mechanics lecture

7/23/2019 mechanics lecture

http://slidepdf.com/reader/full/mechanics-lecture 1/27

HKPhO Pre-Training 2005

Cover Page

The materials in this note serve as a guideline for the mechanics topics of HKPhO in addition

to the S4-5 syllaus before 200!" #ideos of the lecture and the tutorial $ill e availale in

http%&&h'pho"phys"ust"h'&" (hile the 'ey issues are addressed e)plicitly* details of secondary

importance are mostly left out" Students are encouraged to spend additional time to further digest the contents" Small gaps are left for the students to $or' out the details* and learn

physics through the process" +dditional references should e sought to further understand thetopics li'e vectors* elementary calculus* etc"* most of $hich can e found in relevant ,-.

te)too's"

(e hope that the pre-training $ill give a 'ic' start for those $ho hope to do $ell in HKPhO

200 and eyond" (e also encourage the trainees to share the materials $ith their classmates

ac' in schools"

/" angHKPhO 1ommittee 1hairman

http://hkpho.phys.ust.hk/

http://hkpho.phys.ust.hk/




Mechanics

1 Vector

"

+ny vector 000 z A y A x A A z y x ++= 3"

+ddition of t$o vectors%

000 333 z B A y B A x B A B AC z z y y x x +++++=+= 3"2

ot product of t$o vectors

z z y y x x B A B A B A AB B A ++==• θ cos 3"!

6t is also referred to as the pro7ection of A on B * or vise versa"

Amplitude of the vector

A A A A A A z y x •=++≡ 222

88 3"4

"2Position vector of a particle% 000 z z y y x xr ++= 3"5"

6f the particle is moving* then x, y, and z are function of time t "

#elocity%

000000 z v yv xv z

dt

dz y

dt

dy x

dt

dx

dt

r d v

z y x

++=++=≡ 3"

+cceleration 00002

2

02

2

02

2

000 z a ya xa z dt

z d y

dt

yd x

dt

xd z

dt

dv y

dt

dv x

dt

dv

dt

vd a z y x

z y x ++=++=++=≡

3".

"! 9niform circular motion

Ta'e the circle in the :- plane* so z ; 0*cos3 Φ+= t R x ω * sin3 Φ+= t R y ω 3"<

ω is the angular speed" Φ is the initial phase" =oth are constants"

9sing the aove definition of velocity 3"*

00 cos3sin3 yt R xt Rv Φ++Φ+−≡ ω ω ω ω 3">*

v is al$ays perpendicular to r "

6ts amplitude is ω Rv = 3"0

The acceleration is%

r yt R xt Ra 2

0

2

0

2cos3sin3 ω ω ω ω ω −=Φ+−Φ+−≡ 3""

6ts amplitude is R

v Ra

22 == ω 3"2

2 Relative Motion

2




+ reference frame is needed to descrie any motion of an o7ect"

1onsider t$o such reference frames S and S? $ith their origins atO and O?* respectively" The :--/ a)es in S are parallel to the :?-

?-/? a)es in S?" One is moving relative the other"

@ote% Rr r += A " 32"So the velocity is% uv

dt

Rd

dt

r d

dt

r d v

A

A+=+== 32"2

Similar for acceleration% Aaa += A 32"!

This is the classic theory of relativity" 6f u is constant* then Aaa = * i" e"* @e$ton?s Ba$s $or' in all inertia reference frames"

Properly choosing a reference frame can sometimes greatly simplify the prolems"

! Forces

Pull through a rope* push* contact forces 3elastic force and friction force* air resistance* fluid

viscosity* surface tension of liCuid and elastic memrane* gravity, electric and magnetic, strong interaction, weak interaction" Only the last four are fundamental" +ll the others are the

net effect of the electric and magnetic force"

!" Tension

Pulling force 3tension in a thin and light rope%

T$o forces* one on each end* act along the rope direction" Thet$o forces are of eCual amplitude and opposite signs ecause

the rope is massless" 6t is also true for massless stic's"

!"2 Dlastics

Dlastic contact forces are due to the deformation of solids"9sually the deformation is so small that it is not noticed" The

contact force is al$ays perpendicular to the contact surface" 6n

the e)ample oth 2 F and 2

F are pointing at the center of the

sphere"

!"! ,rictionThe friction force et$een t$o contact surfaces is caused y the relative motion or the

tendency of relative motion" 6ts amplitude is proportional to the elastic contact force* so a

friction coefficient µ can e defined"

(hen there is relative motion* the friction force is given y f ;µ' N * $here µ' is the 'inetic friction coefficient" The direction

of the friction is al$ays opposite to the direction of therelative motion"

(hen there is no relative motion ut a tendency for suchmotion* li'e a loc' is eing pushed y a force F * the

amplitude of f is eCual to F until it reaches the limiting value

!




of µs N if F 'eeps increasing* $here µs is the static friction

coefficient"

Once the loc' starts to move* the friction ecomes f ; µ' N " 9sually µ' < µs"

!"4 #iscosity

+ir resistance and fluid viscous forces are proportional to the relative motion speed and thecontact area" + coefficient called viscosity γ is used in these cases"

!"5 6nertial force

6nertial force is a Efa'e? force $hich is present in a reference frame 3say S? $hich itself is

accelerating" Fecall that Aaa += A " +ssume frame-S is not accelerating* then according to

@e$ton?s Second Ba$* 4A3 Aamam F +== " So in the S?-frame* if one $ants to correctly

apply @e$ton?s Ba$* she $ill get Aam Am F =− * i" e"* there seems to e an additional force Am F −=int 3!"

acting upon the o7ect"

Example 3.1

+ loc' is attached y a spring to the $all and placed on thesmooth surface of a cart $hich is accelerating" +ccording to the

ground frame* the force F acting on the loc' y the spring is

'eeping the loc' accelerating $ith the cart* so am F = " 6n thereference frame on the cart* one sees the loc' at rest ut there

is a force on the loc' y the spring" This force is Ealanced? y

the inertial force am F −=int

!" Gravity

=et$een t$o point masses M and m* the force is

r r

Mm F

2−= 3!"2

The gravitation field due to M is r r

M g

2−= 3!"!"

The field due to a sphere at any position outside the

sphere is eCual to that as if all the mass is concentrated at

the sphere center" 3@e$ton spent nearly 0 years trying to

proof it"Potential energy

r

M !

−= 3!"4"

+pplication of superposition% 9niform density

larger sphere $ith a smaller spherical hole"

4




@ear Darth surface* ecause the large radius

of Darth* the gravitation field of Darth can e

ta'en as constant* and its amplitude is

2

"

"

R

M g = ; >"< m&s2 3!"5"

6ts direction is pointing to$ards the center of Darth* $hich in practice can e regarded as

Edo$n$ards? in most cases"

Example 3.2

The E$eight?* or the force of the ground on a person ondifferent places on Darth" Ta'e the radius of Darth as R* and

the rotation speed eing ω 3; 2π&<400 s-"

On the DCuator* $e have mg # N $ ma $ mω % R*

so N $ mg # mω % R $ m&g # ω % R'

+t the South Pole 3or @orth Pole* $e have N $ mg

+t latitude θ * y rea'ing do$n the forces along

the :-direction and -direction* $e have

ma f N mg =−− θ θ sincos3 * and0cossin3 =+− θ θ f N mg "

Solving the t$o eCuations* $e get θ cos3 ma N mg =− * θ sinma f −= * $here θ ω cos2 Ra = "

The negative sign of f means that its direction is the opposite of $hat $e have guessed"

One can also rea' do$n the forces along the tangential and radial directions to otain the

same ans$ers" One can also ta'e Darth as reference frame and introduce the inertia force toaccount for the rotational acceleration"

!". =uoyancy

6n a fluid 3liCuid or gas of mass density ρ at depth ( *

consider a column of it $ith cross section area A* then the

total mass of the column is ρ A( * and the gravity on it is

ρ A(g " The gravity must e alanced y the supporting forcefrom elo$* so the force of the column on the rest of the

liCuid is F ; ρ A(g and pointing do$n$ards" The pressure

) $ F*A $ ρ (g 3!""

@o$ consider a very small cuic of fluid $ith all si) side area of A at depth ( " The force on

its upper surface is ρ A(g and pointing do$n* the force on its lo$er surface is ρ A(g ut

5




pointing up$ards so the cuic is at rest" Ho$ever* for the cuic not to e deformed y the t$o

forces on its upper and lo$er surfaces* the forces on its side surfaces must e of the same

magnitude" This leads to the conclusion that the pressure on any surface at depth ( is ρ A(g *

and its direction is perpendicular to the surface" One can then easily prove that the net force of

the fluid 3buoyancy on a sumerged ody of volume + is eCual ρ +g " 3See the HKPhO 200!

paper" The uoyancy force is acting on the center of mass of the submerged ortion of theo7ect"

!"< TorCue

(hen t$o forces of eCual amplitude and opposite directions

acting upon the t$o ends of a rod* the center of the rod remains

stationary ut the rod $ill spin around the center" The torCue 3ofa force is introduced to descrie its effect on the rotational

motion of the o7ect upon $hich the force is acting" ,irst* an

origin 3pivot point O should e chosen" The amplitude of the

torCue of force F is

τ $ rF 3!".*

$here r is the distance et$een F and the origin O" The

direction of the torCue 3a vector as $ell is point out of the paper

surface using the right hand rule" -ne can c.oose any oint as

origin* so the torCue of a force depends on the choice of origin"

Ho$ever* for t$o forces of eCual amplitude and opposite

directions* the total torCue is independent of the origin"

The general form of torCue is defined as

F r ×≡τ 3!"<*

$hich involves the cross-product of t$o vectors"

4 scillations

4" Simple Harmonic otion




,reCuency f and Period /0

f /

=

4cos343 φ ω += t xt xm

34"

3a Dffects of different amplitudes

3 Dffects of different periods

3c Dffects of different phases

Since the motion returns to its initialvalue after one period / *

"2π ω =/

.




Thus "22

f /

π π

ω == 34"2

Velocit!

I*cos3J3 φ ω +== t xdt

d

dt

dxt v

m34"!a

4I"sin343 φ ω ω +−= t xt vm

34"!

#elocity amplitude% mm xv ω = 34"4"

Acceleration

I*sin3J3 φ ω ω +−== t xdt

d

dt

dvt a

m

"4Icos343 2 φ ω ω +−= t xt am

34"5

+cceleration amplitudemm

xa 2ω = 34""

This eCuation of motion $ill e very useful in identifying simple harmonic motion and itsfreCuency"

4"2 The ,orce Ba$ for Simple Harmonic otion

1onsider the simple harmonic motion of a loc' of mass m su7ect to the elastic force of a

spring

kx F −= 3Hoo'?s Ba$ 34"."

@e$ton?s la$%

<




"makx F =−=

"02

2

=+ kxdt

xd

m

"02

2

=+ xm

k

dt

xd 34"<

1omparing $ith the eCuation of motion for simple harmonic motion*

ω 2 = k

m" 34">

Simple harmonic motion is the motion e)ecuted y an o7ect of mass m su7ect to a force that

is proportional to the displacement of the o7ect ut opposite in sign"

+ngular freCuency%

m

k =ω 34"0

Period%

k

m

/ π 2=34"

Examples 4.1

+ loc' $hose mass m is <0 g is fastened to a spring $hose spring constant k is 5 @m-" The

loc' is pulled a distance x ; cm from its eCuilirium position at x ; 0 on a frictionlesssurface and released from rest at t ; 0"

3a (hat force does the spring e)ert on the loc' 7ust efore the loc' is released

3 (hat are the angular freCuency* the freCuency* and the period of the resulting oscillation3c (hat is the amplitude of the oscillation

3d (hat is the ma)imum speed of the oscillating loc'3e (hat is the magnitude of the ma)imum acceleration of the loc'

3f (hat is the phase constant φ for the motion

+ns$ers%

3a @2"."0-5 −=×−=−= kx F

>




3 -srad.<">

<"0

5 ===m

k ω

HL5"2

==π

ω f

s4!"0==

f

/

3c cm=m x

3d -

ms0<"== mm xv ω

3e 2-22 ms0"50".<"> =×== mm xa ω

3f +t t ; 0* "0cos03 == φ m x x 3

0sin03 =−= φ ω m xv 32

32% 0sin =φ ⇒ 0=φ

Example 4.2

+t t $ 0* the displacement of x30 of the loc' in a linear oscillator is M <"50 cm" 6ts velocityv30 then is M 0">20 ms-* and its acceleration a30 is 4."0 ms-2"

3a (hat are the angular freCuency ω and the freCuency f of this system

3 (hat is the phase constant φ

3c (hat is the amplitude xm of the motion

3a +t t ; 0*

"0<5"0coscos33 −==+= φ φ ω mm xt xt x 3

">20"0sinsin33 −=−=+−= φ ω φ ω ω mm xt xt v 32

"0"4.coscos33 22

+=−=+−= φ ω φ ω ω mm xt xt a 3!

3! ÷ 3% "03

03 2ω −= x

a

"srad5"2!0<50"0

0"4.

03

03 -=−

−=−= x

aω 3ans$er

3 32 ÷ 3% "tancos

sin

03

03φ ω

φ

φ ω −=−=

x

v

"40!"00<5"035"2!3

>20"0

03

03tan −=

−−−=−=

x

v

ω φ

φ ; M24".o or <0o M 24".o ; 55o"

One of these 2 ans$ers $ill e chosen in 3c"

0




3c 3% "cos

03

φ

x xm =

,or φ ; M24".o* "cm4>m

."24cos3

0<5"0 1 x

om −=−

−=

,or φ ; 55o* "cm4>m 55cos

0<5"0 1 x

om =−

=

Since xm is positive* φ ; 55o and xm ; >"4 cm"

3ans$er

Example 4.3

+ uniform ar $ith mass m lies symmetrically across t$o rapidly rotating* fi)ed rollers* A and

B* $ith distance 2 ; 2"0 cm et$een the ar?s centre of mass and each roller" The rollers slip

against the ar $ith coefficient of 'inetic friction µ k

; 0"40" Suppose the ar is displacedhoriLontally y a distance x* and then released" (hat is the angular freCuency ω of the

resulting horiLontal simple harmonic 3ac' and forth motion of the ar

@e$ton?s la$%

"0=−+=∑ mg F F F B A y 3

"makB f kA f x F =∑ += 32

or "ma

B

F

k A

F

k

=− µ µ

1onsidering torCues aout +*

"000320 =⋅+⋅++−⋅+⋅=∑kB f

kA f x 2mg 2

B F

A F

z τ

3!

or "32 x 2mg 2 F B

+=⋅

3!% "2

3

2

x 2mg F B

+=

3% "2

3

2

x 2mg F mg F B A

−=−=

32% "I33J2

ma x 2mg x 2mg 2

k =−−+ µ




"0=+ x 2

k g

a µ

1omparing $ith 02

2

2

=+ xdt

xd ω for simple harmonic motion* 2

g k µ ω =

2

*

"srad402"0

<">340"03 -=== 2

g k µ ω 3ans$er

4"! Dnergy in Simple Harmonic otion

Potential energy%

Since *cos33 φ ω += t xt xm

! t kx kx t m

3 "= = +

2

2

2

2 cos 32 ω φ 34"2

Kinetic energy%

Since *sin33 φ ω ω +−= t xt vm

"3sin3 222

2

2

2

φ ω ω +== t xmmvt 3

m34"!

Since ω 2 = k m& *

3 t kx t m

3 "= +

2

2 sin 32 ω φ

echanical energy%

" ! 3 kx t kx t

kx t t

m m

m

= + = + + +

= + + +

2

2

2

2

2

2

cos 3 sin 3

Jcos 3 sin 3

2 2

2 2

ω φ ω φ

ω φ ω φ

I"

2




Since cos 3 sin 32 2ω φ ω φ t t + + + = *

" ! 3 kxm

= + =

2

2 " 34"4

3a The potential energy ! 3t * 'inetic energy 3 3t * and mechanical energy " as functions of

time* for a linear harmonic oscillator" @ote that all energies are positive and that the potential

energy and 'inetic energy pea' t$ice during every period" 3 The potential energy ! 3t *'inetic energy 3 3t * and mechanical energy " as functions of position* for a linear harmonic

oscillator $ith amplitude 4 m" ,or x ; 0 the energy is all 'inetic* and for x ; ± 4 m it is all

potential"

The mechanical energy is conserved"

4.4 "he #imple Pen$%l%m

1onsider the tangential motion acting on the mass"9sing @e$ton?s la$ of motion*

*sin2

2

dt

d m2mamg

θ θ ==−

d

dt

g

2

2

2

0θ

θ + =sin " 34"4

(hen the pendulum s$ings through a small angle*

sinθ θ ≈ " Therefore

d

dt

g

2

2

2 0

θ θ + = " 34"5

1omparing $ith the eCuation of motion for simple harmonic motion*

2 g =2ω and

/ 2

g = 2π "

& "he Centre o' Mass

!




5" efinition

The centre of mass of a ody or a system of

odies is t.e oint that moves as though all of

the mass $ere concentrated there and alle)ternal forces $ere applied there"

,or t$o particles*

"22

2

22

M

xm xm

mm

xm xm x

cm

+=

++

=

,or n particles*

"

M

xm xm x nn

cm

++=

6n general and in vector form*

"

∑==

n

i

cm ir im

M

r

35"

6f the particles are in a uniform gravity field* then the

total torCue relative to the center of mass is Lero" Thesame applies for the inertia force $hen the particles

are in an accelerating reference frame"

Proof%

0333

=×−=×∑−∑=×∑ −====

g r M r M g mr r m g r r m cmcmcm

n

ii

n

iii

n

icmii

τ

5"2 Figid =odies

4




∫ ∫

∫

∫ ∫ d+ z M

dm z M

z

d+ y M

dm y M

y

d+ x M

dm x M

x

cm

cm

cm

22

*22

*22

ρ

ρ

ρ

==

⌡⌠ ==

==

35"2

$here ρ is the mass density"

6f the o7ect has uniform density*

"+

M

d+

dm== ρ 35"!

Fe$riting d+ dm ρ = and + m ρ = * $e otain

∫

∫

∫ "2

*2

*2

d+ z

+

z

d+ y+

y

d+ x+

x

cm

cm

cm

=

=

=

35"4

Similar to a system of particles* if the rigid ody is in a uniform gravity field* then the total

torCue relative to its center of mass is Lero" This is true even $hen the density of the o7ect isnon-uniform" The same applies to the inertia force" The proof is very much the same as in the

case for particles" One only needs to replace the summation y integration operations"

5"! @e$ton?s Second Ba$ for a System of Particles

6n terms of :--/ components*

"

*

*

**

**

**

z cm z ext

ycm yext

xcm xext

Ma F

Ma F

Ma F

=∑

=∑

=∑

35"5

5




5"4 Binear omentum

,or a single particle* the linear momentum is

"vm , = 35"

@e$ton?s Ba$%

"3

dt

,d vm

dt

d

dt

vd mam F

===

∑ = 35".

This is the most general form of @e$ton?s Second Ba$" 6t accounts for the change of mass as$ell"

"i f

, ,dt F 5 −=∑= ∫ 35"<

The change of momentum is eCual to the time integration of the force* or imulse"

,or a system of particles* the total linear momentum is

"222 nnn vmvm , , ) ++=++= 35">

ifferentiating the position of the centre of mass*

" nncm vmvmv M ++=




"cm

v M ) = 35"0

The linear momentum of a system of particles is eCual to the product of the total mass M of

the system and the velocity of the centre of mass"

+pply @e$ton?s Ba$?s to the particle system*

*3 ∑+=≠i 6

i6

ext

iii F F t am

∑

∑+∑ ==⇒

∑=⇒

∑ ==∑=

≠i i 6

i6

ext

i

i

ii

iii

ii

iii

F F

M

am

M

A

vm M

+

masstotal m M xm M

4

3*

+ccording to the Third Ba$* 6ii6 F F −= " So

0=∑≠i 6

i6 F 35"

and

M

F F

M A

ext

tot

i

ext

i

=

+∑= 03

35"2

@e$ton?s la$%

"cm

a M dt

cmvd M

dt

) d == 35"!

Hence

"

dt

) d F ext

∑ = 35"4

6f 0∑ =ext F * then const vm+ M i

ii =∑= 35"5

The total momentum of a system is conserved if the total e)ternal force is Lero"

5"5 Figid ody at rest

.




The necessary and sufficient conditions for the alance of a rigid ody is that net e)ternal

force ; 0* and the net torCue due to these e)ternal forces ; 0* relative to any origin 3pivot"

1hoosing an appropriate origin can sometimes greatly simplify the prolems" + common tric' is choosing the origin at the point $here an un'no$n e)ternal force is acting upon"

Example &.1+ uniform rod of length 2l and mass m is fi)ed on one end y

a thin and horiLontal rope* and on the $all on the other end"

,ind the tension in the rope and the force of $all acting uponthe lo$er end of the rod"

+ns$er%

The force diagram is sho$n" 1hoose the lo$er end as the

origin* so the torCue of the un'no$n force F is Lero" =y alance of the torCue due to gravity and the tension* $e get

mgl sinθ - /l cosθ ; 0* or / $ mg tanθ

=rea'ing F along the :- 3horiLontal-vertical directions* $e get F ) ; / * and F y ; mg "6t is interesting to e)plore further" Bet us choose another point of origin for the consideration

of torCue alance" One can easily verify that $ith the aove ans$ers the total torCue is alanced relative to any point of origin* li'e the center of the rod* or the upper end of the rod"

1an you prove the follo$ing6f a rigid ody is at rest* the total torCue relative to any pivot point is Lero"

5" 1onservation of Binear omentum

6f the system of particles is isolated 3i"e" there are no e)ternal forces and closed 3i"e" no

particles leave or enter the system* then

"constant= ) 35"

Ba$ of conservation of linear momentum%

" f i

) ) = 35".

Example &.2

6magine a spaceship and cargo module* of total mass M * traveling in deep space $ith velocity

vi ; 200 'm&h relative to the Sun" (ith a small e)plosion* the ship e7ects the cargo module*of mass 0"20 M " The ship then travels 500 'm&h faster than the moduleN that is* the relative

speed vrel et$een the module and the ship is 500 'm&h" (hat then is the velocity v f of the ship

relative to the Sun

9sing conservation of linear momentum*

<




f i ) ) =

f rel f i Mvvv M M+ <"032"0 +−=

rel f i vvv 2"0−=

rel i f vvv 2"0+=

; 200 30"23500

; 2200 'm&h 3ans$er

Example &.3

T$o loc's are connected y an ideal spring and are free to slide on a frictionless horiLontal

surface" =loc' has mass m and loc' 2 has mass m2" The loc's are pulled in oppositedirections 3stretching the spring and then released from rest"

3a (hat is the ratio v&v2 of the velocity of loc' to the velocity of loc' 2 as the separation

et$een the loc's decreases3 (hat is the ratio 3 & 3 2 of the 'inetic energies of the loc's as their separation decreases

+ns$er 3a 9sing conservation of linear momentum*

f i ) ) =

220 vmvm +=

2

2

m

m

v

v−=

3

2

2

2

2

2

2

2

2

22

2

2

2

2

m

m

m

m

m

m

v

v

m

m

vm

vm

3

3 =

−=

==

Example &.4

+ firecrac'er placed inside a coconut of mass M * initially at rest on a frictionless floor* lo$s

the fruit into three pieces and sends them sliding across the floor" +n overhead vie$ is sho$nin the figure" Piece C * $ith mass 0"!0 M * has final speed v fc;5"0ms-"

3a (hat is the speed of piece B* $ith mass 0"20 M

3 (hat is the speed of piece A

+ns$er%

3a 9sing conservation of linear momentum*

3 fxix ) ) =

fyiy ) ) =

050cos<0cos oo =−+ fA A fB B fC C vmvmvm 3

050sin<0sin oo =− fB B fC C

vmvm 32

>




m A ; 0"5 M * m B ; 0"2 M * mC ; 0"! M "

32% 050sin2"0<0sin!"0 oo =− fB fC Mv Mv

--

o

o

ms">ms4">50sin2"0

<0sin53!"03≈== fBv 3ans$er

3 3% fA fB fC Mv Mv Mv 5"050cos2"0<0cos!"0 oo =+

-oo

ms0"!5"0

50cos4">32"03<0cos53!"03=

+= fAv

3ans$er

5". Dlastic 1ollisions in One imension

6n an elastic collision* the 'inetic energy of each colliding ody can change* ut the total'inetic energy of the system does not change"

6n a closed* isolated system* the linear momentum of each colliding ody can change* ut the

net linear momentum cannot change* regardless of $hether the collision is elastic"

6n the case of stationary target* conservation of linear momentum%

"22 f f i

vmvmvm +=

1onservation of 'inetic energy%

"222

2

2 2

2

2

f f i

vmvmvm +=

Fe$riting these eCuations as

*4322 f f i

vmvvm =−

"3 222

2

2

f f ivmvvm =−

ividing*

"2 f f i

vvv =+

(e have t$o linear eCuations for vf and v%f " Solution%

20




"2

*

2

2

2

2

i f

i f

vmm

mv

vmm

mmv

+=

+−

=

otion of the centre of mass% "

2

2icm

vmm

m

mm

) v

+=

+=

Example &.&

6n a nuclear reactor* ne$ly produced fast neutrons must e slo$ed do$n efore they can

participate effectively in the chain-reaction process" This is done y allo$ing them to collide$ith the nuclei of atoms in a moderator "

3a =y $hat fraction is the 'inetic energy of a neutron 3of mass m reduced in a head-on

elastic collision $ith a nucleus of mass m2* initially at rest

3 Dvaluate the fraction for lead* caron* and hydrogen" The ratios of the mass of a nucleus tothe mass of a neutron 3; m2&m for these nuclei are 20 for lead* 2 for caron and aout

for hydrogen"

+ns$er 3a 1onservation of momentum

f if i vmvmvm 22 +=

,or elastic collisions*

2

22

2

2

2

2

2

f f i vmvmvm +=

f f i vmvvm 22 43 =− 3

2

22

2

2

3 f f i vmvvm =− 32

ividing 3 over 32* f f i vvv 2 =+ 3!

3% 4343 2 f i f i vvmvvm +=−

i f vmm

mmv

2

2 +

−=

,raction of 'inetic energy reduction

2




2

2

2

2

2

2

2

2

2

2

2

i

f

i

f i

i

f i

i

f i

v

v

v

vv

vm

vmvm

3

3 3 −=

−=

−=

−=

22

2

2

2

2

3

4

mm

mm

mm

mm

+=

+

−

−= 3ans$er

3 ,or lead* m2 ; 20m*

,raction P>"20.

20-34

20-3

20-3422

==+

=mm

mm 3ans$er

,or caron* m2 ; 2m*

,raction P2<!

234

23

23422

==+

=mm

mm 3ans$er

,or hydrogen* m2 ; m*

,raction P003

342

=+

=mm

mm 3ans$er

6n practice* $ater is preferred"

5"< 6nelastic 1ollisions in One imension

6n an inelastic collision* the 'inetic energy of the

system of colliding odies is not conserved"

6n a completely inelastic collision* the colliding

odies stic' together after the collision"

Ho$ever* t.e conservation of linear momentum still .olds"

*32

+ mmvm += or "2

vmm

m+

+=

Examples &.(

The ballistic endulum $as used to measure the speeds of ullets efore electronic timing

devices $ere developed" Here it consists of a large loc' of $ood of mass M ; 5"4 'g*

hanging from t$o long cords" + ullet of mass m ; >"5 g is fired into the loc'* coming

Cuic'ly to rest" The block 7 bullet then s$ing up$ard* their centre of mass rising a verticaldistance . ; "! cm efore the pendulum comes momentarily to rest at the end of its arc"

3a (hat $as the speed v of the ullet 7ust prior to the collision

22




3 (hat is the initial 'inetic energy of the ullet Ho$ much of this energy remains as

mechanical energy of the s$inging pendulum

+ns$er

3a 9sing conservation of momentum during collision*

+ m M mv 3 +=9sing conservation ofenergy after collision*

g.m M + m M 332

2 +=+

g.+ 2=

+ m

m M v

+=

-ms!00!"03<">32

00>5"0

00>5"04"52 =

+=

+= g.

m

m M

3 6nitial 'inetic energy

Q>00!000>5"032

2 22 === mv 3

,inal mechanical energyQ!"!0!"03<">300>5"04"533 =+=+= g.m M "

3only 0"2 3ans$er

Example &.)

3The Physics of Karate + 'arate e)pert stri'es do$n$ard $ith his fist 3of mass m ; 0".0 'g*

rea'ing a 0"4 'g $ooden oard" He then does the same to a !"2 'g concrete loc'" The

spring constants k for ending are 4" × 04 @m- for the oard and 2" × 0 @m- for the

loc'" =rea'ing occurs at a deflection d of mm for the oard and " mm for the loc'"3a Qust efore the oard and loc' rea'* $hat is the energy stored in each

3 (hat fist speed v is reCuired to rea' the oard and the loc' +ssume that mechanicalenergy is conserved during the ending* that the fist and struc' o7ect stop 7ust efore the

rea'* and that the fist-o7ect collision at the onset of ending is completely inelastic"

+ns$er

3a ,or the oard* Q24<"50"00"432

2

242 =×== kd ! ≈ 5"2 Q 3ans$er

,or the loc'* Q5.!"00"00"232

2

22 =×== kd ! ≈ " Q 3ans$er

3 ,or the oard* first the fist and the oard undergoesan inelastic collision" 1onservation of momentum%

+ mmvm 3 2 += 3

2!




Then the 'inetic energy of the fist and the oard is

converted to the ending energy of the $ooden oard"

1onservation of energy%

32%2

2

mm

! +

+= -

ms5!4"!

4"0."0

24<"532=

+=

3% + m

mmv

2 += 5!4"!."0

4"0."0

+

= ≈ 4"2 ms- 3ans$er

,or the concrete loc'*

2

2

mm

! +

+= -

ms<><"02"!."0

5.!"32=

+= "

+ m

mmv

2 += <><"0."0

2"!."0

+= ≈ 5"0 ms- 3ans$er

The energy to rea' the concrete loc' is &! of that for the $ooden oard* ut the fist speed

reCuired to rea' the concrete loc' is 20 fasterR This is ecause the larger mass of the loc'

ma'es the transfer of energy to the loc' more difficult"

5"> 1ollisions in T$o imensions

1onservation of linear momentum%

x component% *222

coscos θ θ f f i

vmvmvm +=

y component% "222

sinsin0 θ θ f f

vmvm +−=

1onservation of 'inetic energy%

"2

2

2

2

2

2

2

2

2222 f f ii

vmvmvmvm +=+

Typically* $e 'no$

m *2

m * iv and θ" Then $e can solve for f v * f

v2 and θ2"

Examples &.*

24




T$o particles of eCual masses have an elastic collision* the target particle eing initially at

rest" Sho$ that 3unless the collision is head-on the t$o particles $ill al$ays move off

perpendicular to each other after the collision"

9sing conservation of momentum*

f f i vmvmvm 2 +=

f f i vvv

2 +=

The three vector form a triangle"

6n this triangle* cosine la$%

φ cos2 2

2

2

2

2

f f f f i vvvvv ++= 3

9sing conservation of energy%

2

2

2

2

2

2

2

f f i mvmvmv +=

2

2

2

2

f f i vvv += 32

3 M 32% 0cos2 2 =φ f f vv

o>0=φ 3ans$er

Example &.+

T$o s'aters collide and emrace* in a completely inelastic collision" That is* they stic' together after impact" +lfred* $hose mass m+ is <! 'g* is originally moving east $ith speed v+

; "2 'm&h" =arara* $hose mass m= is 55 'g* is originally moving north $ith speed v= ; ."<'m&h"

3a (hat is the velocity+ of the couple after impact

3 (hat is the velocity of the centre of mass of the t$o s'aters efore and after the

collision3c (hat is the fractional change in the 'inetic

energy of the s'aters ecause of the

collision

3a 1onservation of

momentum% θ cos3 + mmvm B A A A += 3

θ sin3 + mmvm B A B B += 32

32 ÷ 3%

A A

B B

vm

vm=θ tan <!4"0

2"3<!3

<".3553 == * so θ ; !>"<o ≈ 40o 3ans$er

25




3% o<"!>cos55<!3

2"3<!3

cos3 +=

+=

θ B A

A A

mm

vm+ ; 4"< 'm&h ≈ 4"> 'm&h 3ans$er

3 #elocity of the centre of mass is not changed y the collision" Therefore + ; 4"> 'm&h and

θ ; 40o oth efore and after the collision" 3ans$er

3c 6nitial 'inetic energy22

2

2

B B A Ai vmvm 3 += ; !2.0 'g 'm

2

&h2

"

,inal 'inetic energy23

2

+ mm 3

B A f += ; !0 'g 'm2&h2"

,raction P50!2.0

!2.0!0−=

−=

−=

i

f i

3

3 3 3ans$er

( ,eneral e-%ations o' motion in the /0 plane ith constant 'orces

" General formulae

t m

F vt v

t m

F vt v

y

yo y

xo x x

+=

+=

3

3

3"

2

2

2

3

2

3

t m

F t v yt y

t m

F t v xt x

y

yoo

xo xo

++=

++=

3"2

These general formulae can e applied to any situations* once the initial conditions 3 x8 , y8 , v x8 ,

v y8 are given"

+ $ord of caution% ,riction forces are not al$ays Econstant?" Pay attention to their directions

ecause they change $ith the direction of the velocity"

"2 Pro7ectile motion near Darth surface

The only force on the o7ect is the gravity $hich is along the M direction" +ccordingly* $e

have

gt vt m

F vt v

vt m

F vt v

yo

y

yo y

o x x

o x x

−=

+=

=

+=

3

3

3"!

for the velocity and

2




22

2

2

2

3

2

3

gt t v yt m

F t v yt y

t v xt m

F t v xt x

yoo

y

yoo

o xo x

o xo

−+=

++=

+=

++=

3"4

for the position"

These general formulae can e applied to any situations* once the initial conditions 3 x8 , y8 , v x8 ,

v y8 are given"

"end"

mechanics lecture

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