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Physics 7B - AB Lecture 10 June 5 Overview Practice Final ProblemsPowerPoint Presentation

Physics 7B - AB Lecture 10 June 5 Overview Practice Final Problems

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Physics 7B - AB Lecture 10 June 5 Overview Practice Final Problems

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Physics 7B - AB Lecture 10 June 5 Overview Practice Final Problems

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

Quiz 6 Rubrics on the website

Review session starts TODAY. Schedule on the course website

7B Final June 9 Mon 1- 3pm

Review session starts TODAY.

Haring Hall, Rm. 1227 A through D

Haring Hall, Rm. 2205 E through Q

Surge 3 Bldg., Rm. 1309 R through Z

Separated by familyname:

- 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.

Formulas will be provided with the final

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

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

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

- 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

Energy density model

No net change in energy density/around a circuit

Forces and its relation to change in motion

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

Then we have learned four techniques

- Vectors (PF5, Q3, Q4)
- Components, use of trigonometry (PF5, Q5)
- Force diagrams (PF5&7, Q6),
Extended force diagrams (PF8, Q6,CDQ6)

- Linear/Angular Momentum Charts (Q4, CDQ5)

Today’s lecture

Overview 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.

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

Wait a while…

Permeable, semipermeable membrane

Forces, Force diagram, Vectors, Components Practice Final 4

Students (m =100kg) in hammocks

= 25

= 45

Hammmock + Students + Strings = single object

What are the contact/non contact forces exerted on the system?

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

Fstring on ball

Fy, string on ball = 100N

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

FEarth on ball

Forces, Force diagram, Vectors, Components Practice Final 4

Students (m =100kg) in hammocks

= 25

= 45

Hammmock + Students + Strings = single object

What are the contact/non contact forces exerted on the system?

Forces, Force diagram, Vectors, Components Practice Final 4

Students (m =100kg) in hammocks

FPost on hammock

FEarth on hammock = 1000N

Forces, Force diagram, Vectors, Components Practice Final 4

Students (m =100kg) in hammocks

FPost on hammock

FEarth on hammock = 1000N

Forces, Force diagram, Vectors, Components Practice Final 4

Students (m =100kg) in hammocks

= 25

= 45

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.

Torque, Extended force diagram Practice Final 7

Another static problem This one is harder.

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

Attached at 15cm from the shoulder joint

40N = FEarth on arm

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

Torque, Extended force diagram Practice Final 7

Another static problem

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

Attached at 15cm from the shoulder joint

40N = FEarth on arm

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

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)

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

40N = FEarth on arm

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

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

Draw extended force diagram

Torque, Extended force diagram Practice Final 7

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

80N =Ftangential

FMuscle on arm

40N = FEarth on arm

Torque, Extended force diagram Practice Final 7

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

80N =Ftangential

FMuscle on arm

Fshoulder joint on arm

40N = FEarth on arm

Torque, Extended force diagram Practice Final 7

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

80N =Ftangential

FMuscle on arm

Fshoulder joint on arm

40N = FEarth on arm

Now both torque and forces are balanced!

Torque, Extended force diagram Practice Final 7

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

What is its after 0.1s?

40N = FEarth on arm

Torque, Extended force diagram Practice Final 7

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

What is its after 0.1s?

40N = FEarth on arm

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

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

Recipes for torque

- 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

Pivot

FEarth on pentagon

Pivot

r

FEarth on pentagon

Magnitude of torque is

rFEarth 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)

Forces, Force diagram, Acceleratin/Velocity Practice Final 7

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

Cushion

Position y (m)

Velocity v (m/s)

Acceleration a (m/s2)

Forces, Force diagram, Acceleration/Velocity Practice Final 7

Compact shaped person jumping off a window of a burning building

(1) Force diagram during the free fall

FEarth on person

Cushion

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

FCushion on person

FEarth on person

Forces, Force diagram, Acceleration/Velocity Practice Final 7

Compact shaped person jumping off a window of a burning building

(1) Force diagram during the free fall

FEarth on person

Cushion

Acceleration a (m/s2)

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

Check whether

the net force, i.e.Fon person,

is consistent with a person

Remember Fon person = m a person!

FCushion on person

FEarth on person

Forces, Momentum Practice Final 6

Initial

mToyota = 1000kg 20m/s to Left

mCadillac = 2000kg 20m/s to Right

Final

mCadillac+Toyota = 3000kg

Stuck together

Traveling either to Right or to Left, or remain stationary

We don’t know the direction/speed of travel after the collision. What we do know is: pC+Tinitial = pC+Tfinal

Forces, Momentum Practice Final 6

Initial

mToyota = 1000kg 20m/s to Left

mCadillac = 2000kg 20m/s to Right

Total

| pC+Tinitial | = 20000kgm/s

pC+Tinitial

pC

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

pT

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

Final

mCadillac+Toyota = 3000kg

Stuck together

Forces, Momentum Practice Final 6

Initial

mToyota = 1000kg 20m/s to Left

mCadillac = 2000kg 20m/s to Right

Total

| pC+Tinitial | = 20000kgm/s

pC+Tinitial

pC

40000kgm/s

pT

20000kgm/s

Final

mCadillac+Toyota = 3000kg

Stuck together

Travelling to Right at 6.66m/s

pC+Tfinal

Forces, Momentum Practice Final 6

Initial

mToyota = 1000kg 20m/s to Left

mCadillac = 2000kg 20m/s to Right

Then find ∆p of each car from:

∆p = m ∆v = m (vfinal–vinitial)

pC

40000kgm/s

pT

20000kgm/s

Pay attention to the direction of vectors when adding/subtracting

Final

mCadillac+Toyota = 3000kg

Stuck together

Travelling to Right at 6.66m/s

pC+Tfinal

Forces, Momentum Practice Final 6

Initial

mToyota = 1000kg 20m/s to Left

mCadillac = 2000kg 20m/s to Right

Then find ∆p of each car from:

∆p = m ∆v = m (vfinal– vinitial)

pC

40000kgm/s

pT

20000kgm/s

Pay attention to the direction of vectors when adding/subtracting

Final

mCadillac+Toyota = 3000kg

Stuck together

Travelling to Right at 6.66m/s

∆pToyota comes out to be equal to ∆pCadillac even with its smaller mass because Toyota changes its direction after the collision

pC+Tfinal

Forces, Momentum Practice Final 6

Initial

mToyota = 1000kg 20m/s to Left

mCadillac = 2000kg 20m/s to Right

Then find ∆v of each car:

pC

40000kgm/s

pT

20000kgm/s

Pay attention to the direction of vectors when adding/subtracting

Final

mCadillac+Toyota = 3000kg

Stuck together

Travelling to Right at 6.66m/s

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

pC+Tfinal

Forces, Momentum Practice Final 6

Initial

mToyota = 1000kg 20m/s to Left

mCadillac = 2000kg 20m/s to Right

What about Fave experienced by each car during the collision?

pC

40000kgm/s

pT

SAME !

20000kgm/s

Final

One approach (Newton’s 1st law):

Fave ∆t = ∆p

mCadillac+Toyota = 3000kg

Stuck together

Travelling to Right at 6.66m/s

Alternative approach (Newton’s 3rd law):

FC on T = – FT on C !!!

pC+Tfinal

Fluids/Circuits

Basic rules for looking at fluids/circuits :

1. Energy (density) conservation

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

OR (the same as)

∆V = – IR

A1 v1 = A2 v2

2. Current entering = current leaving

Junction rule

3. Pressures where two fluids systems touch are equal

Voltages that are connected by wire (no circuit

element in between) are equal

P1 = 200kPa

v1 = 10m/s

Fluids, Circuit Practice Final 1

w = 1000kg/m3

A2 = 0.5A1

v1 vs v2 ??

P1 = 200kPa

v1 = 10m/s

Fluids, Circuit Practice Final 1

w = 1000kg/m3

A2 = 0.5A1

A1 v1 = A2 v2

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

P1 = 200kPa

v1 = 10m/s

Fluids, Circuit Practice Final 1

w = 1000kg/m3

v2 = 20m/s

Keep raising the end 2,

∆P + (1/2)∆(v2) + g∆h = 0

At what h, P2 is equal to zero?

P1 = 200kPa

v1 = 10m/s

Fluids, Circuit Practice Final 1

w = 1000kg/m3

v2 = 20m/s

∆P + (1/2)∆(v2) + g∆h = 0

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

h2 = 5m.

Fluids, Circuit Practice Final 2

Why does a hose have a nozzle at the end?

Fluids, Circuit, Forces Practice Final 2

v2

v1

v2 > v1

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.

Fluids, Circuit, Forces Practice Final 2

v2

v1

v2 > v1

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

Example of a problem that combines concepts from different models.

Fluids, Circuit, Forces Practice Final 2

0

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.

Fluids/Circuits

32

20

12

24

Oops… One more technique to remember :

Know how to find equivalent resistance,

this is an essential technique for analyzing circuits

32

20

12

24

Circuit

Practice Final 3

Req of the whole circuit = 40

∆V3 = 3V

VB ??

32

20

12

24

Circuit

Practice Final 3

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!

Req of the whole circuit = 40

∆V3 = 3V

VB ??

32

20

12

24

Circuit

Practice Final 3

20V

Close the switch… The circuit becomes simpler

What happens to:

∆V1 I2 ∆V3 ∆V4 ??

Two things that don’t change : VB , resistor values

Req decreases, and so Ieq increases

Exponential decay Practice Final 3

Circuit A and B are similar but slightly different RC circuits

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

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

Exponential decay Practice Final 3

What was Capacitor C again??

Capacitor:

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

Exponential decay

±

(Amount of stuff at t = 0 ) e ±kt

Stuff (t)

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 :

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

Exponential decay Practice Final 3

Cgarged to V

Uncharged

Vc = Ve ±(1/RC)t

C

Direction of

Good luck!

Rotation this way