kinematics in one dimension - urząd miasta...

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

lecture based on © 2016 Pearson Education, Ltd.

• 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

VECTORS

• A vector has magnitude as well as direction.

• Examples: displacement, velocity, acceleration,

force, momentum

• A scalar has only magnitude

• Examples: time, mass, temperature, energy

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 segment sense

D

VECTOR ADDITION – ONE DIMENSION

A person walks 8 km East

and then 6 km East.

Displacement =14 km East

A person walks 8 km East

and then 6 km West.

Displacement = 2 km East

VECTOR ADDITION

• A person walks 10 km East

and 5.0 km North

• Order doesn’t matter

21 DDDR

2

2

2

1 DDDR

kmkmkmDR 2.11)5()10( 22

RD

D2sin

0121 5.26)2.11

5(sin)(sin

km

km

D

D

R

GRAPHICAL METHOD OF VECTOR ADDITION TAIL TO TIP METHOD

1V

2V

3V

RV

GRAPHICAL METHOD OF VECTOR ADDITION TAIL TO TIP METHOD

1V

2V

3V

RV

1V

2V

3V

PARALLELOGRAM METHOD

SUBTRACTION OF VECTORS

• Negative of vector has

same magnitude but points

in the opposite direction

• For subtraction, we add the

negative vector.

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 direction

but a magnitude c•V.

• If c is negative, the resultant vector points in the

opposite direction.

ADDING VECTORS BY COMPONENTS

• Any vector can be expressed as the sum of two

other vectors, which are called its components.

Usually the other vectors are chosen so that they

are perpendicular to each other.

TRIGONOMETRY REVIEW

Opp

osi

te

Adjacent

Hypotenuse

Hypotenuse

Oppositesin

Hypotenuse

Adjacentcos

cos

sin

Adjacent

Oppositetan

ADDING VECTORS BY COMPONENTS

• If the components are perpendicular, they can be

found using trigonometric functions.

sinVVy

cosVVx

V

Vy

Hypotenuse

Oppositesin

V

VxHypotenuse

Adjacentcos

cos

sin

Adj

Opptan

ADDING VECTORS BY COMPONENTS

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 and

cosines.

5. Add the components in each direction.

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

V

Vysin

REFERENCE FRAMES

Any measurement of position, distance, or speed

must be made with respect to a reference frame.

REFERENCE FRAMES

Coordinate axes

REFERENCE FRAMES

COORDINATE SYSTEM

DISPLACEMENT

• distance

• displacement

• The displacement is written:

• Displacement is positive

• Δx=30m-10m=20m

• Displacement is negative

• Δx=10m-30m=-20m

DISPLACEMENT

VELOCITY

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

interval

• Velocity includes directional information:

VELOCITY AS A VECTOR

𝑣 =∆𝑥

∆𝑡

AVERAGE VELOCITY

𝑣 =𝑥2 − 𝑥1

𝑡2 − 𝑡1=

Δ𝑥

Δ𝑡

INSTANTANEOUS VELOCITY

constant velocity varying velocity

𝑣 = lim∆𝑡→0

Δ𝑥

Δ𝑡=

𝛿𝑥

𝛿𝑡

ACCELERATION

ACCELERATION

VECTOR

ACCELERATION AS A VECTOR

𝑎 =∆𝑣

∆𝑡

ACCELERATION

There is a difference between negative acceleration

and deceleration

ACCELERATION

Negative acceleration is acceleration in the negative

direction as defined by the coordinate system.

ACCELERATION

Deceleration occurs when the acceleration is

opposite in direction to the velocity.

ACCELERATION

The instantaneous acceleration is the average

acceleration, in the limit as the time interval becomes

infinitesimally short.

𝑎 = lim∆𝑡→0

Δ𝑣

Δ𝑡=

𝛿𝑣

𝛿𝑡=

𝛿2𝑥

𝛿𝑡

MOTION AT CONSTANT ACCELERATION

• The average velocity of an object during a time

interval t is

• The acceleration, assumed constant, is

MOTION AT CONSTANT ACCELERATION

• In addition, as the velocity is increasing at a

constant rate, we know that

• Combining these last three equations, we find:

MOTION AT CONSTANT ACCELERATION

• We can also combine these equations so as to

eliminate t:

• We now have all the equations we need to solve

constant-acceleration problems.

FALLING OBJECTS

The same acceleration

FALLING OBJECTS

FALLING OBJECTS

9.80 m/s2

FREE FALL—HOW FAST?

• The velocity acquired by an object

starting from rest is

• So, under free fall, when

acceleration 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

GRAPHICAL ANALYSIS OF LINEAR MOTION

constant velocity

GRAPHICAL ANALYSIS OF LINEAR MOTION

VARYING VELOCITY

GRAPHICAL ANALYSIS OF LINEAR MOTION

Displacement

Displacement

SUMMARY

• Kinematics is the description of how objects move

with respect to a defined reference frame.

• Displacement is the change in position of an object.

• Average speed is the distance traveled divided by

the time it took; average velocity is the

displacement divided by the time.

• Instantaneous velocity is the limit as the time

becomes infinitesimally short.

SUMMARY

• Average acceleration is the change in velocity

divided 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 of

which requires a different set of quantities.

• Objects falling (or having been projected) near the

surface of the Earth experience a gravitational

acceleration of 9.80 m/s2.

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

lecture based on © 2016 Pearson Education, Ltd.

PROJECTILE MOTION

• two dimensions

• parabola

PROJECTILE MOTION

ay=o

tgrounded=tdropped vertically

EQUATIONS FOR PROJECTILE MOTION

• Horizontal X Vertical Y

• ax=0 ay = - g

• vx= constant

00 xvv tgvv yy 0

2

002

1tgtvyy y tvxx x00

)(2 0

2

0

2 yygvv yy

INITIAL VELOCITY

• For y=0 𝑣0 = 𝑣

• v = vx0 = constans

sin00 vvy

cos00 vvx

sin00 vvy sin00 vvy

PROBLEM SOLVING—A GENERAL APPROACH

• 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.

EXAMPLE

• A football is kicked at an angle of 50˚ above the

horizontal 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=?

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 that the ball was kicked at ground level and lands at ground level.

cos0 vvx sin0 vvy

tgvv yy 0 0

g

vt

y

up

0

280.9

8.13

sms

m

s41.12

0maxmax2

1gttvyyH yo

2

0max

sin

2

1sin0

g

vg

g

vvH y

mH 7.9max

at top:

sms

m /6.11)50)(cos18(

sms

m /8.13)50)(sin18(

g

v sin

22max )41.1)(8.9(

2

1)41.1)(8.13(0 s

sms

smH

LEVEL HORIZONTAL RANGE

• Range is determined by time it takes for ball to return to

ground level or perhaps some other vertical value.

• If ball hits something a fixed distance away, then time is

determined 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

1tgtvtgtvyy yoy

g

vt

y02

g

v

g

vv

g

vvtvxR

yoxy

xox00

2

000

0

cossin22)

2(

LEVEL HORIZONTAL RANGE

• We can use a trig identity

• Greatest range: θ= 450

• θ = 300 and 600 have same

range.

2sincossin2

g

vR

2sin2

0)1545( 00

• Caution– the range formula has limited usefulness. It is

only valid when the projectile returns to the same

vertical position.

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 that the 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

tvxxR x00 )82.2()6.11(0 ss

m m33

g

vR

2sin2

0

msm

sm33

/8.9

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

02

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

on a rocky ridge 235 m below. If the plane is traveling horizontally with a 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

2

1

2

10235 tgtgmy

tvxx xo 0

g

yt

)()2(

mx

ssm

481

)93.6()/4.69(0

s93.6

2/8.9

)235()2(

sm

m

smvv

my

x /4.69

235

0

?x 00 yv

PROJECTILE MOTION IS PARABOLIC

• In order to demonstrate that

projectile motion is parabolic, the

book derives y as a function of x.

When we do, we find that it has

the form:

• This is the equation for a parabola

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

lecture based on © 2016 Pearson Education, Ltd.

• 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 FORCE

• Momentum is a vector symbolized by the symbol p,

and is defined as

• The rate of change of momentum is equal to the

net force:

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

CONSERVATION OF MOMENTUM

• During a collision, measurements show that the

TOTAL MOMENTUM DOES NOT CHANGE:

CONSERVATION OF MOMENTUM

More formally, the law of conservation of momentum

states:

• The total momentum of an isolated system of

objects remains constant.

CONSERVATION OF MOMENTUM EXPERIMENT

Momentum conservation works for a rocket as long

as we consider the rocket and its fuel to be one

system, and account for the mass loss of the rocket.

CONSERVATION OF MOMENTUM 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)

ELASTIC COLLISION: BILLIARD BALLS

• Whack stationary ball with identical ball moving at velocity vcue

• To conserve both energy and momentum, cue ball stops

dead, and 8-ball takes off with vcue • Momentum conservation: mvcue = mvcue, after + mv8-ball • Energy conservation: ½mv2

cue = ½mv2cue, after + ½mv2

8-ball

• The only way v0 = v1 + v2 and v20 = v2

1 + v22 is if either v1 or

v2 is 0. • Since cue ball can’t move through 8-ball, cue ball gets

stopped.

8

8

DESK TOY PHYSICS

• The same principle applies to the suspended-ball

desk toy, which eerily “knows” how many balls you

let go…

• Only way to simultaneously satisfy energy and

momentum conservation

• Relies on balls to all have same mass

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

COLLISIONS AND IMPULSE

• During a collision, objects are deformed due to the

large forces involved.

• Since , we can

• Write

• The definition of impulse:

COLLISIONS AND IMPULSE

• The impulse tells us that we

can get the same change

in momentum with a large

force acting for a short time,

or a small force acting for a

longer time.

• This is why you should bend

your knees when you land;

why airbags work; and why

landing on a pillow hurts less

than landing on concrete.

CONSERVATION OF ENERGY AND MOMENTUM IN COLLISIONS

• Momentum is conserved

in all collisions.

• Collisions in which kinetic

energy is conserved as

well are called elastic

collisions,

and those in which it is not

are called inelastic.

ELASTIC COLLISIONS IN ONE DIMENSION

• Here we have two objects

colliding elastically. We know

the masses and the initial

speeds.

• Since both momentum and

kinetic energy are conserved,

we can write two equations.

This allows us to solve for the

two unknown final speeds.

INELASTIC COLLISIONS

• With inelastic collisions, some of the

initial kinetic energy is lost to

thermal or potential energy. It may

also be gained during explosions,

as there is the addition of chemical

or nuclear energy.

• A completely inelastic collision is

one where the objects stick

together afterwards, so there is only

one final velocity.

COLLISIONS IN TWO OR THREE DIMENSIONS

• Conservation of energy and momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy.

• Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities

COLLISIONS IN TWO OR THREE DIMENSIONS PROBLEM SOLVING

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

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

• Draw diagrams of the initial and final situations, with momentum 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

• Momentum of an object:

• Newton’s second law:

• Total momentum of an isolated system of objects is

conserved.

• During a collision, the colliding objects can be

considered 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

• In an elastic collision, total kinetic energy is also

conserved.

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

• In a completely inelastic collision, the two objects

stick together after the collision.

•