1 physics 7b - ab lecture 10 june 5 overview practice final problems good news everyone! in four...

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Physics 7B - ABLecture 10

June 5Overview

Practice Final Problems

Good news Good news everyone! In four everyone! In four days you will be days you will be sitting physics sitting physics exam. Oooh exam. Oooh

yes.....yes.....

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Quiz 4 Re-evaluation Request Due TODAY

Quiz 5 & 6

Due June 9 at the time of Final

Quiz 6 Rubrics on the website

Review session starts TODAY. Schedule on the course website

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7B Final June 9 Mon 1- 3pmReview session starts TODAY.

Day Date Time Location Topic Instructor

Thursday 6-Jun Lecture 10 Roessler 66 Practice Final Problems Yu Sato

Thursday 6-Jun 10:30am to noon Walker Annex 114 Overview of the Material Dan (Lec A&B)

Thursday 6-Jun 2:10pm to 3:40pm Walker Annex 114 Quizzes from Lec. A&B Abby (Lec A&B)

Thursday 6-Jun 4:40 to 6:10pm Walker Annex 114 Free form / Q & A Marshall (Lec A&B)

Thursday 6-Jun 7:10pm to 8:40pm Walker Annex 114 Quizzes from Lec. A&B Ryan (Lec A&B)

Friday 7-Jun 10:30am to noon Walker Annex 114 Quizzes from Lec. A&B Chris (Lec A&B)

Friday 7-Jun 2:10pm to 3:40pm Walker Annex 114 Free form / Q & A Brandon (Lec A&B)

Saturday 8-Jun 10:00am to 11:00am Roessler 166 Free form / Q & A Rion (Lec C&D)

Saturday 8-Jun 11:00am to noon Roessler 166 Free form / Q & A Mickey (Lec C&D)

Saturday 8-Jun 2:00pm to 3:00pm Roessler 166 Quizzes from Lec. C&D Adam (Lec C&D)

Sunday 9-Jun 10:00am to 11:00am Roessler 166 Free form / Q & A Mike (Lec C&D)

Sunday 9-Jun 2:00pm to 3:00pm Roessler 166 Quizzes from Lec. C&D Owen (Lec C&D)

Sunday 9-Jun 3:00pm to 4:00pm Roessler 166 Free form / Q & A Matt (Lec C&D)Sunday 9-Jun 4:00pm to 5:00pm Roessler 166 Free form / Q & A Jim (Lec C&D)

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Final location

Haring Hall, Rm. 1227 A through DHaring Hall, Rm. 2205 E through QSurge 3 Bldg., Rm. 1309 R through Z

Separated by family name:

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Final checklist

• Pens and pencils

• CalculatorWe will not have spare calculators, make sure you bring yours

• Photo ID (Student or Government ID)

Without it you can’t sit the final, and you will fail the course.

The pages will be separated -- write your name on every single page when you first get your final

Formulas will be provided with the final

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• Momentum (PF6, Q5, CDQ5, CDQ7)• Forces: how to change momentum (PF6&7, Q4)• Angular momentum (Q5,CDQ7)• Torque: how to change ang mom (PF8,Q5,CDQ6)• Simple Harmonic Motion: A specific type of net force

• Fluid (PF1&2, Q1, CDQ1)• Circuits (PF4, Q2, CDQ2)

Physics breaks into two separate “parts” in 7B:

Then we have learned four techniques

• Vectors (PF5, Q3, Q4)Vectors (PF5, Q3, Q4)• Components, use of trigonometry (PF5, Q5)Components, use of trigonometry (PF5, Q5)• Force diagrams (PF5&7, Q6),Force diagrams (PF5&7, Q6), Extended force diagrams (PF8, Q6,CDQ6)Extended force diagrams (PF8, Q6,CDQ6)•Linear/Angular Momentum Charts (Q4, CDQ5)Linear/Angular Momentum Charts (Q4, CDQ5)

Energy density modelNo net change in energy density/around a circuit

•Exponential decay (PF3, Q3,CDQ4)Exponential decay (PF3, Q3,CDQ4)• Osmosis (Q3)Osmosis (Q3)

Forces and its relation to change in motion

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Today’s lectureOverview of the material,

using practice final problems as examples• These review notes are supposed to go over the course, but as you have seen everything before, pieces of the course can be “mixed up” -- this is good practice for the final.

• These notes cover a lot of the class, but not all.For example, practice final does not have any problem on diffusion. See CD Quiz 3 for an example.

Diffusion occurs when there is concentration gradient of a specie of particles.

Wait a while…

Permeable, semipermeable membrane

What flows can be particels or water depending on the membrane property

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Forces, Force diagram, Vectors, Components Practice Final 4

Students (m =100kg) in hammocks

= 45 = 25

Hammmock + Students + Strings = single objectWhat are the contact/non contact forces exerted on the system?

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Identifying forces

There are contact forces and non contact forces

The only non contact force we worry in 7B is gravitational pull of the Earth exerted on all objects, i.e. FEarth on ball

Contact force can be exerted by anything that is in contact with your object, i.e. Fstring on ball

FEarth on ball

Fy, string on ball = 100N

Fx, string on ball = m|a| = 10kg(1.5m/s) = 15N

Fstring on ball

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= 45 = 25

Hammmock + Students + Strings = single objectWhat are the contact/non contact forces exerted on the system?

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FEarth on hammock = 1000N

FPost on hammock

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FEarth on hammock = 1000N

FPost on hammock

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= 45 = 25

A static problem, i.e., torques as well as forces are all balanced (another way of saying this is, net torque is zero & net force is zero)

They have to be balanced compnents wise.

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Torque, Extended force diagram Practice Final 7

Deltoid muscle is attached to the man’s arm and pulls on his arm in this direction

Attached at 15cm from the shoulder joint

Center of Mass of the arm at 30cm from the shoulder joint

40N = FEarth on arm

Another static problem This one is harder.

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40N = FEarth on arm

A static problem, i.e., torques as well as forces are all balanced (another way of saying this is, net torque is zero & net force is zero)

Another static problem

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40N = FEarth on arm

(a) What’s tangential component of the force of Deltoid muscle on the arm?

(b) Draw extended force diagram

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40N = FEarth on arm

80N = Ftangential

FMuscle on arm

Note: An arrow is a vector, a dotted arrow is a component of a vector

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40N = FEarth on arm

80N = Ftangential

FMuscle on arm

Fshoulder joint on arm


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40N = FEarth on arm

80N = Ftangential

FMuscle on arm

Fshoulder joint on arm

Now both torque and forces are balanced!


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40N = FEarth on arm

Muscle goes limp and the arm starts to swing (Now torque is not balanced! )

What is its after 0.1s?

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40N = FEarth on arm

Muscle goes limp and the arm starts to swing (Now torque is not balanced! )

What is its after 0.1s?

|| ∆ t = |∆L| = (0.3m)(40N)(0.1sec) = 1.2Nms Lf = I = 1.2Nms

So then if we knew Iarm, we can figure out !

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• We know how to find each force

• Every force in the problem also contributes a torque.(This torque may turn out to be zero)

• The magnitude of the torque is on obj = (Ftangential) rwhere r is the distance between where the force is applied and the pivot point

FEarth on pentagon

Pivot

Recipes for torque

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FEarth on pentagon

Pivot

r

Magnitude of torque is

r FEarth on obj, perp

Direction is into the screen (RHR)

Do this for each force, then add all the torques up to find the net torque.

(Some forces are easy: applied either at or through the pivot)

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Forces, Force diagram, Acceleratin/Velocity Practice Final 7

Cushion

Compact shaped person jumping off a window of a burning building

He falls freely for 1.5 sec, then the cushion exerts a constant force to bring the person to rest in 0.3sec Hint : Assume |vPerson| = 14.7m/s right before he hits the cushion

Position y (m) Velocity v (m/s) Acceleration a (m/s2)

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Forces, Force diagram, Acceleration/Velocity Practice Final 7

Cushion


(1) Force diagram during the free fall

FEarth on person

2) Force diagram while the cushion is bringing the person to rest

FEarth on person

FCushion on person

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Forces, Force diagram, Acceleration/Velocity Practice Final 7

Cushion


Acceleration a (m/s2)

(1) Force diagram during the free fall

FEarth on person

2) Force diagram while the cushion is bringing the person to rest

FEarth on person

FCushion on personCheck whether the net force, i.e.Fon person,is consistent with a person

Remember Fon person = m a person!

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Forces, Momentum Practice Final 6 mCadillac = 2000kg

20m/s to Right

mToyota = 1000kg 20m/s to Left

Initial

FinalmCadillac+Toyota = 3000kg

Traveling either to Right or to Left, or remain stationary

Stuck together

We don’t know the direction/speed of travel after the collision. What we do know is:

pC+Tinitial = pC+T

final

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20m/s to Right


Initial


Stuck together

|pC | = mC | vC | = 40000kgm/s

| pT | = mT | vT | = 20000kgm/s

pC

pT

| pC+Tinitial | = 20000kgm/s

pC+Tinitial

Total

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20m/s to Right


Initial


Stuck together

| pC+Tinitial | = 20000kgm/s

pC+Tinitial

Total

pC+Tfinal

Travelling to Right at 6.66m/s

40000kgm/s

20000kgm/s

pC

pT

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20m/s to Right


Initial


Stuck together

Then find ∆p of each car from:

∆p = m ∆v = m (vfinal–vinitial)

pC+Tfinal


Pay attention to the direction of vectors when adding/subtracting

40000kgm/s

20000kgm/s

pC

pT

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20m/s to Right


Initial


Stuck together

Then find ∆p of each car from:

∆p = m ∆v = m (vfinal– vinitial)

pC+Tfinal



∆pToyota comes out to be equal to ∆pCadillac

even with its smaller mass because Toyota changes its direction after the collision

40000kgm/s

20000kgm/s

pC

pT

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20m/s to Right


Initial


Stuck together

40000kgm/s

20000kgm/s

pC

pT

Then find ∆v of each car:

pC+Tfinal



∆vToyota comes out to be greater than ∆vCadillac because Toyota changes its direction after the collision

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20m/s to Right


Initial


Stuck together

40000kgm/s

20000kgm/s

What about Fave experienced by each car during the collision?

pC+Tfinal


One approach (Newton’s 1st law):

Fave ∆t = ∆p

Alternative approach (Newton’s 3rd law):

FC on T = – FT on C !!!

pC

pT

SAME !

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Fluids/Circuits

Basic rules for looking at fluids/circuits :

1. Energy (density) conservation

OR (the same as)

2. Current entering = current leaving

3. Pressures where two fluids systems touch are equal Voltages that are connected by wire (no circuit element in between) are equal

∆V = – IR

∆P + (1/2)∆(v2) + g∆h = Epump/volume – IR

Junction rule

A1 v1 = A2 v2

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Fluids, Circuit Practice Final 1

P1 = 200kPa

v1 = 10m/s

w = 1000kg/m3

v1 vs v2 ??

A2 = 0.5A1

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P1 = 200kPa

v1 = 10m/s

w = 1000kg/m3

A1 v1 = A2 v2

v2 = ( A1 /A2) v1 = 2 v1 = 20m/s !

A2 = 0.5A1

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P1 = 200kPa

v1 = 10m/s

w = 1000kg/m3

Keep raising the end 2,

At what h, P2 is equal to zero?

v2 = 20m/s

∆P + (1/2)∆(v2) + g∆h = 0

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P1 = 200kPa

v1 = 10m/s

w = 1000kg/m3

v2 = 20m/s

∆P + (1/2)∆(v2) + g∆h = 0

Substitute P2 = 0, to find h2 (say h1 = 0)

h2 = 5m.

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Why does a hose have a nozzle at the end?

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Fluids, Circuit, Forces Practice Final 2

Why does a hose have a nozzle at the end?

A1 v1 = A2 v2 !!In order to keep the flow rate constant throughout the fluid circuit, fluid velocity will increase at a narrowed nozzle. With greater velocity of the fluid coming out of the nozzle, the water will reach farther, allowing the firefighter to fight the fire far from the fire.

v2 > v1

v2

v1

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What is the direction of the net force on the small amount of water at three different locations?

v2 > v1

v2

v1

Example of a problem that combines concepts from different models.

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These are the direction of the net force on the small amount of water at three different locations. Think about how fluid velocity v is changing at each location, as net force is in the same direction as ∆ v.

0

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Fluids/Circuits

Oops… One more technique to remember :

Know how to find equivalent resistance, this is an essential technique for analyzing circuits

3220 12 24

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Req of the whole circuit = 40

Circuit Practice Final 3

3220 12 24

VB ??

∆V3 = 3V

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Req of the whole circuit = 40

Circuit Practice Final 3

3220 12 24

VB ??

∆V3 = 3V

Find I3 = 0.25A Find ∆V2 = 5V Find ∆V3 = 8V Find I1 = 0.25A Find I = 0.5A Find ∆V4 = 12V Find ∆VB 20V Finally!

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Close the switch… The circuit becomes simpler

Circuit Practice Final 3 32

20 12 24

20V

What happens to:

∆V1 I2 ∆V3 ∆V4 ??

Two things that don’t change : VB , resistor valuesReq decreases, and so Ieq increases

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Exponential decay Practice Final 3

Circuit B has a battery having voltage V and the capacitor is uncharged.

Circuit A has no battery and the capacitor is charged to a voltage V.

Circuit A and B are similar but slightly different RC circuits

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Capacitor:

A capacitor stores electrical energy by accumulating charge on two conducting plates. It can also release the stored energy very quickly.

What was Capacitor C again??

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Exponential decay

Anytime we see change in something is directly proportional to the amount of that something, we have exponential decay (growth)Ex. Microorganisms in a culture dish, a virus of sufficient infectivity, human population, nuclear chain reaction, charge/discharging capacitor

The same statement mathematically:

Solution :

±

Stuff (t) (Amount of stuff at t = 0 ) e ±kt

In each time interval of 1/k (time constant), amount of stuff reduces to 1/e of each previous value/ or grows by a factor of e

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Cgarged to V

Uncharged

C

Vc = Ve ±(1/RC)t

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Good luck!

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