtain the displacement from the graph by simply using the formula we learned in grade 11 physics when d=vt.

Ms. Kozoriz had then given another diagram to use as an example showing the same idea explaining that when finding two different displacements and combining them together, it gives us one of our derived equations where each half represents the shape of the diagram.

An overview of the diagram above:

Finally, Ms. Kozoriz taught us how to derive one of the last equations.

That concludes the lesson of Thursday, September 9, 2010.

HOMEWORK

Ah, the most important part of the day. Today Ms. Kozoriz had handed out 2 sheets of questions with a total of 20 (10 on each sheet). In case you did not get one (for some reason..) here are the the links for the 2 homework sheets.

Homework sheet 1 & Homework sheet 2 (with answers)

In that case, this is an end of my scribe post. Thank you to those who've taken the time to read what I have to share. Please leave some comments :)!!!