1. kinematics-1

8/20/2019 1. Kinematics-1

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INTRODUCTION TO

BIOPHYSICS


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Welcome to the Cytobiology andProteomics Unitat the Department of Biomedical Sciences


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SEMINARS

• Compulsory

• In the case of absence from the seminars thestudent is obligated to justify it within a week.

• The student is entitled to TWO justified absences.Only sick leaves and official certificates arehonored.

• The exam consists of two parts


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1. Introduction prof. dr hab. C. Watała

2. Motion

3. Force dr M. Stasiak

4. Work and Energy/ Gravity

5. Solids and Fluids

6. Vibrations and Waves

7. Electric Fields8. Current

9. TEST middle

SEMINARS

dr R. Bednarek

dr M. Stasiak


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10.Magnetism

11.Properties of Light

12.Quantum Physics13.Atomic and Nuclear Physics

14.Thermodynamics

15.TEST

SEMINARS

dr K. Czarnecka


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TEST

With open questions

• Short answer

• Could be calculation

• Conversion of units

The first part of the exam is held 7th December 3 and4 groups (30 Nov- groups 8 and 9)

The second part – last classes (25th January 2016)

The second and third term exam will be during thesession


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• For more information visit our website:

http://zdn.umed.lodz.pl/~biophysics

(password: bFk4std)

Biophysic’s coordinator:(the teacher of the 4th module):Radoslaw Bednarek, Ph.D. (room 145)[email protected]


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D R M A R T A S T A S I A KD E P A R T M E N T O F C Y T O B I O L O G Y A N D P R O T E O M I C S

KINEMATICS

lecture based on © 2016 Pearson Education, Ltd.


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• Vectors

• Reference Frames

• Displacement

•

Average Velocity• Instantaneous Velocity

• Acceleration

• Motion at Constant Acceleration

•Solving Problems

• Falling Objects

• Graphical Analysis of Linear Motion

• Projectile Motion

•Linear Momentum


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VECTORS

• A vector has magnitude as well as direction.

• Examples: displacement, velocity, acceleration,force, momentum

• A scalar has only magnitude• Examples: time, mass, temperature, enargy


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VECTORS

• Magnitude

• Direction – the line of action (line segment) and

sense (orientation).• Origin (tail) of the vector - point of application,

initial point

A

B

Origin

line segmentsense

D


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VECTOR ADDITION – ONE DIMENSION

A person walks 8 km Eastand then 6 km East.

Displacement =14 km East

A person walks 8 km East

and then 6 km West.

Displacement = 2 km East


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VECTOR ADDITION

• A person walks 10 km Eastand 5.0 km North

•

Order doesn’t matter

21 D D D R

2

2

2

1 D D D R

kmkmkm D R 2.11)5()10( 22

R D D2sin

0121 5.26)2.11

5(sin)(sin

km

km

D

D

R


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GRAPHICAL METHOD OF VECTOR

ADDITION TAIL TO TIP METHOD

1V

2V

3V

RV


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GRAPHICAL METHOD OF VECTOR

ADDITION TAIL TO TIP METHOD

1V

2V

3V

RV

1V

2V

3V


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PARALLELOGRAM METHOD


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SUBTRACTION OF VECTORS

• Negative of vector hassame magnitude but pointsin the opposite direction

•

For subtraction, we add thenegative vector.


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MULTIPLICATION BY A SCALAR

• A vector V can be multiplied by a scalar c;the result is a vector c•V that has the same directionbut a magnitude c•V.

•

If c is negative, the resultant vector points in theopposite direction.


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ADDING VECTORS BY COMPONENTS

• Any vector can be expressed as the sum of twoother vectors, which are called its components.Usually the other vectors are chosen so that they

are perpendicular to each other.


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TRIGONOMETRY REVIEW

O p p o s i t e

Adjacent

Hypotenuse

Hypotenuse

Oppositesin

HypotenuseAdjacentcos

cos

sin

Adjacent

Oppositetan


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• If the components are perpendicular, they can befound using trigonometric functions.

sinV V y

cosV V x

V

V yHypotenuse

Oppositesin

VVx

Hypotenuse

Adjacentcos

cos

sin

Adj

Opptan


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Adding vectors:

1. Draw a diagram; add the vectors graphically.

2. Choose x and y axes.

3. Resolve each vector into x and y components.

4. Calculate each component using sines andcosines.

5. Add the components in each direction.

6. To find the length and direction of the vector, use:

V

V y sin


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REFERENCE FRAMES

Any measurement of position, distance, or speedmust be made with respect to a reference frame.


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REFERENCE FRAMES

Coordinate axes


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REFERENCE FRAMES

COORDINATE SYSTEM


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DISPLACEMENT

• distance

• displacement


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• The displacement is written:

• Displacement is positive

• Δ x=30m-10m=20m

• Displacement is negative

• Δ x=10m-30m=-20m

DISPLACEMENT


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VELOCITY

• Speed: how far an object travels in a given timeinterval

• Velocity includes directional information:


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VELOCITY

CHECK YOUR NEIGHBOR

The average speed of driving 30 km in 1 hour is thesame as the average speed of driving

A. 30 km in 1/2 hour.B. 30 km in 2 hours.

C. 60 km in 1/2 hour.

D. 60 km in 2 hours.


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VELOCITY AS A VECTOR

=∆

∆


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AVERAGE VELOCITY

=2−

1

2− 1=Δ

Δ


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INSTANTANEOUS VELOCITY

constant velocity varying velocity

= lim∆→

Δ

Δ=


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ACCELERATION


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ACCELERATION

VECTOR


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ACCELERATION AS A VECTOR

=∆

∆


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ACCELERATION

CHECK YOUR NEIGHBOR

An automobile is accelerating when it is

A. slowing down to a stop.

B. rounding a curve at a steady speed.C. Both of the above.

D. Neither of the above.


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ACCELERATION

CHECK YOUR ANSWER

An automobile is accelerating when it is

A. slowing down to a stop.

B. rounding a curve at a steady speed.C. Both of the above.

D. Neither of the above.

•

Explanation:• Change in speed (increase or decrease) per time is

acceleration, so slowing is acceleration.

• Change in direction is acceleration (even if speed staysthe same), so rounding a curve is acceleration.


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ACCELERATION

There is a difference between negative acceleration and deceleration


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ACCELERATION

Negative acceleration is acceleration in the negativedirection as defined by the coordinate system.


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ACCELERATION

Deceleration occurs when the acceleration isopposite in direction to the velocity.


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ACCELERATION

The instantaneous acceleration is the averageacceleration, in the limit as the time interval becomesinfinitesimally short.

= lim∆→

Δ

Δ=

=


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MOTION AT CONSTANT

ACCELERATION

• The average velocity of an object during a timeinterval t is

• The acceleration, assumed constant, is


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MOTION AT CONSTANT

ACCELERATION

• In addition, as the velocity is increasing at aconstant rate, we know that

• Combining these last three equations, we find:


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FALLING OBJECTS

The same acceleration


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FALLING OBJECTS


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FALLING OBJECTS

9.80 m/s2


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FREE FALL—HOW FAST?

• The velocity acquired by an objectstarting from rest is

• So, under free fall, whenacceleration is 10 m/s2, the speed is

• 10 m/s after 1 s.

•

20 m/s after 2 s.• 30 m/s after 3 s.

And so on.

© 2015 Pearson Education, Inc.

Velocity = acceleration ´ time


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FREE FALL—HOW FAST?

CHECK YOUR NEIGHBOR

At a particular instant a free-falling object has aspeed of 30 m/s. Exactly 1 s later its speed will be

A. the same.B. 35 m/s.

C. more than 35 m/s.

D. 60 m/s.


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GRAPHICAL ANALYSIS OF LINEAR

MOTION

constant velocity


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MOTION

VARYING VELOCITY


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MOTION

Displacement

Displacement


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SUMMARY

• Kinematics is the description of how objects movewith respect to a defined reference frame.

• Displacement is the change in position of an object.

•

Average speed is the distance traveled divided bythe time it took; average velocity is thedisplacement divided by the time.

• Instantaneous velocity is the limit as the time

becomes infinitesimally short.


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SUMMARY

• Average acceleration is the change in velocitydivided by the time.

• Instantaneous acceleration is the limit as the time

interval becomes infinitesimally small.• The equations of motion for constant acceleration

are given in the text; there are four, each one ofwhich requires a different set of quantities.

•

Objects falling (or having been projected) near thesurface of the Earth experience a gravitationalacceleration of 9.80 m/s2.


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D R M A R T A S T A S I A K

D E P A R T M E N T O F C Y T O B I O L O G Y A N D P R O T E O M I C S

KINEMATICS IN TWO

DIMENSION



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PROJECTILE MOTION

• two dimensions

• parabola


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PROJECTILE MOTION

ay=o

tgrounded=tdropped vertically


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EQUATIONS FOR PROJECTILE MOTION

• Horizontal X Vertical Y

• a x=0 ay = - g

• v x= constant

00 xvv t g vv y y 0

2

002

1t g t v y y y t v x x x00

)(2 02

0

2 y y g vv

y y


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INITIAL VELOCITY

• For y=0 0 =

• v = v x0 = constans

sin00 vv y

cos00 vv x

sin00 vv y sin00 vv y


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PROBLEM SOLVING A GENERAL


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PROBLEM SOLVING—A GENERALAPPROACH

• Read the problem carefully; then read it again.

• Draw a sketch, and then a free-body diagram.

• Choose a convenient coordinate system.

•

List the known and unknown quantities• Find relationships between the knowns and the unknowns.

• Estimate the answer.

• Solve the problem without putting in any numbers(algebraically); once you are satisfied, put the numbers in.

• Keep track of dimensions.

• Make sure your answer is reasonable.


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EXAMPLE

• A football is kicked at an angle of 50˚ above thehorizontal with a velocity of 18 m/s.

• Calculate the maximum height and the range

as well as how long it is in the air• Assume that the ball was kicked at ground level

and lands at ground level.

Θ=50˚ V0=18m/s

Hmax, R , t=?


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EXAMPLE• A football is kicked at an angle of 50˚ above the horizontal with a

velocity of 18.0 m / s. Calculate the maximum height. Assume thatthe ball was kicked at ground level and lands at ground level.

cos0 vv x sin0 vv y

t g vv y y 0 0

g

vt

y

up

0

280.9

8.13

sm s

m

s41.12

0maxmax2

1 gt t v y y H yo

2

0maxsin

21sin0

g v g

g vv H y

m H 7.9max

at top:

sm s

m /6.11)50)(cos18(

sm s

m /8.13)50)(sin18(

g

v sin

22max )41.1)(8.9(

2

1)41.1)(8.13(0 s

sm s

sm H


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LEVEL HORIZONTAL RANGE

• Range is determined by time it takes for ball to return toground level or perhaps some other vertical value.

• If ball hits something a fixed distance away, then time isdetermined by x motion

• If the motion is on a level field, when it hits: y = 0

• Solving we find

• We can substitute this in the x equation to find the range R

22

002

100

2

1t g t vt g t v y y yo y

g

vt

y02

g

v

g

vv

g

vvt v x R

yo x y

xo x

00

2

000

0

cossin22)

2(


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LEVEL HORIZONTAL RANGE

• We can use a trig identity

• Greatest range: θ= 450

• θ = 300 and 600 have same

range.

2sincossin2

g

v R

2sin20 )1545( 00

• Caution – the range formula has limited usefulness. It isonly valid when the projectile returns to the samevertical position.


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EXAMPLE• A football is kicked at an angle of 50˚ above the horizontal with a

velocity of 18.0 m / s. Calculate the maximum height. Assume thatthe ball was kicked at ground level and lands at ground level.

• Assume time down = time up

•

For Range:

• Could also use range formula

)41.1)(2( st s82.2

t v x x R x00 )82.2()6.11(0 s sm m33

g

v R

2sin20 m sm

sm33

/8.9

)50()2(sin)/18(2

02


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EXAMPLE-VERTICAL PROJECTIONA rescue plane wants to drop supplies to isolated mountain climbers

on a rocky ridge 235 m below. If the plane is traveling horizontally witha speed of 250 km/h (69.4 m/s) how far in advance of the recipients

(horizontal distance) must the goods be dropped?

Coordinate system is 235 m below plane

22

21

210235 t g t g m y

t v x x xo 0

g

yt

)()2(

m x

sm

481

93.6()/4.69(0

s93.6

2/8.9

)235()2(

sm

m

smvv

m y

x /4.69

235

0

? x 00 yv


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PROJECTILE MOTION IS PARABOLIC

• In order to demonstrate thatprojectile motion is parabolic, thebook derives y as a function of x.When we do, we find that it has

the form:

• This is the equation for a parabola

PROJECTILE MOTION


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PROJECTILE MOTION

CHECK YOUR NEIGHBOR

The velocity of a typical projectile can berepresented by horizontal and vertical components.Assuming negligible air resistance, the horizontalcomponent along the path of the projectile

A. increases.

B. decreases.

C. remains the same.

D. Not enough information.

PROJECTILE MOTION


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PROJECTILE MOTION

CHECK YOUR ANSWER

The velocity of a typical projectile can berepresented by horizontal and vertical components.Assuming negligible air resistance, the horizontalcomponent along the path of the projectile

A. increases.

B. decreases.

C. remains the same.

D. Not enough information.Explanation:

Since there is no force horizontally, no horizontalacceleration occurs.

PROJECTILE MOTION


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When no air resistance acts on a fast-movingbaseball, its acceleration is

A. downward, g.

B. a combination of constant horizontal motion andaccelerated downward motion.

C. opposite to the force of gravity.

D. centripetal.

PROJECTILE MOTION

CHECK YOUR NEIGHBOR

PROJECTILE MOTION


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PROJECTILE MOTION

CHECK YOUR ANSWER

When no air resistance acts on a fast-movingbaseball, its acceleration is

A. downward, g.

B. a combination of constant horizontal motion andaccelerated downward motion.

C. opposite to the force of gravity.

D. centripetal.

PROJECTILE MOTION


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Neglecting air drag, a ball tossed at an angle of 30°with the horizontal will go as far downrange as onethat is tossed at the same speed at an angle of

A. 45°.

B. 60°.

C. 75°.

D. None of the above.

PROJECTILE MOTION

CHECK YOUR NEIGHBOR

PROJECTILE MOTION


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PROJECTILE MOTION

CHECK YOUR ANSWER

Neglecting air drag, a ball tossed at an angle of 30° with the horizontal will go as far downrange as onethat is tossed at the same speed at an angle of

A. 45°.

B. 60 °.

C. 75°.

D. None of the above.

Explanation:

Same initial-speed projectiles have the same range when their

launching angles add up to 90°. Why this is true involves a bit of

trigonometry — which, in the interest of time, we'll not pursue here.


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D R M A R T A S T A S I A K

D E P A R T M E N T O F C Y T O B I O L O G Y A N D P R O T E O M I C S

LINEAR MOMENTUM



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• Momentum and Its Relation to Force

• Conservation of Momentum

• Collisions and Impulse

• Conservation of Energy and Momentum in Collisions

• Elastic Collisions in One Dimension

• Inelastic Collisions

• Collisions in Two or Three Dimensions

• Center of Mass (CM)

• CM for the Human Body

• Center of Mass and Translational Motion

MOMENTUM AND ITS RELATION TO


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MOMENTUM AND ITS RELATION TO

FORCE

• Momentum is a vector symbolized by the symbol p,and is defined as

•

The rate of change of momentum is equal to thenet force:

• This can be shown using Newton’s second law.


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CONSERVATION OF MOMENTUM

More formally, the law of conservation of momentumstates:

• The total momentum of an isolated system ofobjects remains constant.



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EXPERIMENT

Momentum conservation works for a rocket as longas we consider the rocket and its fuel to be onesystem, and account for the mass loss of the rocket.



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EXPERIMENT

A Wad of Clay Hits Unsuspecting Sled

1 kg clay ball strikes 5 kg sled at 12 m/s and sticks

Momentum before collision:

(1 kg)(12 m/s) + (5 kg)(0 m/s)

Momentum after

= 12 kg·m/s → (6 kg)·(2 m/s)

MOMENTUM


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MOMENTUM

CHECK YOUR NEIGHBOR

When the speed of an object is doubled, itsmomentum

A. remains unchanged in accord with theconservation of momentum.

B. doubles.

C. quadruples.

D. decreases.

MOMENTUM


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MOMENTUM

CHECK YOUR NEIGHBOR

When the speed of an object is doubled, itsmomentum

A. remains unchanged in accord with theconservation of momentum.

B. doubles.

C. quadruples.

D. decreases.

IMPULSE CHANGES MOMENTUM


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CHECK YOUR NEIGHBOR

When the force that produces an impulse acts fortwice as much time, the impulse is

A. not changed.B. doubled.

C. quadrupled.

D. halved.



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CHECK YOUR NEIGHBOR

When the force that produces an impulse acts fortwice as much time, the impulse is

A. not changed.B. doubled.

C. quadrupled.

D. halved.


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DESK TOY PHYSICS

• The same principle applies to the suspended-balldesk toy, which eerily “knows” how many balls youlet go…

• Only way to simultaneously satisfy energy and

momentum conservation

• Relies on balls to all have same mass


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INELASTIC COLLISION

• Energy not conserved (absorbed into other paths)

• Non-bouncy: hacky sack, velcro ball, ball of clay

Momentum before = m1vinitial

Momentum after = (m1 + m2)vfinal = m1vinitial (because conserved)

Energy before = ½m1v2

initial Energy after = ½ (m1 + m2)v

2final + heat energy

CO S O S A P S


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COLLISIONS AND IMPULSE

• During a collision, objects are deformed due to thelarge forces involved.

• Since , we can

• Write

• The definition of impulse:



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• The impulse tells us that wecan get the same changein momentum with a largeforce acting for a short time,

or a small force acting for alonger time.

• This is why you should bendyour knees when you land;why airbags work; and why

landing on a pillow hurts lessthan landing on concrete.

CONSERVATION OF ENERGY AND


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CONSERVATION OF ENERGY AND

MOMENTUM IN COLLISIONS

• Momentum is conservedin all collisions.

• Collisions in which kineticenergy is conserved aswell are called elasticcollisions,and those in which it is notare called inelastic.

ELASTIC COLLISIONS IN ONE


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ELASTIC COLLISIONS IN ONE

DIMENSION

• Here we have two objectscolliding elastically. We knowthe masses and the initialspeeds.

• Since both momentum andkinetic energy are conserved,we can write two equations.This allows us to solve for the

two unknown final speeds.

INELASTIC COLLISIONS


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INELASTIC COLLISIONS

• With inelastic collisions, some of theinitial kinetic energy is lost tothermal or potential energy. It mayalso be gained during explosions,

as there is the addition of chemical or nuclear energy.

• A completely inelastic collision isone where the objects stick

together afterwards, so there is onlyone final velocity.

COLLISIONS IN TWO OR THREE


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DIMENSIONS

• Conservation of energyand momentum can alsobe used to analyzecollisions in two or threedimensions, but unless the

situation is very simple,the math quicklybecomes unwieldy.

• Here, a moving objectcollides with an object

initially at rest. Knowingthe masses and initialvelocities is not enough;we need to know theangles as well in order to

find the final velocities



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DIMENSIONS PROBLEM SOLVING

• Choose the system. If it is complex, subsystems may bechosen where one or more conservation laws apply.

• Is there an external force? If so, is the collision time shortenough that you can ignore it?

• Draw diagrams of the initial and final situations, withmomentum vectors labeled.

• Choose a coordinate system.

• Apply momentum conservation; there will be one

equation for each dimension.• If the collision is elastic, apply conservation of kinetic

energy as well.

• Solve.

• Check units and magnitudes of result.

SUMMARY


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SUMMARY

• Momentum of an object:

• Newton’s second law:

•

Total momentum of an isolated system of objects isconserved.

• During a collision, the colliding objects can beconsidered to be an isolated system even if external

forces exist, as long as they are not too large.• Momentum will therefore be conserved during

collisions.

SUMMARY


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SUMMARY

• In an elastic collision, total kinetic energy is alsoconserved.

• In an inelastic collision, some kinetic energy is lost.

• In a completely inelastic collision, the two objectsstick together after the collision.

•

1. kinematics-1

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