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KKINETICSINETICS OFOF PPARTICLESARTICLESImpulse and MomentumImpulse and Momentum

동역학 및 응용

LLINEARINEAR IIMPULSEMPULSE ANDAND MMOMENTUMOMENTUM OFOF AA PPARTICLEARTICLE

mddtmdtd

dtdmm vRvvaR

if

mv

mv

t

tmmmddt f

i

f

i

vvvR F t t f ti f ti l

▶ Linear Momentum : s]Nm/s;[kg vL m

Forces : constant or functions of time only

s]Nm/s;[kgvL m

▶ Linear Impulse :

Linear Impulse of the Force Rs][N

ft

tdtR

it

for Constant Force Rc

tt ff

ifc

t

tc

t

t c ttdtdt f

i

f

i

RRR


Time Average Impulse Force Ravg

ifavg

t

tavg

t

t avg

t

tttdtdtdt fff RRRR

ftdtRR 1 Equivalent constant force that

gives the same linear impulse as

fgtgt gt iii

itifavg dt

ttRR gives the same linear impulse as

the original time-varying force R(t)

for Varying Force (Magnitude & Direction)

ffff t

z

t

y

t

x

tdtRdtRdtRdt kjiR

iiii t zt yt xtj

The work done by a force is a scalar quantity; The impulse of a force is a vector.

The work done by a constant force can be zero; The impulse of a constant force is never zero.


▶ Principle of Linear Impulse and Momentum

xf

t

xxi mvdtRmv f

f

t

tif dt LRL yf

t

t yyi

xft xxi

mvdtRmv f

i

i

fti

zf

t

t zzi mvdtRmv f

i

The force is known as a function of time.

f fmL vf

i

t

tdt R

i imL v


▶ Conservation of Linear Momentum

0RLL whenfi

Conservation of Linear Momentum is not related to conservation of kinetic energy.


A 10-kg box is resting on a horizontalA 10 kg box is resting on a horizontal surface when a horizontal force P is applied to it. The magnitude of P varies with time as shown If the static and kineticwith time as shown. If the static and kinetic coefficients of friction are 0.4 and 0.3, respectively, determinea. The velocity of the box at t = 10 s.b. The velocity of the box at t = 15 s.c. The time tf at which the box stopsc. The time tf at which the box stops

sliding.

f

t

ti mvRdtmv f

i

N198N


N1.98N

Static friction force Ki ti f i ti f

N432930

N24.394.0 NN

Kinetic friction force N43.293.0 N

a) Linear Impulse due to the force P from t = 0 to t = 10 :110 sN25010502110

0 Pdt

Linear Impulse due to the friction force :Linear Impulse due to the friction force :

848.71043.29848.724.392110

0 Fdt

sN31.217

tmdtm f vRv fti mdtm

i

vRv 101031.2172500 v

sm27.310 v


b) from t = 10 to t = 15 :

15 sN25010155015

10 Pdt

15

sN15.147101543.2915

10 Fdt

101514725027310 v 151015.14725027.310 v

sm55.1315 v

c) between t = 15 and t = tf :

152515

f

ttPdtf

15 f

1543.2915

f

ttFdtf

043.292555.1310 ft s6.45ft


The 46-g golf ball is struck by the five-iron and acquires the velocity shown in a time period of 0.001 sec. Determine the magnitude R of the average g gforce exerted by the club on the ball. What acceleration magnitude a does this force cause, and what is the distance d over which the launch velocity

150 km hwhat is the distance d over which the launch velocity is achieved, assuming constant acceleration.

ft

lb101916 7 NR

avg 0.001 0.046 150 3.6f

i

t

i ftmv Rdt mv R

241,667 m sa mg

avg 1916.7 NR

avg :1916.7 0.047R ma a

41.67 2

0 041.67 2 41,667

dvdv adr vdv a dr d R

25

0 0 0.021 md

IINTERACTINGNTERACTING SSYSTEMSYSTEMS OFOF PPARTICLESARTICLES

f

t

tif

i

dt LfRL11211

t

f

t

ti

f

f

i

dt LfRL22122

ff l

lfl

t ll

lif

i

dt LRL

Rl : Resultant external force acting on particle l

2112 ff

▶ Motion of the Mass Center

Rl : Resultant external force acting on particle l

m mm Mass Center rG : calculated using

the first moment of massN N N r

1m

2m

21 mm

1 1 1

, ,N N N

G l l G l l ll l l

m m m m m m

r r v v 1r2r

Gr

f

i

t

G l Gi ftl

m dt m v R v


▶ Conservation of Linear Momentum for a System of Particles y

Impulses of all the external forces acting on a system of particles is zero :

l

lfl

li mm vv

t f

fGiG

ll

mm vv 0R

lt l

f

i

dt

GfGi vv

The impulses of all the external forces acting on a system of particles is zero. The velocity of the mass center of the particles is constant.


▶ Impulsive and Nonimpulsive Force

Impulsive forces have large magnitudes and can produce a significantImpulsive forces have large magnitudes and can produce a significant change in momentum even over very short time periods.The forces generated when one body strikes another

Nonimpulsive forces have smaller magnitude than impulsive forces.Weight, Spring Forces, Friction

t

dt

0Impulse F

Impulsive forces : Large forces that have a significant impulse even over very short time interval (t → 0).

Ex) Forces of collision between two elastic bodies 12 vvF mdt

N i l i f W i ht F i tiNonimpulsive forces : Weight, Friction, …


• What is the difference between impulsive pforces and nonimpulsive forces?

• How can one be sure that a force is nonimpulsive and can be neglected?nonimpulsive and can be neglected?

•Hammer : 0.5 kg

•Steel surface : Fp = 1500 N, Td = 0.00044 s

NF

p , d

•Hand : Fp = 250 N, Td = 0.01 s

F : maximum forceFp : maximum force

Td : impact duration


I l d f ti h NFImpulse : area under force-time graph

Area of half sine wave

2impulse p dF T

dT

dp dtt

TF

0sin

dp

T

d

dp

TFt

TTF

d 2cos

0

0


Hammer : 0.5 kg, Steel surface : Fp = 1500 N, Td = 0.00044 s

2impulse 1500 0.00044 0.42 N s

g, p , d

I l f th i ht f th h 0 0022 NW tImpulse of the weight of the hammer : 0.0022 N sW t

The error introduced by neglecting the impulse of the weight of the hammer : 0.52%

Velocity of the hammer immediately before the impact : 0.5 m/s

0.5 0.5 0.42 0.5 0.34 m sf fv v


Hammer : 0 5 kg Hand : F = 250 N T = 0 01 s

2impulse 250 0.01 1.59 N s

Hammer : 0.5 kg, Hand : Fp = 250 N, Td = 0.01 s

Impulse of the weight of the hammer : 0.049 N sW t

The error introduced by neglecting the impulse of the weight of the hammer : 3.1%

IINTERACTINGNTERACTING SSYSTEMSYSTEMS OFOF PPARTICLESARTICLESA 3 kg cannonball is fired with an initial velocity of v = 150 m/s and = 60 asA 3-kg cannonball is fired with an initial velocity of v0 = 150 m/s and 0 = 60 as shown. At the peak of its trajectory, the ball explodes and splits into two pieces. The 1-kg piece hits the ground at x = 500 m, and y = 2500 m when t = 35 s.a. Determine when and where the 2-kg piece hits the ground.b. Determine the average magnitude of the explosive force Favg if the duration

of the explosion is t = 0.005 s.of the explosion is t 0.005 s.

sm1500 v kg1

kg2sm9.129750 kjv

60

s35t


Before the explosion :

sm9.129750 kjv sm9.129750 kjv

2sm81.9 ka G

sm81.99.1297581.900 kjkjv ttvv zyG

m905.49.12975 2 kjkjr tttzy GGG

At the top of the trajectory ( Explosion) :

s241308199129 ttdtdzG s24.13081.99.129 ttdtdzG

m1.8602.993 kjr t

sm75jv t


Kinematics of 1 kg-block after the explosion :

Let initial velocity of the 1kg-block right after the explosion :

024.13' tt

Let initial velocity of the 1kg-block right after the explosion :

kjiv zyx vvv 10101010

2819 k kji '819 t21 sm81.9 ka kjiv '81.91010101 tvvv zyx

kjir 21010101010101 '905.4''' ttvztvytvx zyx

kji 2101010 '905.4'1.860'2.993' ttvtvtv zyx

At t = 35 s (t’ = 35 – 13.24 = 21.76) :

kjir 76.21905.476.211.86076.212.99376.21 21010101 zyx vvv

kji 02500500 y

sm216726699822 kjikjiv vvv sm21.6726.6998.2210101010 kjikjiv zyx vvv

000


2010 LLLLRL Gtl

lfl

t

t ll

lif

i

dt

During the explosion : s005.0t

lll i

jv 7533 t

tavgF 202v

t avgF

101v

20221.6726.6998.221753 vkjij

kjiv 61.3387.7749.1120 j20

for 1 kg-block : kjiFj 21.6726.6998.221005.0751 avg

N1344211484596avg kjiF


Kinematics of 2 kg-block after the explosion : 024.13' tt

Initial velocity of the 2kg-block right after the explosion :kjiv 61.3387.7749.1120

22 sm81.9 ka

kjiv '819613387774911 t kjiv 81.961.3387.7749.112 t

kjir 22020202 '905.4'61.33'87.77'49.11 ttztytx

kji 2'905.4'61.331.860'87.772.993'49.11 tttt

The 2 kg block hits the ground when z = 0 :The 2 kg-block hits the ground when z2 = 0 :

s49.2324.13's25.10' ttt

m179180.1172 jir


A 20-kg box A slides down a frictionless ramp and strikes a 10-kg box B. As a result of the impact, the two boxes become hooked together and slide as a single unit on the rough surface (k = 0.6). Determine a. The velocity of the boxes immediately after impact.b The distance d that the boxes will slide before coming to restb. The distance d that the boxes will slide before coming to rest.

20 kg

15

10 kg

5 m

2 3 4k = 0.6

2 3 4


a. Work-Energy from 1 → 2 :

VTUVT o 2AAvm 0Bm

222111 VTUVT

22

10 20 5 0 20 02 Ag v 2

2 9.905 m sAv tF

Linear Impulse-Momentum from 2 → 3 :

tF

fxix mvmv

20 9 905 10 0 20 10 v 3vmm BA 320 9.905 10 0 20 10 v

6 603 m sv

3vmm BA

3 6.603 m sv


b. Newton’s 2nd Law :

30 0F N 30 0yF N g 30g

0.6 30 176.58 NkF N g

Work-Energy from 3 → 4 :

222111 VTUVT o

NF k

N 1 2 176.58oU Fd d

21 30 6.603 0 176.58 0 02

d

3.704 md

AANGULARNGULAR IIMPULSEMPULSE ANDAND AANGULARNGULAR MMOMENTUMOMENTUM OFOFAA PPARTICLEARTICLE

▶ Angular Momentum of a particle P about a fixed point O : t f Lmoment of L vrH mOPO s]m[N

keerH mvrvvm OPrrOPO

▶ Angular Impulse of the resultant force about a fixed point O :

fimpulse of moment

s]m[N ff t

O

t

OP dtdt MRr s]m[N ii t Ot OP dtdt MRr


▶ Principle of Angular Impulse and Momentum

vrH ddd

vrH mOPO

R

vrvrH

OP

OPO

dtdmm

dtd

dtd

OO

dtd MH

Rrarvv OPOP mm

Principle of Angular Impulse and Momentum : Principle of Angular Impulse and Momentum :

Of

t

t OOif

i

dt HMH ti

▶ Conservation of Angular Momentum

0MHH OOfOi ,


The vertical shaft shown in the figure is rotating with an initial angular velocity of 20 rad/s when the 1-kg collar A starts to slide slowly outward along the g y glightweight horizontal arm. Determine the decrease in the angular velocity of the shaft as the collar A slides from 0.1 m out to 1 m from the axis of the shaft.

22 0.1 20 0.2 N m sOiH rmv rmr mr m m

21 N m sOf f fH m m 0.1 m

0.9 m

Of

t

t OOif

i

dt HMH

0.2 0 fm m

0.2 rad sf


A 0.6-kg mass slides on a smooth horizontal surface at the end of an inextensible string. The other end of the string passes through a hole in theinextensible string. The other end of the string passes through a hole in the surface and is attached to a spring having k = 100 N/m. The spring is un-stretched when l = 0. If v = 10 m/s and l = 0.5 m at the instant shown, d t i th i i d i l f l i th lti ti

t f dt HMH

determine the minimum and maximum values of l in the resulting motion.

Oft OOii

dt HMH W, N, T have no moment O m50 ,about the vertical axis through O.

O m5.0

OfOi HH

0M O0.6kg

OfOi


sin 60 0 5 0 6 10sin 60H lm v

Angular momentum at the instant shown :

sin 60 0.5 0.6 10sin 602.5981 N m s

OiH lm v

Wh th t i i t it i i i l th

vllmvHOf 6.0

When the string is at its minimum or maximum length, the velocity is perpendicular to the string.

Of

Conservation of angular momentum : 5981.26.0 vl

Work-Energy Equation : 222111 VTUVT

2222 11011 kmvkmv 22

022 ffii kmvkmv

2222 100160150100110601 lv m405.0m828.0

min

max

ll

1002

6.02

5.01002

106.02

lv

AANGULARNGULAR IIMPULSEMPULSE ANDAND AANGULARNGULAR MMOMENTUMOMENTUM OFOFSSYSTEMYSTEM OFOF PPARTICLESARTICLES

ijjiij f ff'OkOk

d MH

iOO mdtd ffRrvr 11211111 frfr jijiji

ijjiij fffOkOkdt

iOO mdtddt

ffRrvr 22122222sin

sin

k

k

ijjj

ijii

jijiji

fr

fr

kikkOkkkOk mdtd ffRrvr 1

d

0ijjj f

212211

frfr

frfrRrH

OO

OOk

kOkk

Okdtd

313131 frfr OO

Nd MH

N

HHTotal angular momentum about O :

k

OkOdt 1MH

k

OkO1HHTotal angular momentum about O :


Principle of Angular Impulse and Momentum

fO

t

t

N

lOliO

f

i

dt HMH 1

1

N

O l Ol

H H

llGlGlG m vrHH

Angular Momentum about G

llGlGlG m vrHH

GlGOll vvrv

GlkGlGGllGlGlGlG mmm vrvrvvrH

GllGlllGlG mm vrvrH


Time Derivative of Angular Momentum

l

lGlllGlG

dtdmm

dtd

dtd vrv

rH llGlG m vrH

llllll

llGlllGl

mm

mm

arvvv

arvv

llGlGllGlGGll

llGlGlGlGl

mmm

mm

arvvvv

arvvv

GlGd MH Gldt

M

t f

fG

t

t GliGf

i

dt HMH


A pendulum consists of two 3.2-kg concentrated 3.2 kg

masses positioned as shown on a light but rigid bar. The pendulum is swinging through the vertical position with a clockwise angular p gvelocity = 6 rad/s when a 50-g bullet traveling with velocity v = 300 m/s in the direction shown strikes the lower mass and becomes embeddedstrikes the lower mass and becomes embedded in it. Calculate the angular velocity ’ which the pendulum has immediately after impact and find the maximum angular reflection of the pendulum.

6 rad si

1

f

i

Nt

O l O O O Oi f i ftl

dt

H M H H H 300 m s0.05 kgb

vm


1 1 1 1 2 2 2O b bi i iim m m H r v r v r v

0.4 0.05 300 cos 20 sin 20

0.4 3.2 0.4 6 0.2 3.2 0.2 6

j i j

j i j i

2r

1r

1 1 1 2 2 2

0.4 0.05 3.2 0.4 ' 0.2 3.2 0.2 'O b f ff

m m m

H r v r v

j i j i 0.4 0.05 3.2 0.4 0.2 3.2 0.2 j i j i

' 2.77 rad s

i i f fT V T V

2 21 1

y

2 21 10.05 3.2 0.4 2.77 3.2 0.2 2.772 2

3.2 0.2 0.05 3.2 0.4g g

52.1

x

0 3.2 0.2cos 0.05 3.2 0.4cosg g

IIMPACTMPACT

▶ Impact (collision between two bodies) is an event that usually i b i f i t l f tioccurs in a very brief interval of time

Relatively large reaction forces between the two bodies Relatively large change of velocity Relatively large change of velocity Considerable deformation

Conversion of mechanical energy into sound, heatgy ,

▶ Line of Impact : Straight line normal to the contacting surface at the point of impactthe point of impact

IIMPACTMPACT

▶ Categorization : relative location of the mass centers, relative velocity of the mass centers, line of impact

Central impact : mass t f b th b dicenters of both bodies are

on L.O.I. Eccentric impact : mass p

center of one or both bodies does not lie on the L.O.I.

Direct impact : initial Direct impact : initial velocities of the impacting bodies are along the L.O.I.

Obli i t Oblique impact

IIMPACTMPACT

Deformation or Compression phase• Instant of contact → Instant of maximum deformation• Two bodies are compressed by the large interaction force• At the instant of maximum deformation : the bodies are neither coming closer

together nor moving apart: the relative velocity along the L.O.I. is zero.

:itt Aiv Biv

:ctt cv cv

IIMPACTMPACT

Restoration or Restitution phase• Instant of maximum deformation → Instant of separationp• Generally, not all of the deformation is recoverable. • Permanent deformation, sound vibrations → some of initial mechanical energy

is dissipatedis dissipated.

:ctt cv cv

:ftt Afv Bfv

IIMPACTMPACT

▶ Direct Central Impact

During the brief impact interval : t = tf - t During the brief impact interval : t tf - ti

1. Velocity of one or both particles may change greatly. 2. Positions of the particles do not change significantly. p g g y3. Nonimpulsive forces may be neglected. 4. Friction forces b/w two bodies are negligible.

During the collision : no impulsive, external forces → Linear Momentum is conserved

BfBAfABiBAiA mmmm vvvv

If the particles are examined individually, the internal force is impulsive and must be included in the linear impulse-momentumincluded in the linear impulse momentum equation

IIMPACTMPACT

During the deformation phase : td = tc - tit

vmdtFvm c

:itt Aiv Biv

cAt dAiA vmdtFvmi

cB

t

dBiB vmdtFvm c

During the restoration phase : t = tf - t

cBt dBiBi

:ctt cv cvDuring the restoration phase : tr tf tc

AfA

t

t rcA vmdtFvm f

c

AfB

t

t rcB vmdtFvm f

c

:ftt Afv Bfv

Fd, Fr : interaction forces on particles during the deformation and

vc : common velocity of the two particles at the end of the deformation phase

t : time at the end of the deformationdu g t e de o at o a drestoration phases, respectively.

tc : time at the end of the deformation phase

IIMPACTMPACTR i i I l

ftF dt

Coefficient of restitution Restitution Impulse

Deformation Impulsec

c

i

rtt

dt

F dte

F dt

AfcAfAcA

t

t r vvvmvmdtFe

f

c cAAiA

t

t d vmvmdtFc

i

for particle A :

cAi

f

cAAiA

ft

t dvvvmvmdtF

ec

i

c

t

dFf

BiBcB

t

t d vmvmdtFc

i

for particle B :

cBfcBBfBt

t

t r vvvmvmdtFe

f

c

AfAcAt r vmvmdtFf

c

cBAfB

t

r vmvmdtFf

for particle B :

BicBiBcBt

t dvvvmvmdtFc

i

cBAfBt rc

fABAfBf

v

v

vvvv

e

Coefficient of restitution : The negative of the ratio of the relative velocity of the two particles after impact and the relative velocity

iABAiBi vvv of the two particles before impact.

IIMPACTMPACT

Coefficient of restitution : a measure of the elastic properties of the particles

AB

AfBf

vvvv

e

Perfectly Elastic Impact : e = 1vvvv

AiBi vv

BfBiAfAi vvvv

BiBfBAfAiABfBAfABiBAiA vvmvvmvmvmvmvm BiBfBAfAiABfBAfABiBAiA

BfBiBiBfBAfAiAfAiA vvvvmvvvvm

2222BiBfBAfAiA vvmvvm

2222 1111

Kinetic Energy is conserved !

2222

21

21

21

21

BfBAfABiBAiA vmvmvmvm

Kinetic Energy is conserved !

Perfectly Plastic Impact : e = 0 AfBf vv

IIMPACTMPACT

Two spheres are hanging from cords as shown. The distance from the ceiling to the center of each sphere is 2 m, and the coefficient of restitution is e = 0.75. If sphere A (mA = 2 kg) is drawn back 60 and released from rest, determinea. The maximum angle B that sphere B (mB = 3 kg) will swing as a result ofa. The maximum angle B that sphere B (mB 3 kg) will swing as a result of

the impact.b. The angle A that sphere A will rebound as a result of the impact.

222111 VTUVT o

datum location

281.92221

060cos281.920

2

Aiv m2Bcos2

2

sm4295.4Aivkg2 kg3

75.0e

IIMPACTMPACT

Across the impact :

vv 3203429542 vmvmvmvm BfAf vv 32034295.42

75.0

AfBfAfBf vvvve

BfBAfABiBAiA vmvmvmvm

4295.4075.0

AiBi vve

sm1013sm22150 fAf vv sm101.3,sm2215.0 BfAf vv

for B : cos281930028193101331 2 for B : Bcos281.9300281.93101.332

97.40B

for A : Acos281.9200281.922215.0221 2

87.2A

IIMPACTMPACT

The figure shows n spheres of equal mass m suspended in a line by wires of equal length so that the spheres are almost touching each other. If sphere 1 is released from the dashed position and strikes sphere 2 with a velocity v1, write an expression for the velocity vn of the nth sphere immediately after being struck by the one adjacent to it The common coefficient of restitution is eby the one adjacent to it. The common coefficient of restitution is e.

IIMPACTMPACT

Before:

1iv

1 2

1 fv 2 fv

2

3 2 11 1

2 2f f ie ev v v

After: 1 2

1 f 2 f 2 2 3

4 3 11 1

2 2f f ie ev v v

1 1 2 2 1 1i f f f i fmv mv mv v v v 1

2 2

1 ne

2 1f fv ve v ev v

1 1 2 2 1 1i f f f i fmv mv mv v v v 1

12nf i

ev v

1 1 210 f i fi

e v ev vv

1 ev v ev v v2 1 1 2 12f i i f iv v ev v v

IIMPACTMPACT

▶ Oblique Central Impact

Superposition ofSuperposition of a motion in the direction perpendicular to the L O I andL.O.I and a direct central impact along the L.O.I.

No impulsive, external forces o pu s e, e te a o ces

BfBAfABiBAiA mmmm vvvv

t-Direction : tBfBtAfAtBiBtAiA vmvmvmvm

vmvmvmvm n-Direction : nBfBnAfAnBiBnAiA vmvmvmvm n-Direction :

IIMPACTMPACT

t-Direction (no impulsive, internal force) vmvm

tAftAi

tAfAtAiA

vv

vmvm

,f

i

t

A Ai t A Aft ttm v F dt m v

BfBi

tBfBtBiB

vv

vmvm

f

i

t

B Bi t B Bft ttm v F dt m v

tBftBi vv

IIMPACTMPACT

n-Direction (Direct central impact)

nBfBnAfAnBiBnAiA vmvmvmvm

( p )

fnABnAfnBf vvve

inABnAinBi vvv

IIMPACTMPACT

Two pucks of equal radius sliding on a smooth horizontal surface collide obliquely Puck A (5 kg) is traveling to the right at 6 m/s whereas puck B (2 kg)obliquely. Puck A (5 kg) is traveling to the right at 6 m/s, whereas puck B (2 kg) is traveling to the left at 3 m/s. If the coefficient of restitution for the collision is 0.7 and the duration of the contact is 0.001 s, determinea The velocities of the pucks immediately after they collidea. The velocities of the pucks immediately after they collide.b. The percentage energy loss due to the collision.c. The average interaction force of pucks B on A.

5kg

6m s3m s

2kg

IIMPACTMPACT

L.O.I.

t

AivBiv

n30

1sin 302rr

6sin 30 3m sAi tv

2r

6cos30 5.1962m s

3sin 30 1 5m sAi n

v

v

3sin 30 1.5m s

3cos30 -2.5981m sBi t

Bi n

v

v

IIMPACTMPACT

3m s, 1.5m sAf Ai Bf Bit tt tv v v v

A Ai B Bi A Af B Bfn n n nm v m v m v m v f fn n n n

5 5.196 2 -2.598 5 2Af Bfn nv v

Bf Afn nv v

e

0.7 Bf Afn nv v

Bi Ain nv v

0.72.598 5.196

1 410 6 866 1.410m s, 6.866m sAf Bfn nv v

IIMPACTMPACT

2 21.410 3 3.31m sAfv

tan 30 3 1 410 34 8 tan 30 3 1.410 34.8Af Af

2 21.5 6.866 7.03m sBfv

tan 30 1.5 6.866 42.3Bf Bf

IIMPACTMPACT

2 21 15 6 2 3 99JiT T T 2 2i

2 21 15 3.31 2 7.03 76.81JfT 100 22.4%i f

i

T TT

2 2f

A Ai A Afm t m v F vi f

5 6 0.001 5 3.31 cos34.8 sin 34.8 i F i j

4-1.641 9.445 10 N F i j

IIMPACTMPACT

In a pool shot, the cue ball knocks the 1-ball into the corner pocket as shown. If the coefficient of restitution is e = 0 95 determine

f1vtn

coefficient of restitution is e = 0.95, determine the velocity of the cue ball after the collision. 1

L.O.I

civ

00 vv 0

,00

1

11

ni

tfti

v

vv 01iv

sm5230sin5 v

sm330.430cos5 tcftci vv

sm5.230sin5 nciv

IIMPACTMPACT

f1vtn

nfncfninci mmmm 11 vvvv

1

520

95.0 11

ncfnfncfnfe

vvvvvv

L.O.I

5.201 ncini vv

sm43752v

civ sm0625.0ncfv

sm4375.21 nfv

83.6033.4cfv

sm3304v tcfv sm330.4

tcfv

ncfv

3060

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