solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion cap 13

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 13, Solution 1.

Given: Weight of satellite, 1000 lbW =

Speed of satellite, 14,000 mi/hv =

Find: Kinetic energy, T

( )( ) h14,000 mi/h 5280 ft/mi 20,533 ft/s3600 s

v = =

( )( )

22

1000 lbMass of satellite 31.0559 lbs /ft

32.2 ft/s= =

( )( )22 91 1 31.0559 20,533 6.5466 10 lb ft2 2

T mv= = = × ⋅

96.55 10 lb ftT = × ⋅

Note: Acceleration of gravity has no effect on the mass of the satellite.

96.55 10 lb ftT = × ⋅ !




CircumferenceTime

v

=

( )( )( )( )( ) ( )( )2 6370 km 35,800 km 1000 m/km

3075.2 m/s23 hr 3600 s/hr 56 min 60 s/hr

vπ +

= =+

3075.2 m/sv =

( ) ( )221 1Kinetic energy, 500 kg 3075.2 m/s2 2

T mv= =

2.36 GJT = !




Given: Mass of stone, 2 kgm =

Velocity of stone, 24 m/sv =

Acceleration of gravity on the moon, 21.62 m/smg =

Find:

(a) Kinetic energy, T

Height h, from which the stone was dropped

(b) T and h on the Moon

(a) On the Earth ( )( )221 1 2 kg 24 m/s 576 N m2 2

T mv= = = ⋅ 576 JT = !

( )( )22 kg 9.81 m/s 19.62 NW mg= = =

1 1 2 2 1 1 2 2 0 576 JT U T T U Wh T− −= = = = =

( )( )

22

576 N m 29.36 m

19.62 NTWh T hW

⋅= = = =

29.4 mh = !

(b) On the Moon

Mass is unchanged. 2 kgm =

Thus T is unchanged. 576 JT = !

Weight on the moon is, ( )( )22 kg 1.62 m/sm mW mg= =

N24.3=mW

( )576 N m177.8 m

3.24 Nmm

ThW

⋅= = =

177.8 mmh = !




21 lb 11.6203

16 oz 32.2 ft/sm

=

(a) 2 21 1 1.62 (160 ft/s)2 2 16(32.2)

T mv

= =

40.2 ft-lbT = !

At maximum height,

(160 ft/s) cos 25xv v= = °

(b) 21 (160cos 25 )2

T m= °

33.1 ft-lbT = !



Chapter 13, Solution 5. Use work and energy with position 1 at A and position 2 at C.

At 1

0yFΣ = 1

1

cos30 0cos30

N mgN mg

⇒ − °== °

0yFΣ = 2

2

= 0N mgN mg

⇒ −=

Work and energy

1 1 2 2T V T→+ = (1)

Where

2 21 1

1 1 (4ft/s) 82 2

T mv m m+ = =

1 2 1 2 (20) ( sin 30 )k kV N d N mg dµ µ→ = − − + °

2 22 2

1 1 (8) 322 2

T mv m m= = =

Into (1)

8 cos30 (20) sin 30 32k km mgd mg mgd mµ µ− ° − + °=

Solve for 32 8 20 32 8 (0.25)(32.2)(20) 20.3 ftcos30 sin 30 ( 0.25) (32.2) (0.866 32.2(0.5))

k

k

gdg g

µµ

− + − += = =− °+ ° − +

!

At 2



Chapter 13, Solution 6. (a) Use work and energy from A to B. 1 1 2 2T V T→+ =

2 21 1

1 1 50 (40) 1242.24 ft lb2 2 32.2

T mv = = = ⋅

2 0T = (Stops at top)

1 2 sin 20U Nx mg xµ→ = − − °

N is needed

y 0FΣ = cos 20 (50 lb)cos 20 46.985 lbN W⇒ = °= °=

So

1 2 0.15(46.985) 50sin 2024.149

U x xx

→ = − − °= −

Substitute

1242.24 24.149 0x− =

51.44 ftx = 51.4 ftx = !

(b) Package returns to A � use work and energy from B to A

2 2 3 3T U T→+ =

Where 2 0T = (At B)

2 3 sin 20 kU W x Nxµ→ = ° − (50) sin 20 (51.44) 0.15(46.985)(51.44)= ° − 517.13 ft lb= ⋅



2 2 23 3 3 3

1 1 50 0.77642 2 32.2

T mv v v = = =

Substitute 2

30 517.13 0.7764v+ =

3 25.81 ft/sv = 3 25.8 ft/sv = 20°!

(c) Energy dissipated is equal to change of kinetic energy

2 21 3 1 2

1 12 2

T T mv mv− = −

2 21 50 (40 25.81 )2 32.2 = −

725 ft lb= ⋅

Energy dissipated 725 ft lb= ⋅ !



Chapter 13, Solution 7. Given: Automobile Weight W = mg = (2000 kg) (9.81) 19,620 NW =

Initial Velocity A, 0m/sAv =

Incline Angle, 6α = °

Vehicle brakes at impending slip for 20 m from B to C

0Cv =

Find; speed of automobile at point B, vB

Coefficient of static friction, µ

(a) (19620 N) (150 m)sin 6A B A BU Wh→ →= − ° 3307.63 10 N m= × ⋅

21 02A B B AU T T mv→ = − = −

3 21307.63 10 N m (2000kg) 02 Bv× ⋅ = −

17.54 m/sBv = !

(b) 0A C A C B C C AU Wh Fd T T→ → →= − = − = 20 mB Cd → = NF µ= Where coefficient of static frictionµ = (19620 N)(sin 6 )(170m) (20m)A CU F→ = ° − (19620 N) cos6F µ= ° (19620 N)(sin 6 )(170m) (19620 N)(cos6 ) (20m) 0µ° − ° =

170 tan 6 0.89320

µ = °= 0.893µ = !



Chapter 13, Solution 8. Given: Automobile weight, (2000) (9.81) 19620 NW = =

Initial velocity at A, 0 m/sAv =

Incline Angle, 6α = °

Vehicle costs 150 m from A to B

Vehicle skids 20 m from B to C

Dynamic friction coefficient, 0.75µ =

Find: Work done on automobile by air resistance and rolling resistance between points A and C.

(20 m) 0A C R A C C AU U Wh F T T→ →= + − = − =

N 0.75 (19620 N) cos6F µ= = °

UR = Resistance work 0.75 (19620 N)cos6 (20 m) (19620 N)sin 6 (170 m)= ° − °

356.0 10 N mRU = − × ⋅ 356.0 10 Jor − × !




90cos 20 sin 50 0NF N P= − °− °=∑

90cos 20 sin 50N P= °+ °

1 2 [ cos50 90sin 20 0.35 N](3 ft)U P→ = °− °−

22 2

1 90 lb (2 ft/s)2 32.2 ft/s

T

=

2( cos50 ) 3 (90sin 20 ) (3) 0.35 (90cos 20 sin 50 )3P P T° − ° − °+ ° =

21 90(3 cos50 0.35(3) sin 50 ) 90sin 20 (3) 0.35 (90cos 20 ) (3) (2)2 32.2

P °− ° = ° + ° +

186.736 166.1 lb1.12402

P = = !




(a) First stage: 1 2 2(5.5 3)(50) 125 lb ft =U T→ = − = ⋅

22 2

1 32 32.2

T v =

2 51.8 ft/sv = ! (b) At the top: 2 3 3( 50) 0 125U h→ = − − = −

275 ,3

h∴ = 91.7 fth = !

(c) At the return: 23 4 4 4

1 33(91.6667)2 32.2

U T v→ = + = =

4 76.8 ft/sv = !




WA = 7(9.81) = 68.67 N

Given: Block A is released from rest and moves up incline 0.6 m.

Friction and other masses are neglected

Find: Velocity of the block after 0.6 m, v

From the Law of Cosines

2 2 2(1.2) (0.6) 2(1.2) (0.6)cos 15d = + − °

2 20.4091 md = 0.63958 md =

C CU W= (Distance pulley C lowered)

1140 N (1.2 0.63958) m 39.229 N m2 = − = ⋅

68.67 N (sin15 ) (0.6 m) 10.6639 N mAU = − ° = − ⋅

2 1 C AU T T U U= − = −

21 02 A C Am v U U− = −

21 (7 kg) (39.229 10.6639) N m2

v = − ⋅

2 8.1615v = 2.857 m/sv = 2.86m/sv = 15°!




WA = 7(9.81) = 68.67 N

Given: Block A is released at the position shown at a velocity of

1.5 m/s up.

After moving 0.6 m the velocity is 3 m/s.

Find: work done by friction force on the block, Vf J

From the Law of Cosines

2 2 2(1.2) (0.6) 2(1.2) (0.6)(cos 15 )d = + − °

2 20.4091md = 0.63958md =

UC 1140 N (1.2 0.63958) m 39.229 N m2 = − = ⋅

68.67 N (sin15 ) (0.6 m) 10.664 N mAU = − ° = − ⋅

2 22 1 2 1

1 [ ]2C A friction AU U U T T m v v+ − = − = −

2 2 2139.229 10.664 (7 kg)[(3) (1.5) ] m2frictionU− − = −

39.229 10.664 23.625frictionU− = − + +

4.94 J= − 4 94 JfrictionU .= − !




Given: At A, 0 v v=

For AB, 0.40kµ =

At B, 2 m/sv =

Find: 0v

( )22 20

1 1 1 2 m/s2 2 2A B BT mv T mv m= = =

2 mBT =

( )( )sin15 N 6 mA B kU W µ− = ° −

0 cos15 0F N WΣ = − ° =

cos15N W= °

( )( )sin15 0.40cos15 6 mA BU W− = ° − °

( )0.76531 0.76531A BU W mg− = − = −

A A B BT U T−+ =

20

1 0.76531 2 m2

mv mg− =

( ) ( ) ( )( )2 20 2 2 0.76531 9.81 m/sv = +

20 19.0154v =

0 4.36 m/sv = !




Given: At A, 0v v=

At B, 0v =

For AB, 0.40kµ =

Find: 0v

20

1 02A BT mv T= =

( )( )sin15 6 mA B kU W Nµ− = ° −

0 cos15 0F N WΣ = − ° =

cos15N W= °

( )( )sin15 0.40cos15 6 mA BU W− = ° − °

( )0.76531 0.76531A BU W mg− = − = −

A A B BT U T−+ =

20

1 0.76531 02

mv mg− =

( ) ( )( )2 20 2 0.76531 9.81 m/sv =

20 15.015v =

0 3.87 m/sv = ! Down to the left.




Car C 3 2 30 0 (35 10 kg) (9.81 m/s ) 343.35 10 Ny C C CF N M g NΣ − ⇒ − = ⇒ = × = ×

So, 3 3(0.35) (343.35 10 ) 120.173 10 NCF = × = ×

Car B 3 2 30 0 (45 10 kg) (9.81m/s ) 441.45 10 Ny B B BF N M g NΣ = ⇒ − = ⇒ = × = ×

So, 3 30.35(441.45 10 ) 154.508 10 NBF = × = ×

Also,

1 (54 km/h)(1h/3600s) (1000 m/km) 15 m/sv = =

(a) Work and energy for the train

1 1 2 2T U T→+ =

3 3 3 2 3 31 (35 10 45 10 35 10 )(15) (120.173 10 154.508 10 ) 02

x× + × + × − × + × =

47.10 mx =

47.1 mx = !



(b) Force in each coupling

Car A

Car C

1 1 2 2T U T→+ =

( ) ( ) ( )231 35 10 15 47.10 02 ABF× − =

383.599 10 NABF = ×

83.6 kNABF = ! Tension

1 1 2xT U T→+ =

( ) ( ) ( ) ( )23 31 35 10 15 120.173 10 47.10 02 BCF× + − × =

Solve for FBC 336.6 10 NBCF = ×

36.6 kNBCF = ! Tension




0 0y A AF N M gΣ = ⇒ − =

( ) ( )3 335 10 9.81 343.35 10 NAN = × = ×

so,

( ) ( )3 30.35 343.35 10 120.173 10 NAF = × = ×

( )1 54 km/h 15 m/sv = =

(a) Work - energy for the entire train

1 1 2 2T U T→+ =

( ) ( ) ( ) ( )23 3 3 31 35 10 45 10 35 10 15 120.173 10 02

x × + × + × − × =

107.66 mx =

107.7 mx = !

(b) Force in each coupling

Car A

1 1 2 2T U T→+ =

( )( ) ( ) ( )23 31 35 10 15 120.173 10 107.66 02 ABF× − + × =

383.60 10 NABF = − ×

83.6 kNABF = ! Compression

1 1 2 2T U T−+ =

( ) ( ) ( )231 35 10 15 107.66 02 BCF× + =

336.57 10 NBCF = − ×

36.6 kNBCF = ! Compression

Car A

Car C




Car B:

Car A:

Given: Car B towing car A uphill at A constant speed of 30 ft/s

Car B skids to a stop. µk = 0.9

Car A strikes rear of car B.

Find: Speed of car A before collision, vA

Let d = Distance traveled by car B after braking.

1 2 2 1U T T− = − ( )21sin 5

2 Bm v mg F d− = − °−

( )2130

2sin 5 0.9 cos5

md

mg mg= −

°+ °

( ) ( ) ( )450 450

32.2 sin 5 0.9cos5 32.2 0.9837d = =

°+ °

14.206 ft traveled by d B=

For car A, travel to contact

2 21 1 1

1 1

2 2C A AU T T mv mv→ = − = −

( )( ) ( )221 1sin 5 15 30

2 2Amg d mv m− ° + = −

( )( )21450 32.2sin 5 14.206 15

2 Av − = − ° +

21368.036

2 Av =

27.13Av = 27.1 ft/sAv = !




F = 0.8 NA

Given: Car B tows car A at 30 ft/s uphill.

Car A brakes for 4 wheels skid µk = 0.8

Car B continues in same gear and throttle setting.

Find: (a) distance, d, traveled to stop

(b) tension in cable

(a) 1 Traction force (from equilibrium)F =

( ) ( )1 3000 sin 5 2500 sin 5F = ° + °

5500sin 5= °

For system A + B

( )1 2 1 3000 sin 5 2500sin 5U F F d→ = − °− ° −

( )222 1

1 1 55000 30

2 2 32.2A BT T m v+ − = − = −

Since ( )1 3000 sin 5 2500 sin 5 0F − ° − ° =

( )0.8 3000cos5 76863 ft lbFd d− = − ° = − ⋅

32.1 ftd = !

(b) cable tension, T

( )( )1 2 2 10.8 sin 5 32.149A AU T N W T T→ = − − ° = −

( )( )( ) ( ) ( )( )

23000 30

0.8 3000 cos5 3000 sin 5 32.1492 32.2

T − ° − ° =

( )2652.3 1304T − = −

1348 lb=

1348 lbT = !

NA = 3000 cos 5° NB = 2500 cos 5°




Given: Blocks A, B released from rest and friction and masses of pulleys neglected.

Find: (a) Velocity of block A, vA, after moving down dA = 1.5 ft. (b) The tension in the cable.

(a) constraint 3 0A Bv v+ = 13B Av v=

Also, 13B Ad d=

( ) ( )1 2 sin 30 sin 30A A B BU W d W d→ = ° − °

( )( ) ( ) 1.520 sin 30 1.5 16 sin 303

= ° − °

11 ft lb= ⋅

2 21 2

10,2 2A A B BT T m v m v1= = +

2

2 21 20 1 16 0.338162 32.2 2 32.2 3

AA A

vv v = + =

21 2 2 1 ; 11 0.33816 AU T T v→ = − =

5.703Av = 5.70 ft/sAv = 30°!

(b) For A alone

( ) ( ) ( )21 2

1sin 302A A A A AU W d T d m v→ = ° − =

( )( ) ( ) ( )21 2020 0.5 1.5 1.5 5.703 10.1022 32.2

T − = =

3.265 ft lbT = ⋅

3.27 ft lbT = ⋅ !




Given: System at rest when 500 N force is applied to collar A. No friction. Ignore pulleys mass.

Find: (a) Velocity, Av of A just before it hits C.

(b) Av If counter weight B is replaced by a 98.1 N downward force.

Kinematics

2B AX X=

2B Av v=

(a) Blocks A and B

2 21 2

1 10 2 2B B A AT T m v m v= = +

( )( ) ( )( )2 22

1 110 kg 2 20 kg2 2A AT v v= +

( )( )22 30 kg AT v=

( ) ( )( ) ( )( )1 2 500 A A A B BU X W X W X− = + −

( )( ) ( )( )21 2 500 N 0.6 m 20 kg 9.81 m/s 0.6 mU − = + ×

( )( )210 kg 9.81 m/s 1.2 m− ×

1 2 300 117.72 117.72 300 JU − = + − =



( ) 21 1 2 2 0 300 J 30 kg AT U T v−+ = + =

2 10Av =

3.16 m/sAv = ! (b) Since the 10 kg mass at B is replaced by a 98.1 N force, kinetic energy at 2 is,

( )2 22 1

1 1 20 kg 02 2A A AT m v v T= = =

The work done is the same as in part (a)

1 2 300 JU − =

( ) 21 1 2 2 0 300 J 10 kg AT U T v−+ = + =

2 30Av =

5.48 m/sAv = !




Given: System at rest when 500 N force is applied to collar A. No friction. Ignore pulleys mass.

Find: (a) Velocity, Av of A just before it hits C.

(b) Av If counter weight B is replaced by a 98.1 N downward force.

Kinematics

2B AX X=

2B Av v=

(a) Blocks A and B

2 21 2

1 10 2 2B B A AT T m v m v= = +

( )( ) ( )( )2 22

1 110 kg 2 20 kg2 2A AT v v= +

( )( )22 30 kg AT v=

( ) ( )( ) ( )( )1 2 500 A A A B BU X W X W X− = + −

( )( ) ( )( )21 2 500 N 0.6 m 20 kg 9.81 m/s 0.6 mU − = + ×

( )( )210 kg 9.81 m/s 1.2 m− ×

1 2 300 117.72 117.72 300 JU − = + − =



PROBLEM 13.21 CONTINUED

( ) 21 1 2 2 0 300 J 30 kg AT U T v−+ = + =

2 10Av =

3.16 m/sAv = ! (b) Since the 10 kg mass at B is replaced by a 98.1 N force, kinetic energy at 2 is,

( )2 22 1

1 1 20 kg 02 2A A AT m v v T= = =

The work done is the same as in part (a)

1 2 300 JU − =

( ) 21 1 2 2 0 300 J 10 kg AT U T v−+ = + =

2 30Av =

5.48 m/sAv = !




Given: 10 kg; 4 kg; 0.5 mA Bm m h= = =

System released from rest.

Block A hits the ground without rebound. Block B reaches a height of 1.18 m.

Find: (a) Av just before block A hits the ground.

(b) Energy, ,PE dissipated by the pulley friction.

(a) Bv at 2 Av= at 2 just before impact.

from 2 to 3; Block B

( )2 2 23 2

1 10 4 22 2B B B BT T m v v v= = = =

Tension in the cord is zero, thus

( )( )( )22 3 4 kg 9.81 m/s 0.18 m 7.0632 JU − = − = −

2 22 2 3 3; 2 7.0632; 3.5316B BT U T v v−+ = = =

2 2 3.5316 1.8793B A B Av v v v= = = = 1.879 m/sAv = ! (b) From 1 to 2 Blocks A and B,

( ) 21 2 2

10 2 A BT T m m v= = +

Just before impact 2 1.793 m/sB Av v v= = =

( )( )22

1 10 4 1.8793 24.722 J2

T = + =

( ) ( )1 2 0.5 0.5 ;A B PU W W E− = − −

Energy dissipated by pulleyPE =

( )( ) ( )21 2 9.81 m/s 10 4 kg 0.5 m 29.43P PU E E−

= − − = −

1 1 2 2; 0 29.43 24.722PT U T E−+ = + − =

4.708PE = 4.71 JPE = !




Given: 8 kg; 10 kg; 6 kgA B Cm m m= = =

System released from rest. Collar C removed after blocks move 1.8 m.

Find: Av , just before it strikes the ground.

Position 1 to position 2

1 10 0v T= =

At 2, before C is removed from the system

( ) 22 2

12 A B CT m m m v= + +

( ) 2 22 2 2

1 24 kg 122

T v v= =

( ) ( )1 2 1.8 mA C BU m m m g− = + −

( ) ( )1 2 8 6 10 g 1.8 m 70.632 JU − = + − =

21 1 2 2 2; 0 70.632 12T U T v−+ = + =

22 5.886v =


( ) ( )22 2

1 18 5.886 52.9742 2A BT m m v′ = + = =

( ) 2 23 3 3

1 92 A BT m m v v= + =

( ) ( ) ( )( )( )22 3 2 0.6 2 kg 9.81 m/s 1.4 mA BU m m g′− = − − = −

2 3 27.468 JU ′− = −

22 2 3 3 352.974 27.468 9T U T v′−′ + = = − =

23 32.834 1.68345v v= = 1.683 m/sAv = !




Given: 8 kg; 10 kg; 6 kgA B Cm m m= = =

System released from rest.

Collar C removed after blocks move 1.8 m.

Find: Av , just before it strikes the ground.


1 10 0v T= =

At 2, before C is removed from the system

( ) 22 2

1

2 A B CT m m m v= + +

( ) 2 22 2 2

124 kg 12

2T v v= =

( ) ( )1 2 1.8 mA C BU m m m g− = + −

( ) ( )1 2 8 6 10 g 1.8 m 70.632 JU − = + − =

21 1 2 2 2; 0 70.632 12T U T v−+ = + =

22 5.886v =


( ) ( )22 2

1 185.886 52.974

2 2A BT m m v′ = + = =

( ) 2 23 3 3

19

2 A BT m m v v= + =

( ) ( ) ( )( )( )22 3 2 0.6 2 kg 9.81 m/s 1.4 mA BU m m g′− = − − = −

2 3 27.468 JU ′− = −

22 2 3 3 352.974 27.468 9T U T v′−′ + = = − =

23 32.834 1.68345v v= = 1.683 m/sAv = !




Given: Conveyor is disengaged, packages held by friction and system is released from rest. Neglect mass of belt and rollers. Package 1 leaves the belt as package 4 comes onto the belt.

Find: (a) Velocity of package 2 as it leaves the belt at A.

(b) Velocity of package 3 as it leaves the belt at A.

(a) Package 1 falls off the belt, and 2, 3, 4 move down.

2.4

0.8 m3

=

22 2

13

2T mv

=

( ) 22 2

33 kg

2T v=

22 24.5T v=

( )( )( ) ( )( ) ( )21 2 3 0.8 3 3 kg 9.81 m/s 0.8U W− = = ×

1 2 70.632 JU − =

21 1 2 2 2 0 70.632 4.5T U T v−+ = + =

22 15.696v =

2 3.9618v = 2 3.96 m/sv = !




Work and energy 1 1 2 2T U T→+ = (1)

Where 1 20; 0T T= =

Work

Outer spring ( )221 2 1

1 1 N3000 1.5 m2 2 m

V k x− = − = −

33.75 J= −

Inner spring ( )221 2 2

1 1 N10,000 0.06 m2 2 m

U k x− = − = −

18 J= −

Gravity ( )1 2 0.15U mg h− = +

( ) ( ) ( )8 9.81 0.15 78.48 11.722h h= + = +

Total work 1 2 33.75 18 78.48 11.772U h− = − − + +

39.978 78.48 h= − +

Substituting into (1)

0 39.978 78.48 0h− + =

0.5094 mh =

509 mmh = !




Work and energy

Assume 0.09 mx >

1 1 2 2T U T→+ = (1)

Where 1 20; 0T T= =

Work

Outer spring ( )12 2 2

1 2 11 1 3000 15002 2

U k x x x→ = − = = −

Inner spring ( ) ( )2

2 21 2 2

1 0.09 5000 0.092

U k x x→ = − − = − −

Gravity ( )1 2 0.6U mg x→ = +

( )( )( )8 9.81 0.6 78.48 47.09x x= + = +

Total work

( )221 2 1500 5000 0.09 78.48 47.09U x x x→ = − − − + +

Substitute into (1)

26500 978.48 6.588 0x x− + + =

Solve

0.1570 mx = or � 0.00646

Reject negative solution 157.0 mmx = !




(a)

Given: A 0.7 lb block rests on a 0.5 lb block which is not attached to a spring of constant 9 lb/ft; upper block is suddenly removed.

Find: (a) maxv of 0.5 lb block

(b) maximum height reached by the 0.5 lb block

At the initial position (1), the force in the spring equals the weight of both blocks, i.e., 1.2 lb.

Thus at a distance x, the force in the spring is,

1.2sF kx= −

1.2 9sF x= −

Max velocity of the 0.5 lb block occurs while the spring is still in contact with the block.

2 2 21 2

1 1 0.5 0.250

2 2T T mv v v

g g

= = = =

( ) 21 2 0

91.2 9 0.5 0.7

2x

U x dx x x x− = − − = −∫

2 21 1 2 2

9 0.250.7

2T U T x x v

g−+ = = − =

2 294 0.7

2v g x x

= −

max when 0 0.7 9 0.077778 ftdv

V x xdx

= = − ⇒ =

( ) ( )22max

94 0.7 0.077778 0.077778

2v g

= −

2max 3.5063v =

max 1.87249v =

max 1.872 ft/sv = !



(b)

0 Initial compressionx =

01.2 lb

0.133333 ft9 lb/ft

x = =

1.2 9sF x= −

1 30, 0T T= =

01 3 0

0.5x

sU F dx h− = −∫

( )01 3 0

1.2 9 0.5x

U x dx h− = − −∫

20 0

91.2 0.5

2x x h= − −

( ) ( )291.2 0.133333 0.133333 0.5

2h= − −

0.08 0.5h= −

( )1 1 3 3 : 0 0.08 0.5 0T U T h−+ = + − =

0.16 fth =

1.920 in.h = !




(a) Same as 13.25 solution for Part (a) max 1.872 ft/sv = !

(b) With 0.5 lb block attached to the spring, refer to figure in (b) of Problem 13.27.

( )1 3 1 3 00 0 1.2 9 0.5hT T U x dx h−= = = − −∫

Since the spring remains attached to the 0.5 lb block,

the integration must be carried out for the total distance, h.

21 1 3 3

9 0 0.7 02

T U T h h−+ = + − =

( )2 0.7 lb 0.155556 ft9 lb/ft

h = =

1.86667 in.h =

1.867 in.h = !




Position 1, initial condition Position 2, spring deflected 5 inches Position 3, initial contact of spring with collar

( )2

1 218 5 1 5 18 560 + 7.5 sin 30

12 2 12 12(Friction) (Spring) (Gravity)

U F→+ + = − − °

1 2 1 20, 0T T U −= = ∴ =

( ) ( ) ( ) ( )223 1 5 230 7.5 0.866 60 7.5 0.5

12 2 12 12µ = − − +

(a) 0.1590µ = !

(b) Max speed occurs just before contact with the spring

( ) ( ) ( ) 21 3 3 max

18 18 1 7.57.5 0.866 7.5 0.512 12 2 32.2

U T vµ→ = − + = =

max 5.92 ft/sv = !




(a) W = Weight of the block ( )10 9.81 98.1 N= =

1

2B Ax x=

( ) ( ) ( )2 21 2

1 1

2 2A A A B BU W x k x k x− = − −

(Gravity) (Spring A) (Spring B)

( )( ) ( )( )21 2

198.1 N 0.05 m 2000 N/m 0.05 m

2U − = −

( ) ( )212000 N/m 0.025 m

2−

( ) ( )2 21 2

1 110 kg

2 2U m v v− = =

( ) 214.905 2.5 0.625 10

2v− − =

0.597 m/sv = !

(b) Let Distance moved down by the 10 kg blockx =

( ) ( ) ( )2

2 21 2

1 1 1

2 2 2 2A Bx

U W x k x k m v− = − − =

( ) ( ) ( )210 2

2 8B

Ad k

m v W k x xdx

= = − −



( ) ( ) ( )2000

0 98.1 2000 2 98.1 2000 2508

x x x= − − = − +

( )0.0436 m 43.6 mmx =

For ( ) 210.0436, 4.2772 1.9010 0.4752 10

2x U v= = − − =

max 0.6166 m/sv =

max 0.617 m/sv = !




(a) W = Weight of the block = (10)(9.81) = 98.1 N

1

2B Ax x=

Block moves down at release after spring A is stretched 25 mm

2 21 2

1 1( ) ( ) ( )

2 2A A A B BU W x k x k x− = + −

(Gravity) (Spring A) (Spring B)

21 2

1(98.1 N)(0.025 m) (2000 N/m)(0.025 m)

2U − = +

21(2000 N/m)(0.0125 m)

2−

2 21 1( ) (10 kg)

2 2m v v= =

21 2

12.4525 0.625 0.15625 (10)

2U v− = + − =

0.764 m/sv = 0.764 m/sv = !

(b) Let x = Distance moved down by the 10 kg block

(for x > 25 mm)

21 2

1 1(0.025) ( 0.025)

2 2A AU Wx k k x− = + − −

221 1

( )2 2 2B

xk M v − =

22 2

1 21 1 1

98.1 (2000)(0.025) (2000)( 0.025) (2000)2 2 2 2

xU x x−

= + − − −

21(10)

2v=



210 (10) 98.1 2000( 0.025) 2000

2 4

d xv x

dx = = − − −

98.1 2000 50 500x x= − + −

148.10.05924 m ( 59.24 mm)

2500x = = =

For 0.05924 mx =

21 2 98.1(0.05924) 0.625 1000(0.03424) 0.87734U − = + − −

215.8114 0.625 1.1724 0.87734 (10)

2v+ − − =

max 0.937 m/sv =

max 0.937 m/sv = !




(a)

Assume auto stops in 1.5 4 m.d< <

( )( )221 1 1

1 127.778 m/s 1000 kg 27.778 m/s

2 2v T mv= = =

385809 J 385.81 kJ= =

2 20 0T v= =

( )( ) ( )( )1 2 80 kN 1.5 m 120 kN 1.5 120 120 180 120 60U d d d− = + − = + − = −

1 1 2 2 385.81 120 60 3.715 mT U T d d−+ = = − =

3.72 md = !

Assumption that 4 md < is O.K.

(b) Maximum deceleration occurs when F is largest.

For 3.3401 m, 120 kN, thus Dd F F ma= = =

( ) ( )( )120,000 N 1000 kg Da=

2120 m/sDa =

2120.0 m/sDa = !




Pressures vary inversely as the volume

LL

P Aa PaPP Ax x

= =

( ) ( ) 2 2

RR

P Aa PaPP A a x a x

= =− −

Initially at 1 0 2av x= =

1 0T =

At 2, 22

1, 2

x a T mv= =

( )2 2

1 21 1

2a aa a

L Ru P P Adx PaA dxx a x− = − = − −

∫ ∫

( )1 22

ln ln 2aau paA x a x− = + −

1 23ln ln ln ln

2 2a au paA a a−

= + − −

2

21 2

3 4ln ln ln4 3au paA a paA−

= − =

21 1 2 2

4 10 ln3 2

T U T paA mv− + = + =

2

42 ln3 0.5754

paApaAv

m m

= = 0.759 paAv

m= !




( ) ( )2

2 2/

1E E

hhR

GM m GM m RF mgh R

= = =+ +

At earths� surface ( )0h = 02EGM m mg

R=

( )2

02 2 1

EGME R

hhR

GM g gR

= =+

Thus ( )

02

1h

hR

gg =+

6370 kmR =

At altitude h, �true� weight h TF mg W= =

Assume weight 0 0W mg=

0 0 0

0 0 0 Error T h hW W mg mg g gE

W mg g− − −= = = =

( )( )

( )

020

102 2

0

1 11 1

hR

g

hh hR R

ggg E

g+

− = = = − + +

(a) ( )21

6370

11 km: 100 100 11

h P E = = = − +

0.0314%P = !

(b) ( )21000

6370

11000 km: 100 100 11

h P E = = = − +

25.3%P = !




Newtons law of gravitation

2 21 0 2

1 1 2 2

T mv T mv= =

( )2

1 2 2 m n

m

R h m mn nR

mg Ru F dr Fr

+− = − =∫

21 2 2

m n

m

R hm m R

dru mg Rr

+− = − ∫

21 2

1 1m m

m m nu mg R

R R h−

= − +

1 1 2 2T U T−+ =

2 20

1 12 2

mm m

m n

Rmv mg R mvR h

+ − = +

( )

( )2 20

2 20

22

mn v vm

m gm

v v Rhg

R−

−

= −

(1)

Uniform gravitational field

2 21 0 2

1 2

T mv T mv= =

( ) ( )1 2m n

m

R hu m m u m uRu F dr mg R h R mgh+

− = − = − + − = −∫

2 21 1 2 2 0

1 1 2 2m uT u T mv mg h mv−+ = − =

( )2 2

0

2um

v vh

g

−= (2)

Divide (1) by (2) ( )( )

2 20

2

1

1m m

nv vug R

hh −

=−

!




6(3960 mi)(5280 ft/mi) 20.9088 10 ftR = = ×

1 2 2 12 1

GMm GMmU T Tr r− = − = −

Since 1r is very large,

11

0 thus 0GMm Tr

≈ ≈

2 22

2 2

1 , 2 2

GMm v Rmv gr r

= =

( )( )22 622

2 2

2 32.2 ft/s 20.9088 10 ft2gRvr r

×= =

(a) For 62 20.9088 10 ft ((620 mi)(5280 ft/mi))r = × +

624.1824 10 ft= ×

34,121 ft/sv = 6.46 mi/sv = !

(b) For 62 20.9088 10 ft ((30 mi)(5280 ft/mi))r = × +

621.0672 10 ft= ×

36,557 ft/sv = 6.92 mi/sv = !

(c) For 62 20.9088 10 ft/sr = ×

36,695 ft/sv = 6.95 mi/sv = !




0 620 mi 3960 mi 4580 mi 24,182,400 ftr = + = =

3960 mi 5200 mi 9160 mi = 48,364,800 ftBr = + =

Parabola: 2y Kx=

At B: ( )220 24,182,400 ft 48,364,800 ftBr Kr K= =

9 110.3381 10 ftK − −= ×

At A: ( ) ( )20sin 45 , cos 45A A A A Ax r y Kx r r= ° = = − °

( ) ( )220 cos 45 sin 45A Ar r Kr− ° = °

20 0A AKx x r+ − =

0 (6.5)(5280) 34,320 ft/sv = =

( )01 1 1 4

2Ax KrK

= − + +

6 620.0334 10 ft, 14.1657 10 ft.A Ax r= × = ×

(a) 2 20 0

0

1 12 2A A

A

GMm GMmU mv mvr r→ = − = −

( )26 2 20

0

1 132.2 20.9088 10 , 2AA

GM v v GMr r

= × = + −

( )226 6

1 134320 214.1657 10 24.1824 10Av GM

= + − × ×

44734 ft/sAv = 8.47 mi/sAv = !

(b) 2 20

0

1 12BB

v v GMr r

= + −

( )226 6

1 134320 248.3648 10 28.1824 10Bv GM

= + −

× ×

24408 ft/sBv = 4.62 mi/sBv = !




(a) 21.62 m/sg = Use work and energy 1 1 2 2T U T→+ = (1)

where 2 21 1

1 1 (600) 180,0002 2

T mv m m= = =

2 0T = (maximum elevation)

1 2U mgh→ = −

(1.62) 1.62m h mh= − = −

Substituting into (1) 180,000 1.62 0m mh− =

3111.11 10 mh = ×

111.1 kmh = !

(b) 12 so 180,000 (same as above)GMmF T mR

= =

22

GMmW mg GM gRR

= = ⇒ =

At some elevation r 2GMmF

r= −

so,

1 2 2

R h

R

GMmU drr

+→ = −∫

22

2

1 1 1r

R

GMm gR mr r R

= = −

6 26

2

1 1(1.62)(1.740 10 )r 1.740 10

m

= × − ×




126

2

1 1180,000 4.9047 10 m 01.74 10

mr

+ × − = ×

Solve for r2 6

2 1.8587 10 mr = ×

1858.7 km=

2so, 1858.7 1740h r R= − = −

118.7 km=

118.7 kmh = !




Assume the blocks move as one: 2 1 1 2T T U −− = 2 2

1 2friction1 1( )2 2A Bm m v kx U −+ = −

2 21 1(3 kg) (180 N/m)(0.1 m) (3)(9.81)(0.1)(0.1 m)2 2

v = −

2 0.4038 0.63545 m/sv v= = Check assumption at release, s18 N > (3)(9.81) 4.41 Nµ = ∴ Slips at the horizontal surface At release

18 NsF = x xF maΣ = 18 2.94 3a− = 25.02 m/sa =

For A alone:

x xF maΣ =

18 (1.5)(5.02)fF− =

10.47 NfF = s18 7.53 10.47 N < (1.5)(9.81) (0.95)(14.175) 13.98 NfF µ= − = = =

∴ A and B move as one, thus (a) 0.635 m/sv = !

(b) vmax is max at a = 0, 0 180 2.943, 0.01635 ms fF F x x= = = − =

2 2 2max 0 0

1 1( ) ( ) ( )2 2A B fm m v k x x F x x+ = − − −

2 2 2max

1 1(3) (180)[(0.1) (0.01635) ] 2.943(0.1 0.01635)2 2

v = − − −

max 0.648 m/sv = !

0




From problem 13.39 assuming the blocks move together, 25.02 m/s at release.a =

2(1.5 kg)(9.81 m/s ) 14.715 NA BW W= = =

s18 7.53 10.47 N < (1.5)(9.81) 0.35(14.715) 5.15 NfF µ= − = = =

∴ Block A slides on Block B A alone:

(a) 22 1 1 2 1 2 friction

1 1, ( )2 2AT T U m v kx U− −− = = −

2 21 1(1.5)( (180 N/m)(0.1 m ) 14.175(0.3)(0.1 m)2 2

v = −

2 0.6114v = 0.782 m/sv = !

(b) max at acceleration 0,v v= =

0s f k AF F kx Wµ− = = −

180 (0.30)(14.715) 4.4145, 0.0245 mx x= = =

2 22 1 0 0

1 ( ) ( )2 A kT T k x x W x xµ− = − − −

2max

1 (1.5) 0.84598 0.33329 0.512682

v = − =

max 0.827 m/sv = !

0

0




21 0

12

T mv=

22

12

T mv=

1 2U mgl− = −

2 21 1 2 2 0

1 1 2 2

T U T mv mgl mv−+ = − =

2 20 2v v gl= +

Newtons� law at 2

(a) For minimum v, tension in the cord must be zero.

Thus 2v gl=

2 20 2 3v v gl gl= + =

0 3v gl= !

(b) Force in the rod can support the weight so that v can be zero.

Thus 20 0 2v gl= +

0 2v gl= !




( )221 0

1 1 16 128 m2 2

T mv m= = =

22

12

T mv=

( )1 2 6sinU mg θ− =

21 1 2 2

1: 128 6 sin2

T U T m mg mvθ−+ = + =

2256 12 sing vθ+ = (a)

Newton�s law

2

: 2 sin6n

mvF ma mg mg θΣ = − =

Using (a) ( )12 6sin 256 12 sing gθ θ− = +

18 sin 12 256g gθ = −

( )( )

12 32.2 256sin 0.22498

18 32.2θ

−= = 13.00θ = °!




Use work - energy : position 1 is bottom and position 2 is the top

1 1 2 2T U T→+ = (1)

where,

21 0

1

2T mv=

22 2

1

2T mv=

1 2 (0.5)U mgh mg→ = − = −


2 20 2

1 1(0.5)

2 2mv mg mv− =

so

2 20 2v v g= + (2)

(a) Slender rod 2 00v v g= ⇒ =

0 3.13 m/sv = !

(b) Cord, so the critical condition is tension = 0 at the top

2222n

mvF ma mg v gρ

ρΣ = ⇒ = ⇒ =


20 9.81(0.25 1)v g gρ= + = +

12.2625=

0 3.502 m/sv =

0 3.50 m/sv = !




Use work - energy : position 1 is at A, position 2 is at B.

1 1 2 2T U T→+ = (1)

Where 21 1 2 2

10; sin ;2 BT U mg l T mvθ→= = =

Substitute

210 sin2 Bmg l mvθ+ =

2 2 sinBv g l θ= (2)

For T = 2 W use Newtons 2nd law. 2

2 sin Bn n

mvF ma W Wl

θΣ = ⇒ − = (3)

Substitute (2) into (3)

sin2 sin 2 lmg mg mgl

θθ− =

2 3sinθ= 2or sin 41.813

θ θ= ⇒ = °

41.8θ = °!




( )2 2 21 10 0 250 kg 1252 2A A B B B Bv T T mv v v= = = = =

( ) ( )21250 kg 9.81 m/sW = ×

( )( )27 1 cos 40A BU W− = − °

( )( )( )2250 kg 9.81 m/s 27 m 0.234A BU − = ×

15495 JA BU − =

20 15495 125A A B A BT U T v−+ = + =

( )( )

2 15495 J125 kgBv =

2 2 2124.0 m /sBv = Newtons Law at B

2

2 2 2cos 40 ; 124.0 m /sBB

mvN W vR

−− ° = =

( )( )( )( )2 2

2250 kg 124.0 m /s

250 kg 9.81 m/s cos 4027 m

N = × ° −

1879 1148 731 NN = − = 731 NN = !




Normal force at B

See solution to Problem 13.45, 731.0 NBN =

Newtons Law

From B to C (car moves in a straight line)

cos 40 0BN W′ − ° =

( )2250 kg 9.81 m/s cos 40 1878.7 NBN ′ = × ° =

At C and D (car in the curve at C)

At C

2

cos CC

W vN W

g Rθ− =

( )2

2250 kg 9.81 m/s cos CC

vN

gRθ

= × +

At D

2D

DW v

N Wg R

− = +

continued



( )2

250 9.81 1 DD

vN

gR

= × +

Since and cos 1,D C D Cv v N Nθ> < >

Work and energy from A to D

2 210, 0 125

2A A D D Dv T T mv v= = = =

( ) ( )( )( )227 18 250 kg 9.81 m/s 45 m 110362.5A DU W− = + = =

20 110362.5 125A A D D DT U T v−+ = + =

2 882.90Dv =

( ) ( )2 882.90

250 1 250 9.81 1 5518.1 N72 72 9.81

DD

vN g

g

= + = + =

min max731 N; 5520 NB DN N N N= = = = !




Kinematics: 21constant, ,2

a v at x at= = =

( )2 21110 5.4 , 7.5446 ft/s2

x a a= = =

( )22

150 lbconstant 7.5446 ft/s 35.1456 lb32.2 ft/s

F ma

= = = =

7.5446v t=

( )2150Power 7.544632.2

Fv mav t = = =

( )( ) ( )2 25.4

0

150 / 32.2 7.5446 5.41Average power 05.4 5.4 2

Fv dt = = −

∫

Average power 715.93 ft lb/s= ⋅

Average power 1.302 hp= !




3tan100

θ =

1.718θ = °

( ) 27 60 kg 9.81 m/sB WW W W= + = + ×

657.3 NW =

( )( )sinWP W v W vθ= ⋅ =

( )( )( )657.3 sin1.718 2WP = °

39.41 WWP =

39.4 WWP = !

( ) 29 90 kg 9.81 m/sB mW W W= + = + ×

971.2 NW =

Brake must dissipate the power generated by the bike and the man going down the slope at 6 m/s.

( )( )sinBP W v W vθ= ⋅ =

( )( )( )971.2 sin1.718 6 174.701BP = ° =

174.7 WBP = !




(a) ( ) ( )( ) ( )( ) ( )2800 650p A A L A AAP F v W W v v= = + = +

6.5 ft

s/t 0.40625 ft/s16 sAv = = =

( ) ( )( )3450 lb 0.40625 ft/s 1401.56 lb ft/sp AP = = ⋅

( )1 hp 550 ft lb/s, 2.548 hpp AP= ⋅ =

( ) 2.55 hpp AP = !

(b) ( ) ( ) 2.55

0.82

p AE A

PP

η= =

( ) 3.11 hpE AP = !




(a) Material is lifted to a height b at a rate, ( )( ) ( )2kg/h m/s N/hm g mg =

Thus,

( ) ( )( )N/h

N m/s3600 s/h 3600

mg b mU mgb

t

∆ = = ⋅ ∆

1000 N m/s 1 kw⋅ =

Thus, including motor efficiency, η

( ) ( )( ) ( )

N m/skw

1000 N m/s3600

kw

mgbP

η

⋅=

⋅

( ) 6kw 0.278 10mgb

Pη

−= × !

(b) ( )( ) ( )tons/h 2000 lb/ton ft

3600 s/h

W bU

t

∆ =∆

ft lb/s; 1hp 550 ft lb/s1.8

Wb= ⋅ = ⋅

With ,η ( ) 1hp 1ft lb/s

1.8 550 ft lb/s

Wbhp

η = ⋅ ⋅

31.010 10 Wbhp

η

−×= !




For constant power, P:

dv

P Fv mav m vdt

= = =

Separate variables

12 21 0

0 0 2 2t v m dv m v vdt t

P v P

= ⇒ = −

∫ ∫ (1)

Distance

2dv dx dvP mv mv

dx dt dx= =

Separate variables

1

0

3 32 1 0

0 3 3x v

v

m m v vdx v dv x

P P

= ⇒ = −

∫ ∫ (2)

with numbers

(a) 0 136 km/h 10 m/s; 54 km/h 15 m/s, so,v v= = = =

( ) ( ) ( )3

2 2

3

15 10 kg15 m/s 10 m/s

2 50 × 10 Wt

× = − 18.75 st⇒ = !

( ) ( )

33 3

3

15 1015 10 237.5 m

3 50 10x

× = − = × 238 mx⇒ = !

(b) 0 154 km/h 15 m/s; 72 km/h 20 m/sv v= = = =

( ) ( )

32 2

3

15 1020 15 26.25 s

2 50 10t

× = − = × 26.2 st⇒ = !

( ) ( )3

3 33

15 1020 15 462.5 m

3 50 10x

× = − = × 462 mx⇒ = !




(a) constantdv

P F v m vdt

= ⋅ = =

4.3 5

2.0 0m v dv P dt∴ =∫ ∫

( ) ( )2 24.3 m/s 20 m/s

60 kg 5 , 86.94 W2

P P − = =

86.9 WP = !

(b) constantdv

P F v mv vdx

= ⋅ = =

4.3 22.0 0

xm v dv P dx∴ =∫ ∫

( ) ( )3 34.3 2.0

60 86.943

x − =

16.45 mx = !




Motion is determined as a function of time as,

( )12000 ln cosh 0.03x t=

Velocity ( )( )112000 sinh 0.03 0.03cosh 0.03

dxv tdt t

= =

360 sinh 0.03cosh 0.03

tvt

=

Power dissipated ( )2 30.01 0.01P Dv v v v= = =

( )33 0.03 0.03

3 30.03 0.03

sinh 0.030.01 360 466.56 10cosh 0.03

t t

t tt e ePt e e

−

−

−= = × +

(a) 10 s,t = 11534 ft lb/sP = ⋅ 21.0 hpP = !

(b) 15 s,t = 35037 ft lb/sP = ⋅ 63.7 hpP = !




Motion is defined by the following function:

0.000511 x dva e vdx

−= =

0.00050.00050 0 0

11110.0005

v x xx uvdv e dx e du−− −= =∫ ∫ ∫

( )2

0.000522000 12

xv e−= −

( )2 0.000544000 1 xv e−= −

( )1

0.0005 2209.76 1 xv e−= −

3Power dissipated 0.01P Dv v= =

3

0.0005 292295 1 xP e− = −

(a) 600 ftx = , 12178 ft lb/sP = ⋅ 22.1 hpP = !

(b) 1200 ft,x = 27971 ft lb/sP = ⋅ 50.9 hpP = !




System is in equilibrium in deflected 0x position.

Case (a) Force in both springs is the same P=

0 1 2x x x= +

0e

Pxk

=

1 21 2

P Px xk k

= =

Thus 1 2e

P P Pk k k

= +

1 2

1 1 1

ek k k= +

1 2

1 2e

k kkk k

=+

!

Case (b) Deflection in both springs is the same 0x=

1 0 2 0P k x k x= +

( )1 2 0P k k x= +

0eP k x=

Equating the two expressions for

( )1 2 0 0eP k k x k x= + =

1 2ek k k= + !




Use conservation of energy Let position 1 be where A is compressed 0.1 m; position 2 when B is compressed a maximum distance So 1 1 2 2T V T V+ = + (1)

Where 1 0;T = ( )( )221 1

1 1 1600 N/m 0.1 m 8 J2 2AV k x= = =

2 0;T = ( )2 2 22 2 2 2

1 1 2800 N/m 14002 2BV k x x x= = =


22 20 8 0 1400 0.07559 mx x+ = + ⇒ =

This answer is independent of mass Distance traveled 0.5 m 0.05 m 0.07559 m 0.526 m= − + =

The maximum velocity will occur when the mass is between the two springs where 1 1 2 2T V T V+ = +

1 0;T = ( )1 8 J same as beforeV =

22 max

1 ;2

T mv= 2 0V =


2max

10 8 0;2

mv+ = + 2max

16vm

=

For 1 kgm = 2max 16v = (a) max 4 m/sv = !

For 2.5 kgm = 2max

16 6.42.5

v = = (b) max 2.53 m/sv = !




2 21 6 9 10.817 in.= + =l

( ) ( )2 20 6 8 10 in. 0.8333 ft= + = =l

Stretch 10.817 10 0.817 in.= − =

1 0.06805 ftS =

( ) ( )2 22 7 6 9.215 in.= + =l

Stretch 9.2195 10 0.7805 in.= − = −

2 0.06504 ftS =

1 20, 0T V= =

2 22 2 2

1 1 42 2 32.2

T mv v = =

( )( )2 21 1 2

1 33,600 lb/ft2

V S S= +

( )( )1 16,800 0.008861 148.86 ft lbV = = ⋅

1 1 2 2T V T V+ = +

22 2396.7v = 2 49.0 ft/sv = !

0 0




2800 lb/in. 33,600 lb/ftk = =

( ) ( )2

21 1 1

1 1 10 33,6002 2 12

T V k l = = ∆ =

116.667 ft lb= ⋅

2 21 6 9 10.817 in. 0.9014 ft= + = =l

1 Stretch 10.817 10 0.817 in. 0.06808 ftS = = − = =

2 22 6 7 9.2195 in.= + =l

2 Stretch 9.2195 9 0.2195 in. 0.018295 ftS = = − = =

2 2 22 2 2 2

1 1 4 0.06212 2 32.2

T mv v v = = =

( )( )2 22 1 2

1 33,6002

V S S= +

( ) ( )2 216800 0.06808 0.018295 = +

83.489 ft lb= ⋅

1 1 2 2T V T V+ = +

22116.667 0.06211 83.489v= +

2 23.1 ft/sv = !

0




Use conservation of energy 1 1 2 2T V T V+ = + (1)

(a) Position 1 is at A and position 2 is at B

1 0;T = 21 1

12

V kx= where 1 0x = −l l

1

2 2 2 2500 400 350 729.726 mm = + + = l

So 0 429.726 mm− =l l

( )( )21

1 1500 N/m 0.429726 m 13.8498 J2

V = =

At B 1

2 2 20350 400 531.507 mm 231.507 mm = + = ⇒ − = l l l

( )2 2 22 2 2 2

1 1 0.75 0.3752 2

T mv v v= = =

( )( )22

1 150 0.231507 4.01966 J2

V = =

Substituting into (1) 220 13.8498 0.375 4.01966v+ = +

2 5.12 m/sBv v= = !

(b) At E ( )1 10; 13.8498 same as beforeT V= =

At E 1

2 2 20350 500 610.328 mm 310.328 mm = + = ⇒ − = l l l

2 23

1 0.3752 E ET mv v= =

( )( )221 1 150 0.310328 7.223 J2 2Ev kx= = =


20 13.8498 0.375 7.223 4.20 m/sE Ev v+ = + ⇒ = !




Conservation of energy 1 1 2 2T V T V+ = +

At A 1

2 2 2 20500 400 350 729.726 mm 729.726 450 279.726 mm = + + = ⇒ − = − = l l l

So ( )( )221 10; 150 0.279726 5.8685 J2 2A AT V kx= = = =

At B 1

2 2 20350 400 531.507 81.507 mm = + = ⇒ − = l l l

( )( )22 2 21 1 10.375 ; 150 0.081507 0.49825 J2 2 2B B B BT mv v V kx= = = = =

Substituting into (1) 20 5.8685 0.375 0.49825Bv+ = +

3.78 m/sBv = !

At E ( ) ( )1

2 2 20350 500 610.328 mm 160.328 mm = + = ⇒ − =

l l l

So ( )( )22 2 21 1 10.375 ; 150 0.160328 1.9279 J2 2 2E E E ET mv v V kx= = = = =

Substituting into (1) 20 5.8685 0.375 1.9279EV+ = +

3.24 m/sEv = !

The fact the cord becomes slack doesn�t matter.




(a) Maximum height when 2 0v =

1 2 0T T∴ = =

g eV V V= +

Position 1 ( )1 0gV =

16 lb 6 in. 0.4 6 6.4 in.

15 lb/in.x = + = + =

( ) ( )( )2211

1 1 15 lb/in. 6.4 in.2 2eV kx= =

307.2 lb in. 25.6 lb ft= ⋅ = ⋅

Position 2 ( ) ( )2

6 6 0.512gV mg h h = + = +

( )2 0eV =

( ) ( ) ( ) ( )1 1 2 2 1 21 2: g e g eT V T V V V V V+ = + + = +

( )25.6 6 0.5 h= +

3.767 fth = 45.2 in.h = !

(b) Maximum velocity occurs when acceleration is 0, equilibrium

position

2 2 23 3 3 3

1 1 6 0.0931672 2 32.2

T mv v v = = =

( ) ( ) ( ) ( ) ( )2 23 133

16 6 6 36 7.5 6.4 62g eV V V k x= + = + − = + −

37.2 lb in. 3.1 lb ft= ⋅ = ⋅

21 1 3 3 3: 25.6 0.093167 3.1T V T V v+ = + = +

max 15.54 ft/sv = !




(a) Collar is in equilibrium.

( )15 lb/in. 6 lbF δΣ = −

( )( )

6 lb0.4 in.

15 lb/in.δ = =

max 0.4 in.δ = !

(b) Maximum compression occurs when velocity at 2 is zero.

1 10 0T V= =

22 2 max max

102

T V W kδ δ= = − +

max12

W kδ=

( )( )max

2 6 lb0.8 in.

15 lb/in.δ = =

max 0.8 in.δ = !




30 rad6πθ = ° =

0.3 mR =

(a) Maximum height

Above B is reached when the velocity at E is zero

0CT =

0ET =

e gV V V= +

Point C

( )0.3 m rad6BCL π ∆ =

m20BCL π∆ =

( ) ( ) ( )2

21 1 40 N/m m 0.4935 J2 2 20C BCeV k L π = ∆ = =

( ) ( ) ( )( )( )21 cos 0.2 kg 9.81 m/s 0.3 m 1 cos30C gV WR θ= − = × − °

( ) 0.07886 JC gV =

( ) 0 (spring is unattached)E eV =

( ) ( )( ) ( )0.2 9.81 1.962 JE gV WH H H= = × =

C C E ET V T V+ = +

0 0.4935 0.07886 0 0 1.962H+ + = + +

0.292 mH = !

continued



(b) The maximum velocity is at B where the potential energy is zero, maxBv v=

0 0.4935 0.07886 0.5724 JC CT V= = + =

( )2 2max

1 1 0.2 kg2 2B BT mv v= =

2max0.1BT v=

0BV =

( ) 2max0 0.5724 0.1C C B BT V T V v+ = + + =

2 2 2max 5.72 m /sv =

max 2.39 m/sv = !




0.3 mR =

( ) ( )20.2 kg 9.81 m/sW = ×

1.962 N=

(a) Smallest angleθ occurs when the velocity at D is close to zero

0 0C Dv v= =

0 0C DT T= =

e gV V V= +

Point C

( )0.3 m 0.3 mBCL θ θ∆ = =

( ) ( )212C BCeV k L= ∆

( ) 21.8C eV θ=

( ) ( )1 cosC gV WR θ= −

( ) ( )( )( )1.962 N 0.3 m 1 cosC gV θ= −

( ) ( ) ( )21.8 0.5886 1 cosC C Ce gV V V θ θ= + = + −

Point D

( ) 0 (spring is unattached)D eV =

( ) ( ) ( )( )( )2 2 1.962 N 0.3 m 1.1772 JD gV W R= = =

( )2; 0 1.8 0.5586 1 cos 1.1772 JC C D DT V T V θ θ+ = + + + − =

( ) ( )21.8 0.5886 cos 0.5886θ θ− =

By trial 0.7522 radθ =

43.1θ = ° !

continued



(b) Velocity at A Point D

( )0 0 1.1772 J see Part ( )D D DV T V a= = =

Point A

( )2 21 1 0.2 kg2 2A A AT mv v= =

20.1A AT v=

( ) ( ) ( )( )1.962 N 0.3 m 0.5886 JA A gV V W R= = = =

A A D DT V T V+ = +

20.1 0.5886 0 1.1772Av + = +

2 2 25.886 m /sAv =

2.43 m/sAv = !




Conservation of energy

Position (1) is at the top of the incline; position (2) is when the spring has maximum deformation

1500 lb/ftk =

Where 1 1 2 2T V T V+ = +

At (1) ( )221 1

1 1 200 8 198.76 ft lb2 2 32.2

T mv + = = ⋅

( )21 1 1 1 1

1 datum at point 22g eV V V mgz k x= + = =

( ) ( ) ( )21200 25 sin 20 1500 0.52

x= − ° +

Deformation of the springx =

1 1710.1 68.404 187.5V x= + +

At (2) ( )( )222 2 2 2 2

1 10; 1500 0.52 2g eT V V V k x x= = + = = +

Substituting into (1) ( )2198.78 1710.1 68.404 187.5 750 0.5x x+ + + = +

Solve 2750 681.596 1908.9 0x x+ − =

2.11 or +1.2044 ftx = −

1.204 ftx = !

14.45 in.= !

0




Spring length 2 214 28= +

31.305 in.=

Stretch 31.305 in. 14 in.12 in./ft

−=

1.44208 ft=

( )( )20 0

10 0 48 lb/ft 1.44208 ft 49.910 lb ft2

T V+ = + + = ⋅

(a) At A: ( )2

22

1 10 lb 1 14 2 1449.910 48 lb/ft2 2 1232.2 ft/s Av

−= +

16.89 ft/sAv = !

(b) At B: ( )2

21 10 1 14 1449.910 48 102 32.2 2 12 12Bv = + −

13.64 ft/sBv = !




( ) ( )1 1 10, 0 Constraint: 2e g B AT V V y x= = = ↓ = →

2 22

1 12 2A A B BT M v M v= +

( ) ( )2

2 21 14 kg 1.5 kg 1.252 2 2

BB B

v v v = + =

(a) ( )( )22

10.15 m, 0.075 m, 300 N/m 0.075 m 0.84375 N m2B A ey x V= = = = ⋅

( ) ( ) ( )2 1.5 9.81 0.15 m 2.2073 JgV = − = −

21 1 2 2 ; 0 1.25 0.84375 2.20725BT V T V v+ = + = + −

1.044 m/sBv = ! (b) Maximum velocity when acceleration 0=

; 2 0; Cord tensile forcex x AF m a k x T T= − = =∑

( )2 14.7152 ; 0.0981 m; 2 0.1962 m300A A B A

Tx x x xk

= = = = =

( )( )2 14.715 N 0.1962 m 2.8871gV = − = −

( )( )22

1 300 N/m 0.0981 m 1.44352eV = =

21 1 2 2 ; 0 1.25 1.44354BT V T V v+ = + = − 1.075 m/sBv = !

(c) ( )2

2 210; 0 300 14.7152 2

BB

yT V y = = = − 0.392 m 392 mmBy = = !




(a) Calculate spring lengths after deflection.

Original spring length 0.75 m,= collar moved 100 mm 0.1 m=

( ) 2 21 1 2

10, 4 22

T V T v v= = = =

( )( )( )22 4 kg 9.81 m/s 0.1 m 3.924 JgV = − = −

( ) ( ) ( )2 22

1 300 0.8322 0.75 0.6727 0.752eV = − + −

1.9098 J=

21 1 2 2 : 0 2 3.924 1.9098T V T V v+ = + = − +

1.0035 m/sv = 1.004 m/sv = !

(b) Calculate spring lengths after deflection, collar moved190 mm 0.19 m=

( ) ( ) ( )22 4 kg 9.81 m/s 0.19 m 7.4556 JgV = − = −

( ) ( ) ( )2 22

1 300 0.90918 0.75 0.60877 0.752eV = − + −

6.7927 J=

21 1 2 2 : 0 2 7.4556 6.7927T V T V v+ = + = − +

0.576 m/sv = 0.576 m/sv = !




(a) 0, 0C Cv T= =

21

2B BT mv=

( ) 210.2 kg

2B BT v=

20.1B BT v= ( ) ( )C C Ce gV V V= +

arc BCBC L Rθ= ∆ =

( )( ) ( )0.3 m 30

180BCLπ

∆ = °°

0.15708 mBCL∆ =

( ) ( ) ( )( )2 21 140 N/m 0.15708 m 0.49348 J

2 2C BCeV k L= ∆ = =

( ) ( ) ( )( )( )( )21 cos 0.2 kg 9.81 m/s 0.3 m 1 cos30C gV WR θ= − = − °

( ) 0.078857 JC gV =

( ) ( ) 0.49348 J 0.078857 J 0.57234 JC C Ce gV V V= + = + =

( ) ( ) 0 0 0B B Be gV V V= + = + =

2; 0 0.57234 0.1C C B B BT V T V v+ = + + =

2 2 25.7234 m /sBv = 2.39 m/sBv = !

(b) 2B

Rmv

F F WR

Σ = − =

( ) ( )( )

2 25.7234 m /s1.962 N 0.2 kg

0.3 mRF = +

1.962 N 3.8156 N 5.7776 NRF = + = 5.78 NRF = !




(a) Speed at C

( ) ( ) ( )2 2 2300 150 75 343.69318 mmABL = + + =

320 N/mk =

At B 0 0B Bv T= =

( ) ( )B B Be gV V V= +

343.69318 mm 200 mmABL∆ = −

143.69318 mm 0.14369318 mABL∆ = =

( ) ( ) ( )( )2 21 1 320 N/m 0.1436932 m2 2B ABeV k L= ∆ =

( ) 3.303637 JB eV =

( ) ( )( )( )20.5 kg 9.81 m/s 0.15 m 0.73575 JB gV Wr= = =

( ) ( ) 3.303637 J 0.73575 J 4.03939 JB B Be gV V V= + = + =

At C ( )( )2 21 1 0.5 kg2 2C C CT mv v= =

20.25C CT v=

( ) ( )212C ACeV k L= ∆

309.23 mm 200 mm 109.23 mm 0.10923 mACL∆ = − = =

( ) ( )( )21 320 N/m 0.10923 m 1.90909 J2C eV = =



B B C CT V T V+ = +

20 4.0394 0.25 1.90909Cv+ = +

2 2 24.0394 1.90909 8.5212 m /s0.25Cv −= = 2.92 m/sCv = !

(b) Force of rod on collar AC

0zF = (no friction)

x yF F= +F i j

1 75tan 14.04300

θ −= = °

( )( )cos sinACk L θ θ= ∆ +eF i k

( )( )( )320 0.10923 cos14.04 sin14.04= ° + °eF i k

33.909 8.4797 (N)= +eF i k

( ) ( )2

33.909 4.905 8.4797x ymvF F mg

rΣ = + + − + = +F i j k j k

( ) ( )2 28.5212 m /s33.909 N 0 4.905 N 0.5

0.15 mx yF F+ = = +

33.909 NxF = −

33.309 NyF =

33.9 N 33.3 N= − +F i j !




Datum at point C. 2

22.5 lb 1b s0.07764

ft32.2 ft/sm

⋅= =

( ) ( ) 21 2.50, 0, 0,2 32.2C C C A Ag eT V V T v = = = =

( ) ( )( )2.5 lb 7/12 ft 1.4583A gV = − = −

(a) ( ) ( ) ( )21 20 lb/ft 0.63465 ft 0.3333 ft 0.908122A CV = − =

From conservation of energy:

( ) 210 0.07764 1.4583 0.90812 2 Av= − +

3.7646 ft/sAV = 3.76 ft/sAv = !

At point A, ( ) ( )20 lb/ft 0.63465 ft 0.3333 ft 6.0263 lbS CAF k L= ∆ = − =

( ) ( )

( )

222 2.5 lb/32.2 ft/s 3.76551 ft/s1.8872 lb

7 /12 ftAmv

r= =



( )2 7; 6.0263 1.8872

7.61577A

AmvF N

r= − =∑

3.65 lbAN = !

(b) Datum at C.

( ) ( )( )2.5 lb 14 /12 ft 2.9167 ft lbB gV = − = − ⋅

( ) ( ) ( )( )2 21 1 20 lb/ft 0.5 ft 2.5 ft lb2 2B CBeV k L= ∆ = = ⋅

From Conservation of energy:

21 2.50 2.9167 2.52 32.2 Bv + − +

3.2762 ft/sBv = 3.28 ft/sBv = !

( ) ( ) ( )20 0.5 10 lbS CBF k L= ∆ = =

( )22.5 1.4286 lb

32.2 7 /12Bv =

10 2.5 1.4286y BF N= − + − =∑

6.07 lbBN = !




(a) For maximum velocity,

7.61577 4 3.61577 in. 0.30131 ftL∆ = − = =

0ta s= =&& ( )sin 3/ 7.61577θ =

2.5 lbW =

( )0 0.30131 3/ 7.61577 2.5 0yF k= = − =∑

21.063 lb/ftk = 21.1 lb/ftk = !

(b) Put datum at C

( ) ( ) 0,C C Cg eT V V= = = 21 2.5

2 32.2 AAT v =

( ) ( )2.5 7 /12 1.4583A gV = − = −

( ) ( ) ( )21 21.063 0.30131 0.95612A eV = =

Conservation of energy: 21 2.50 1.4583 0.95612 32.2 Av = − +

3.597Av = 3.60 ft/sAv = !




Loop 1

(a) The smallest velocity at B will occur when the force exerted by the tube on the package is zero.

2 0 BmvF mg

rΣ = + =

( )2 2 1.5 ft 32.2 ft/sBv rg= =

2 48.30Bv =

At A 20

12AT m v=

A0.50 8 oz 0.5 lb 0.01553

32.2V = = ⇒ = =

At B ( )21 1 48.30 24.15 m2 2B BT mv m= = =

( ) ( )7.5 1.5 9 9 0.5 4.5 lb ftBV mg mg= + = = = ⋅

( ) ( )20

1: 0.01553 24.15 0.01553 4.52A A B BT V T V v+ = + = +

20 0 627.82 25.056v v= = 0 25.1 ft/sv = "

At C

( )2 21 0.007765 7.5 7.5 0.5 3.752C C C CT mv v v mg= = = = =

2 20: 0.007765 0.007765 3.75A A C C CT V T V v v+ = + = +

( )2 20.007765 25.056 3.75 0.007765 Cv− =

2 144.87Cv =

continued



Loop 2

(b)

( )144.87 : 0.01553

1.5nF ma NΣ = =

1.49989N =

{Package in tube} 1.500 lbCN = "

(a) At B, tube supports the package so,

0Bv ≈

( )0, 0 7.5 1.5B B Bv T V mg= = = +

4.5 lb ft= ⋅

A A B BT V T V+ = +

( ) 21 0.01553 4.5 24.0732 A Av v= ⇒ =

24.1 ft/sAv = "

(b) At C 20.007765 , 7.5 3.75C C CT v V mg= = =

( )2 2: 0.007765 24.073 0.007765 3.75A A C C CT V T V v+ = + = +

2 96.573Cv =

96.5730.01553 0.999851.5CN = =

{Package on tube} 1.000 lb CN = "




(a) Loop 1

From 13.75, at B

2 2 2 48.3 ft /s 6.9498 ft/sB Bv gr v= = ⇒ =

( )( )21 1 0.01553 48.3 0.375052 2B BT mv= = =

( ) ( )( )7.5 1.5 0.5 9 4.5 lb ftBV mg= + = = ⋅

( )2 2 21 1 0.01553 0.007765 2 2C C C CT mv v v= = =

( )7.5 0.5 3.75 lb ftCV = = ⋅

2: 0.37505 4.5 0.007765 3.75B B C C CT V T V v+ = + + = +

2 144.887 12.039 ft/sC Cv v= ⇒ = 12.04 ft/s 10 ft/s> ⇒ Loop (1) does not work !

(b) Loop 2 at A 2 20 0

1 0.007765 2AT mv v= =

0AV = At C assume 10 ft/sCv =

( )221 0.007765 10 0.77652C CT mv= = =

( )7.5 0.5 3.75Cv = =

20: 0.007765 0.7765 3.75A A C CT V T V v+ = + = +

0 24.144v = 0 24.1 ft/sv = !




Use conservation of energy from the point of release (A) and the top of the circle.

1 1 2 2T V T V+ = + (1) (datum at lowest point)

where

1 10;T V mg= = l

At 2 ( )( )22 2

1 ; 22

T mv V mgz mg a= = = −l

Substituting into (1) ( )210 22

mg mv mg a+ = + −l l (2)

We need another equation � use Newton�s 2nd law at the top. ( )0Tension, 0 at topT =

n nF ma m= ⇒∑mg =

2vρ

( )2v g g aρ= = −l


( ) ( )1 22

mg mg a mg a= − + −l l l

2 4 45 3

a aa

= − + −=

l l l

l

35

a = l!




( ) ( )( )( )2 21 0, 70 , 70 kg 9.81 m/s 40 m 1 0.7071

2A A B B BT V T v V= = = = − −

( ) 2170 ,

2C CT v= 2C BV V= −

Conservation of energy: 0, 15.161 m/sB B BT V v+ = =

0 , 21.441 m/sC C CT V v+ = =

(a)

(b)

(c)

( )270 /40 + 70

402.26 686.7

BN v g

N

=

= +

1 1089 NBN = !

( )270 /40 70

402.26 686.7

BN v g

N

= − +

= − +

2 = 284 NBN !

( ) 270 kg 0.7071 70 /40 CN v− = −

1 804.5 485.6 0CN = − + <

Skier airborne? Yes!

,

,

,




For a conservative force, Equation (13.22) must be satisfied

x y zV V VF F Fx y z

∂ ∂ ∂= − = − = −∂ ∂ ∂

We now write 2 2

yx FF V Vy x y x y x

∂∂ ∂ ∂= − = −∂ ∂ ∂ ∂ ∂ ∂

Since 2 2

: yx FV V Fx y y x y x

∂∂ ∂ ∂= =∂ ∂ ∂ ∂ ∂ ∂

!

We obtain in a similar way

y z z xF F F Fz y x z

∂ ∂ ∂ ∂= =∂ ∂ ∂ ∂

!




(a) x yyz zxF Fxyz xyz

= =

( ) ( )11

0 0yyxx FFy y x x

∂∂ ∂∂ = = = =∂ ∂ ∂ ∂

Thus yx FFy x

∂∂ =∂ ∂

The other two equations derived in Problem 13.80 are checked in a similar way.

(b) Recall that , , x y zv v vF F Fx y z

∂ ∂ ∂= − = − = −∂ ∂ ∂

( )1 ln ,xvF V x f y z

x x∂= = − = − +∂

(1)

( )1 ln ,yvF V y g z x

y y∂= = − = − +∂

(2)

( )1 ln ,zvF V z h x y

z z∂= = − = − +∂

(3)

Equating (1) and (2)

( ) ( )ln , ln ,x f y z y g z x− + = − +

Thus ( ) ( ), lnf y z y k z= − + (4)

( ) ( ), lng z x x k z= − + (5)

Equating (2) and (3)

( ) ( )ln , ln ,z h x y y g z x− + = − +

( ) ( ), lng z x z l x= − +

From (5)

( ) ( ), lng z x x k z= − +

Thus ( ) lnk z z= −

( ) lnl x x= −

continued



From (4)

( ), ln lnf y z y z= − −

Substitute for ( ),f y z in (1)

ln ln lnV x y z= − − −

lnV xyz= − "




(a)

( ) ( )3 1

2 2 2 2 2 22 2

x yx yF F

x y z x y z= =

+ + + +

( )( )

( )( )

( )

3 32 2

5 52 2 2 2 2 22 2

2 2 yx

x y y yFFy x

x y z x y z

− −∂∂ = =∂ ∂

+ + + +

Thus yx FFy x

∂∂ =∂ ∂

The other two equations derived in Problem 13.79 are checked in a similar fashion

(b) Recalling that , , x y zV V VF F Fx y z

∂ ∂ ∂= − = − = −∂ ∂ ∂

( )3

2 2 2 2

xV xF V dxx

x y z

∂= − = −∂

+ +∫

( ) ( )1

2 2 2 2 ,V x y z f y z−

= + + +

Similarly integrating Vy

∂∂

and Vz

∂∂

shows that the unknown function ( ),f x y is a constant.

( )1

2 2 2 2

1Vx y z

=+ +

!




(a)

2

0 2a

ABaU kxdx k= =∫

, x yF F F= ∴ is normal to BC, 0BCU =

( )2

0 2a

CAaU a u du −= − − =∫

( )2

1 ,2ABCAaU k= − not conservative !

(b) From Problem 13.77, 1 yx FFy x

∂∂ = =∂ ∂

Conservative, 0ABCAU = !




(a) ( )2 2

1 1

31 2 1 2

x x

x xU Fdx k x k x dx→ = − = − +∫ ∫

( ) ( )2 2 4 41 22 1 2 12 4

k kx x x x= − − −

1 2 1 2e eU V V→ = −

2 41 2

1 1

2 4eV k x k x= + !

(b) Conservation of energy: 21 2

10,

2T T mv= =

2 41 1 0 2 0 2

1 1, 0

2 4e eV k x k x V= + =

2 2 41 0 2 0

1 1 1

2 2 4mv k x k x= +

2 41 20 02

k kv x x

m m = +

!




(a) ( )2 2

1 1

31 2 1 2

x x

x xU Fdx k x k x dx→ = − = − +∫ ∫

( ) ( )2 2 4 41 22 1 2 12 4

k kx x x x= − − + −

1 2 1 2:e eU V V→ = −

2 41 2

1 1

2 4eV k x k x= − !

(b) Conservation of energy: 21 2

10,

2T T mv= =

2 41 1 0 2 0 2

1 1, 0

2 4e eV k x k x V= − =

2 2 41 0 2 0

1 1 1

2 2 4mv k x k x= −

2 41 20 02

k kv x x

m m = +

!

10

2

2Requires

kx

k

<




Circular orbit velocity 63960 mi = 20.9088 10 ftR = ×

6930 mi = 4.9104 10 ftAor = ×

2

22 ,Cv GM GM gR

r r= =

( )( )

( )262

26 6

32.2 ft/s 20.9088 10 ft

20.9088 10 ft + 4.9104 10 ftC

GM gRvr r

×= = =

× ×

2 6 2 2545.22 10 ft /sCv = ×

23350 ft/sCv =

Velocity reduced to 60% of 14010 ft/sCv =

Conservation of energy:

A A B BT V T V+ = +

2 21 12 2A B

A B

GMm GMmmv mvr r

− = −

( ) ( )( )

( )( )

2 26 622

6 6

32.2 20.9088 10 32.2 20.9088 101 140102 225.819 10 20.9088 10

Bv× ×− = −

× ×

21269 ft/sBv = 4.03 mi/sBv = !




Distance 6OA 3960 mi + 376 mi = 4336 mi = 22.894 10 ft= ×

Distance 6EOO 10,840 mi 4336 mi = 6504 mi = 34.341 10 ft= − ×

( ) ( )2 22 8670 6504Cr = +

610,838.4 mi = 57.2268 10 ftCr = ×

( )2

21= 15681.62 C

gR mT V mr

+ −

( ) 6 2 21constant = 123.032 10 ft /sT Vm

+ = − ×

63960 mi = 20.9088 10 ft; 2.97 mi/s = 15681.6 ft/sCR v= × =

(a) At point A, 2

26

1=2 22.894 10 ftA

T V gRvm+ −

×

( ) ( )( )( )

22 66 2

6

32.2 ft/s 20.9088 10 ft1123.032 102 22.894 10 ft

Av×

− × = −×

31364 ft/sAv = 5.94 mi/sAv = !

(b) At point B, ( ) ( )2 10,840 mi 4336 miBr = −

617344 mi = 91.5763 10 ft= ×

( ) ( )( )( )

266 2

6

32.2 20.9088 101123.032 102 91.5763 10

Bv×

− × = −×

7834.3 ft/sBv = 1.484 mi/sBv = !




- At A, 32.5 /h 9028 m/sAv Mm= =

( )2 61 9028 m/s 40.752 102AT m m= = ×

2

AA A

GMm gR mVr r

−= − =

610.67 M 10.67 10 mAr m= = ×

66370 km 6.37 10 mR = = ×

( ) ( )

( )22 6

66

9.81 m/s 6.37 10 mm 37.306 10 m

10.67 10 mAV

×= − = − ×

×

At B 2

21 ;2B B B

B B

GMm gR mT mv Vr r

−= = − =

619.07 M 19.07 10 mBr m= = ×

( ) ( )( )

22 66

6

9.81 m/s 6.37 10 m20.874 10 m

19.07 10 mB

mV

×= − = − ×

×

6 6 2 61; 40.752 10 m 37.306 10 m 20.874 10 m2A A B B BT V T V mv+ = + × − × = − ×

2 6 6 62 40.752 10 37.306 10 20.874 10Bv = × − × + ×

2 6 2 248.64 10 m /sBv = ×

36.9742 10 m/s 25.107 Mm/hBv = × = 25.1 Mm/hBv = !

4.3 6.37A Ar h R Mm Mm= + = +

10 67Ar Mm=

72.7 6.37B Br h R Mm Mm= + = +

19.07Br Mm=




Note: moon earth0.0123GM GM=

By Equation 12.30 2moon 0.0123 EGM gR=

At ∞ distance from moon: 2 2, assume 0r v= ∞ =

lem2 2 2 0 0 0 0mGM mE T V= + = − = − =

∞

(a) On surface of moon: 61740 km 1.74 10 mMR = = ×

61 10, 0 6370 km 6.37 10 mEv T R= = = = ×

2lem lem

1 1 1 10.01230m E

M m

GM m gR mV E T VR R

−= = + = −

( )( )( )( )

22 6lem

1 6

0.0123 9.81 m/s 6.37 10 m

1.740 10 m

mE

×= −

×

Where lem mass of the lemm =

( )6 2 21 lem2.814 10 m /sE m= − ×

( )6 2 22 1 lem0 2.814 10 m /sE E E m∆ = − = + ×

Energy per kilogram: lem

2810 kj/kgEm∆ = !

(b) 1 80 kmmr R= +

( ) 61 1740 km 80 km 1820 km 1.82 10 mr = + = = ×

Newton�s second law:

2

lem lem 1lem lem 2

11: mGM m m vF m a

rr= =

2 2 lem1 1 lem 1

1 1

1 12 2

m mGM m GMv T m vr r

= = =



lem lem lem1 1 1 1

1 1 1

12

m m mGM m GM m GM mV E T Vr r r

−= = + = −

( ) 2lemlem

11 1

0.01231 12 2

Em gR mGM mEr r

= − = −

( )( )( )2 3

lem1 6

0.0123 9.81 m/s 6.37 10 m12 1.82 10 m

mE

×= −

×

( )6 2 21 lem1.345 10 m /sE m= − ×

( )6 2 22 1 lem0 1.345 10 m /sE E E m∆ = − = + ×

Energy per kilogram: lem

1345 kJ/kgEm∆ = !




Total energy per unit weight

2 20 0

1 1;2 2A A p p A p

A p

GMm GMmE T V T V E mv mvr r

= + = + = − = −

Unit weightW mg=

2 20 2 2A p

A p

W W GM W W GME v vg g r g g r

= − = −

22

0

2 2pA

A p

vE v GM GMW g gr g gr

= − = − (1)

22

21 11

2pA

A pA

vv GMg g r rv

− = −

2

221 2p p A

AA pA

v r rv GM

r rv

− − =

( )givenpA

p A

rvv r

=

2

221 2 p AA

AA pp

r rrv GMr rr

− − =

2 2

22 2p A p A

AA pp

r r r rv GM

r rr

− − =

2 12 pA

A p A

rv GM

r r r

= +

(2)



Substitute Av from (2) into (1)

0 1 1 1 122

p p

A p A A A p A A

r rE GM GMGMW g r r r gr g r r r r

= − = − + +

( )

1 p p Ap

A p A A p A

r r rrGM GMr g r r r g r r

− + = − = + +

( )A B

GMg r r

−=+

2

2 0 EE

A p

E RGM gRW r r

= ⇒ = −+

( )( )

260 3960 mi 5280 ft/mi

1.65598 10 ft lb/lb50,000 mi 5280 ft/mi

EW

×= − = − × ⋅

×

On earth: , 0, 0,E E E E E EE

WGME T V V T VgR

= + = = = −

2

620.9088 10 ft lb/lbE EE

E E

E GM gR RW gR gR

= − = − = − = − × ⋅

For propulsion: 0p EE E EW W W

= −

( )6 61.65598 10 20.9088 10= − × − − ×

619.25 10 ft lb/lbpEW

= × ⋅ "




Geosynchronous orbit

61 3960 200 4160 mi 21.965 10 ftr = + = = ×

62 3960 22,000 25,960 mi 137.07 10 ftr = + = = ×

Total energy 212

GMmE T V mvr

= + = −

mass of earthM = mass of satellitem =

Newton�s second law 2

22;n

GMm mv GMF ma vr rr

= = ⇒ =

212 2

GM GMmT mv m Vr r

= = = −

1 12 2

GMm GMm GMmE T Vr r r

= + = − = −

( )2 2

2 1 1 where 2 2

E EE

gR m R WGM gR E W mgr r

= = − = − =

( )( )26 186000 20.9088 10 ft1 1.3115 10 lb ft2

Er r

× ×= − = − ⋅

Geosynchronous orbit at 62 137.07 10 ftr = ×

189

61.3115 10 9.5681 10 lb ft137.07 10GsE − ×= = − × ⋅

×

(a) At 200 mi, 61 21.965 10 ftr = × 18

10200 6

1.3115 10 5.9709 1021.965 10

E ×= − = − ××

10300 200 5.0141 10GsE E E∆ = − = ×

9300 50.1 10 ft lbE∆ = × ⋅ !



(b) Launch from earth

At launch pad 2E

E EE E

GMm gR mE WRR R

−= − = = −

( ) 116000 3960 5280 1.25453 10EE = − × = − ×

9 99.5681 10 125.453 10E Gs EE E E∆ = − = − × + ×

9115.9 10 ft lbE

E∆ = × ⋅ !




We know

Geosynchronous orbit: 2 35780 6370 42,150 kmr = + =

Orbit of shuttle: 1 6370 km + 296 km = 6666 kmr =

Radius of Earth: 26370km also =R GM = gR=

For any circular orbit 2

22n

GMm mv GMF ma v

r rr= ⇒ = ⇒ =

Energy

2

2

1 1;

2 2

1 1 1

2 2 2

GMm GMmT mv v

r r

GMm GMm GMm gR mE T V

r r r r

= = = −

= + = − = − = −

For Geosynchronous orbit

( )( ) ( )22 69

2 6

9.81 m/s 6.370 10 m 3600 kg116.999 10 J

2 42.150 10 mE

×= − = − ×

×

For orbit of shuttle

( )( ) ( )266

1 6

9.81 6.370 10 36001107.487 10 J

2 6.666 10E

×= − = − ×

×

On the launching pad vo = 0

( )( )( )6 93600 9.81 6.370 10 224.963 10 JoGMm

E mgRR

= − = − = − × = − ×

(a) From shuttle to orbit

( )9 92 1 16.999 10 J 107.487 10 JE E E∆ = − = − × − − ×

990.5 10 JE∆ = × " (b) From surface to orbit

( )9 92 0 16.999 10 J 224.963 10 JE E E∆ = − = − × − − ×

9208 10 JE∆ = × "




(a) Potential energy 2

constantGMm gR mVr r

= − = − +

(cf. Equation 13.17)

Choosing the constant so that 0 for :V r R= =

1 RV mgRr

= −

!

(b) Kinetic energy

Newton�s second law 2

2:nGMm vF ma m

rr= =

22 GM gRv

r r= =

212

T mv= 21

2mgRT

r= !

Energy

(c) Total energy

21 12

gR RE T V m mgr r

= + = + −

12RE mgRr

= −

!




In a circular orbit, 2

2Nmv GMmF

r r= =

mass of Venus, mass of Sunm M= =

2 ,GMvr

∴ = 212 2

GMmT mvr

= =

,2

GMm GMmV T Vr r

= − + = −

(a)

23 6

2

19

88(78.3 10 ) (67.2 10 )(5280)60

34.4 10v rMG −

× × = =×

27 2136.0 10 lb s /ftM = × ⋅ !

(b) 212 2

GMmT V mvr

+ = − = −

227

31 136.029 10 8878.3 102 60407 10

T V 3

× + = − × ×

332.20 10 lb ftT V+ = − × ⋅ !




2 2setting :

1g g yR

WR WR WRV r R y Vr R y

= − = + = − = −+ +

( ) ( )( )1 21 1 21 1

1 1 2gy y yV WR WRR R R

− − − − = − + = − + + + ⋅ !

We add the constant WR, which is equivalent to changing the datum from to :r r R= ∞ =

2

gy yV WRR R

= − +

!

(a) First order approximation:

gyV WR WyR

= =

!

[ ]Equation 13.16

(b) Second order approximation: 2

gy yV WRR R

= −

2

gWyV Wy

R= − !




Use Conservation of energy and Conservation of angular momentum.

Conservation of angular momentum.

( )( )2

1 11 1 2 2

2

0.2 60.5

= ⇒ = =r vr mv r mv vr

θθ θ θ

22.4 m/s=vθ !


1 1 2 2T V T V+ = + (1)

At 1 ( ) ( )221 1

1 1 4 kg 6 m/s 72 J2 2

T mv= = =

( )( )221 1

1 1 1500 N/m 0.2 m 0.4 m 72 J2 2

V kx= = − =

At 2 ( )( ) ( )22 2 22 2 2 2

1 1 1 14 2.4 42 2 2 2r rT mv mv v= + = +θ

2211.52 2 rv= +

( )( )222 2

1 1 1500 N/m 0.5 m 0.4 m 7.5 J2 2

V kx= = − =

Substituting into (1) 2272 30 11.52 2 7.5rv+ = + +

222 82.98rv =

2 6.44 m/srv = !




Conservation of angular momentum.

1r m 1 maxv r mθ = 2v θ

so

( )( )1 12

max max max

0.2 6 1.2r vvr r r

= = =θθ

Energy

1 1 2 2T V T V+ = + (1)

where,

At 1 ( ) ( )221 1

1 1 4 kg 6 m/s 72 J2 2

T mv= = =

( )( )221 1

1 1 1500 0.2 0.4 30 J2 2

V kx= = − =

At 2 ( )2

2 22 2 2 2

max max

1 1 1 1.2 28842 2 2rT mv mv

r r

= + = =

θ

( )( )222 2 max

1 1 1500 0.42 2

V kx r= = −

Substituting into (1) ( )2max2

max

2.8872 30 750 0.4rr

+ = + −

Solve by trial for maxr ⇒ max 0.760 mr = !

Solve for 2v θ

2max

1.2 1.20.760

vr

= = ⇒ 2 1.580 m/sv = !

0




Initial state

F maΣ =

22

0 0, AA

mr θkx mr θ xk

= =!!

( )( )( )2

04 / 32.2 3 ft 5 rad/s

1.331 ft7 lb/ft

= =x

Unstretched length = 4.5 ft + 1.331 ft = 5.831 ft

Conservation of angular momentum

(A) ( ) ( )22 = 3 5 45= =!A Ah r θ (1)

(B) ( ) ( ) ( )2 22 = 9 = 7.5 5 , = 3.4722= ! ! !B Bh r θ θ θ (2)

(a) Substitute into (1) gives 3.6 ftAr = !

(b) = 3.47 rad/sθ! !

Conservation of energy ( )( )20 0 0

1, 7 lb/ft 1.331 ft 6.200 ft lb2

T V T V V+ = + = = ⋅

( ) ( )2 20

1 4 1 45 3 5 7.5 101.320 ft lb2 32.2 2 32.2

T = + = ⋅

( ) ( ) 21 7 9 3.6 5.831 0.6501 ft lb2

V = − − = ⋅

0 0= + −T T V V

101.32 6.200 0.6501= + −

(c) T = 106.9 ft ⋅ lb!




Initial state

F maΣ =

20 Akx mr θ= !

( ) ( ) ( )22

04 / 32.2 3 5

1.331 ft7

Amr θxk

= = =!

Unstretched length = 4.5 + 1.331 = 5.831 ft


( ) ( )2 20

1 4 1 45 3 5 7.52 32.2 2 32.2

T = +

101.320 ft lb= ⋅

( )( )20

1 7 1.331 6.200 ft lb2

V = = ⋅

( ) 21

1100 7 101.32 6.2002

T V x+ = + = +

1 1.4658 ftx = ±

For compression:

( ) 5.831 1.4658B Ar r− − = −


(A) ( ) ( )22 = 3 5 45A Ah r θ= =!

(B) ( ) ( )22 = 7.5 5 = 281.25B Bh r θ= !

continued



( )22 22.5 , 2.5B A B Ar r r r= =

5.831 2.5 5.831 1.5 5.831B A A A Ar r r r r− − = − − = −

(a) 1.5 5.831 1.4658Ar∴ − = − 2.91 ftAr = !

(b) 2.5 7.2753 ftB Ar r= = 7.28 ftBr = !

(c) 2 = 45, = 5.31 rad/sAr θ θ! ! = 5.31 rad/sθ! !




6370 kmR =

0 500 km 6370 kmr = +

0 6870 kmr =

66.87 10 m= ×

0 36,900 km/hv =

6

336.9 10 m3.6 10 s

×=×

310.25 10 m/s= ×


0 0 1 0 min 1 max, ,Ar mv r mv r r r r= = =

( )6

300

1 1

6.870 10 10.25 10ArV vr r′

×= = ×

9

1

70.418 10AV

r′×= (1)


Point A

30 10.25 10 m/sv = ×

( )22 30

1 1 10.25 102 2AT mv m= = ×

( )( )( )652.53 10 JAT m= ×

( )( )22 2 6

09.81 m/s 6.37 10 mA

GMmV GM gRr

= − = = ×

12 3 2398 10 m /sGM = ×

60 6.87 10 mr = ×

continued



( )

( ) ( )12 3 2

66

398 10 m /s57.93 10 m J

6.87 10 mA

mV

×= − = − ×

×

Point A′

212A AT mv′ ′=

( )12

1 1

398 10 m JAGMmV

r r′×= − = −

A A A AT V T V′ ′+ = +

652.53 10 m× 657.93 10 m− × 12

m=12

2 398 10A

mv ′×−

1r

Substituting for Av ′ from (1)

( )

( )( )

29 126

211

70.418 10 398 105.402 102 rr

× ×− × = −

( )21 12

62

11

2.4793 10 398 105.402 10rr

× ×− × = −

( ) ( )6 2 12 211 15.402 10 398 10 2.4793 10 0r r× − × + × =

6 61 66.7 10 m, 6.87 10 mr = × ×

max 66,700 kmr = !




The cord will not go slack if 2v is perpendicular to the undeformed cord length, 0,L at 2


1 2 2 1 00.80.8 0.6 1.3330.6

v v v v v= = =


Point 1 2 21 0 1 0 0

1 0.352

v v T mv v= = =

( ) ( )( )2 21 0

1 1 150 N/m 0.8 m 0.6 m2 2

V k L L= − = −

1 3JV =

Point 2 2 22 2 2

1 0.352

T mv v= =

2 21 1 2 2 20 0 : 0.35 3 0.35 0BL V T V T V v v∆ = = + = + + = +

From conservation of angular momentum 2 1.3158 Bv v=

( )2200.35 1.3158 1 3v − =

( )( )( )

2 2 20

3J11.72 m /s

0.35 kg 0.7313v = =

0 3.42 m/sv = "

continued



The ball travels in a straight line after the cord goes slack.


( )( )0.8 1.71 dv=

1.368dv

=


1 1.71 m/sv =

Point 1

( )( )221 1

1 1 0.7 kg 1.71 m/s 1.0234J2 2

T mv= = =

( ) ( )( )2 21 0

1 1 150 N/m 0.8 m 0.6 m 3J2 2

V k L L= − = − =

Point 3 2 23 3

1 0.352

T mv v= =

3 0V =

21 1 3 3 : 1.0234 3 0.35 0T V T V v+ = + + = +

3.39 m/sv =

From conservation of momentum

1.368 1.368 404 mm3.39

dv

= = =

404 mmd = "




(a) Conservation of angular momentum: About O

00.8 0.27v v=

02.963v v=


Point 1 2 21 0 1 0 00.35

2v v T mv v

1= = =

( ) ( )( )2 21 1 0

1 1150 N/m 0.8 m 0.6 m

2 2V k L L= − = −

1 3 JV =

Point 2 2 22 2

10.35

2v v T mv v= = =

( )2 0 cord is slackV =

2 21 1 2 2 0: 0.35 3 0.35 0T V T V v v+ = + + = +

From conservation of angular momentum, 03.125v v=

( )2200.35 3.125 1 3v − =

( )

( )( )20

3J

0.35 kg 8.7656v =

2 2 20 0.9779 m /sv =

0 0.989 m/sv = "

(b) Maximum velocity occurs when the ball is at its minimum distance from O, (when 0.27 m)d =

( )( )03.125 3.125 0.9889 3.09 m/smv v= = =

3.09 m/smv = "





AA A B B B A

B

rmr v mr v v vr

= ⇒ = (1)


2 21 12 2A B

A B

GMm GMmmv mvr r

− = − (2)

Put (1) into (2) and solve for Av

( )2 2 BA

A B A

GMrvr r r

=+

(3)

Given data 31740 8 1748 km = 1.748 10 mA Ar R h= + = + = ×

At B 31740 140 1880 km = 1.880 10 mB Br R h= + = + = ×

moon earth0.0123M M=

( ) ( ) 2moon earth0.0123 0.0123GM GM gR= =

( )( )26 12 3 20.0123 9.81 6.370 10 4.8961 10 m /s= × = ×

(a) Speed at A

( )( )( )

12 62

6 6 6

2 4.8961 10 1.88 102.9029

1.748 10 1.880 10 1.748 10Av

× ×= =

× × + ×

1703.8 m/sAv = 1704 m/sAv = !



(b) At B

( )1748 1703.8 1584.2 m/s1880

AB A

B

rv vr

= = =

The command module is in a circular orbit

At 61.880 10 mBr = ×

112 2

circ 64.8961 10 1613.8 m/s1.880 10B

GMvr

×= = = ×

( )Relative velocity 1613.8 1584.2 29.6 m/scirc Bv v= − = − =

Relative velocity 29.6 m/s=




From 13.100 AB A

B

rv vr

= (1) and 2 2( )

BA

A A B

GMrvr r r

=+

(2)

Given data 264 km;Ah = 6370 km + 264 km 6634 kmAr = =

66.634 10 m= ×

35780 km;Bh = 6370 km 35780 km 42,150 kmBr = + =

642.150 10 m= ×

( )( )22 6 129.81 6.37 10 398.06 10GM gR= = × = ×

Substitute into (2) ( ) ( )

( ) ( )12 6

26 6 6

2 398.06 10 42.150 10

6.634 10 42.150 10 6.634 10Av

× ×=

× × + ×

6 2 2103.69 10 m /s= ×

10,183 m/sAv =

Substitute into (1) ( )6634 10,183 1602.7 m/s42,150Bv = =

At A we have a circular orbit 12

6398.06 10 7746.2 m/s6.634 10circ

A

GMvr

×= = =×

Relative velocity ( )10,183 7746.2A circv v− = −

Relative velocity 2.44 km/s= "

At B

112 2

6398.06 10 3073.1 m/s42.15 10circ

B

GMvr

×= = = ×

Increase in velocity 3073.1 1602.7 1470 m/scirc Bv v= − = − = Increase in 1.470 km/sv = "




M = Mass of the sun; 2;EGM gR= R = 3960 mi 620.9088 10 ft= ×

( ) ( ) ( )22 2 6 21 3 2= 332.8 10 32.2 ft/s 20.9088 10 ft 4.6849 10 ft /sGM × = ×

Circular orbits Earth 6 97,677 ft/s(93 10 )(5280)E

GMv = =×

Mars 6 79187 ft/s(141.5 10 )(5280)M

GMv = =×


Elliptical orbit ( ) ( )93 141.5A Bv v=


( ) ( ) ( ) ( )2 2

6 61 12 293 10 5280 141.5 10 5280A B

GM GMv v− = −× ×

141.5 1.521593A B Bv v v = =

( ) ( ) ( ) ( ) ( )21 21

2 2 26 6

1 4.6849 10 1 4.6849 101.52152 293 10 5280 141.5 10 5280

B Bv v× ×− = −× ×

2 90.6575 3.270 10Bv = ×

70524 ft/s;Bv = 107,303 ft/sAv =

(a) Increase at A, 107303 97677 9626 ft/s = 1.823 mi/sA Ev v− = − = !

(b) Increase at B, 79187 70524 8663 ft/s = 1.641 mi/sB Mv v− = − = !




R = planet radius

; A A B Br R h r R h= + = +


( ) ( )A A B Bv R h v R h+ = +

( ) ( )5 1200 mi 1.2 16300 mi+ = +R R

5 6000 1.2 19560; 3.8 13560+ = + =R R R

(a) 3568.4 miR = = 3570 miR !

Conservation of energy mass of spacecraft=sm

( )( )( )

( )( )( )( )

2 25 5280 1.2 5280

2 3568.4 1200 5280 2 3568.4 16300 5280

GM GM− = −+ +

6 66 6348.48 10 20.072 10

25.177 10 104.905 10× − = × −

× ×GM GM

15 3 210.879 10 ft /s= ×GM

Using 9 4 434.4 10 ft /lb sG −= × ⋅

(b) 21316.26 10 slugs= ×M 21316 10 slugs= ×M ! (planet is Venus)

( ) ( )2 21 1

;2 2

+ = + − = −+ +

s sA A B B s A s B

A B

GMm m GMT V T V m v m v

R h R h




Elliptical orbit between A and B


A A B Bmr v mr v=

7.1706.690

BA B B

A

rv v vr

= =

66370 km 320 km 6690 km, 6.690 10 mA Ar r= + = = ×

1.0718A Bv v= (1)

66370 km 800 km 7170 km, 7.170 10 mB Br r= + = = ×

( ) 66370 km 6.37 10 mR = = ×


( )( )22 2 6 12 3 29.81 m/s 6.37 10 m 398.060 10 m /sGM gR= = × = ×

Point A

( )( )

122

6

398.060 101 2 6.690 10

A A AA

mGMmT mv Vr

×= = − = −

×

659.501 10 mAV = ×

Point B

( )( )

122

6

398.060 101 2 7.170 10

B B BB

mGMmT mv Vr

×= = − = −

×

655.5 10 mBV = ×

A A B BT V T V+ = +

continued



2 6 2 61 159.501 10 m 55.5 10 m2 2A Bmv mv− × = − ×

2 2 68.002 10A Bv v− = ×

From (1) ( )22 61.0718 1.0718 1 8.002 10A B Bv v v = − = ×

2 6 2 253.79 10 m /s , 7334 m/sB Bv v= × =

( )( )1.0718 7334 m/s 7861 m/sAv = =

Circular orbit at A and B

(Equation 12.44)

( )12

6398.060 10 7714 m/s

6.690 10A CA

GMvr

×= = =×

( )12

6398.060 10 7451 m/s

7.170 10B CB

GMvr

×= = =×

(a) Increases in speed at A and B

( ) 7861 7714 147 m/sA A A Cv v v∆ = − = − = !

( ) 7451 7334 117 m/sB B BCv v v∆ = − = − = !

(b) Total energy per unit mass

( ) ( ) ( ) ( )2 2 2 21/2 A A B BC CE m v v v v = − + −

( ) ( ) ( ) ( )2 2 2 21/ 7861 7714 7451 73342

E m = − + −

6/ 2.01 10 J/kgE m = × !




(a) 6185 10 mAr = ×

6295 10 mBr = ×

Speed of spacecraft in the elliptical orbit after its speedAv′ = has been decreased

Elliptical orbit between A and B conservation of energy

Point A 2 sat1 , 2A A A

A

GM mT mv Vr

= = −

sat mass of saturnM =

Determine satGM from the speed of Tethys in its circular orbit.

(Equation 12.44) 2satcirc sat circ B

GMv GM r vr

= =

( )( )26 3 15 3 2sat 295 10 m 11.3 10 m/s 37.67 10 m /sGM = × × = ×

( )( )

15 3 29

6

37.67 10 m /s0.2036 10 m

185 10 mA

mV

×= − = − ×

×

Point B

( )( )

15 3 22 sat

6

37.67 10 m /s1 2 295 10 m

B B BB

mGM mT mv Vr

×= = − = −

×

90.1277 10BV = − ×

;A A B BT V T V+ = +

2 9 2 91 10.2036 10 0.1277 102 2A Bmv m mv m− × = − ×

2 2 90.1518 10A Bv v′ − = ×

continued



Conservation of angular momentum 6

6185 10 0.6271295 10

AA A B B B A A A

B

rr mv r mv v v v vr

×= = = =′ ′ ′ ′×

( )22 91 0.6271 0.1518 10 , 15817 m/sA Av v ′ ′− = × =

21000 15817 5183 m/s 5.18 km/sA A Av v v′∆ = − = − = = !

(b) ( )( )0.6271 15817 9919 m/s,AB A

B

rv vr

′= = =

9.92 km/sBv = !





2 20

1 1

2 2 AB A

GMm GMmmv mv

r r− = −

So 2 20

21 B

AB A

GM rv v

r r

= − −

(1)

Given

66370 km = 6.37 10 mR = ×

( )( )22 6 129.81 6.37 10 398 10GM gR= = × = ×

66370 360 6730 km 6.73 10 mAr = + = = ×

66370 60 6430 km 6.43 10 mBr = + = = ×

Substitute into (1)

( )12 6

2 20 6 6

2 398 10 6.43 101

6.43 10 6.73 10Av v

× ×= − − × ×

2 2 60 5.518 10Av v= − × (2)

We need another equation → conservation of angular momentum

sinB A Ar mv r mvφ =

6

006

sin 6.43 10sin 50

6.73 10B

AA

r vv v

r

φ ×= = ° ×

00.7319Av v= Substitute into (2)

( )2 2 60 00.7319 5.518 10v v= − ×

2 600.46433 5.518 10v = ×

0 3477 m/sv =

0 3450 m/sv = "




21Let constant2 A

A

T V GME vm r+= = = −

km m 130,000 1000hr km 3600 s/hrAv

=

8333.33 m/sAv =

( )( )22 2 69.81 m/s 6.37 10 mGM gR= = ×

12 3 2398.059 10 m /sGM = ×

6 6 64.3 10 m 4.3 10 m + 6.37 10 mAr R= × + = × ×

610.67 10 mAr = ×

( )12

26

1 398.059 108333.32 10.67 10

E ×∴ = −×

3 2 22584.19 10 m /sE = − ×

/ constant = sin 60A Ah m v r= °

( ) ( ) ( )6/ 8333.33 m/s 10.67 10 m 3/2h m = ×

9 2/ 77.0041 10 m /sh m = ×

At min or max altitude, / ,h m r v= /h mvr

∴ =

Eliminate

2

2/ 1 1 /:2 2

h m GM h m GMv E vr R r r

= = − = −

Multiply by r2: 2

2 12

hEr GMrm

= −



And rearrange ( )2

2 1 02

hEr GM rm

+ − =

Quadratic formula for minimum r: ( ) ( )2 2

min2 /

2GM GM E h m

rE

− + +=

( ) ( )( )( )

2 212 24 3 18

min 3

398.06 10 398.06 10 2 2584.2 10 77.004 10

2 2584.2 10r

− × + × + − × ×=

− ×

( )12 12

min 3398.06 10 357.50 10

2 2584.2 10r − × + ×=

− ×

6min 7.848 10 mr = ×

Minimum altitude 67.848 10 R= × −

6 67.848 10 m 6.37 10 m= × − ×

61.478 10 m= ×

Minimum altitude 1478 km= !




At A: ( )( ) ( )( )6.5 5280 3960 mi + 567 mi 5280 ft/miAh v r = =

9 2820.336 10 ft /sAh = ×

( )( ) 63960 mi 5280 ft/mi 20.9088 10 ftR = = ×

( ) 21 1

2A AGM

T V vm r

+ = −

( )( )( )

( )( )

262 32.2 20.9088 101

6.5 5280 02 3960 567 5280

× = − ≅ +

Parabolic orbit

At B: ( ) 21 10

2B B BB

GMT V v

m r+ = − =

( )( )

262

32.2 20.9088 101

2 3960 5190 5280Bv

× = +

2 6582.76 10 ; 24140 ft/sB Bv v= × =

(a) 4.57 mi/sBv = !

9sin 820.336 10B B B Bh v rφ= = ×

( )( )

9820.336 10sin

24140 3960 5190 5280Bφ ×=

+

0.7034=

(b) 44.7Bφ = ° !




6, 3960 mi 377 mi = 4337 mi = 22.8994 10 ftA A A Ah v r r= = + ×

( )( )2.97 mi/s 5280 ft/mi 15681.6 ft/sCv = =

( )( ) 63960 5280 20.9088 10 ftR = = ×

615681.6 22.8994 10C C Ah v b b v = = = × (1)

( )2

61

2 22.8994 10A

A Av GMT V

m+ = −

×

2632.2 20.9088 10GM = ×

2

62 22.8994 10Av GM= −

× (2)

Two equations in two unknowns: , .Av b

Solve ;

31361 ft/sAv =

645.8 10 ftb = ×

(a) 5.94 mi/sAv = "

(b) 8670 mib = "




61080 87 1167 mi 6.1618 10 ftAr = + = = × 61080 mi 5.7024 10 ftCr R= = = ×

2moon 0.0123 0.0123E EGM GM gR= =

( )( )20.0123 32.2 3960 5280= ×

14 3 21.7315 10 ft /s= ×

At 87 mi: mooncirc 5301.0 ft/s

A

GMvr

= =

(a) An elliptic trajectory between A and C, where the lem is just tangent to the surface of the moon, will give the smallest reduction of speed at A which will cause impact.

2 61 28.101 10 m2

mA A A

A

GM mT mv Vr

= = − = − ×

2 61 30.364 10 m2

mC C C

C

GM mT mv Vr

= = − = − ×

2 61: 28.101 10 m2A A C C AT V T V mv+ = + − ×

2 61 30.364 10 m2 Cmv= − ×

2 2 64.526 10A Cv v= − × (1)



Conservation of angular momentum: A A C Cr mv r mv=

6.1618 1.08065.7024

AC A A A

C

rv v v vr

= = =

( )22 61.0806 4.526 10 5195.1 ft/sA A Av v v= − × ⇒ =

( )circ 5343.9 5195.1 148.8A A Av v v∆ = − = − =

148.8 ft/sAv∆ = !

(b) Conservation of energy (A and B)

Since B Cr r= conservation of energy is the same as between A and C

Conservation of angular momentum:

sin ; 45A A B Br mv r mv φ φ= = °

6.1618 1.5281sin 45 5.7024 0.70711

A A AB A

B

r v Vv vr

= = = °

From (1)

( )22 61.5281 4.526 10 1841.4 ft/sA A Av v v= − × ⇒ =

( )circ 5343.9 1841.4 3487.3A A Av v v∆ = − = − =

3503 ft/sAv∆ = !




For circular orbit of radius 0r

20

200

nGMm vF ma m

rr= =

20

0

GMvr

=

But 0v forms an angle α with the intended circular path

For elliptic orbit Conservation of angular momentum

0 0 cos A Ar mv r mvα =

00cosA

A

rv vr

α

=

(1)


2 20

0

1 12 2 A

A

GMm GMmmv mvr r

= = =

2 2 00

0

2 1AA

GM rv vr r

− = −

Substitute for Av from (1) 2

2 20 00

0

21 cos 1A A

r GM rvr r r

α − = −

But 20

0

GMvr

= thus 2

20 01 cos 2 1A A

r rr r

α

− = −

22 0 0cos 2 1 0

A A

r rr r

α

− + =



Solving for 0

A

rr

20

2 22 4 4cos 1 sin

2cos 1 sinA

rr

α αα α

+ ± − ±= =−

( )( ) ( )0 01 sin 1 sin

1 sin1 sinAr r r

α αα

α+ −

= =±

∓

also valid for point A′

Thus

( )max 01 sinr rα= + ( )min 01 sinr rα= − !




6370 km 362 km 6732 kmAr = + =

66.732 10 mAr = ×

6370 km 64.4 km 6434.4 kmBr = + =

66.4344 10 mBr = ×

66370 km 6.37 10 mR = = ×

( )( )22 2 69.81 m/s 6.37 10 mGM gR= = ×

9 3 2398.06 10 m /sGM = ×

Conservation of energy 9

2 66

1 398.06 10 59.130 10 m2 6.732 10A A A

A

GMm mT mv Vr

×= = − = − = − ××

92 6

61 398.06 10 61.86 10 m2 6.4344 10B B B

B

GMm mT mv Vr

×= = − = − = − ××

2 6 2 61 1: 59.130 10 61.86 102 2A A B B A BT V T V mv m mv m+ = + − × = − ×

2 2 65.4609 10A Bv v= − × (1)




sinA A B B Br mv r mv φ=

( )( )( )

6732 1 1.208sin 6434.4 sin 60

A AB A B A

B B

r vv v v v

r φ = = = °

(2)

Substitute Bv from (2) in (1)

( ) ( )2 22 6 2 61.208 5.4609 10 ; 1.208 1 5.4609 10A A Av v v = − × − = ×

2 6 2 211.8905 10 m /sAv = ×

3.448 km/sAv =

(a) 3.45 km/sAv = !

From (2) ( )1.208 1.208 3.45 km/s 4.1655 km/sB Av v= = =

(b) 4.17 km/sBv = !




6370 km 362 kmAr = +

6732 km=

6370 km 64.4 kmBr = +

6434.4 km=

( ) 22 2 69.81 m/s 6.37 10 mGM gR = = ×

15 3 20.39806 10 m /sGM = ×

Velocity in circular orbit at 362 km altitude

2A

GMmFr

=

( )2circ

2A

nA

m va

r=

Newton�s second law

( )2circ

2 An

AA

m vGMmF marr

= =

( )15

36circ

0.39806 10 7.69 10 m/s6.732 10A

A

GMvr

×= = = ××

Energy expenditure

From Problem 13.112, 33.448 10 m/sAv = ×



Energy, ( )2 2112 circ

1 12 2A AE m v mv∆ = −

( ) ( )2 23 3112

1 17.690 10 3.448 102 2

E m m∆ = × − ×

6112 23.624 10 m JE∆ = ×

( )6

113 112

23.624 10 m0.50 J

2E E

×∆ = ∆ =

Thus, additional kinetic energy at A is,

( ) ( )62

113

23.624 10 m12 2Am v E

×∆ = ∆ = (1)

Conservation of energy between A and B

( ) ( )2 2circ

1 2A A A A

A

GMmT m v v Vr

= + ∆ = −

21 2B B B

B

GMmT mv Vr

= = −

A A B BT V T V+ = +

( )6 1523

61 23.624 10 m 0.3980 10 m7.690 102 2 6.732 10

m × ×× + −×

15

26

1 0.39806 10 m2 6.434 10Bmv ×= −

×

2 6 6 6 659.136 10 23.624 10 118.26 10 123.74 10Bv = × + × − × + ×

2 688.240 10Bv = ×

9.39 km/sBv = !

Conservation of angular momentum between A and B

( )circ sinA A B B Br m v r mv φ=

( )( )

( )( )

( )( )

3circ

3

7.690 106732sin 0.8565

6434.4 9.394 10AA

BB B

vrr v

φ×

= = = ×

58.9Bφ = °!





A A P Pr mv r mv=

PA P

A

rv v

r= (1)


2 21 1

2 2P AP A

GMm GMmmv mv

r r− = − (2)

Substituting for Av from (1) into (2)

2

2 22 2PP P

P A A

GM r GMv v

r r r

− = −

22 1 1

1 2PP

A P A

rv GM

r r r

− = −

2 2

22 2A P A P

PA PA

r r r rv GM

r rr

− −=

With ( )( )2 2A P A P A Pr r r r r r− = − +

2 2 AP

A P P

GM rv

r r r

= +

(3) !

Exchanging subscripts P and A

( )2 2 QEDP

AA P A

GM rv

r r r

= +

!




See solution to Problem 13.113 (above) for derivation of Equation (3)

( )2 2 AP

A P P

GM rvr r r

=+

Total energy at point P is

212P P P

P

GMmE T V mvr

= + = −

( )0

1 22

A

A P P

GMm r GMmr r r r

= −

+

( )1A

P A P P

rGMmr r r r

= − +

( )( )

A A P

P A P

r r rGMm

r r r− −

=+

A P

GMmEr r

= −+

"

Note: Recall that gravitational potential of a satellite is defined as being zero at an infinite distance from the earth.




(a) For a circular orbit of radius r

2

2:nGMm vF ma m

rr= =

2 GMvr

=

21 12 2

GMm GMmE T V mvr r

= + = − = − (1)

Thus E∆ required to pass from circular orbit of radius 1r to circular orbit of radius 2r is

1 21 2

1 12 2

GMm GMmE E Er r

∆ = − = − +

( )2 1

1 22GMm r r

Er r

−∆ = (2) (Q.E.D.)

(b) For an elliptic orbit we recall Equation (3) derived in

Problem 13.113 ( )1with Pv v=

( )2 21

1 2 1

2Gm rvr r r

=+

At point A: Initially spacecraft is in a circular orbit of radius 1r

2circ

1

GMvr

=

2circ circ

1

1 12 2

GMT mv mr

= =



After the spacecraft engines are fired and it is placed on a semi-elliptic path AB, we recall

( )2 21

1 2 1

2GM rvr r r

= ⋅+

And ( )2 2

1 11 1 2

1 1 22 2

GMrT mv mr r r

= =+

At point A, the increase in energy is

( )2

1 circ1 1 2 1

1 2 12 2A

GMr GME T T m mr r r r

∆ = − = −+

( )( )

( )( )

2 1 2 2 1

1 1 2 1 1 2

22 2A

GMm r r r GMm r rE

r r r r r r− − −

∆ = =+ +

( )2 12

1 2 1 22AGMm r rrE

r r r r −

∆ = +

Recall Equation (2): ( ) ( )2

1 2 Q.E.DA

rE Er r

∆ = ∆+

A similar derivation at point B yields,

( ) ( )1

1 2 Q.E.DB

rE Er r

∆ = ∆+




If the point of intersection 0P of the circular and elliptic orbits is at an end of the minor axis, then 0v is parallel to the major axis. This will be the case only if 090 ,α θ+ ° = that is if 0cos sinθ α= − we must therefore prove that

0cos sinθ α= − (1)

We recall from Equation (12.39):

21 cosGM Cr h

θ= + (2)

When 0,θ = ( )min min 0 and 1 sinr r r r α= = −

( ) 20

11 sin

GM Cr hα

= +−

(3)

For 180 ,θ = ° ( )max 0 1 sinr r r α= = +

( ) 20

11 sin

GM Cr hα

= −+

(4)

Adding (3) and (4) and dividing by 2:

2 20 0

1 1 1 12 1 sin 1 sin cos

GMrh rα α α = + = − +

Subtracting (4) from (3) and dividing by 2:

20 0

1 1 1 1 2sin2 1 sin 1 sin 2 1 sin

Cr r

= − = − + −

αα α α

20

sincos

Cr

αα

=

Substitute for 2GMh

and C into Equation (2)

( )20

1 1 1 sin coscosr r

α θα

= + (5)



Letting 0 0 and r r θ θ= = in Equation (5), we have

20cos 1 sin cosα α θ= +

2 2

0cos 1 sincos sin

sin sinα αθ αα α

−= = − = −

This proves the validity of Equation (1) and thus 0P is an end of the minor axis of the elliptic orbit.




(a) Ar R=


0 0sin B BRmv r mvφ =

( )1Br R R Rα α= + = +

( ) ( )0 0 0 0sin sin

1 1BRv vv

Rφ φ

α α= =

+ + (1)

Conservation of Energy

( )2 20

1 1 2 2 1A A B B B

GMm GMmT V T V mv mvR Rα

+ = + − = −+

2 20

2 1 211 1B

GMm GMmv vR R

αα α

− = − = + +

Substitute for Bv from (1)

( )2

2 00 2

sin 2111

GMmvR

φ ααα

− = + +

From Equation (12.43): 2esc

2GMvR

=

( )2

2 200 esc2

sin111

v vφ ααα

− = + +

( )

220 esc2

0

sin 111

vv

φ ααα

= − ++

(2)

( )2

esc0

0sin 1 1 Q.E.D.

1vv

αφ αα

= + − +



(b) Allowable values of 0v ( )for which maximum altitude Rα=

200 sin 1φ≤ ≤

For 0sin 0,φ = from (2)

2esc

00 1

1vv

αα

= − +

0 esc 1v v α

α=

+

For 0sin 1,φ = from (2)

( )

2esc

20

1 111

vv

ααα

= − ++

( )

2 2esc

0

1 1 1 2 1 211 1 1

vv

α α ααα α α α α

+ + − + = + − = = + + +

0 esc12

v v αα

+=+

esc 0 esc1

1 2v v vα α

α α+≤ ≤

+ +!




6

0 60m dt m+ =∫v F v

( ) ( )660

30 20sin 2 24cos 2

32.2t t dt + + = ∫ i j v

[ ]660

30 10cos 2 12sin 2

32.2t t+ − + =i j v

[ ] [ ] 610 cos12 cos0 12 sin12 sin 0 0.09317− − + − =i j v

61.5615 6.4389 0.09317− =i j v

[ ]6 16.759 69.109 ft/s= −v i j

6 71.1 ft/sv = !

76.4°!




0dt m m= −∫F v v

( )( ) ( )2

04 8 2 4 2 8 2

tt dt t t t− − = = − −∫ i j v i j

( )20.5 2 0.5 m/st t t= − −v i j

( ) ( ) ( )22 2 22speed 0.5 2 0.5v t t t= = − +

4 3 2 20.25 2 4 0.25t t t t= − + +

2 20.25 2 4.25t t t = − +

( ) [ ]2 2 2

3 2

speed 2 0.25 2 4.25 0.5 2

6 8.5 0

d t t t t tdt

t t t

= − + + −

= − + =

Roots: ( )6 36 4 8.5

0, 0,2

t v t± −

= = =

( )6 2 2.2929 s, 3.7071 s outside interval2

t ±= =

At 2.2929 s, 1.9571 1.1464t v= = − −i j

2.27 m/s, maxv = "

0




1 1 212 km/h 3.33 m/s 10 sv t −= = =

2 18 km/h 5.00 m/sv = =

1 1 2 2impulsemv mv−+ =

( ) ( ) ( )333 m/s 10 s 5.00 m/sNm F m+ =

( )( )440 kg 5.00 m/s 3.3333 m/s73.33 N

10 sNF−

= =

73.3 NNF = !

Note: NF is the net force provided by the sails. The force on the sails is actually greater and includes the force needed to overcome the water resistance of the hull.




(a) 0θ = °

k kFt Wt mgtµ µ= =

A k Bmv mgt mvµ− =

( )9 0.30 9.81 0t− =

3.06 st = !

(b) 20θ = °

cos 20 cos 20Nt Wt mgt= ° = °

kFt Ntµ=

cos 20 sin 20 0A kmv mg t mgtµ− ° − ° =

( )9 0.65 9.81 cos20 9.81sin 20 0t − ° + ° =

( )9 9.81 0.9528 0t− =

0.963 st = !




20,000 lbW =

220,000 621.118 lb s ft32.2

m = = ⋅

Momentum in the x direction

( )0 1: sin15x mv F mg t mv− + ° =

( ) ( ) ( )( )621.118 108 sin15 6 621.118 36F mg− + ° =

sin15 7453.4F mg+ ° =

(a) 7453.4 20,000 sin15 2277 lbF = − ° =

2280 lbF = !

(b) ( )0 sin15 0mv F mg t− + ° = t = total time

( )621.118 108 7453.4 0;t− = t = 9.00 s

Additional time = 9 � 6 = 3 s !




20,000 lbW =

220,000 621.118 lb s ft32.2

m = = ⋅

Momentum, ( )0 1: sin15x mv F mg t mv− + ° =

( ) ( )( ) 00621.118 sin15 5.5 621.118

2vv F mg − + ° =

( )( )0310.559 sin15 5.5v F mg= + ° (1)


( )2 20 1

1 1sin152 2

mv F mg x mv− + ° =

( ) ( )2

2 00

310.559310.559 sin15 5404

vv F mg− + ° =

( ) ( ) ( )

203 310.559

sin15 5404

vF mg= + °

Using (1) eliminate ( )sin15 :F mg+ °

( )2

00

3 310.559 5.5310.5594 180

vv =

(a) ( )( )( )0

4 540130.909 ft s 89.3 mi h

3 5.5v = = = !

(b) ( ) ( )( )0 310.559 130.909310.559+ sin15 7391.85.5 5.5

vF mg ° = = =

( )7391.8 20000 sin15 2215 lbF = − ° =

F = 2220 lb !




100 km h 27.777 m s2v = =

80 km h 22.222 m s1v = =

(a) ( ) ( ) ( )82 10

10008 100 803600

F dt F m v v m = = − = −

∫

0.69444F m= on the level

on the up grade

( ) ( )10100: sin 6 27.777x F mg dt m v− ° = −∫

( ) ( ) ( ) ( )100.69444 10 9.81 sin 6 10 27.777m m m v− ° = −

(a) 10 1027.777 3.3098, 24.468 m sv v− = − =

10 88.1 km hv = !

(b) ( ) ( )01000sin 6 60 1003600

t F mg dt m − ° = −

∫

0.69444 m m− ( )9.81 sin 6 t m ° = ( )11.111−

t = 33.6 s !

F = 0.69444 m




Impulse diagonally (assume sliding)

[ ]( ) ( ): cos 20 196.2 sin 20 0.3 N 6 20 15x P ° − ° − =

[ ]( ): N 196.2 cos 20 sin 20 6 0y P− ° − ° =

[ ] 15cos 20 196.2 sin 20 0.3 196.2 cos 20 + sin 20 20 506

P P ° − ° − ° ° = =

( )cos 20 0.3 sin 20 50 67.104 55.310P ° − ° = + +

205.97 NP = 206 NP = !

Check sµ

Static value ( )0.4 N = 0.4 196.2 cos 20 sin 20P° + °

cos 20 196.2 sin 20 0.4 N = 0P ° − ° −

( )cos 20 0.4 sin 20 196.2 sin 20 78.48 cos 20 140.85P ° − ° = ° + ° =

static = 175.4 N 206 NP <

W = (20)(9.81) = 196.2 N




Use impulse momentum 1 55mi h 80.667 ft sv = =

x-Direction

m 1 sv µ m− 0gt =

( )( )

12

s

80.667 ft s0.4 32.2 ft s

vtµ g

= =

6.26 st = !

Since this is the shortest time the load can be brought to rest and the load does not slide it is also the shortest time the rig can be brought to rest.




( )42 10

tt F dt m v v=

= = −∫

( ) ( ) ( )( )40 5 40 40 32.2 2 3P dt− = − −∫

( ) ( )( )20 40 4 40 32.2 5P − =

(a) ( )= 6.2112 + 160 20P

8.31 lbP = !

(b) ( ) ( ) ( )( )0 5 40 40 32.2 0 3t P dt− = − −∫

( )( )5 t 40 40 32.2 3P t =−

2.4 st = !




(a) Combined

90km h 25 m sv = =

( )( ) ( )( )1 26500 9.81 63765 N; 3600 9.81 35316 NW W= = = =

1 1 2 2 1; 0.75N W N W F N= = =

+⊗← Impulse = 0 � mv0

( )10.75 10,100 kg 25 m sN t− = −

( )( )( )

10,100 255.2798 s

0.75 63765t = = 5.28 st = !

(b) Second trailer alone

+⊗← Impulse 2= C t m v− =

( ) ( )5.2798 3600 kg 25 m sC− = −

= 17046 NC

17.05 kNC = !

Compression




(a) Entire train 1 72 km/h 20 m/sv = =

318 13 31 Mg 31 10 kgA Bm m+ = + = = ×

( ) ( )( )31 20 19000 N 19000 N 31 10 kg 20 m/st −= − + + ×

( )( )3

1 2

31 10 kg 20 m/s16.3158 s

38000 Nt −

×= =

1 2 16.32 st − = !

(b) Car A 31 218 Mg 18 10 kg; 16.32 sAm t −= = × =

( ) [ ] ( )( )0 19000 N 16.32 s 18000 kg 20 m/sCF = + +

3058.8 NCF =

3.06 kN TCF = !




(a) Entire train 1 72 km/h 20 m/sv = =

18 13 31 Mg 31000 kgA Bm m+ = + = =

( ) ( )( )1 20 19000 N 31000 kg 20 m/st −= − +

1 2 32.63 st − =

1 2 32.6 st − = !

(b) Car A

1 2 1 1 20 ; 32.63 sC AF t m v t− −= − + =

( )( )( )

18000 kg 20 m/s11033 N

32.63 sCF = =

11.03 kN TCF = !




Impulse diagrams

Constraint: Av 3 Bv=

A: x ( ) ( ) 2020 0.5 sin 30 0.532.2 AT v° − =

B: x ( ) ( ) ( )16 163 0.5 16 0.5 sin 30

32.2 32.2 3A

BvT v− ° = =

Substituting for T(0.5) from the equation for A into the equation for B

From A:

( )0.5 5 0.62112 AT v= −

A0.496915 1.8634 4

3Avv− − =

2.029 11Av =

vA = 5.4214

(a) 5.42 ft/sAv = 30° !

( ) ( )0.5 5 0.62112 5.4214T = −

3.2653 lbT =

(b) 3.27 lbT = !




( ) ( ) ( )3 C A C B Bl x x x x d x≅ − + − + −

Constraint: 4 2 3 0C B Av v v− − = (1)

For , andA B Cv v v +

At 0,t = ( ) ( )4 15ft/s 2 3 9ft/s 0, 16.5 ft/sB Bv v− − = =

(2) ( )203 4 9

32.2 ATt t v− = − (3) ( )204 4 15

32.2 CTt t v− − = − (4) ( )102 2 16.5

32.2 BTt t v− = −

Given 0.5 s 4t = ⇒ Equations in T, vA, vB, vC

3 2 4 0A B Cv v v− − + = (1)

1.5 0.6211 AT v− 3.5901= − (2)

2T− 0.6211 7.3168Cv− = − (3)

0.31056 BT v− 4.1242= − (4)

From the solution of the above equations.

(a) 6.07 ft/sAv = "

13.7 ft/sBv = "

11.4 ft/sCv = "

(b) 0.1212 lbT = "




Kinematics

Length of cable is constant.

2 A BL X X= +

2 0A BdL v vdt

= + =

2B Av v= −

( )2 0.6 m/sAv =

Collar A

15 kgAm =

( ) ( )( ) ( )1 2 1 21 22A A A Am v T t W t m v− −+ − =

( ) ( )( )1 20 2 15 9.81 15 0.6T t − + − × =

( ) 1 273.575 4.5T t −− = (1)

Collar B 10 kgBm =

( ) ( )2 22 1.2 m/sB Av v= =

( ) ( ) ( ) ( )1 2 1 21 2B A B B Bm v T t W t m v− −− + =

( ) ( ) ( )1 20 10 9.81 10 1.2T t − + × − = (2)

Add Equation (1) and Equation (2) (eliminating T)

( ) 1 298.1 73.575 4.5 12t −− = +

1 216.5 0.673 s24.52

t − = = 0.673 st = !




Lets find out if they slide � assume they don�t slide and find the required angle for impending motion

0; cos 0; cosy A A i A A iF N m g N m gθ θ= ⇒ − = ⇒ =∑

0; sin 0x s A A iF N m gµ θ= ⇒ − =∑

s mµ A g cos i mθ = A g sin iθ

tan 0.3i sθ µ= =

so 16.7iθ = ° so they slide

Assume they slide at the same velocity (remain in contact) impulse � momentum

-x dir

( ) ( ) ( )0 sinA B kA A kB B A Bm g m g t N N t m m vθ µ µ+ + − + = +

(a) Solve for v

( ) [ ]sin cos cosA B k A A k B B

A B

m g m g t m g m g tv

m mθ µ θ µ θ+ − +

=+

( )( )( ) ( )( ) ( ) ( )6 9 9.81 3 sin 20 0.25 6 0.15 9 9.81 cos 20 3

6 9 + ° − + ° =

+

4.811 m/sv = 4.81 m/sA Bv v= = !



(b) Just look at AB

x 0 + sinB AB k B B Bm gt F t N t m vθ µ− − =

So ( )sin cosB k B B BAB

m gt m g t m vF

tθ µ θ− −

=

( )( )( ) ( )( ) ( ) ( )( )9 9.81 3 sin 20 0.15 9 9.81 cos 20 3 4.81 93

°− ° −=

= 3.319 N

3.32 NABF = !

Since this is positive our assumption that the blocks stay in contact is correct




The block does not move until 22 kg 9.81 m/s 19.62 NP = × =

From 0 to 2 st t= = 20P t=

Thus, the block starts to move when 19.62

0.981s20

t = =

(a) For 0 2 st< <

20P t=

1 2 10.981s 2 s, 0t t v= = =

( )2

11 2 1 2

t

tmv Pdt W t t mv+ − − =∫

( ) ( )220.981

0 20 2 9.81 2 0.981 2t dt v+ − × − =∫

( ) ( ) ( )( )2 22

1 20 N2 s 0.981s 19.62 N 2 s 0.981s

2 kg 2 sv

= − − −

2 5.1918 m/sv =

2 5.19 m/sv = !

(b) From 2 s to 3 st t= =

2 5.19 m/s,v = from (a)

40 N 2 s 3 sP t= ≤ ≤

2 32 s 3 st t= =

( )3

22 3 2 3

t

tmv Pdt W t t mv+ − − =∫

( ) ( )( )332

2 5.1918 40 19.62 3 2 2dt v+ − − =∫

( ) ( ) ( )( )31

5.1918 m/s 20.38 N 1s2 kg

v = +

3 5.1918 10.19 15.3818 m/sv = + =

3 15.38 m/sv = !




(a) Determine time at which collar starts to move.

20 , 0 2 sP t t= < <

Collar moves when

2 19.622 kg 9.81 m/s , or 0.981s20 20PP t= × = = =

1 20.981 0.981t tmv Pdt Wdt mv+ − =∫ ∫

For ( )2 s 20 Nt P t< =

2 s 3 s 40 Nt P< < =

3 s 0t P> =

For 23 s 2 kg 9.81 m/st W< = ×

The maximum velocity occurs when the total impulse is maximum.

area maximum impulseABCD =

( )( ) ( )( )1 20.38 N 1.019 s 20.38 N 1s2

= +

area 30.76 N sABCD = ⋅

( ) max0 30.76 N s 2 kg v+ ⋅ =

max 15.38 m/sv = !

(b) Velocity is zero when total impulse is zero at .t t+ ∆

For ( )0.981s 3 s, impulse 19.62 N st t< < = − ∆ ⋅

Thus, total impulse 0 30.76 19.62 t= = − ∆

1.57 st∆ =

Time to zero velocity� 3 s 1.57 s 4.57 st = + = !




(a)

20 4P t= −

1 20

sin 30t

mv W t Pdt mv− + =∫

( ) ( )5

20

1212sin 30 5 20 4

32.2t dt v− ° + − =∫

52

2030 20 2 0.37267t t v − + − =

2 53.7 ft/sv = !

(b) After 5 s,t = 50 53.7 ft/s,P v t′= = is time after 5 s

5 12sin 30 0mv t′− =

( )1253.7 6 3.34 s; 5

32.2t t t t′ ′ ′= ⇒ = = +

8.34 st = !

0




20

1 2 1 0 3, where ,

1.6 10

pp C C t C p C −= − = =

×

16 oz1 lb

0.70 oz

0.04375m

g g

= =

( )2 20.4 0.12566 m4

Aπ= =

( ) ( )( )

31.6 101 20 2

0.04375 2100 ft/s0.12566 2.85326

32.2 ft/sC C t dt

−× − = =∫

31.6 102

1 20

122.706

2C t C t

−× − =

( ) ( )233 0

0 3

1.6 10 s1.6 10 s 22.706

2 1.6 10 s

pp

−−

−

×× − =

×

( )3

01.6 10 s22.706

2

p−×=

20 28.383 lb/inp =

0 28.4 ksip = !




( ) 22

12 oz 1 lb/16 oz 0.003882 lb s /ft32.2 ft/s

m

= = ⋅

Conservation of energy (before impact)

( )2 2 21 1 1

1 12 2 Aym v mgh m v v+ = +

( ) ( ) ( ) ( )2 2 21 154 32.2 4.5 542 2 Aym m m v+ = +

17.0235 ft/sAyv = (Just before impact)

Conservation of energy (after impact)

( ) ( )22 22 2 2

1 12 2Aym v v m v mg h+ = +′

( ) ( ) ( ) ( )2 221 130 30 32.2 32 2Aym v m m + = +′

13.8996 ft/sAyv =′ (Just after impact)

( ) ( ) ( ): 0.003882 54 0.004 0.003882 30 , 23.292 lbH Hx F F− = =

( ) ( ) ( ): 0.003882 17.0235 0.004 0.003882 13.8996 , 30.011 lbv vy F F− + = =

impulsive forceIF =



38.0 lbIF = 52.2° !




1 230 ft/s 36 ft/s 0.18 sv v t= = ∆ =

( )1 2mv P W t mv+ − ∆ =

Vertical components

( )( ) ( )0 0.18 36sin 50VW

P Wg

+ − = °

4.758VP W W− =

5.76VP W= !




Use impulse–momentum for bullet

Knowns: 0.028 kg,m = 1 650 m/s,v = 2 500 m/s,v =

-dirx

1 2cos 20 cos10xmv F t mv°− ∆ = °

So, 1 2cos 20 cos10xF t mv mv∆ = °− °

( ) ( )0.028 650 cos 20 0.028 500 cos10 3.3151 N s= °− °= ⋅

-diry

1 2sin 20 sin10ymv F t mv− °+ ∆ = °

So, 2 1sin10 sin 20yF t mv mv∆ = °+ °

( ) ( )0.028 500 sin10 0.028 650 sin 20 8.6558 N s= °+ °= ⋅

We need ∆t. The average velocity is 600 m/s

6ave

ave

0.05 m; 83.33 10 s

600 m/s

xx v t t

v−∆∆ = ∆ ∆ = = = ×

So

6

6

3.315139.78 kN

83.33 10111.2 kN

8.6558F 103.87 kN

83.33 10

x

y

F

F−

−

= = × == =×

!





( ) ( ) ( ) ( ) ( )2

2

0.15 18 0.15 0.5 0.5

1.62 rad/s

m m θ

θ

=

=

&

&


( ) ( )( )22 21 10.15 18 0.5 1.622 2

m m R = + &

2.5756 m/sR =&

Motion relative to the rod:

( ) ( )1.5 2.5756 0F t− ∆ =

3.86 N sF t∆ = ⋅ !




2 38 m/sv =

01 20

tmv Fdt mv+ =∫

( )( )30.5 10

300 sin 0.045 kg 38 m/s0.5 10m

tF dtπ−×−+ =

×∫

5.37 kNmF = !




(a) Force on the belt is opposite to the direction shown.

1 72 km/h 20 m/s, 100 kgv m= = =

1 2 avemv Fdt mv Fdt F t− = = ∆∫ ∫

( )( ) ( )ave100 kg 20 m/s 0.110 s 0 0.110 sF t− = ∆ =

( )( )( )ave100 20

18182 N0.110

F = =

ave 18.18 kNF = !

(b) Impulse area under diagramF t= −

( )10.110 s

2 mF=

From (a) aveImpulse F t= ∆

( )( )18182 N 0.110 s=

( ) ( )10.110 18182 0.110

2 mF =

36.4 kNmF = !




( ) ( ) 21100 2000 6211.18 lb s /ft

32.2Bm = = ⋅

( ) ( ) 21120 2000 7453.42 lb s /ft

32.2Tm = = ⋅

( )0 6211.18B B BF t m v v+ ∆ = =

( )6T T Tm F t m v− ∆ =

( ) ( ) ( )7453.42 6 7453.42 TF t v− ∆ =

constraint: / 5.4 ft/sT B T Bv v v= − =

Solving; ( ) ( ) [ ]7453.42 6 6211.18 5.4 7453.42T Tv v= − +

( )13664.6 78260.9Tv =

5.7273 ft/sTv =

(a) 5.73 ft/sTv = !

5.7273 ft/s 5.4 ft/s 0.3273 ft/sBv = − =

( ) ( )6211.18 0.3273 2032 lb sB BF t m v∆ = = = ⋅

(b) 2030 lb sF t∆ = ⋅ !




21 1 1lb 0.0625 lb; 0.001941 lb s /ft16 16 32.2B BW m = = = = ⋅

2block 8 lb; 0.248447 lb s /ftBW m= = ⋅

Initial impact (Bullet + Block)

( )0 block: cos30 0B Bx m v m m v°+ = + ′ (1)

0: sin 30 0By m v F t− °+ ∆ = (2)

After impact

( ) ( )( ) ( )block block: 32.2 1.2 sin15 0B Bx m m v m m′+ − + °=

sin15 10.001 ft/sv gt− °=′

From (1)

( )( ) ( )

block

08.0625 10.0010.0625cos30

cos30

B

B

W Wv

gv

Wg

+′

= = °

°

(a) 0 1489.7 ft/sv = 0 1490 ft/sv = !



(b) Bullet alone

( ) ( ) ( ) ( )0 0.001941 1489.7 2.8915; 0.001941 10.001 0.01941B Bm v m v′= = = =

x : 0 cos15 cos15B x Bm v F t m v° + ∆ = °′

y : 0 sin15 sin15B y Bm v F t m v− ° + ∆ = °′

Solve: 2.7742xF t∆ = −

0.7534yF t∆ = 2.87 lb sF t∆ = ⋅ !




/30

:3.59 m/sA A B B A

°= + = +v v v v Bv

[ ] ( )0 2 3.59cos30 10 0A Ax B Bx B Bm v m v v v+ = °− + − =

0.518 m/sBv = Just before impact

After impact, 0A Bv v= =

(a) A AF t m v∴ ∆ = − =

"

(b) ( ) ( )2 21 1Loss , just before impact 2 10

2 2A BT v v= = +

( )2

26.30385 0.51817 11.28 J

2T

= + = "




( )75 kg, 50 kg, 200 kg BoatA B Cm m m= = =

(a) Swimmers dive simultaneously

( ) 20 C C A Bm v m m v= + + (1)

Relative velocity of swimmers with respect to the boat is 3 m/s

2 23 m/s 3C Cv v v v− = ⇒ = +

Substitute into (1)

( ) ( )0 3C C A B Cm v m m v= + + +

Solve

( ) ( )

( )3 3 75 50

75 50 200A B

CA B C

m mv

m m m− + − +

= =+ + + +

1.154 m/sCv = !

(b) A dives first and then B

x-dir

( ) 2 20 C B C Am m v m v= + − (2)

continued

mC vC



Relative velocity

2 223 3C C Cv v v v− = ⇒ = +

Substitute into (2)

( ) ( )2 20 3C B C A Cm m v m v= + − +

Solve for 2Cv

2

3 AC

A B C

mvm m m

−=+ +

(3)

Now look at C and B.

-dir.x ( ) 2 3 3C B C C C Bm m v m v m v+ = + (4)

Relative velocity

2 33 33 3C Cv v v v− = ⇒ = +

Substitute into (4)

( ) ( )2 3 23C B C C C B Cm m v m v m v+ = + +

so 3 2

3C B BC C

C B C B

m m mv vm m m m

+= −+ +

(5)

Substituting (3) into (5)

3

3 3A BC

A B C C B

m mvm m m m m

−= −+ + +

(6)

with numbers

3

75 503 1.292375 50 200 200 50Cv

= − + = − + + +

3

1.292 m/sCv = !

(c) Swimmer B dives first � solution is the same as for (b) except switch mA and mB

3

3 3B AC

A B C C A

m mvm m m m m

−= −+ + +

50 753 1.28075 50 200 200 75

= − + = − + + +

3

1.280 m/sCv = !




ball3 1

0.00582316 32.2

m = =

plate14 1

0.02717416 32.2

m = =

( )2 4.8 17.582 ft/syv g= = ( )2 1.8 10.7666 ft/syv g= =′

(a) Conservation of momentum

ball ball plate plate0y ym v m v m v+ = − +′ ′

( ) ( ) ( ) ( ) ( ) plate0.005823 17.582 0 0.005823 10.7666 0.027174 v+ = − + ′

plate 6.0747 ft/sv =′ plate 6.07 ft/sv =′ !

(b) Energy loss

Initial energy ( ) ( ) ( ) ( ) ( )2

1

10.005823 6 0.005823 4.8 1.0048

2T V g+ = + =

Final energy ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2

2

1 10.005823 6 0.005823 1.8 0.027174 6.0747

2 2T V g+ = + +

0.9437=

Energy lost ( )1.0048 0.9437 ft lb= − ⋅ = 0.0611 ft lb⋅ !




Before impact

( )( )( )21 10, 30 kg 9.81 m/s 2 m 588.6 NT V mgh= = = =

22 2

1 , 02

T mv V= =

( ) 21 1 2 2

1: 588.6 30 6.2642 m/s2

T V T V v v+ = + = ⇒ =

(a) Rigid columns

0mv F t− + ∆ =

( )30 6.2642 F t= ∆

187.93 N sF t∆ = ⋅ on the block

187.9 N sF t∆ = ⋅ !

All of the kinetic energy of the block is absorbed by the chain.

( )( )21 30 6.2642 588.6J2

T = =

589JE = !



(b) Elastic columns

Momentum of system of block and beam is conserved

( ) ( )30 6.2642 1.2528 m/s150

mmv M m v v vm M

′ ′= + = − = =+

Referring to figure in Part (a) mv F t mv′− + ∆ = −

( ) ( )30 6.2642 1.2528 150.34F t m v v′∆ = − = − =

150.3 N sF t∆ = ⋅ !

( ) ( ) ( )2 2 22 21 1 30 1206.2642 1.2528 1.25282 2 2 2

E mv mv ′= − = − −

565.06 94.170 470.89= − =

471JE = !




Before impact 2 21 1 (0.75) (6) 13.5J2 2H HT m v= = =

(a) For ,Am = ∞ 2 0T = So,

Energy absorbed = 13.5 J!

Impulse 2= ( ) (0.75) (6) 4.5 N sH Hm v v− = = ⋅ !

(b) 4kgAm =

y-dir : 2( )H H A Hm v m m v= +

2(0.75) (6) 0.9474 m/s(4 0.75)

H H

A H

m vvm m

= = =+ +

So 2 22 2

1 1( ) (4 0.75)(0.9474) 2.1316 J2 2A HT m m v= + = + =

Energy absorbed 1 2 13.5 2.1316T T= − = −

11.37 JE∆ = !

System = hammer

0



y-dir : 2H H Hm v F t m v− ∆ =

So 2( )H HF t m v v∆ = − 0.75(6 0.9474)= − 3.79= 3.79 N sF t∆ = ⋅ !




96 mi/h 140.8 ft/sv = =

5 / 0.3125/16

m g g= =

AVE

812 0.02667 s25

dtv

= = =

( ) ( )AVE0.3125 140.8 0.02667 0F

g+ − =

AVE 51.2 lbF = !




For the sphere at A′ immediately before and after the cord becomes taut

0 Amv F t mv ′+ ∆ =

0 sin 0 0.8 lb smv F t F tθ − ∆ = ∆ = ⋅

4mg

=

( ) 04 sin 65.38 0.8vg

° =

0 7.08 ft/sv = !




(a)

Conservation of total momentum

2A Bmv mv mv′− =

( )12 A Bv v v′ = − !

(b) Energy loss

( )L A B A BE T T T T′ ′= + − +

( ) ( )2 2 2 21 12 2L A BE m v v m v v′ ′= + − +

From (a)

( )12 A Bv v v′ = −

( ) ( )22 21 1 12 2 2L A B A BE m v v m v v = + − −

( ) ( )2 2 2 21 1 22 4L A B A A B BE m v v m v v v v= + − − +

( )22 21 124 4L A A B B A BE m v v v v m v v = + + = + !




Before impact After impact

A B A Bm v m v m v m v′ ′+ = + (1)

( )B A A Bv v e v v′ ′− = − (2)

From (1) and (2) solve for ,A Bv v′ ′ ( ) 0.5 ( )

2A B A B

Av v v v

v− − −′ =

( ) 0.5( )

2A B A B

Bv v v v

v+ + −′ =

( ) ( 3 ) / 4A A Ba v v v′ = + !

(3 ) / 4B A Bv v v′ = + !

(b) Loss of energy 2 2 2 2( ) ( )2 2A B A Bm m

v v v v′ ′= + − +

Loss of energy = 2 2 2 2 2 21( 6 9 9 6 )

2 16A B A A B B A A B Bm

v v v v v v v v v v + − + + + + +

Loss of energy 23( )

16 A Bm

v v= − !




System = A + B

(a) x-dir

A A B B A A B Bm v m v m v m v′ ′− = + (1)

Unknowns ,B Av v′ ′

Coefficient of restitution

( )Br Ar Ar Brv v e v v′ ′− = −

For our problem

( )B A A Bv v e v v′ ′− = + (2)

With numbers 0.6; 0.9; 4 m/s; 2 m/sA B A Bm m v v= = = =

Solve 2 equations and 2 unknowns

2.3 m/s; 2.2 m/sA Bv v′ ′= − = 2.3 m/sAv =′ !

2.2 m/sBv =′ !

(b) Energy lost

2 2 2 21

1 1 1 1(0.6) (4) (0.9) (2) 6.6 J

2 2 2 2A A B BT m v m v= + = + =

( ) ( )2 2 2 22

1 1 1 1(0.6)(2.3) (0.9) (2.2) 3.765 J

2 2 2 2A A B BT m v m v= + = + =′ ′

1 2 2.835JE T T∆ = − =

2.84 JE∆ = !




System = A + B

x-dir

A A B B A A B Bm v m v m v m v′ ′− = + (1)

Unknowns , Ae v′


( )B A A Bv v e v v′ ′− = + (2)

Where, 4 m/s; 2 m/s; 2.5 m/sA B Bv v v′= = =

0.6 kg; 0.9 kgA Bm m= =

Solve 2 equations and 2 unknowns

2.75 m/s; 0.875Av e= − =′

0.875e = !





A A B B A A B Bm v m v m v m v′ ′+ = +

1.2 2.4 1.2 0A Agv g gv

− = − +′

g�s cancel (1)

From restitution

0.8 , 0.8 14.418

AA A

A

v v vv

′ ′= = ++

(2)

(a) Velocity of A before impact from equations (1) and (2)

1.2 43.2 1.2(0.8 14.4) 0.96 17.28A A Av v v− = − + = − −

2.16 25.92Av =

12 ft/sAv = !

(b) Velocity of A after impact

0.8(12) 14.4Av = +′

24 ft/sAv′ = !





A A B B A B B Bm v m v m v m v′ ′+ = + g’s cancel

1.2 2.4 2.4

(24 ft/s) 0B Bv vg g g

′− = +

(1)

From restitution

0.2 , 4.8 0.224

BB B

B

vv v

v

′ ′= = ++

(2)

(a) Velocity of B before impact from equations (1) and (2)

28.8 2.4 2.4(4.8 0.2 ) 11.52 0.48B B Bv v v− = + = +

2.88 17.28Bv = 6 ft/sBv = !

(b) Velocity of B after impact

4.8 0.2(6)Bv′ = + 6 ft/sBv′ = !




(a) Total momentum conserved

A A B B A A Bm v m v m v m v′ ′+ = +

( ) ( )0 0 066 kg 0

1B BB

v m v m v v vm

′ ′+ − = + ⇒ = −

(1)

Relative velocities

( ) 02A B B Av v e v v v v e′ ′ ′− = − ⇒ = (2)

From equations (1) and (2)

( )0 0 0 06 62 2 0.5

1 1B Bv e v v v

m m

= ⇒ = − −

3 kgBm = !

(b) Using 0 062

1Bv e v

m

= −

Gives, 6 62 12 1B

Be m

m e+ = ⇒ =

+

0, 6 kgBe m= =

1, 2 kgBe m= =

2 kg 6 kgBm≤ ≤ !




(a) First collision (between A and B)

The total momentum is conserved A B A Bmv mv mv mv′ ′+ = +

0 A Bv v v′ ′= + (1)

Relative velocities

( ) ( )A B B Av v e v v′ ′− = −

0 B Av e v v′ ′= − (2)

Solving equations (1) and (2) simultaneously

( )0 12A

v ev

−′ = !

( )0 1

2Bv e

v+

′ = !

(b) Second collision (Between B and C) The total momentum is conserved.

B C B Cmv mv mv mv′ ′′ ′+ = +

Using the result from (a) for Bv′

( )0 10

2 B Cv e

v v+

′′ ′+ = + (3)

Relative velocities

( )0B C Bv e v v′ ′ ′′− = −

Substituting again for Bv′ from (a)

( ) ( )01

2 C Be

v e v v+

′ ′′= − (4)

continued



Solving equations (3) and (4) simultaneously

( ) ( ) ( )00

11 12 2 2C

v e ev v e

+′ = + +

( )20 1

4Cv e

v+

′ = !

( )2

0 14B

v ev

−′′ = !

(c) For n spheres

n Balls

1 collisionn th−

We note from the answer to part (b), with 3n =

( )20

31

4n Cv e

v v v+

′ ′ ′= = =

or ( )( )

( )

3 10

3 3 1

1

2

v ev

−

−

+′ =

Thus for n balls

( )( )

( )

10

1

1

2

n

n n

v ev

−

−

+′ = !

(d) For 8, 0.90n e= =

From the answer to part (c) with 8n =

( )( )

( )( )( )

8 1 70 0

78 1

1 0.9 1.9

2 2B

v vv

−

−

+′ = =

8 00.698v v′ = !




(a) Packages A and B

Total momentum conserved


( )16 16 86 16 8 96A B A Bv v v vg g g

′ ′ ′ ′= + ⇒ + =

2 12A Bv v′ ′+ = (1)

Relative velocities ( ) ( )0.3 6 1.8A B B A B Av v e v v v v′ ′ ′ ′− = − ⇒ − = = (2)

Solving Equations (1) and (2) simultaneously

3.4 ft/sAv′ =

5.2 ft/sBv′ =

Packages B and C

B B C C B B C Cm v m v m v m v′ ′′ ′′+ = +

( )8 8 125.2 B Cv vg g g

′′ ′= + 4 6 20.8B Cv v′′ ′+ = (3)

Relative velocities

( ) ( )0.3 5.2 1.56B C C B C Bv v e v v v v′ ′ ′′ ′ ′′− = − ⇒ − = = (4)

continued



Solving (3) and (4) simultaneously 2.70 ft/sCv′ = !

1.144 ft/sBv′′ =

(b) Packages A and B (second time)

A A B B A A B Bm v m v m v m v′ ′′ ′′ ′′′+ = +

( ) ( )16 8 16 83.4 1.144 ; 2 7.944A B A Bv v v vg g g g

′′ ′′ ′′ ′′′+ = + + = (5)

( )A B B Av v e v v′ ′′ ′′′ ′− = −

( )( )3.4 1.144 0.3 0.6768 ; 0.6768B A A Bv v v v′′′− = = − − + = (6)

Solving Equations (5) and (6) simultaneously

2.42 ft/sAv′′ = !




Impact 2.5 cos 40 1.915 m/sAnv = − °= −

2.5 sin 40 1.607 m/sAtv = − °= −

2 m/sBnv =

0Btv =

Impulse-momentum

Unknowns , , ,An Bn At Btv v v v′ ′ ′ ′

System A + B n-dir A An B Bn A An B Bnm v m v m v m v′ ′+ = + (1)

Coefficient of restitution ( )Bn An An Bnv v e v v′ ′− = − (2)

Solve (1) and (2) for An Bnv v′ ′+

0.7493 m/s; 2.1870 m/sAn Bnv v′ ′= = −

System A t-dir

1.607 m/sA At A At Atm v m v v= ⇒ = −′ ′

continued



System B t-dir 0 m/sB Bt B Bt Btm v m v v′ ′= ⇒ =

Resolve into components

Ball A

2 2(0.7493) (1.607) 1.773 m/sAv′ = + =

1 0.7493tan 25.01.607

β − = = °

40 25 15.0θ = ° − °= °

So 1.773 m/sAv′ = !

Ball B

2.19m/sBv′ = !




Ball A t-dir 0 0sin sinAt Atmv mv v vθ θ′ ′= ⇒ =

Ball B t-dir 0 0B Bt Btm v v′ ′= ⇒ =

Ball A + B n-dir 0 cos 0 An Bnmv m v m vθ ′ ′+ = + (1)

Coefficient of restitution ( )Bn An An Bnv v e v v− = −′ ′

0( cos 0)Bn Anv v e v θ′ ′− = − (2)

Solve (1) and (2)

0 01 1cos ; cos

2 2An Bne ev v v vθ θ− + ′ ′= =

With numbers 0.8; 45e θ= = °

0 0sin 45 0.707Atv v v= °=′

0 01 0.8 cos 45 0.0707

2Anv v v− ′ = ° =

0Btv′ =

0 01 0.8 cos 45 0.6364

2Bnv v v+ ′ = °=

continued



(A)

1

2 2 20 0(0.707 ) (0.0707 )Av" v v = +

00.711v=

1 0.0707tan 5.71060.707

β − = = °

So 45 5.7106 39.3θ = − = °

(B)

00.711Av v′ = "

00.636Bv v′ = "




Angle of impulse force from geometry

1 6cos 22.626.5

θ −= = °

Total momentum conserved Ball A:

x : ( )cos 0A A A Am v F t m vθ ′− ∆ + = (1) Ball B: B BF t m v′∆ = Restitution ( )0Av v=

continued



Approach Separation

( ) ( )0 0cos cos ; cosB A B Av v e v v v evθ θ θ′ ′ ′ ′− = = +

Using equations (1) and (2) x : cosA A B B A Am v m v m vθ′ ′= +

( )( ) ( ) ( )617.5 / 6 ft/s 1.6 / 17.5 / v6.5B Ag g v g ′ ′= +

g�s cancel

Substituting for 6 6 6; 105 1.6 4.8 17.56.5 6.5 6.5B A Av v v ′ ′ ′= + +

5.22 ft/sAv′ = !

9.25 ft/sBv′ = 22.6°!




Angle of impulse force from geometry

1 6cos 22.626.5

θ −= = °


Ball A:

x : ( )cos 0A A A Am v F t m vθ ′− ∆ + = (1)

Ball B:

x : cos B BF t m vθ ′∆ = (2)

Restitution ( )0Av v−

continued



Approach Separation

( )cos cos cos ; 4.8B A A B Av v e v v vθ θ θ′ ′ ′ ′− = − = (3)

Using Equations (1) and (2) x : A A B B A Am v m v m v′ ′= +

( )( ) ( ) ( )17.5 / 6 1.6 / 17.5 /B Ag g v g v′ ′= + g�s cancel

Substituting for Bv′ from (3) ( )105 1.6 4.8 17.5A Av v′ ′= + +

5.10 ft/sAv′ = !

9.90 ft/sBv′ = !




Angle of impulse force from geometry of A and B

1 6cos 22.62

6.5θ − = = °


Ball A:

Ball B:

Restitution

Approach

Separation

( ) 6

cos 22.66.5

16

66.5

A BBn An

An Bn

v vv v

e ev v

θ ′ ′ ′− + ° + ′ ′− = ⇒ = = −

(1)

A: ( )sin sin 22.6A A A Am v m vθ θ′ ′= + °

( )2.56 sin 22.6

6.5 Av θ ′ ′= + °

(2)

continued



:A B+ cosA A A A B Bm v m v m vθ′ ′ ′= +

( )( ) ( ) ( )58 / 6 58 / cos 5.3 /A Bg g v g vθ′ ′ ′= + g’s cancel

Equations (1), (2), and (3) in , and A Bv v θ′ ′ ′

5.027 ft/s; 10.838 ft/s; 0.08218 rad 4.71A Bv v θ′ ′ ′= = = = °

5.03 ft/sAv′ = 4.71° !

10.84 ft/sBv′ = !




(a)

Before After

t-Direction

Momentum of A is conserved.

( )sinA A tmv m vθ ′=

( ) sinA Atv v θ′ =

Momentum of B is conserved.

( )cosB B tmv m vθ ′=

( ) cosB Btv v θ′ =

n-Direction Total momentum is conserved.

( ) ( )cos sinA B A Bn nmv mv m v m vθ θ ′ ′− = +

( ) ( ) cos sinA B A Bn nv v v vθ θ′ ′+ = − (1)

Relative velocities (coefficient of restitution)

1e = ( ) ( ) ( )( )1 cos sinB A A Bn nv v v vθ θ′ ′− = + (2)

Adding Equation (1) and (2)

( ) cosB Anv v θ′ =

(1) (2)− ( ) sinA Bnv v θ′ = −

continued

A Bm m m= =



Thus, after impact

tan tanA A

B B

v vv v

α β= Thus and A Bv vα β ′ ′= ⊥ !

(b) Using the results from (a)

( ) ( )2 2 2 2 2 2sin sinA A A A Bt nv v v v vθ θ′ ′ ′= + = +

( ) ( )2 2sin 30 30 40 25 ft/sAv′ = ° + = !

( ) ( )2 2 2 2 2 2cos cosB B B B At nv v v v vθ θ′ ′ ′= + = +

( ) ( )2 2cos30 40 30 43.3 ft/sBv′ = ° + = !

1 1 30tan tan 36.940

A

B

vv

α β − −= = = = °

( )180 90 90γ α β = − + − = ° !




(a) Since Bv′ is in the x-direction and (assuming no friction), the common tangent between A and B at impact must be parallel to the y-axis

Thus 250tan150 D

θ =−

1 250tan 70.20150 60

θ −= = °−

70.2θ = °!

(b) Conservation of momentum in x(n) direction

( ) ( )cosA B A Bn nmv m v m v mvθ ′ ′+ = +

( ) ( ) ( )1 cos 70.20 0 A Bnv v′ ′+ = +

( ) ( )0.3387 A Bnv v′ ′= + (1)

Relative velocities in the n direction

( )( ) ( )0.9 cosA B B An ne v v e v vθ ′ ′= − = −

( )( ) ( )0.3387 0 0.9 B A nv v′ ′− = − (2)

(1) + (2)

( )2 0.3387 1.9 0.322 m/sB Bv v′ ′= = !




Momentum: cos 0A Ax Bv v vθ ′+ = +

Restitution: 0.9 cosB Ax Av v v θ′ ′− =

( )cos 0.9 cosA Ax Ax Av v v vθ θ′ ′− − =

( )10.1 1 cos 70.2 0.016936 m/s

2Axv′ = ° =

( )1 sin 70.2 0.94089, 0.32178Ay Bv v′ ′= ° = =

(a) 0.941 m/sAv′ = !

(b) Fraction of Initial Energy loss = F. L.

12F. L. =

( )2 11

2m − ( )2 1

2Bm v′ − ( )2

12

Am v′

( )21m

F. L. 1 0.1035 0.8856 0.01090= − − = !




Rebound at A

Projectile motion between A and B

Conservation of momentum – t-direction

( ) ( )0 sin 30 5sin 30A At tmv m v v′ ′° = ⇒ = °

( ) 2.5 m/sA tv′ =

Relative velocities in the n-direction

( ) ( ) ( ) ( )( )0 cos30 0 0 5cos30 0.8A An nv e v v′ ′− ° − = − ⇒ =

( ) 3.464 m/sA nv′ =

After rebound ( ) ( ) ( )0cos30 sin 30x A At n

v v v′ ′= ° + °

2.5cos30 3.464sin 30= ° + °

( )03.897 m/sxv =

( ) ( ) ( )0

sin 30 cos30y A At nv v v′ ′= − ° + °

2.5sin 30 3.464cos30= − ° + °

( )0

1.75 m/syv =

x-direction: ( ) ( )0 03.897 , 3.897 m/sx x xx v t t v v= = = =

y-direction: ( ) 2 2

0

11.75 4.905

2yy v t gt t t= − = −

( )0

1.75 9.81y yv v gt t= − = −

At B: 0 1.75 9.81 0.17839 sy A B A Bv t t− −= = − ⇒ =

( ) ( ) ( )22

04.905 1.75 0.17839 4.905 0.17839y A B A By h v t t− −= = − = −

0.15609 mh =

156.1 mmh = !




22 2

09.817.5 m, 0.6 m, m/s 0.6 m 0.34975 s

2 2A

A A A Agtx v t t t= = = = ⇒ =

0 21.444 m/sv =

(a) First bounce:

0 1.5 m, 0.06995 sB Bv t t= =

( )2

29.81 m/s0.12 m 3.431 m2A B Be t t= −

( )( ) ( )( )20.12 3.431 0.06995 4.905 0.06995Ae= − 0.24 0.024Ae= −

0.600Ae = !



(b) Second bounce:

Before

After

( )0.6 3.431 9.81By Bv t= −

1.3724 m/s=

( )0 21.444 m/sB Be v e=

2 20.12 4.905 0 0.12 1.3724 4.905C By C C C Cy v t t t t= + − = = + −

0.3497 sCt =

( )( )06.75 21.444 0.3497C B C Bx e v t e= = =

0.900Be = !




Momentum in t direction is conserved

sin 30 tmv mv′° =

( )( )25 sin 30 tv′° =

12.5 ft/stv′ =

Coefficient of restitution in n-direction

( )cos30 nv e v′° =

( )( )( )25 cos30 0.9 19.49 ft/sn nv v′ ′° = =

Write v′ in terms of x and y components

( ) ( ) ( ) ( ) ( )0 cos30 sin 30 19.49 cos30 12.5 sin 30x n tv v v′ ′ ′= ° − ° = ° − °

10.63 ft/s=

( ) ( ) ( ) ( ) ( )0

sin 30 cos30 19.49 sin 30 12.5 cos30y n tv v v′ ′ ′= ° + ° = ° + °

20.57 ft/s=



Projectile motion

( ) ( ) ( )2

2 20 0

1 3 ft 20.57 ft/s 32.2 ft/s2 2y

ty y v t gt t′= + − = + −

At B, 20 3 20.57 16.1 ; 1.4098 sB B By t t t= = + − =

( ) ( )0 0 0 10.63 1.4098 ; 14.986 ftB x B Bx x v t x′= + = + =

( ) ( )3cos60 14.986 ft 3 ft cot 60 13.254 ftBd x= − ° = − ° =

13.25 ftd = !




Find x and y components of v

x yv v v= −i j

After the first impact x component is multiplied by e and the y component is unchanged

x yv ev v= − −′ i j

After rebound at C the y component is multiplied by e and the x component is unchanged

( )x y x yv ev ev e v v= − + = − −′′ i j i j

so v ev′′ = − And the final velocity is parallel to the original velocity !




Velocities just after impact

Total momentum in the horizontal direction is conserved

( ) ( )1.5 2.5 1.5 2.5; 0 6A A B B A A B B A Bm v m v m v m v v vg g g g

′ ′ ′ ′+ = + + = +

15 1.5 2.5A Bv v′ ′= + (1) Relative velocities ( ) ( )( ): 0 6 0.8 4.8A B B A B A B Av v e v v v v v v′ ′ ′ ′ ′ ′− = − − = − ⇒ − = − (2)

Solving (1) and (2) simultaneously 6.75 ft/s 1.95 ft/sA Bv v′ ′= =

(a)


21 1

1 02 A AT m v V= =

( )( )2

1 21.5 lb 6.75 ft/s1 1.06124

2 32.2 ft/sT = =

continued



2 0T =

2 1.5AV m gh h= =

1 1 2 2 : 1.06124 1.5T V T V h+ = + =

0.70749 ft 8.9899 in.h = =

8.49 in.h = !

(b) Work and energy

2 0T =

( )221 1 2.5 1.95 0.147612 2B BT m v

g ′= = =

( )1 2 0.6 2.5 1.5f k BU F x W x x xµ− = − = − = − = −

1 1 2 2 : 0.14761 1.5 0T U T x−+ = − =

0.0984 ft 1.1808 in.x = =

1.181 in.x = !




(a) Impact between A and B


80A Bm m

g= =

30Cm

g=

15A A B B A A B B A Bm v m v m v m v v v′ ′ ′ ′+ = + ⇒ + = (1)

Relative velocities

( ) ( )( )15 0 0.8 : 12A B AB B A B A B Av v e v v v v v v′ ′ ′ ′ ′ ′− = − ⇒ − = − − = (2)

Solving (1) & (2) 13.5 ft/sBv′ =

Impact between B and C (after A hits B)


( ) ( )80 30 80 30: 13.5 0B B C C B B C C B Cm v m v m v m v v v

g g g g′ ′ ′′ ′′ ′′ ′′+ = + + = +

1080 80 30B Cv v′′ ′′= + (3)

continued



Relative velocities

( ) ( )( ): 13.5 0 0.3B C BC C B C Bv v e v v v v′ ′ ′′ ′′ ′′ ′′− = − − = −

4.05 C Bv v′′ ′′= − (4)

Solving (3) and (4) 8.7136 ft/s 12.7636 ft/sB Cv v′′ ′′= =

8.71 ft/sBv′′ = !

(b) ( ) ( )L B C B CT T T T T′ ′ ′′ ′′∆ = + − +

( ) ( )2 2

2

1 1 80 lb13.5 ft/s 226.39 lb ft

2 2 32.2 ft/sB B BT m v

′ ′= = = ⋅

0CT ′ =

( ) ( )2 2

2

1 1 80 lb8.7136 ft/s 94.319 lb ft

2 2 32.2 ft/sB B BT m v

′′ ′′= = = ⋅

( ) ( )2 2

2

1 1 30 lb12.764 ft/s 75.894 lb ft

2 2 32.2 ft/sC C CT m v

′′ ′′= = = ⋅

( ) ( )226.39 0 94.319 75.894 56.177LT∆ = + − + =

56.2 lb ftLT∆ = ⋅ !




(a)

Before After

A Bm m m= =

5 km/h 1.3889 m/s= Conservation of total momentum


1.3889 1.3889B A B A B Bv v v v v v′ ′ ′ ′− = − − + = − (1)

Work and energy � car A (after impact)

2

112 A AT m v′=

2 0T =

( )1 2 4fU F− =

( )4k Am gµ= −

21 1 2 2

1; 4 02 A A k AT U T m v m gµ− ′+ = − =

continued



( )( )( )2 2 2 22 4 m 0.3 9.81 m/s 23.544 m /sAv′ = =

4.852 m/sAv′ =

Car B � (after impact)

21 2

1 , 02 B BT m v T′= =

( )1 2 1k BU m gµ− = −

( )21 1 2 2

1: 1 02 B B k BT U T m v m gµ− ′+ = − =

( )( )( )2 2 2 22 0.3 1 m 9.81 m/s 5.886 m /s ; 2.426 m/sB Bv v′ ′= = =

From (1) 1.3889 4.852 2.426 1.38B A Bv v v′ ′= + + = + +

31.2 km/hBv = !

(b) Relative velocities

( )A B B Av v e v v′ ′− − = −

( )1.3889 8.667 2.426 4.852e− − = −

( )10.0559 2.426e− = −

0.241e = !




(a) Work and energy

Velocity of A just before impact with B

( )2 21 0 2 2

1 1

2 2A A AT m v T m v= =

( )1 2 0.3 mk AU m gµ− = −

f k k AF N m gµ µ= =

( )( ) ( )( )( ) ( )( )2 2 21 1 2 2 2

1 1: 0.4 kg 3 m/s 0.3 0.4 kg 9.81 m/s 0.3 m 0.4 kg

2 2 AT U T v−+ = − =

( ) ( )222

7.2342 2.6896 m/sA Av v= =

Velocity of A after impact with B ( )2Av′

( )22 32

1 0

2 A AT m v T′= =

( )2 3 0.075 mk AU m gµ− = −

continued



2 2 3 3T U T−+ =

( )( ) ( )( )( )2 2

2

10.4 kg 0.3 0.4 kg 9.81 m/s 0.075 m 0

2 Av′ − =

( )20.6644 m/sAv′ =

Conservation of momentum as A hits B

( )22.6896 m/sAv =

( )20.6644 m/sAv′ =

( ) ( ) ( ) ( )2 2 A A B B A A B B A Bm v m v m v m v m m′ ′+ = + =

2.6896 0 0.6644 2.0252 m/sB Bv v′ ′+ = + =

Relative velocities (A and B)

( ) ( )2 2A B AB B Av v e v v ′ ′− = −

( )2.6896 0 2.0252 0.6644ABe− = − 0.506ABe = !

Work and energy

Velocity of B just before impact with C

( ) ( )2 22 2

1 0.42.0252 0.8203

2 2B BT m v′= = =

( ) ( )2 24 4 4

1 0.4

2 2B B BT m v v′ ′= =

( )2 4 0.30 0.35316k BU m gµ− = − = −

( )21 2 4 4 4

: 0.8203 0.35316 0.2 BT U T v− ′+ = − =

( )41.5283 m/sBv′ =



Conservation of momentum as B hits C

1.2 kg, 0.4 kgC Bm m= =

( ) ( )4 4B B C C B B C Cm v m v m v m v′ ′′ ′+ = +

( ) ( )40.4 1.5283 0 0.4 1.2B Cv v′′ ′+ = +

Velocity of B after B hits C, ( )40Bv′′ =

With ( )40; 0.61132 1.2 0.5094 m/sB C Cv v v′′ ′ ′= = ⇒ =

Relative velocities (B and C)

( ) ( ) ( )4 4; 1.5283 0 0.5094 0B C BC C B BCv v e v v e ′ ′ ′′− = − − = −

0.333BCe = !

(b) Work and energy – Block C

5 0T =

( )( )24

11.2 0.5094 0.15569

2T = =

( )( )4 5 0.3 1.2 9.81 3.5316kU mgx x xµ− = − = − = −

4 4 5 5 : 0.15569 3.5361 0 0.044 mT U T x x−+ = − = ⇒ =

44.0 mmx = !




Conservation of energy before impact

( )( ) ( )2 20

10 8 kg 9.81 m/s 0.15 m 8 kg

2T V v+ = + =

0 1.7155 m/sv =

0 1

1 2

2

0

Cylinder C: 8

Platform A: 0 54 unknowns

Counterweight B: 0 5

Restitution: 0.8

C

A

A

A C

v F dt v

F dt F dt v

F dt v

v v v

′− =

′+ − =

′+ = ′ ′− =

∫∫ ∫

∫

Simultaneous solution 4 Equations and 4 unknowns

( ) 0 0, 1.372 m/sAa v v′ ′= = !

( ) 2 6.86 N sb F dt = ⋅∫ !




(a) A and B after the first impact

2 21

2 22 2 2

1 2

1 1(5 kg)(1.372 m/s) (5 kg)(1.372 m/s)

2 21 1

(5) (5)2 2

0

T

T v v

U Td Td→

= + = +

= + − =

2 Av v∴ =

1.372 m/sAv = !

1 1.372d v t t= =

C after the first impact

2

0 9.81 , 0.2798 s2

td t

= + ∴ =

At which time 9.81Cv t= 2.74 m/sCv = !

(b) Second impact: (As before)

C: 18(2.7448) 8C

F dt v ′′− =∫

A: 1 25(1.3724) 5 AF dt F dt v ′′+ − =∫ ∫

B: 25(1.3724) 5A

F dt v ′′+ =∫

Restitution: 0.8[2.7448 1.3724]A Cv v′′ ′′− = −

With 4 unknowns and four equations, solve for

2.47 m/sAv ′′ = !

1.372 m/sCv ′′ = !




After impact

Velocity of A just before impact, 0v

( )( )20 2 2 32.2 ft/s 3.6 ft sin 30v gh= = °

( ) ( ) ( )2 32.2 3.6 0.5 10.7666 ft/s= =

Conservation of momentum

A B B B A Am v m v m v= −

00.6 0.6(1.8 ) B Av g v vg g

= −

(1)

g�s cancel

Restitution

( ) ( )0 00 0.9A Bv v e v v+ = + = (2)

Substituting for vB from (2) in (1)

0 00.6 1.8(0.9 ) 0.6 ; 2.4 1.02A A A Bv v v v v v= − − =

(a) A moves up the distance d where,

2 2 21 1sin 30 ; (4.5758 ft/s) (32.2 ft/s ) (0.5)2 2A A Am v m gd d= ° =

0.65025 ft 7.80 mAd = = !

(b) Static deflection 0,x= B moves down

Conservation of energy (1) to (2)

Position (1) � spring deflected, 0x

0 sin 30Bkx m g= °



21 1 2 2 1 2

1; , 02 B BT V T V T m v T+ = + = =

21 0

1 sin 302e g B BV V V kx m gd= + = + °

( )0 2 22 0 00

1 22

Bx de g B BV V V kxdx k d d x x+= + = = + +′ ′ ∫

( )2 2 2 20 0 0

1 1 1sin 30 2 0 02 2 2B B B B Bkx mgd m v k d d x x+ ° + = + + + +

2 2 2 21.8; 34 (5.1141)32.2B B B Bkd m v d ∴ = =

0.20737 ftBd =

2.49 in.Bd = !




After impact

Velocity of A just before impact, 0v

( )( )20 2 2 32.2 ft/s 3.6 ft sin 30v gh= = °

0 2(32.2)(3.6)(0.5) 10.7666 ft/sv = =

Conservation of momentum

0 00.6 1.8;A B B A A Bm v m v m v v vg g

= − =

(1)

g�s cancel Restitution ( ) ( )0 00 ;A B Bv v e v v ev+ = + = (2)

From (1) ( )00.6 0.6 10.7666 ft/s 3.5889 ft/s1.8 1.8Bv v = = =

From (2) ( )01/ ,3Be v v e= =

(a) 0.333e = ! (b) Energy loss

( ) 21Energy 3.6 sin 302A B Bm g m v∆ = ° −

( )( )( )0.6 lb 3.6 ft 0.5= ( )21 1.8 3.5889 ft/s2 32.2 −

1.08 0.36 0.72 ft lb= − = ⋅ Loss 0.720 ft lb= ⋅ !

(c) Static deflection 0,x= B moves down Conservation of energy 1 to 2 Position 1-spring deflected, 0x

0 sin 30Bkx m g= °



21 1 2 2 1 2

1; , 02 B BT V T V T m v T+ = + = =

21 0

1 sin 302e g B BV V V kx m gd= + = + °

( )0 2 22 0 00

1 22

Bx de g B BV V V kxdx k d d x x+= + = = + +′ ′ ∫

( )2 2 2 20 0 0

1 1 1sin 30 2 0 02 2 2B B B B Bkx mgd m v k d d x x+ ° + = + + + +

2 2 2 21.8; 34 (3.5889)32.2B B B Bkd m v d ∴ = =

0.1455 ftBd =

1.746 in.Bd = !




Ball A falls

1 1 2 2T V T V+ = + (Put datum at 2)

21 22 A Amgh mv v gh= ⇒ =

(2)(9.81)(0.2) 1.9809 m/s= =

Impact

1sin 302rr

θ −= = °

Impulse�Momentum

Unknowns , ,B At Anv v v′ ′ ′ x-dir

0 0 sin 30 cos30B B A An A Atm v m v m v′ ′ ′+ = + ° + ° (1)

We need more equations

0 0




( )Bn An An Bnv v e v v′ ′− = −

For our problem

sin 30 ( cos30 0)B An Av v e v′ ′° − = ° − (2)

System = A

t-dir ( sin 30 )A A A Atm v m v′− ° = (3)

Solve 3 equations and 3 unknowns (maple) using A Bm m m= =

1.3724 m/sBv′ =

1.029 m/sAnv′ = −

0.9905 m/sAtv′ = −

Now lets look at B after impact.

1 1 2 2T V T V+ = +

21 ( )2 B Bm v mgh′ =

So 2 2( ) (1.3724)

2 (2)(9.81)B

Bvh

g′

= =

0.0960 m=

96.0 mmBh = !

0 0




Momentum: 1 1 20 (0.7071)n nmv mv Mv′ ′+ = +

1 2(2 kg)(5 m/s)(0.7071) 2 kg 9 kg (0.7071)nv v= +′ ′

Restitution: 2 1 1(0.7071) 0.6 0.6(5)(0.7071)n nv v v− = =′ ′

Solve for 2 116 17m/s, (0.7071)11 11nv v = = −′ ′

1 1.092801 m/snv′ = −

"

(b) Conservation of energy � cylinder + spring:

2 2 20 2 2

1 1 1( )2 2 2

kx M v kx′+ =

22 2

220,000 1 16 20,000(0.05) (9) 34.52

2 2 11 2x + = =

2 0.05875 m,x = 2N20,000 (0.0587 m) = 1175 Nm

F kx= = "





1 6cos 22.626.5

θ − = = °

Total momentum conserved Ball A:

Ball B:

(1)

Restitution

( ) ( )

0

cos cos sin6 ft/s

cosB A Ax y

AA

v v ve v v

v

θ θ θ

θ

′ ′ ′− += = =

( ) ( ) ( ) ( ) ( )25

6tan

2B A A B A Ax y x y

A

v v v v v ve

v

θ′ ′ ′ ′ ′ ′− + − += =

continued



( ) ( ) ( ) ( ): sin sin cosA A A A A Ax yA m v m v m vθ θ θ′ ′= +

2.5 2.5tan ( ) tan ( ) ; 6 ( ) ( )6 6A A x A y A x A yv v v v vθ θ ′ ′ ′ ′= + = +

( ) ( )15 2.5 6A Ax yv v′ ′= + (2)

( ) 50 50 4.6: ; (6 ft/s) ( )A A A A B B A x BxA B m v m v m v v vg g g

′ ′ ′ ′+ = + = +

(3)

g�s cancel

From equation (1) 22(32.2 ft/s )(0.75 ft) 6.9498 ft/sBv′ = =

From equation (3) (50)(6) 50( ) 4.6(6.9498)A xv′= +

( ) 5.3606 ft/sA xv′ =

From equation (2) 15 2.5(5.3606) 6( )A yv′= +

( ) 0.2664 ft/sA yv′ =

2.56.9498 5.3606 0.26646 0.2834

6e

− + = =

0.283e = !





1 6cos 22.626.5

θ − = = °

Momentum consideration Ball A:

Ball B:

2B B Bm v m gh′ = (1)

Restitution

Approach

Separation

( ) ( )0

cos cos sin6 ft/s

cosB A Ax y

AA

v v ve v v

v

θ θ θ

θ

′ ′ ′− += = =

( ) ( ) ( ) ( ) ( )2.5

6tan

6B A A B A Ax y x y

A

v v v v v ve

v

θ′ ′ ′ ′ ′ ′− + − += =

continued



( ) ( ) ( ) ( ): sin sin cosA A A A A Ax yA m v m v m vθ θ θ′ ′= +

2 2tan ( ) tan ( ) ; 6 ( ) ( )6 6A A x A y A x A yv v v v vθ θ ′ ′ ′ ′= + = +

( ) ( )12 2.5 6A Ax yv v′ ′= + (2)

( ) 20 20 2: ; (6) ( )A A A A B B A x BxA B m v m v m v v vg g g

+ = + = +′ ′ ′ ′

6 ( )10

BA x

vv′′= + (3)

From the equation for e

2.50; ( ) ( ) 06B A x A ye v v v ′ ′ ′= − + =

(4)

2.51; ( ) ( ) 66B A x A ye v v v ′ ′ ′= − + =

(5)

Simultaneous solution of equations (2), (3) and (4) for e = 0 and equations (2), (3) and (5) for e = 1 yields

0 : ( ) 5.463 ft/s, ( ) 0.224 ft/s, 5.370 ft/sA x A y Be v v v′ ′ ′= = = =

1 : ( ) 4.926 ft/s, ( ) 0.4475 ft/s, 10.740 ft/sA x A y Be v v v′ ′ ′= = = =

2( ) 0.4478 ft, 1.791 ft2(32.2)

Bvh′

= =

5.37 in. 21.5 in.h≤ ≤ !




Ball A alone

Momentum in t-direction conserved

( ) ( )A A A At tm v m v′=

( ) ( )0A At tv v′= =

Thus ( )A Anv v′ ′= 60°

Total momentum in the x-direction is conserved.

( ) ( )sin 60 sin 60A A B B A A B Bxm v m v m v m v′ ′° + = − +

( )0 1.5 m/s 0A B xv v v= = =

( )( ) ( )( )( ) ( )0.17 1.5 sin 60 0 0.17 sin 60 0.34A Bv v′ ′° + = − ° +

0.2208 0.1472 0.34A Bv v′ ′= − + (1)

Relative velocity in the n-direction

( ) cos30 ;A B B Anv v e v v ′ ′− − = − ° −

( )( )1.5 0 1 0.866 B Av v′ ′− − = − − (2)

Solving Equations (1) and (2) simultaneously 0.9446 m/s, 0.6820 m/sB Av v′ ′= =

Conservation of energy ball B

( )21

12 B BT m v′=

( )21 2

1 3.0232 02

BWT Tg

= =

continued



1 20 BV V W h= =

( )21 1 2 2

1; 0.9446 0 ;2

BB

WT V T V W hg

+ = + = +

( )( )( )

20.94460.0455 m

2 9.81h = =

45.5 mmh = !




(a) Momentum of the sphere A alone is conserved in the t-direction.

( ) ( ) ( ) 0A A A A At t tm v m v v′= =

( ) ( )0 A A At nv v v′ ′ ′= = 50°

Total momentum is conserved in the x-direction.

( )cos50 cos50A A B B A A B Bm v m v m v m v′ ′° + = − ° +

0 4 m/sB Av v= =

( ) ( )2 4 cos50 0 2 cos50 6A Bv v′ ′° + = − ° +

5.1423 1.2855 6A Bv v′ ′= − + (1)


( ) ( )cos50 ; 0, 4 m/sA B B A B Av v e v v v v′ ′− = ° + = =

( )4 0.5 0.6428 ; 2 0.6428B A B Av v v v′ ′ ′ ′= + = + (2)

Solving Equation (1) and Equation (2) simultaneously 1.2736 m/s; 1.1299 m/sA Bv v′ ′= =

1.274 m/sAv′ = 50° !

1.130 m/sBv′ = !

(b) ( ) ( )2 22 lost

1 12 2A A A A B BT m v m v m v ′ ′= − +

( )( ) ( )( )2 21 2 kg 4 m/s 2 kg 1.274 m/s2= −

( )( )26 kg 1.130 m/s 10.546 J− =

lost 10.55 JT = !




First 18 m: Since all the cars� weight is on the rear wheels which skid, the force on the car is

( )k kF N Wµ µ= =

( )( )181 hr58 km/h 1000 m/km

3600 sv =

16.1 m/s=

( ) ( )221 2 18

1 10 16.1 m/s 129.62 2

W WT T mvg g

= = = =

( )( ) ( )( )1 2 18 m 18 mkU F Wµ− = =

1 1 2 2T U T−+ =

( )0 18 129.6kWWg

µ + =

( )( )129.6 0.73395

18 9.81kµ = =

For 400 m: Force moving the car is for the first 18 m,

( )( ) ( )1 0.73395kF W Wµ= =

For the remaining 382 m, with 75% of weight on rear drive wheels and impending sliding,

( )( ) ( ) ( ) ( )2 0.75 0.80 0.73395 0.80 0.91744s s kF Wµ µ µ= = = =

( )( )( )2 0.91744 0.75 0.68808F W= =

( )21 2 400

10 2

WT T vg

= =



( ) ( )1 2 1 218 m 382 mU F F− = +

( )( )( ) ( )( )( )0.73395 18 m 0.68808 328 mW W= +

13.21 262.8 276.01W W W= + =

( )21 1 2 2 400

1 0 276.012

WT U T W vg−

+ = + =

( ) ( )( )( )4002 22 276.01 2 9.81 m/s 276.01v g= =

2400 4005415.3 73.6 m/sv v= =

400 265 km/hv = !




(a) Spring constants 3 N/mm 3000 N/m=

2 N/mm 2000 N/m=

Max deflection at 2 when velocity of 0C =

1 1 2 20, 0, 0, 0v T v T= = = =

1 2 e gU U U− = +

( ) ( ) ( )0.051 2 0 01 2 0.15my

e e C mU F dx F dx W y− = − + +∫ ∫

( ) ( ) ( ) ( )221 2

3000 N/m 2000 N/m0.05 m

2 2 mU y− = −

( ) ( ) ( )23 kg 9.81 m/s 0.15 my+ +

( ) ( )23.750 1000 4.4145 29.43m my y= − + +

( ) ( )21 1 2 2 : 0 1000 29.43 8.1645 0m mT U T y y−+ = − + + =

0.10626 mmy = 106.3 mmmy = !

(b) Maximum velocity occurs as the lower spring is compressed a distance y′

( ) ( )2 2 21 2

1 10; 3 kg 1.52 2CT T m v v v= = = =

( ) ( ) ( ) ( )2 21 1 2 2; 0 1000 29.43 8.1645 1.5T U T y y v−+ = − + + =′ ′

Substitute 0.014715 my′ =

( )2

0 2000 29.43 0; 0.014715 mdv y ydy

′ ′= − + = =′

20.21653 0.43306 8.1645 1.5v− + + =

2 5.5873 m/sv = 2.36 m/sv = !



Chapter 13, Solution 192. (a)

Block leaves surface at C when the normal force 0N =

cos nmg maθ =

2

cos Cvgh

θ = (1)

2 cosCv gh gyθ= =

Work-energy principle

( ) ( )212B C B C CT mv U W h y mg h y−= = − = −

B B C CT U T−+ =

Use Equation (1) ( ) 214.52 Cm mg h y mv+ − =

( ) 14.52 Cg h y gy+ − = (2)

34.52 Cgh gy+ =

( )4.532

Cgh

yg

+=

( )( )( )( )

24.5 9.81 1

3 9.812

y+

=

0.97248 my = (3)

continued



0.97248cos cos 0.97248

1 mC

Cyy hh

θ θ= = = =

1cos 0.97248 13.473θ −= = ° 13.47θ = ° !

(b)

From Equations (1) and (3)

( )9.81 0.97248 3.0887 m/sCv gy= = =

At C; ( ) cos 3.0887cos13.47C Cxv v θ= = °

3.0037 m/s=

( ) sin 3.0887sin13.47C Cyv v θ= − = °

0.71947 m/s= −

( ) ( )2 21 10.97248 0.71947 9.812 2C C yy y v t gt t t= + − = − −

At E: 20: 4.905 0.7194 0.97248 0Ey t t= + − =

0.37793 st =

At E: ( ) ( ) ( )cos 1 sin13.47 3.0037 0.37793C xx h v tθ= + = ° +

0.23294 1.3519 1.3681 m= + =

1.368 mx = !




Find unstretched length of the spring

1 0.3tan 71.5650.1

θ θ− = = °

( ) ( )2 20.3 0.1 0.3162 mBDL = + =

= length at equilibrium

Equilibrium: ( )0.1 sin 0.6 10 g 0A sM F θΣ = − =

63.25 gsF =

( )( ): 63.25 g 8000 N/m 0.07756 ms BD BD BDF k L L L= ∆ = ∆ ⇒ ∆ =

Unstretched length 0 0.3162 0.07756BD BDL L L= − ∆ = −

0.23864 m=

Spring elongation, BDL′∆ when 90φ = °

( ) 00.3 m 0.1 m 0.4 0.23864BDL L′∆ = + − = −

0.16136 m=

At 1 1 190 0, 0V Tφ = ° = =

( ) ( )1 1 1e gV V V= +

( ) ( ) ( )2 21

1 8000 0.161362 2BDeV k L′= ∆ =

104.15 N m= ⋅

continued



( ) ( )1 10g 0.6 58.86 N mgV = − = − ⋅

1 104.15 58.86 45.29 N mV = − = ⋅

At 2 ( ) ( ) ( )2 22

1 80000 N/m 0.07756 m2 2BDeV k Lφ = = ∆ =

24.06 N m= ⋅

2 2 22 2 2 2

1 10 kg 52 2

T mv v v = = =

21 1 2 2 2: 0 45.29 5 24.06T V T V v+ = + + = +

22 4.246v = 2 2.06 m/sv = "




63960 1500 5460 mi 28.829 10 ftAr = + = = ×

63960 6000 9960 mi 52.589 10 ftBr = + = = ×

Conservation of momentum A A B Br mv r mv=

28.829 0.5481952.589

AB A A A

B

rv v v vr

= = =

(1)


2 21 1, , , 2 2A A A B B B

A B

GMm GMmT mv V T mv Vr r

= = − = = −

( )( )22 232.2 ft/s 3960 mi 5280 ft/miGM gR= = ×

14 3 2140.77 10 ft /s= × 14

66

140.77 10 488.29 10 m28.829 10A

mV ×= − = − ××

146

6140.77 10 267.68 10 m52.589 10B

mV − ×= = − ××

2 61: 488.29 10 m2A B B B AT T T V mv+ = + − ×

2 61 267.68 10 m2 Bmv= − ×

2 6 21 1220.61 102 2A Bv v− × =

Using (1) ( )22 61 1220.61 10 0.548192 2A Av v− × =

2 60.34974 220.61 10Av = ×

25115.39Av =

325.1 10 ft/sAv = × !




s sFt Nt mgtµ µ= =

1 260 mi/h 88 ft/s 20 mi/h 29.333 ft/sv v= = = =

1 2 0.65s smv mgt mvµ µ− = =

( )88 0.65 32.2 29.333 2.803t t− = ⇒ =

2.80 st = !




0.22t∆ =

( )1 2mv P W t mv+ − ∆ =

Horizontal components

( ) ( )84 9.14cos35 0.22 0HP° − =

22858.69 kg m/sHP = ⋅

2.86 kNHP = !




(a) Total momentum of the two cars is conserved.

( ), : cos30 cos10A A A Amv x m v m m vΣ ° = + ° (1)

( ), : sin 30 sin10A A B B A Bmv y m v m v m m vΣ ° − = + ° (2)

Dividing (1) into (2)

sin 30 sin10

cos30 cos30 cos10B B

A A

m v

m v

° °− =° ° °

( )( )tan 30 tan10 cos30AB

A B

mv

v m

° − ° °=

3600 28000.3473B A

A BA B

v mm m

v m g g= = =

36001.2857

2800A

B

m

m= =

( )( )0.3473 1.2857 0.4465B Av v= =

Car A was going faster !

(b) Since B was the slower car, 30 mi/hBv =

2.2396A Bv v=

67.2 mi/hAv = !




A B Cm m m m= = =

Collision between B and C

The total momentum is conserved.

4.5B C B C B Cmv mv mv mv v v′ ′ ′ ′+ = + ⇒ + = (1)

Relative velocities

( ) ( )0.8 4.5 3.6C B B C B Cv v e v v v v′ ′ ′ ′− = − = − ⇒ − = (2)

Solving (1) and (2) simultaneously

4.05 ft/sBv′ = !

0.450 ft/sCv′ = !

Since ,B Cv v′ ′> Car B collides with Car A

Collision between A and B

4.05A Bv v′ ′′+ = (3)

Relative velocities ( ) ( )0.5 4.05 ; 2.025A B B A B A A Bv v e v v v v v v′ ′′ ′ ′′ ′ ′ ′′− = − ⇒ − = − − = (4)

Solving (3) and (4) simultaneously

1.013 ft/sBv′′ = !

3.04 ft/sAv′ = !

C B Av v v′ ′′ ′< < ⇒ No more collisions




Before

After

( ) ( )18 ft/s; 18cos 40 13.79 ft/s; 18sin 40 11.57 ft/sA A An tv v v= = ° = = − ° = −

( ) ( )12 ft/s; 0B B Bn tv v v= = − =

t-direction


( ) ( ) ( ) ( )A A B B A A B Bt t t tm v m v m v m v′ ′+ = +

( ) ( ) ( ) ( ) ( ) ( )1.5 lb 1.5 lb 2.5 lb11.57 ft/s 0 A At tv v

g g g′ ′− + = +

( ) ( )17.36 1.5 2.5A Bt tv v′ ′− = + (1)

Ball A alone momentum conserved

( ) ( ) ( ) 11.57 ft/sA A A A At t tm v m v v′ ′= ⇒ = − (2)

Replace ( )A tv′ in (2) in equation (1)

( ) ( ) ( )17.36 1.5 11.57 2.5 ; 0B Bt tv v′ ′− = − + =



n-Direction Relative velocities

( ) ( ) ( ) ( )A B B An n nv v e v v ′ ′− = −

( ) ( ) ( )13.79 12 0.8 B An nv v′ ′ − − = −

( ) ( ) 20.632B An nv v′ ′− = (3)


( ) ( ) ( ) ( )A A B B A A B Bn n n nm v m v m v m v′ ′+ = +

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1.5 lb 2.5 lb 1.5 lb 2.5 lb13.79 ft/s 12 ft/s A Bn nv v

g g g g′ ′+ − = +

( ) ( )1.5 2.5 9.315A Bn nv v′ ′+ = − (4)

Solve (3) and (4): ( )4 21.633B nv′ =

( ) 5.408 ft/sB nv′ =

( ) 15.224 ft/sA nv′ = −

A

( ) ( )2 215.224 11.57 19.12 ft/s, 37.23Av θ= + = = °

19.12 ft/sAv = 72.2°!

B

5.41 ft/sBv = 40° !




(a) Rebound at A

Projectile motion between A and B

Conservation of momentum – t direction

( ) ( )0 cos60 6cosA At tmv m v v θ′ ′° = ⇒ =

( ) 3 m/sA tv′ =

Coefficient of restitution in the n direction

( )( ) ( ) ( )( ) ( )0 0 : 6sin 60 0.6A A An n nv e v v′ ′− − = − ° =

( ) 3.12 m/sA nv′ =

After rebound ( ) ( )03 m/sx A t

v v′= − = −

( ) ( )0

3.12 m/sy A nv v′= =

( )03 , 3 m/sx xx v t t v= = − = −

( ) 2 2

0

13.12 4.905 ;

2yy v t gt t t= − = −

( )0

3.12 9.81y yv v gt t= − = −

At B, 0: 3.12 9.81 0 0.318 sy A B A Bv t t− −= − = ⇒ =

2: 3.12 4.905 0.496 mB A B A By h h t t− −= = − =

3B A Bx d t −= − = −

0.496 mh = !

0.953 md = !

(b) ( )03 m/sB xv v= = −

3.00 m/sBv = !




(a) Momentum of sphere A alone is conserved in the t-direction.

0 cos sinA A Am v m vθ θ′=

0 tanAv v θ′= (1)

Total momentum is conserved in the x-direction.

( ) ( )0 0, 0B B A B B A A B Ax xm v m v m v m v v v′ ′ ′+ = + = =

01.5 4.50 0Bv vg g

′+ = +

0

3Bvv′ = (2)


( )0 sin 0 sin cosB Av e v vθ θ θ′ ′− − = − −

( )( )0 0.6 cotB Av v v θ′ ′= + (3)

Substituting Bv′ from (2) into (3)

0 00.6 0.333 cotAv v v θ′= +

00.267 cotAv v θ′= (4)

Divide (4) into (1)

21 tan tan0.267 cot

θ θθ

= =

tan 1.935θ = 62.7θ = °!

(b) From (1) ( )0 tan 1.935A Av v vθ′ ′= =

000.5168 ,

3A Bvv v v′ ′= = (2)

continued



( )( )22 2lost

1 12 2A A A A B BT m v m v m v′= − +

( ) ( )2

2 2 0 lost 0 0

1 1.5 1 1.5 4.50.51682 2 2 3

vT v vg g

= − +

[ ]2 20 00.31.5 0.40 0.50

2v vg g

= − − =

2lost 00.00932 ft lbT v= ⋅ !

(For 0v in ft/s).

solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion cap 13

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t t mv mv

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