solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion cap 14

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 14, Solution 1.

The masses are 1350 kg and 5400 kg.A B C

m m m= = =

Let , , and be the sought after final velocities, positive to the left.A B Cv v v

Initial velocities: ( ) ( ) ( )0 0 0

0, 8 km/h 2.2222 m/sA B Cv v v= = = =

. Truck strikes car . Plastic impact: 0First collision C B e =

( )0

Let be the common velocity of and after impact.BCv B C

:Conservation of momentum for B and C

( ) ( ) ( )0 0B C BC B B C C

m m v m v m v+ = +

( )( )6750 0 5400 2.2222BCv = + 1.77778 m/s

BCv =

Car-truck strikes carSecond collision. BC A.

Elastic impact. 1e =

( ) ( )0 0

1.77778 m/sA BC A BCv v e v v − = − − = − (1)

Conservation of momentum for A, B, and C.

( )( ) ( ) ( )( )0 0A A B C BC A A B C BC

m v m m v m v m m v+ + = + +

( )( )1350 6750 0 6750 1.77778A BCv v+ = + (2)

Solving (1) and (2) simultaneously for and ,A BCv v

2.9630 m/s, 1.18519 m/sA BC B Cv v v v= = = =

10.67 km/hA

=v �

4.27 km/hB

=v �

4.27 km/hC

=v �






Conservation of linear momentum for block, cart, and bullet together.

components : ( )0B A B C fm v m m m v= + +

( )( )

00.028 550

1.7058 m/s5 0.028 4

Bf

A B C

m v

v

m m m

= = =+ + + +

(a) 1.706 m/sfv = �

Consider block and bullet alone.

Principle of impulse and momentum.

components ( ) ( ): N t mg t N mg∆ − ∆ =

components : ( ) ( )0B k A Bm v N t m m v'µ− ∆ = +

Just after impact, t∆ is negligible. The velocity then is

( )( )0

0

0.028 5503.0628 m/s

5 0.028

B

A B

'm v

v

m m

= = =+ +

Also, just after impact, the velocity of the cart is zero.

Accelerations after impact.





Block and bullet: :F ma=∑

( ) ( )k A B A B ABm m g m m aµ + = +

( )( )0.50 9.81AB k

a gµ= =

2

4.905 m/s=

Cart:

:CF ma=∑

( ) :k A B C Cm m g m aµ + =

( ) ( )( )( )0.50 5.028 9.81

4

k A B

C

C

m m ga

m

µ += =

2

6.1656 m/s=

Acceleration of block relative to cart.

( ) 2

/4.905 6.1656 11.0706 m/s

AB Ca = − − =

Motion of the block relative to the cart.

( ) ( ) ( )( )

2 2

/

/ /2

2 2

AB C

AB C AB C

v v'a s− =

In the final position, /

0AB Cv =

( ) ( )

( )( )2 2

/

/C

3.06280.424 m

2 2 11.0706AB C

AB

v's

a= − = − = −

The block moves 0.424 to the left relative to the cart.

(b) This places the block 1.000 0.424 0.576 m− = from the left end of the cart.

0.576 m from left end of cart �






( )( )22 20004000 3700The masses are 124.2 slugs, 114.9 slugs, and 1366.5 slugs

32.2 32.2 32.2A B F

m m m= = = = = =

Let , , and be the sought after velocities in ft/s, positive to the right.A B Fv v v

Initial values: ( ) ( ) ( )0 0 0

0.A B Fv v v= = =

Initial momentum of system: ( ) ( ) ( )0 0 0

0.A A B B F F

m v m v m v+ + =

There are no horizontal external forces acting during the time period under consideration. Momentum is

conserved.

0A A B B F F

m v m v m v= + +

124.2 114.9 1366.5 0A B Fv v v+ + = (1)

The relative velocities are given as

/

/

7 ft/s

3.5 ft/s

A F A F

B F B F

v v v

v v v

= − = −

= − = −

(2)

(3)

Solving (1), (2), and (3) simultaneously,

6.208 ft/s, 2.708 ft/s, 0.7919 ft/s= − = − =A B Fv v v

0.792 ft/sF

=v �






The masses are , , and .A B F

A B F

W W Wm m m

g g g= = =

Let the final velocities be , , and 0.34 ft/s, positive to the right.=A B Fv v v


0A B Fv v v= = =


0A A B B F F

m v m v m v+ + =


conserved.

0A B F

A A B B F F A B F

W W Wm v m v m v v v v

g g g= + + = + +

Solving for ,F

W A A B B

F

F

W v W vW

v

+= − (1)

From the given relative velocities,

/

/

1.02 7.65 6.63 ft/s

1.02 7.5 6.48 ft/s

A F A F

B F B F

v v v

v v v

= + = − = −

= + = − = −

Substituting these values in (1),

( )( ) ( )( )4000 6.63 3700 6.48

49506 lb1.02

FW

− + −= − =

24.8 tonsF

W = �






( ) ( )( )

3 3

3

The masses are the engine 80 10 kg , the load 30 10 kg , and the flat car

20 10 kg .

A B

C

m m

m

= × = ×

= ×

Initial velocities: ( ) ( ) ( )0 0 0

6.5 km/h 1.80556 m/s, 0.A B Cv v v= = = =

No horizontal external forces act on the system during the impact and while the load is sliding relative to the flat

car. Momentum is conserved.

Initial momentum: ( ) ( ) ( ) ( )0 0

0 0A A B C A A

m v m m m v+ + = (1)

(a) Let v′ be the common velocity of the engine and flat car immediately after impact. Assume that the

impact takes place before the load has time to acquire velocity.

Momentum immediately after impact:

( ) ( )0A B C A C

m v m m v m m v′ ′ ′+ + = + (2)

Equating (1) and (2) and solving for ,v′

( ) ( )( )

( )3

0

3

80 10 1.80556

1.44444 m/s

100 10

A A

A C

m v

v

m m

×′ = = =

+ ×

5.20 km/h′ =v �

(b) Let fv be the common velocity of all three masses after the load has slid to a stop relative to the car.

Corresponding momentum:

( )A f B f C f A B C fm v m v m v m m m v+ + = + + (3)

Equating (1) and (3) and solving for ,fv

( ) ( )( )( )

3

0

3

80 10 1.80556

1.11111m/s

130 10

A Af

A B C

m v

v

m m m

×= = =

+ + ×

4.00 km/hf =v �






The masses are m for the bullet and A

m and B

m for the blocks.

(a) The bullet passes through block A and embeds in block B. Momentum is conserved.

( ) ( )0 0Initial momentum: 0 0

A Bmv m m mv+ + =

Final momentum:B A A B B

mv m v m v+ +

0Equating,

B A A B Bmv mv m v m v= + +

( )( ) ( )( ) 3

0

3 3 2.5 543.434 10 kg

500 5

A A B B

B

m v m v

m

v v

−++= = = ×− −

43.4 gm = �

(b) The bullet passes through block A. Momentum is conserved.

( )0 0Initial momentum: 0

Amv m mv+ =

1Final momentum:

A Amv m v+

0 1Equating,

A Amv mv m v= +

( )( ) ( )( )3

0

1 3

43.434 10 500 3 3

292.79 m/s43.434 10

A Amv m v

v

m

−

−

× −−= = =×

1293m/s=v �






(a) Woman dives first.

Conservation of momentum:

( )1 1

120 300 18016 0v v

g g

+− − =

( )( )

1

120 163.20 ft/s

600v = =

Man dives next. Conservation of momentum:

( )1 2 2

300 180 300 18016v v v

g g g

+− = − + −

( )( )1

2

480 180 169.20 ft/s

480

v

v

+= =

29.20 ft/s=v �

(b) Man dives first.


( )1 1

180 300 12016 v v

g g

+′ ′− −

( )( )

1

180 164.80 ft/s

600v′ = =

Woman dives next. Conservation of momentum:

( )1 2 2

300 120 300 12016v v v

g g g

+ ′ ′ ′− = − + −

( )( )1

2

420 120 169.37 ft/s

420

v

v

′ +′ = =

29.37 ft/s′ =v �






(a) Woman dives first.


( )1 1

120 300 18016 0v v

g g

+− − + =

( )( )

1

120 163.20 ft/s

600v = =

Man dives next. Conservation of momentum:

( )1 2 2

300 180 300 18016v v v

g g g

+ = − + −

( )( )1

2

480 180 162.80 ft/s

480

− += =

v

v 2

2.80 ft/s=v �

(b) Man dives first.


( )1 1

180 300 12016 0v v

g g

+′ ′− − =

( )( )

1

180 164.80 ft/s

600v′ = =

Woman dives next. Conservation of momentum:

( )1 2 2

300 120 300 12016v v v

g g g

+ ′ ′ ′− = + −

( )( )1

2

420 120 160.229 ft/s

420

v

v

′− +′ = = −

2

0.229 ft/s′ =v �






The masses are 9 kg.A B C

m m m= = =

Position vectors (m): 0.9 , 0.6 0.6 0.9 , 0.3 1.2A B C

= = + + = +r k r i j k r i j

2In units of kg m /s,⋅ ( ) ( ) ( )O A A A B B B C C Cm m m= × + × + ×H r v r v r v

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

0 0 0.9 0.6 0.6 0.9 0.3 1.2 0

0 9 0 9 0 0 0 0 9

8.1 8.1 5.4 10.8 2.7

8.1 10.8 8.1 2.7 5.4

A B C

A B B C C

A C B C B

v v v

v v v v v

v v v v v

= + +

= − + − + −

= − + + − + −

i j k i j k i j k

i j k i j

i j k

But, O

H is given as ( )21.8 kg m /s− ⋅ k

Equating the two expressions for O

H and resolving into components,

: 8.1 10.8 0

: 8.1 2.7 0

: 5.4 1.8

A C

B C

B

v v

v v

v

− + =− =

− = −

i

j

k

(1)

(2)

(3)

( ) Solving for , , and ,A B C

a v v v 1.333 m/sAv = ( )1.333 m/s

A=v j�

0.333m/sBv = ( )0.333m/s

B=v i�

1.000 m/sCv = ( )1.000 m/s

C=v k�

Coordinates of mass center G in m.

( )( ) ( )( ) ( )( )9 0.9 9 0.6 0.6 0.9 9 0.3 1.2

27

0.3 0.6 0.6

A A B B C C

A B C

m m m

m m m

+ +=+ +

+ + + + +=

= + +

r r rr

k i j k i j

i j k

Position vectors relative to the mass center in m.

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

0.9 0.3 0.6 0.6 0.3 0.6 0.3

0.6 0.6 0.9 0.3 0.6 0.6 0.3 0.3

0.3 1.2 0.3 0.6 0.6 0.6 0.6

A A

B B

C C

′ = − = − + + = − − +

′ = − = + + − + + = +

′ = − = + − + + = −

r r r k i j k i j k

r r r i j k i j k i k

r r r i j i j k j k

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

9 1.333 12 kg m/s

9 0.333 3 kg m/s

9 1 9 kg m/s

A A

B B

C C

m

m

m

= = ⋅

= = ⋅

= = ⋅

v j j

v i i

v k k





( ) ( ) ( )

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

( )

0.3 0.6 0.3 12 0.3 0.3 3 0.6 0.6 9

3.6 3.6 0.9 5.4 1.8 0.9 3.6

G A A A B B B C C Cb m m m′ ′ ′= × + × + ×

= − − + × + + × + − ×

= − − + + = + −

H r v r v r v

i j k j i k i j k k

i k j i i j k

( ) ( ) ( )2 2 21.800 kg m /s 0.900 kg m /s 3.60 kg m /sG

= ⋅ + ⋅ − ⋅H i j k�






The masses are 9 kg.A B C

m m m= = =

Position vectors (m): 0.9 , 0.6 0.6 0.9 , 0.3 1.2A B C

= = + + = +r k r i j k r i j

Coordinates of mass center G expressed in m.

( )( ) ( )( ) ( )( )9 0.9 9 0.6 0.6 0.9 9 0.3 1.2

27

0.3 0.6 0.6

A A B B C C

A B C

m m m

m m m

+ +=+ +

+ + + + +=

= + +

r r rr

k i j k i j

i j k

Position vectors relative to the mass center expressed in m.

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

0.9 0.3 0.6 0.6 0.3 0.6 0.3

0.6 0.6 0.9 0.3 0.6 0.6 0.3 0.3

0.3 1.2 0.3 0.6 0.6 0.6 0.6

A A

B B

C C

′ = − = − + + = − − +

′ = − = + + − + + = +

′ = − = + − + + = −

r r r k i j k i j k

r r r i j k i j k i k

r r r i j i j k j k

Angular momenta.

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

O A A A B B B C C C

G A A A B B B C C C

m m m

m m m

= × + × + ×

′ ′ ′= × + × + ×

H r v r v r v

H r v r v r v

Subtracting,

( ) ( ) ( ) ( )O G A A A A B B B B C C C Cm m m′ ′ ′− = − × + − × + − ×H H r r v r r v r r v

( ) ( ) ( )( )

0A A B B C C

A A B B C C

m m m

m m m

= × + × + ×

= × + + = ×

r v r v r v

r v v v r L

is parallel to .L r 2λ λ= ⋅ = ⋅L r L L r r

( )( )

2

2 2

2

4550 , 50 N s/m

0.9λ λ⋅= = = = ± ⋅

⋅L L

r r

( )( ) ( )( ) ( )( ) ( )9 9 9 50 0.3 0.6 0.6

A A B B C C

A B C

m m m

v v v

v v v r

j i k i j k

λ+ + =

+ + = ± + +





( ) Resolve into components and solve for , , and .A B C

a v v v

3.333 m/sAv = ( )3.33 m/s

A=v j�

1.6667 m/sBv = ( )1.667 m/s

B=v i�

3.333 m/sCv = ( )3.33 m/s

C=v k�

(b) Angular momentum about O expressed in 2kg m /s.⋅

( ) ( ) ( )

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

0 0 0.9 0.6 0.6 0.9 0.3 1.2 0

0 9 0 9 0 0 0 0 9

8.1 8.1 5.4 10.8 2.7

8.1 10.8 8.1 2.7 5.4

9 4.5 9

O A A A B B B C C C

A B C

A B B C C

A C B C B

m m m

v v v

v v v v v

v v v v v

= × + × + ×

= + +

= − + − + −

= − + + − + −

= + −

H r v r v r v

i j k i j k i j k

i j k i j

i j k

i j k

( ) ( ) ( )2 2 29.00 kg m /s 4.50 kg m /s 9.00 kg m /sO

= ⋅ + ⋅ − ⋅H i j k�






Position vectors expressed in ft.

3 6 , 6 3 , 3 3A B C

= + = + = +r i j r j k r i k

Momentum of each particle expressed in lb s.⋅

( ) ( )4 142 63 168 252

A AW

g g g= + = +v

i j i j

( ) ( )4 142 63 168 252

B BW

g g g= − + = − +v

i j i j

( ) ( )28 19 6 252 168

C CW

g g g= − = − −v

j k j k−−−−

Angular momentum of the system about O expressed in ft lb s.⋅ ⋅

( ) ( ) ( ){ }

( )

13 6 0 0 6 3 3 0 3

168 252 0 168 252 0 0 252 168

1252 756 504 1008 756 504 756

10 0 0

A A B B C C

O A B C

W W W

g g g

g

g

g

= × + × + ×

= + + − − −

= − + − − + + + −

= + +

v v vH r r r

i j k i j k i j k

k i j k i j k

i j k

zeroO

=H �






(a) 4 4 28 36 lbA B C

W W W W= + + = + + =

( )( ) ( )( ) ( )( ){ }14 3 6 4 6 3 28 3 3

36

A A B B C C A A B B C Cm m m W W m

m W

+ + + += =

= + + + + +

r r r r r rr

i j j k i k

2.667 1.333 2.667= + +i j k

( ) ( ) ( )2.67 ft 1.333 ft 2.67 ft= + +r i j k� (b) Linear momentum

( )1

A A B B C C A A B B C Cm m m m W W W

g= + + = + +v v v v v v v

( )( ) ( )( ) ( )( )14 42 63 4 42 63 28 9 6

g = + + − + + − − i j i j j k

( )1252 168

32.2= −j k

( ) ( )7.83 lb s 5.22 lb sm = ⋅ − ⋅v j k� (c) Position vectors relative to the mass center G (ft).

( ) ( )

( ) ( )

( ) ( )

3 6 2.667 1.333 2.667

0.333 4.667 2.667

6 3 2.667 1.333 2.667

2.667 4.667 0.333

3 3 2.667 1.333 2.667

0.333 1.333 0.333

A A

B B

C C

′ = − = + − + +

= + −′ = − = + − + +

= − + +′ = − = + − + +

= − +

r r r i j i j k

i j k

r r r j k i j k

i j k

r r r i k i j k

i j k

Angular momentum about the mass center.

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

10.333 4.667 2.667 2.667 4.667 0.333 0.333 1.333 0.333

4 42 4 63 0 4 42 4 63 0 0 28 9 28 6

A A B B C C

G A B C

W W W

g g g

g

′ ′ ′= × + × + ×

= − + − + − − − −

v v vH r r r

i j k i j k i j k

( ) ( ) ( ){ }

( )

1672 448 700 84 56 112 308 56 84

1896 448 672 27.827 13.913 20.870

32.2

g= − − + − − + + + −

= − − = − −

i j k i j k i j k

i j k i j k

( ) ( ) ( )27.8 ft lb s 13.91 ft lb s 20.9 ft lb sG

= ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅H i j k�





From Problem 14.28, O Gm= × +H r v H

2.667 1.333 2.667

0 7.83 5.22

27.826 13.913 20.870 27.827 13.913 20.870

0 0 0

O=

−

= − + + + − −= + +

i j k

H

i j k i j k

i j k

From Prob 14.11,

( ) ( ) ( )

( ) ( ) ( ){ }

( ) ( ) ( ){ }

( )

1

13 6 0 0 6 3 3 0 3

168 252 0 168 252 0 0 252 168

252 756 504 1008 756 504 756

10 0 0

O A A A B B B C C C

A A A B B B C C C

m m m

W W Wg

g

g

g

= × + × + ×

= × + × + ×

= + + − − −

1= − + − − + + + −

= + +

H r v r v r v

r v r v r v

i j k i j k i j k

k i j k i j k

i j k






Linear momentum of each particle expressed in kg m/s.⋅

3 2 4

8 6

6 15 9

A A

B B

C C

m

m

m

= − += += + −

v i j k

v i j

v i j k

Position vectors, (meters): 3 , 3 2.5 , 4 2A B C

= + = + = + +r j k r i k r i j k

( )2Angular momentum about , kg m /s .O ⋅

( ) ( ) ( )

( ) ( ) ( )

0 3 1 3 0 2.5 4 2 1

3 2 4 8 6 0 6 15 9

14 3 9 15 20 18 33 42 48

34 65 57

O A A A B B B C C Cm m m= × + × + ×

= + +− −

= + − + − + + + − + +

= − + +

H r v r v r v

i j k i j k i j k

i j k i j k i j k

i j k

( ) ( ) ( )2 2 234 kg m /s 65 kg m /s 57 kg m /sO

= − ⋅ + ⋅ + ⋅H i j k �






Position vectors, (meters): 3 , 3 2.5 , 4 2A B C

= + = + = + +r j k r i k r i j k

( ) Mass center:a ( )A B C A A B B C Cm m m m m m+ + = + +r r r r

( )( ) ( )( ) ( )( )6 1 3 2 3 2.5 3 4 2

3 1.5 1.5

= + + + + + +

= + +

r j k i k i j k

r i j k

( ) ( ) ( )3.00 m 1.500 m 1.500 m= + +r i j k�

Linear momentum of each particle, ( )kg m/s .⋅

3 2 4

8 6

6 15 9

A A

B B

C C

m

m

m

= − += += + −

v i j k

v i j

v i j k

(b) Linear momentum of the system, ( )kg m/s.⋅

17 19 5A A B B C C

m m m m= + + = + −v v v v i j k

( ) ( ) ( )17.00 kg m/s 19.00 kg m/s 5.00 kg m/sm = ⋅ + ⋅ − ⋅v i j k�

Position vectors relative to the mass center, (meters).

3 1.5 0.5

1.5

0.5 0.5

A A

B B

C C

′ = − = − + −′ = − = − +′ = − = + −

r r r i j k

r r r j k

r r r i j k

(c) Angular momentum about G, ( )2kg m /s .⋅

( ) ( ) ( )

3 1.5 0.5 0 1.5 1 1 0.5 0.5

3 2 4 8 6 0 6 15 9

5 10.5 1.5 6 8 12 3 6 12

2 24.5 25.5

G A A A B B B C C Cm m m′ ′ ′= × + × + ×

= − − + − + −− −

= + + + − + + + + +

= + +

H r v r v r v

i j k i j k i j k

i j k i j k i j k

i j k

( ) ( ) ( )2 2 22.00 kg m /s 24.5 kg m /s 25.5 kg m /sG

= ⋅ + ⋅ + ⋅H i j k�

( ) ( ) ( )2 2 2

3 1.5 1.5

17 19 5

36 kg m /s 40.5 kg m /s 31.5 kg m /s

m× =−

= − ⋅ + ⋅ + ⋅

i j k

r v

i j k

( ) ( ) ( )2 2 234 kg m /s + 65 kg m /s 57 kg m /sG

m+ × = − ⋅ ⋅ + ⋅H r v i j k





Angular momentum about O.

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )2 2 2

0 3 1 3 0 2.5 4 2 1

3 2 4 8 6 0 6 15 9

14 3 9 15 20 18 33 42 48

34 kg m /s + 65 kg m /s 57 kg m /s

O A A A B B B C C Cm m m= × + × + ×

= + +− −

= + − + − + + + − + +

= − ⋅ ⋅ + ⋅

H r v r v r v

i j k i j k i j k

i j k i j k i j k

i j k

Note that

O Gm= + ×H H r v






The mass center moves as if the projectile had not exploded.

( ) ( )( ) ( )( )

( ) ( )

22

0

1 160 2 9.81 2

2 2

120 m 19.62 m

t gt = − = −

= −

r v j i j

i j

( )A B A A B Bm m m m+ = +r r r

( )

( )( ) ( )

1

120 120 19.62 8 120 10 20

12

120 26.033 13.333

B A B A A

B

m m m

m

= + −

= − − − −

= − +

r r r

i j i j k

i j k

( ) ( ) ( )120.0 m 26.0 m 13.33 mB

= − +r i j k�






There are no external forces. The mass center moves as if the explosion had not occurred.

( )( ) ( )0

450 4 1800 mt= = =r v i i

( )A B C A A B B C Cm m m m m m+ + = + +r r r r

( )

( )( ) ( )( )

( )( )

1

1500 1800 300 1200 350 600

50

150 2500 450 900

3300 750 900

C A B C A A B B

C

m m m m m

m

= + + − −

= − − −

− + +

= + +

r r r r

i i j k

i j k

i j k

( ) ( ) ( )3300 m 750 m 900 mC

= + +r i j k�






Mass center at time of first collision.

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

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

1 1 1 1

1 1 1 1

1

1

9600 2800 27.8 3600 38.4 3200 120

40 ft 22.508 ft

A B C A A B B C C

A B C A A B B C C

m m m m m m

W W W W W W

+ + = + +

+ + = + +

= − + − +

= −

r r r r

r r r r

r j j i

r i j

Mass center at time of photo.

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

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

2 2 2 2

2 2 2 2

2

2

9600 2800 30.3 50.7 3600 30.3 61.2

3200 59.4 45.6

40 ft 22.5375 ft

A B C A A B B C C

A B C A A B B C C

m m m m m m

W W W W W W

+ + = + +

+ + = + +

= − + + − +

+ − −

= − +

r r r r

r r r r

r i j i j

i j

r i j

Since no external horizontal forces act, momentum is conserved, and the mass center moves at constant

velocity.

( ) ( ) ( ) ( )1 1 1A B C A A B B C C

m m m m m m+ + = + +v v v v (1)

2 1t− =r r v (2)

Combining (1) and (2), ( )( ) ( ) ( ) ( )2 1 1 1 1A B C A A B B C Cm m m m m m t + + − = + + r r v v v

( )( ) ( ) ( ) ( )2 1 1 1 1A B C A A B B C CW W W W W W t + + − = + + r r v v v

( )( ) ( )( ) ( )( )1

9600 80 45.0455 0 3600 3200 66Bv t − + = + + − i j j i

Components. : 768000 211200t− = −i 3.64 st = �

: 432440 3600Bv t=j

( )

( )( )432440

30.0343600 3.6363

Bv = = 30.0 ft/s

Bv = �






Mass center at time of first collision.

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

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

1 1 1 1

1 1 1 1

1

1

9600 2800 27.8 3600 38.4 3200 120

40 ft 22.508 ft

A B C A A B B C C

A B C A A B B C C

m m m m m m

W W W W W W

+ + = + +

+ + = + +

= − + − +

= −

r r r r

r r r r

r j j i

r i j

Mass center at time of photo.

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

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

2 2 2 2

2 2 2 2

2

2

9600 2800 30.3 50.7 3600 30.3 61.2

3200 59.4 45.6

40 ft 22.5375 ft

A B C A A B B C C

A B C A A B B C C

m m m m m m

W W W W W W

+ + = + +

+ + = + +

= − + + − +

+ − −

= − +

r r r r

r r r r

r i j i j

i j

r i j

Since no external horizontal forces act, momentum is conserved, and the mass center moves at constant

velocity.

( ) ( ) ( ) ( )1 1 1A B C A A B B C C

m m m m m m+ + = + +v v v v (1)

2 1t− =r r v (2)

Combining (1) and (2), ( )( ) ( ) ( ) ( )2 1 1 1 1A B C A A B B C Cm m m m m m t + + − = + + r r v v v

( )( ) ( ) ( ) ( )2 1 1 1 1A B C A A B B C CW W W W W W t + + − = + + r r v v v

( )( ) ( )( ) ( )( ) ( )1 1

9600 80 45.0455 0 3600 3200 3.4B Cv v − + = + + i j j i

Components.

( ) ( )1 1

: 432440 12240 , 35.33 ft/s,B Bv v= =j

24.1mi/hBv = �

( ) ( )1 1

: 768000 10880 , 70.588 ft/s,C Cv v− = − =i

48.1mi/hCv = �






Projectile motion 2

0, 9.81 m/s , 0x y z

a a g a= = − = − =

( ) ( ) ( )00 0

165 m/s, 0, 0x y zv v v= = =

After the chain breaks the mass center continues the original projectile motion.

At 1.5 s,t =

( ) ( )( )0 00 165 1.5 247.5 m

xx x v t= + = + =

( ) ( )( )22

00

1 115 0 9.81 1.5 3.9638 m

2 2y

y y v t gt= + − = + − =

( )0 00

zz z v t= + =

Position of first cannon ball at this time is

1 1 1240 m, 0, 7 mx y z= = =

Definition of mass center: ( )1 2 1 1 2 2m m m m+ = +r r r

( )1 2

2 1

2 2

m m m

m m

1+= −r r r

( ) ( )

( ) ( )

30 15247.5 3.9638 240 7

15 15

255 m . m 7

= + − +

= + 7 9276 −

i j i k

i j k

Position of second cannon ball: 2 2 2

255 m, 7.93 m, 7 mx y z= = = − �






Place the vertical y axis along the initial vertical path of the rocket. Let the x axis be directed to the right (east).

Motion of the mass center: 0, 0, 0x x

a v x= = =

2

9.81 m/sy

a g= − = −

028 9.81

y yv v a t t= + = −

2 2

0 0

160 28 4.905

2

At 5.85 s, 0, 55.939 m

yy y v t a t t t

t x y

= + + = + −

= = =

:A A B B

Definition of mass center m m m= +r r r

component: 3 1 2

0 74.4 2 37.2 m

A B

B B

x x x x

x x

= += − + =

( )( ) component: 3 1 2

3 55.939 0 2 83.9 m

A B

B B

y y y y

y y

= +

= + =

.Position of part B 37.2 m(east), 83.9 m(up)�






Velocities of pieces C and D after impact and fracture.

( ) ( )

( ) ( )

2.13 m/s, 3tan30 m/s

0.7

2.12.333 m/s, 2.3333tan m/s

0.9

CC Cx y

C

DD Dx y

D

xv v

t

xv v

t

θ

′ ′= = = = °

′ ′= = = = −

Assume that during the impact the impulse between spheres A and B is directed along the x axis. Then, the y

component of momentum of sphere A is conserved.

( )0 A ym v′=

Conservation of momentum of system:

( ) ( ) ( )0: 0

A B A A C C D D xxm v m m v m v m v′ ′ ′+ = + +

( ) ( ) ( )4.8 0 3 2.33332 2

A

m m

m mv′+ = + +

( )a 2.13 m/sA′ =v �

( ) ( ) ( ) ( ) ( ): 0 0A B A A C C D Dy yym m m v m v m v′ ′ ′+ = + +

( ) ( )0 0 0 3tan30 2.3333tan2 2

m m θ+ = + ° −

( )b 3

tan tan30 0.74232.3333

θ = ° = 36.6θ = °�

( ) ( ) ( ) ( )22 2 23 3tan30C C Cx y

v v v= + = + o

3.46 m/sCv = �

( ) ( ) ( ) ( )22 2 22.3333 2.3333tan36.6D D Dx y

v v v= + = + o

2.91m/sDv = �






( )( )( )( )

0 0Velocity vectors: cos30 sin 30

sin 7.4 cos7.4

sin 49.3 cos 49.3

cos 45 sin 45

A A

B B

C C

v

v

v

v

= ° + °

= ° + °

= ° − °

= ° + °

v i j

v i j

v i j

v i j


0A A A B B C Cm m m m= + +v v v v

Divide by and substitute data.A B C

m m m= =

( ) ( ) ( )( )

4 cos30 sin30 sin 7.4 cos7.4 sin 49.3 cos49.3

2.1 cos45 sin 45

A Bv v° + ° = ° + ° + ° − °

+ ° + °

i j i j i j

i j

Resolve into components and rearrange.

( ) ( )( ) ( )

: sin 7.4 sin 49.3 4cos30 2.1cos 45

: cos7.4 cos 49.3 4sin 30 2.1sin 45

A B

A B

v v

v v

° + ° = − °

° − ° = − °

i

j

o

o

Solving simultaneously,

(a) 2.01m/sAv = �

(b) 2.27 m/sBv = �






( ) ( )

0190 mi/h east 278.67 ft/s

Place orgin at point of impact.

A= =v i

0 0 00, 0, 0x y z= = =

After impact the motion is projectile motion.

2

0 0

1

2t gt= + −r r v j

0

0

1

2gt

t

−= +r rv j

where ( ) ( ) ( )1600 ft 2400 ft 400 ft= − +r i j k

0

0=r

( )( )

( ) ( ) ( )

0

1600 2400 400 132.2 12

12 12 12 2

133.333 ft/s ft/s 33.333 ft/s

= − + +

= − 6.80 +

v i j k j

i j k

Impact: Conservation of momentum.

( ) ( ) ( ) 00 0A A B B A Bm m m m+ = +v v v

( ) ( )00 0

A B A

B A

B B

m m m

m m

+= −v v v

( ) ( )

( ) ( ) ( )

23000 10000133.333 6.80 33.333 278.67

13000 13000

21.537 ft/s 12.031 ft/s 58.975 ft/s

= − + −

= − +

i j k i

i j k

Components: ( )21.537 ft/s 14.69 mi/h east =i �

( )58.974 ft/s 40.2 mi/h south =k �

( )12.031 ft/s 8.20 mi/h down − =j �






Weight of arrow: 2 oz 0.125 lb.

Weight of bird: 6 lb.

A

B

W

W

= =

=

Conservation of momentum: Let v be velocity immediately after impact.

A B A B

A B

W W W W

g g g

++ =v v v

( )( ) ( )( )0.125 180 240 6 30

6.125

29.388 3.6735 4.8980

A A B B

A B A B

W W

W W W W

+ += + =

+ += + +

j k iv vv

i j k

( ) 2

00

1Vertical motion:

2y

y y v t gt= + −

( ) 2 210 45 3.6735 32.2 or 0.22817 2.7950 0

2t t t t= + − − − =

Solving for , 1.7898 st t =

Horizontal motion: ,x z

x v t z v t= =

( )( )( )( )29.388 1.7898 52.6 ft

4.8980 1.7898 8.77 ft

x

z

= =

= =

( ) ( )52.6 ft 8.77 ftP

= +r i k�






( )( )( )

1 2 1 2

1 2 1 2

1 2 1 2

Position vectors (mm): 80 80 40 120

33 70 10 78.032

48 15 50.289

A A A A

B B B B

C C C C

= + + =

= − + − =

= − =

i j k

i j k

j k

��

��

��

1 2

1 2

1 2

Unit vectors: Along , 0.66667 0.66667 0.33333

Along , 0.42290 0.89707 0.12815

Along , 0.95448 0.29828

A

B

C

A A

B B

C C

= + += − + −= −

i j k

i j k

j k

λλλλλλλλλλλλ

Velocity vectors after the collisions:

A A A B B B C C Cv v v= = =v v vλλλλ λλλλ λλλλ


0 0 04 4 4 4

A B Cm m m m m m+ + = + +u v v v v v

Divide by m and substitute data.

( )600 750 800 2400 2400 4 4A A B B C Cv v v− + − + + = + +i j k j j λλλλ λλλλ λλλλ

Resolving into components,

: 600 0.66667 1.69160

: 5550 0.66667 3.58828 3.81792

: 800 0.33333 0.51260 1.19312

A B

A B C

A B C

v v

v v v

v v v

− = −= + +

− = − −

i

j

k

Solving the three equations simultaneously,

919.26 m/s, 716.98 m/s, 619.30 m/sA B Cv v v= = =

919 m/sAv = �

717 m/sBv = �

619 m/sCv = �






(ft):Position vectors 18D

=r k

/ /

/ /

/ /

7.5 7.5 18 19.5

18 9 18 9 18 27

13.5 13.5 18 22.5

A A D A D

B B D B D

C C D C D

r

r

r

= − = − − =

= + = + − =

= − = − − =

r i r i k

r i j r i j k

r j r j k

( )

( )

( )

/

/

/

1 : Along , 7.5 18

19.5

1Along , 18 9 18

27

1Along , 13.5 18

22.5

A D A

B D B

C D C

Unit vectors = − −

= + −

= − −

r i k

r i j k

r j k

λλλλ

λλλλ

λλλλ

Assume that elevation changes due to gravity may be neglected. Then, the velocity vectors after the

explosition have the directions of the unit vectors.


0Conservation of momentum:

A A B B C Cm m m m= + +v v v v

( ) ( ) ( ) ( )18 8 6 460 45 1800 7.5 18 18 9 18 13.5 18

19.5 27 22.5

A B Cv v v

g g g g

− − = − − + + − + − −

i j k i k i j k j k

Multiply by g and resolve into components.

1080 60 10819.5 27

810 54 5227 22.5

32400 144 108 7219.5 27 22.5

A B

B C

A B C

v v

v v

v v v

= − +

− = −

− = − − −

Solving, 119.94419.5

Av = 2340 ft/s

Av = �

76.63527

Bv = 2070 ft/s

Bv = �

95.16022.5

Cv = 2140 ft/s

Cv = �






(ft):Position vectors 18D

=r k

/ /

/ /

/ /

7.5 7.5 18 19.5

18 9 18 9 18 27

13.5 13.5 18 22.5

A A D A D

B B D B D

C C D C D

r

r

r

= − = − − =

= + = + − =

= − = − − =

r i r i k

r i j r i j k

r j r j k

( )

( )

( )

/

/

/

1 : Along , 7.5 18

19.5

1Along , 18 9 18

27

1Along , 13.5 18

22.5

A D A

B D B

C D C

Unit vectors = − −

= + −

= − −

r i k

r i j k

r j k

λλλλ

λλλλ

λλλλ

Assume that elevation changes due to gravity may be neglected. Then, the velocity vectors after the explosition

have the directions of the unit vectors.


/19.5

where 1950 ft/s0.010

A D

A

A

rv

t

= = =

/

271500 ft/s

0.018

B D

B

B

rv

t

= = =

C/

22.51875 ft/s

0.012

D

C

C

rv

t

= = =

( ) ( )so that 750 ft/s 1800 ft/sA

= − −v i k

( ) ( ) ( )1000 ft/s + 500 ft/s 1000 ft/sB

= −v i j k

( ) ( )1125 ft 1500 ft/sC

= − −v j k

Conservation of momentum: 0 A A B B C C

m m m m= + +v v v v

( ) ( ) ( )0750 1800 1000 500 1000 1125 1500

A B CW W W W

vg g g g

− = − − + + − + − −

k i k i j k j k





Divide by g and resolve into components.

0

: 0 750 1000

: 0 500 1125

: 1800 1000 1500

A B

B C

A B C

W W

W W

Wv W W W

= − += −

− = − − −

i

j

k

(1)

(2)

(3)

Since mass is conserved, 6 lbA B C

W W W W= + + = (4)

Solving equations (1), (2), and (4) simultaneously,

(a) 2.88 lb, 2.16 lb, 0.96 lbA B C

W W W= = = �

substituting into (3),

( )( ) ( )( ) ( )( )06 1800 2.88 1000 2.16 1500 0.96v− = − − −

(b) 0

1464 ft/sv = �






( )

( )

( ) ( )

1

1

1 1

From Eq. (14.7),n

O i i i

i

n

i i i

i

n n

i i i i i

i i

G

m

m

m v r m

m

=

=

= =

= ×

′ = + ×

′= × + ×

= × +

∑

∑

∑ ∑

H r v

r r v

r v

r v H






( )

( )

( )

( )

( )( )

1

1

1 1

1

1

if, and only if, 0

i A i

n

A i i i

i

n

i i A i

i

n n

i i A i i i

i i

n

i i A A

i

n

i i A A A

i

A A A

A A A A

m

m

m m

m

m

m

m

=

=

= =

=

=

′= +

′= ×

′ ′= × +

′ ′= × + ×

′ ′= × +

′= − × +

′= − × +

′= − × =

∑

∑

∑ ∑

∑

∑

v v v

H r v

r v v

r v r v

r v H

r r v H

r r v H

H H r r v

This condition is satisfied if,

( ) 0 Point has zero velocity.

or ( ) Point coincides with the mass center.

or ( ) is parallel to . Velocity is directed along line .

A

A

A A A

a A

b A

c AG

==

−

v

r r

v r r v






From equation (1), ( )1

n

A i i i

i

m

=′ ′ ′= ×∑H r v

( ) ( )1

n

A i A i i A

i

m

=′ = − × − ∑H r r v v

Differentiate with respect to time.

( ) ( ) ( ) ( )1 1

n n

A i A i i A i A i i A

i i

m m

= =′ = − × − + − × − ∑ ∑H r r v v r r v v& & & & &

But , , , andi i i i A A A A

= = = =r v v a r v v a& && &

( ) ( )

( ) ( )

( ) ( )

( )

1

1

1 1

Hence, 0n

A i A i i A

i

n

i A i i A

i

n n

i A i i i A A

i i

A A A

m

m

m

m r

=

=

= =

′ = + − × −

= − × −

= − × − − ×

= − − ×

∑

∑

∑ ∑

H r r a a

r r F a

r r F r r a

M r a

&

( )if, and only if, 0A A A A

M m′ = − × =H r r a&

This condition is satisfied if

( ) 0 The frame is newtonian.

or ( ) Point coincides with the mass center.

or ( ) is parallel to . Acceleration is directed along line .

A

A

A A A

a

b A

c AG

==

−

a

r r

a r r a






The masses are m for the bullet and A

m and B

m for the blocks.

The bullet passes through block A and embeds in block B. Momentum is conserved.

( ) ( )0 0Initial momentum: 0 0

A Bmv m m mv+ + =

Final momentum:B A A B B

mv m v m v+ +

0Equating,

B A A B Bmv mv m v m v= + +

( )( ) ( )( ) 3

0

3 3 2.5 543.434 10 kg

500 5

A A B B

B

m v m v

m

v v

−++= = = ×− −

The bullet passes through block A. Momentum is conserved.

( )0 0Initial momentum: 0

Amv m mv+ =

1Final momentum:

A Amv m v+

0 1Equating,

A Amv mv m v= +

( )( ) ( )( )3

0

1 3

43.434 10 500 3 3

292.79 m/s43.434 10

A Amv m v

v

m

−

−

× −−= = =×

(a) Bullet passes through block A. Kinetic energies.

( )( )22 3

0 0

1 1Before: 43.434 10 500 5429 J

2 2T mv

−= = × =

( )( ) ( )( )2 22 2 3

1 1

1 1 1 1After: 43.434 10 292.79 3 3 1875 J

2 2 2 2A A

T mv m v−= + = × + =

0 1Lost: 5429 1875 3554 JT T− = − = energy lost 3550 J= �

(b) Bullet becomes embedded in block B. Kinetic energies.

( )( )22 3

2 1

1 1Before: 43.434 10 292.79 1861.7 J

2 2T mv

−= = × =

( ) ( )( )22

3

1 1After: 2.54343 5 31.8 J

2 2B B

T m m v= + = =

2 3Lost: 1862 31.8 1830 JT T− = − = energy lost 1830 J= �






Data and results from Prob. 14.1.

Masses: 1350 kg,A B

m m= = 5400 kgc

m =

Initial velocities: 0 0

( ) ( ) 0,A Bv v= =

0( ) 8 km/h = 2.2222 m/s

cv =

Velocities after first collision:

1

( ) 0,Av =

1 1( ) ( ) 1.77778 m/s

B cv v= =

Velocities after second collision

2.9630 m/s,Av = 1.18519 m/s

B cv v= =

Initial kinetic energy: 2 2 2

0 0 0 0

1 1 1( ) ( ) ( )

2 2 2A A B B c B

T m v m v m v= + +

2 3

0

10 0 (5400)(2.2222) 13.3333 10 J

2T = + + = ×

Kinetic energy after the first collision:

( )22 2

1 1 11

1 1 1( ) ( )

2 2 2A A B B c c

T m v m v m v= + +

2 2 31 10 (1350) (1.77778) (5400)(1.77778) 10.6667 10 J

2 2= + + = ×

Kinetic energy after the second collision:

2 2 2

2

1 1 1

2 2 2A A B B c c

T m v m v m v= + +

2 2 2 31 1 1(1350)(2.9630) (1350)(1.18519) (5400)(1.18519) 10.6668 10 J

2 2 2= + + = ×

Kinetic energy lost in first collision: 3

0 12.6667 10 JT T− = ×

2.67 kJ�

Kinetic energies before and after second collision:

2 110.67 kJT T= = �






( )( )22 20004000 3700The masses are 124.2 slugs, 114.9 slugs, and 1366.5 slugs

32.2 32.2 32.2A B F

m m m= = = = = =

Let , , and be the sought after velocities in ft/s, positive to the right.A B Fv v v


0.A B Fv v v= = =


0.A A B B F F

m v m v m v+ + =


conserved.

0A A B B F F

m v m v m v= + +

124.2 114.9 1366.5 0A B Fv v v+ + = (1)

The relative velocities are given as

/

/

7 ft/s

3.5 ft/s

A F A F

B F B F

v v v

v v v

= − = −

= − = −

(2)

(3)


6.208 ft/s, 2.708 ft/s, 0.7919 ft/sA B Fv v v= − = − =

( ) ( ) ( )22 2

1 0 0 0

1 1 1Initial kinetic energy: 0

2 2 2A A B B C

T m v m v v= + + =

2 2 2

2

1 1 1Final kinetic energy:

2 2 2A A B B C C

T m v m v m v= + +

( )( ) ( )( ) ( )( )2 2 2

2

1 1 1124.2 6.208 114.9 2.708 1366.5 0.7919

2 2 2

3243 ft lb

T = + +

= ⋅

Work done by engines: 1 1T U+

2 2T=

1U

2 2 13243 ft lbT T= − = ⋅ 1

U2

3240 ft lb= ⋅ �






From the solution to Prob. 14.27,

Initial velocity of 6-lb shell: 0

1464 ft/sv =

Weights of fragments: 2.88 lb,A

W = 2.16 lb,B

W = 0.96 lbc

W =

Speeds of fragments: 1950 ft/s,Av = 1500 ft/s,

Bv = 1875 ft/s

cv =

Total kinetic energy before the explosion.

( )22 3

0 0

1 1 61464 199.69 10 ft lb

2 2 32.2

WT v

g

= = = × ⋅

Total kinetic energy after the explosion.

2 2 2

1

1 1 1

2 2 2

A B c

A B c

W W WT v v v

g g g= + +

( ) ( ) ( )2 2 21 2.88 1 2.16 1 0.961950 1500 1875

2 32.2 2 32.2 2 32.2

= + +

3297.92 10 ft lb= × ⋅

Increase in kinetic energy. 3

1 098.2 10 ft lbT T− = × ⋅ �






Velocity of mass center: ( )A B A A B Bm m m m+ = +v v v

A A B B

A B

m m

m m

+=+

v v

v

Velocities relative to the mass center:

( )

( )

B A BA A B B

A A A

A B A B

A A BA A B B

B B B

A B A B

mm m

m m m m

mm m

m m m m

−+′ = − = − =+ +

−+′ = − = − =+ +

v vv v

v v v v

v vv v

v v v v

Energies:

( ) ( )( )

( ) ( )( )

2

2

2

2

1

2 2

1

2 2

A B A B A B

A A A A

A B

A B A B A B

B B B B

A B

m mE m

m m

m mE m

m m

− ⋅ −′ ′= ⋅ =

+

− ⋅ −′ ′= ⋅ =

+

v v v v

v v

v v v v

v v

( ) :a Ratio / /A B B A

E E m m= �

( ) 135 km/h 37.5 m/sA

b = =v , 90 km/h 25 m/sB

= =v

62.5 m/sA B

− =v v

( )( ) ( )

( )( )

2 2

3

2

2400 1350 62.5607.5 10 J

2 3750A

E = = × 608 kJA

E = �

( ) ( )( )

( )( )

2 2

6

2

2400 1350 62.51.08 10 J

2 3750B

E = = × 1080 kJB

E = �






( ) :A B A A B B

Velocity of mass center m m m m+ = +v v v

A A B B

A B

m m

m m

+=+

v v

v

Velocities relative to the mass center:

( )

( )

B A BA A B B

A A A

A B A B

A A BA A B B

B B B

A B A B

mm m

m m m m

mm m

m m m m

−+′ = − = − =+ +

−+′ = − = − =+ +

v vv v

v v v v

v vv v

v v v v

Energies:

( ) ( )( )

( ) ( )( )

2

2

2

2

1

2 2

1

2 2

A B A B A B

A A A A

A B

A B A B A B

B B B B

A B

m mE m

m m

m mE m

m m

− ⋅ −′ ′= ⋅ =

+

− ⋅ −′ ′= ⋅ =

+

v v v v

v v

v v v v

v v

( ) ( )2 2

0 00 0

1 1Energies from tests: ,

2 2A A B B

E m v E m v= =

( )( ) ( )

( )

( )( ) ( )

( )

2

2 2

0 0

2

2 2

0 0

Serverities:B A B A BA

A

A A B

A A B A BB

B

B A B

mES

E m m v

mES

E m m v

− ⋅ −= =

+

− ⋅ −= =

+

v v v v

v v v v

:Ratio

2

2

A B

B A

S m

S m= �






(a) A strikes B and C simultaneously.

During the impact, the contact impulses make 30° angles with the velocity 0

v

( )( )

Thus, cos30 sin30

cos30 sin30

B B

C C

v

v

= ° + °

= ° − °

v i j

v i j

By symmetry,A A

v=v i

0Conservation of momentum:

A B Cm m m m= + +v v v v

0

component: 0 0 sin 30 sin 30

component: cos30 cos30

B C C B

A B C

y mv mv v v

x mv mv mv mv

= + ° − ° == + ° + °

( )0 0

0

2,

cos30 3 3

A A

B C A B C

v v v v

v v v v v v

− −+ = = − = =o

Conservation of energy: 2 2 2 2

0

1 1 1 1

2 2 2 2A B C

mv mv mv mv= + +

( )

( )( ) ( )

( )

22 2

0 0

22 2

0 0 0 0

0 0 0 0

0 0

2

3

2

3

2 1 5 1

3 3 3 5

6 2 3

55 3

A A

A A A A

A A A A

B C

v v v v

v v v v v v v v

v v v v v v v v

v v v v

= + −

− = − + = −

+ = − = − = −

= = =

00.200

Av=v �

00.693

Bv=v 30°�





(b) A strikes B before it strikes C.

First impact; A strikes B.

During the impact, the contact impulse makes a 30° angle with the velocity 0.v

( )Thus, cos30 sin30B B

v= ° + °v i j

0Conservation of momentum.

A Bm m m= +v v v

( ) ( )( ) ( )0 0

component: 0 sin30 sin30

component: cos30 cos30

A B A By y

A B A Bx x

y m v mv v v

x v m v mv v v v

′ ′= + ° = − °

′ ′= + ° = − °

Conservation of energy:

( ) ( )

( ) ( )

( )

2 22 2

0

2 2 2

0

2 2 2 2 2 2

0 0

1 1 1 1

2 2 2 2

1 1 1cos30 sin30

2 2 2

12 cos30 cos 30 sin 30

2

A A Bx y

B B B

B B B B

mv m v m v mv

m v v v v

m v v v v v v

′ ′= + +

= − ° + ° +

= − ° + ° + ° +

( )

( )

2

0 0 0 0

0 0

3 1cos30 , sin 30 ,

2 4

3cos30 sin 30

4

B A x

A y

v v v v v v

v v v

′= ° = = ° =

′ = − ° ° = −

Second impact: A strikes C.

During the impact, the contact impulse makes a 30o angle with the velocity 0.v

( )Thus, cos30 sin30C C

v= ° − °v i j

Conservation of momentum:A A C

m m m′ = +v v v





( ) ( )

( ) ( )

( ) ( )

( ) ( )

0

0

component: cos30 ,

1cos30 cos30

4

component: sin30

3sin30 sin30

4

A A Cx x

A A C Cx x

A A Cy y

A A C Cy y

x m v m v mv

v v v v v

y m v m v mv

v v v v v

′ = + °

′= − ° = − °

′ = − °

′= + ° = − + °


( ) ( ) ( ) ( )2 2 2 2 2

22

2 2 2

0 0 0 0

2 2 2

0 0

2 2 2 2

0 0

1 1 1 1 1

2 2 2 2 2

1 1 3 1 1 3cos30 sin30

2 16 16 2 4 4

1 1 1cos30 cos 30

2 16 2

3 3sin30 sin 30

16 2

A A A A Cx y x y

C C C

C C

C C C

m v m v m v m v mv

m v v m v v v v v

m v v v v

v v v v v

′ ′+ = + +

+ = − ° + − + ° +

= − ° + °

+ − ° + ° +

2

0

1 30 cos30 sin30 2

2 2C C

v v v

= − ° + ° +

( )

( )

0 0

0 0 0

0 0 0

1 3 3cos30 sin30

4 4 4

1 3 1cos30

4 4 8

3 3 3sin30

4 4 8

C

A x

A y

v v v

v v v v

v v v v

= ° + ° =

= − ° = −

= − + ° = −

00.250

Av=v 60°�

00.866

Bv=v 30°�

00.433

Cv=v 30°�






(a) Velocity of B at maximum elevation. At maximum elevation ball B is at rest relative to cart A. B A

=v v

Use impulse-momentum principle.

components:x ( )00

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

0A

B

A B

m v

v

m m

=+ �

(b) Conservation of energy:

( ) ( )

2

1 0 1

2 2

2 2 2 0

2

2

1, 0

2

1 1 1

2 2 2 2

A

A

A A B B A B B

A B

B

T m v V

m vT m v m v m m v

m m

V m gh

= =

= + = + =+

=

( )

2 2 1 1

2 2

20

0

1

2 2

A

B A

A B

T V T V

m vm gh m v

m m

+ = +

+ =+

2 2

2 0

0

1

2

A

A

B A B

m vh m v

m g m m

= − +

2

0

2

A

A B

m vh

m m g=

+�






Velocity vectors: ( )0 0cos30 sin30v= ° − °v i j

015 ft/sv =

A A

v= −v j

( )( )sin30 cos30

cos30 sin30

B B

C C

v

v

= ° − °

= ° + °

v i j

v i j


0 A B C

m m m m= + +v v v v

Divide by m and resolve into components.

i: 0cos30 sin 30 cos30

B Cv v v° = ° + °

j: 0sin30 cos30 sin30

A B Cv v v v− ° = − − ° + °

Solving for and ,B Cv v

( ) ( )0 0

3 1

2 2B A C Av v v v v v= − = +


0

1 1 1 1

2 2 2 2A B C

mv mv mv mv= + +

Divide by 1

2m and substitute for and .

B Cv v

( ) ( )2 22 2

0 0 0

2 2

0 0

3 1

4 4

2

A A A

A A

v v v v v v

v v v v

= + − + +

= + −

0

17.5 ft/s

2Av v= = 7.50 ft/s

Av = �

( )315 7.5 6.4952 ft/s

2Bv = − = 6.50 ft/s

Bv = �

( )115 7.5 11.25 ft/s

2Cv = + = 11.25 ft/s

Cv = �






Velocity vectors: ( )0 0cos45 sin 45v= ° + °v i j

015 ft/sv =

A A

v=v j

( )( )sin 60 cos60

cos60 sin 60

B B

C C

v

v

= ° − °

= ° + °

v i j

v i j


0 A B C

m m m m= + +v v v v


i: 0cos45 sin 60 cos60

B Cv v v° = ° + °

j: 0sin 45 cos60 sin 60

A B Cv v v v° = − ° + °

Solving for and ,B Cv v

0 00.25882 0.5 0.96593 0.86603

B A C Av v v v v v= − = −


0

1 1 1 1

2 2 2 2A B C

mv mv mv mv= + +

Divide by m and substitute for and .B Cv v

( ) ( )2 22 2

0 0 0

2 2

0 0

0.25882 0.5 0.96593 0.86603

1.4142 2

A A A

A A

v v v v v v

v v v v

= + + + −

= + +

0

0.70711 10.607 ft/sAv v= = 10.61 ft/s

Av = �

0

0.61237 9.1856 ft/sBv v= = 9.19 ft/s

Bv = �

0

0.35356 5.303 ft/sCv v= = 5.30 ft/s

Cv = �






1sin ,

3θ =

8cos ,

3θ = 19.471θ = °

Velocity vectors 0 0

v= −v j

( )cos sinA A

v θ θ= −v i j

( )/sin cos

B A Bu θ θ= − −v i j

/B A B A

= +v v v

C C

v=v j

Conservation of momentum: 0 /

2A B C A B A C

m m m m m m m= + + = + +v v v v v v v


i: 0 2 cos sinA Bv uθ θ= −

−j : 0

2 sin cosA B C

v v u vθ θ= + + −

Solving for and ,A Bv u ( ) ( )0 0

1 0.94281

6A C B Cv v v u v v= + = +


0

1 1 1 1

2 2 2 2A B C

mv mv mv mv= + +

( )2 2 2 21 1 1

2 2 2A A B C

mv m v u mv= + + +

Divide by 1

2m and substitute for and .

A Bv u

( ) ( ) ( )2

2 222 2

0 0 0

12 0.94281

6C C C

v v v v v v

= + + + +

( )22 2

0 0 00.94445 0.02857

C C Cv v v v v v− = + =

00.0286

Cv=v �

[0 00.17143 0.17143

A Av v v= =v ]19.471° ,

00.1714

Av=v 19.5°�

[0 / 00.96975 0.96975

B B Au v v= v ]19.471°

/B A B A

= +v v v 0.985B

=v 80.1°�






1 8sin , cos , 19.471

3 3θ θ θ= = = °

C strikes B.


0 0

or B C B C

m m m v v v′ ′= + = −v v v


( )22 2

0

1 1 1

2 2 2B C

mv m v mv′= +

( )22 2

0 0 C Cv v v v= − +

0Cv =

0B

v v′ =

Cord becomes taut.

Velocity vectors:

A A

v=v θ

/B A B

u=v θ

Conservation of momentum: /B A A B A

m m m m′ = + +v v v v


+ :θ sin 2B Av vθ′ =

0

1 1sin

2 6A Bv v vθ′= =





(a) 0

6A

v=v 19.5°�

+ :θ cosB Bv uθ′ = 0

cos 83

B B

v

u v θ′= =

0

1 19.471

6B

v = °

v [ 00.94281v+ ]19.471°

[ 00.95743

Bv=v ]80.8°

00.957

Bv=v 80.5°�

0C

=v �

Initial kinetic energy: 2

1 0

1

2T mv=

Final kinetic energy: 2 2 2

2

1 1 1

2 2 2A B C

T mv mv mv= + +

( ) ( )2

22 2

0 0

1 1 1.95743 0 0.94444

2 6 2mv mv

= + + =

(b) Fraction lost: 1 2

1

1 0.944440.05555

1

T T

T

− −= =

Fraction of energy lost = 0.0556 �






(a) Use part (a) of sample Problem 14.4 with A B

m m m.= =

0 0

1

2

m

v v v

m m

= =+

0

1

2A Bv v v= = �

(b) Consider initial position and position when 0θ = again.

Conservation of momentum

0 A A B B

mv m v m v= +

0B Av v v+ = (1)

Conservation of energy

2 2 2

0

1 1 1

2 2 2A B

mv mv mv= + (2)

Substituting (1) into (2),

2 2 21 1 1( )

2 2 2A B A B

m v v mv mv+ = +

0A B

mv v =

Either 0Av = with

0Bv v= �

or 0Bv = with

0Av v= �

(c) Consider positions when max

θ θ= and min

.θ θ=

Since /

0B Av = at these states

B Av v=

Conservation of momentum

0 A B

mv mv mv= +

0

1

2B Av v v= = �





Conservation of energy would show that the elevation of B is the same for max

θ θ= and min

.θ θ=

Both A and B keep moving to the right with A and B stopping intermittently.






Relative velocity and acceleration.

/B A B A= + =v v v [vA ] + [vB/A 30° ]

/B A B A= + =a a a [aA ] + [aB/A 30° ]

Draw free body diagrams and apply Newton’s second law.

Block:

:F ma∑ = 1

cos30 sin30B B A

N m g m a− °= − ° (1)

:F ma∑ = 1 /sin30 cos30

B B A B AN m a m a°= °− (2)

Wedge:

:F ma∑ = 1sin30

A AN m a°= (3)

Rearranging (1), (2), and (3) and applying numerical data,

1

(6sin30 ) (6)(9.81)cos30A

N a+ ° = ° (1)

1 /

(sin30 ) 6 (6cos30 ) 0A B A

N a a° + − ° = (2)

1

(sin30 ) 10 0A

N a° − = (3)


1

44.325N,N = 22.2163 m/s ,

Aa = 2

/6.8243 m/s

B Aa =





Sliding motion of block relative to wedge.

2

/

/ /

( )

2

B A

B A B A

v

a s=

/ / /

2 (2)(6.8243)(1.0) 3.6944 m/sB A B A B Av a s= = =

v 3.6944

0.54136 s6.8243

B/A

B/A

t

a

= = =

Motion of wedge.

(2.2163)(0.54136) 1.1998 m/sA Av a t= = =

(a) Velocity of B relative to A. /

3.69 m/sB A

=v 30° �

(b) Velocity of A . 1.200 m/sA

=v �






There are no external forces. Momentum is conserved.

(a) Moments about D : ( )00.9 1.8 0.9

A C C A B Bm v m v m m v= + +

( ) ( )( ) ( )0

0.90.90.5 12 2.5 3.50

1.8 1.8

A BA

C B

C C

m mm

v v v

m m

+= − = − = 3.50 m/s

Cv = �

Moments about C : ( )00.9 0.9 1.8

A A B B D Dm v m m v m v= + +

( ) ( )( ) ( )( )0

0.90.90.25 12 0.5 2.5 1.750m/s

1.8 1.8

A BA

D B

D D

m mm v

v v

m m

+= − = − = 1.750 m/s

Dv = �

(b) Initial kinetic energy:

( )22

1 0

1 17.5 12 540 N m

2 2A

T m v= = = ⋅

Final kinetic energy:

( ) ( ) ( )

2 2 2

2

2 2 2

1 1 1( )

2 2 2

1 1 115 2.5 7.5 3.5 15 1.750 115.78 N m

2 2 2

A B B C C D DT m m v m v m v= + + +

= + + = ⋅

Energy lost: 540 115.78 424.22 N m− = ⋅

Fraction of energy lost 424.22

0.786540

= =

( )1 2

1

0.786T T

T

−= �






There are no external forces. Momentum is conserved.

(a) Moments about D : 00.9 1.8 0.9

A C C B Bm v m v m v= +

( )( ) ( )( )0

0.9 0.90.5 12 0.5 3.5 4.25

1.8 1.8

A B

C B

C C

m m

v v v

m m

= − = − = 4.25 m/sCv = �

Moments about C : 0

0.9 1.8 0.9A D D B B

m v m v m v= +

( )( ) ( )( )0

0.9 0.90.25 12 0.25 3.5 2.125 m/s

1.8 1.8

A B

D B

D D

m m

v v v

m m

= − = − = 2.13 m/sDv = �

(b) Initial kinetic energy:

( )22

1 0

1 17.5 12 540 N m

2 2A

T m v= = = ⋅

Final kinetic energy:

( ) ( ) ( )

2 2 2

2

2 2 2

1 1 1

2 2 2

1 1 17.5 3.5 7.5 4.25 15 2.125 147.54 N m

2 2 2

B B C C D DT m v m v m v= + +

= + + = ⋅

Energy lost: 540 147.54 392.46 N m− = ⋅

Fraction of energy lost 392.46

0.727540

= =

( )1 2

1

0.727T T

T

−= �






(a) Linear and angular momentum.

00

A A B Bm m mv= + = +L v v i

0mv=L i�

0

2( ) ( ) 0

3 3G

l lmv= × + − ×H j i j i

0

2

3G

lmv= −H k�

Motion of mass center G. Since there is no external force,

0constant

A A B Bm m mv= + = =L v v i

03m mv=v i

0

1

3v=v i

Motion about mass center.

( ) ( )G G i i i A A A B B B

m m m′ ′ ′ ′ ′ ′ ′= = Σ × = × + ×H H r v r v r v

where 2 1

3 3A B

l , l′ ′ ′ ′= = −r j r j

2 1,

3 3A B

l lθ θ′ ′ ′ ′= = −v i v i& &

Thus, 2 2 1 1 1

23 3 3 3 3

Gl ml l m lθ θ ′ ′ ′ ′= × + − × ⋅

H j i j i& &

22

3ml θ= − k&

But HG

is constant.

2 0

0

2 2

3 3

vml lmv

lθ θ− = − =k k &

0

2 2

3 3Av l vθ′ = =&

0

1 1

3 3Bv l vθ′ = =&





(b) After 180º rotation.

0 0

1 2

3 3A A

v v′= + = −v v v i i

0

1

3A

v= −v i�

0 0

1 1

3 3B B

v v′= + = +v v v i i

0

2

3B

v=v i �






Masses: 21253.882 lb s /ft, 2 , 3 .

32.2A B A C A

m m m m m= = ⋅ = =

Conservation of angular momentum about O.

240 240 2160 ( ) ( ) ( )A A A x A y A zv v v= + + = + +r i j k v i j k

600 1320 3240 500 1100 2200B B

= + + = + +r i j k v i j k

480 960 1920 400 ( ) ( )C C C y C zv v= − − + = − + +r i j k v i j k

Since the three parts pass through O, the angular momentum about O is zero. 0

0=H

00

A A A B B B C C Cm m m= × + × + × =H r v r v r v

[ 2 3 ] 0A A A B B C C

m × + × + × =r v r v r v

Dividing by mA and using determinant form,

240 240 2160 1200 2640 6480 1440 2880 5760

( ) ( ) ( ) 500 1100 2200 400 ( ) ( )A x A y A z C y C zv v v v v

+ + − −−

i j k i j k i j k

[240( ) 2160( ) ] [2160( ) 240( ) ]A z A y A x A zv v v v= − + −i j

6 6[240( ) 240( ) ] 1.320 10 0.6 10A y A xv v+ − − × + ×k i j

0 [ 2880( ) 5760( ) ]C z C yv v+ + − −k i

6 6[1440( ) 2.304 10 ] [ 1440( ) 1.152 10 ] 0C z C yv v+ − × + − − × =j k

Dividing by 240 and equating to zero the coefficients of i, j, and k,

: ( ) 9( ) 5500 12( ) 24( ) 0A z A y C z C yv v v vi − − − − = (1)

: 9( ) ( ) 7100 6( ) 0A x A z C zv v vj − − + = (2)

: ( ) ( ) 6( ) 4800 0A y A x C yv v vk − − − = (3)

Conservation of linear momentum.

0( )

A A B C C C A B Cm m m m m m+ + = + +v v v v

0( 2 3 ) 6 ( )

A A C C Am m+ + =v v v v





Dividing by mA and substituting given data,

( ) + ( ) ( ) (2)(500 1100 2200 ) (3)[ 400 + ( ) ( ) ] (6)(1500)A x A y A z C y C zv v v + v + v+ + + + − =i j k i j k i j k k

Separating into components,

: ( ) 1000 1200 0A xvi + − = (4)

: ( ) 2200 3( ) 0A y C yv vj + + = (5)

: ( ) 4400 3( ) 9000A z C zv vk + + = (6)

From (4), ( ) 200 ft/sA xv =

Solving (3) and (5) simultaneously,

( ) 200 ft/s ( ) 800 ft/sA y C yv v= = −

Solving (2) and (6) simultaneously,

( ) 1300 ft/s ( ) 1100 ft/sA z C zv v= =

(200 ft/s) (200 ft/s) (1300 ft/s)A

= + +v i j k �






Let the system consist of spheres A and B.

State 1. Instant cord DC breaks.

( ) 01

3 1

2 2A

m mv

= − −

v i j

( ) 01

3 1

2 2B

m mv

= −

v i j

( ) ( )1 01 1A Bm m mv= + = −L v v j

1

0

1

2 2v

m

= = −Lv j

Mass center lies at point G as shown.

( ) ( ) ( )1 11

0

3 3

2 2

3

2

G A Bl m l m

lmv

= × + − ×

=

H j v j v

k

2 2 2

1 0 0 0

1 1

2 2T mv mv mv= + =

State 2. The cord is taut. Conservation of linear momentum:

(a) 0

1

2D

v= = −v v j

00.500

Dv v= �

Let ( )2

and A A B B

= + = +v v u v v u

2 12

A Bm m m= + + =L v u u L

B A B Au u= − =u u

( )2

2G A B A

lmu lmu lmu= + =H k k k





(b) Conservation of angular momentum:

( ) ( )2 1G G

=H H

0

32

2A

lmu lmv=k k

0

3

4A B

u u v= =

00.750u v= �

( ) 2 2 2

2

2 2

0 0

1 1 12

2 2 2

1 1 9 9 13

2 2 16 16 16

A BT m v mu mu

mv mv

= + +

= + + =

(c) Fraction of energy lost: 13

1 2 16

1

1 3

1 16

T T

T

−− = =

1 2

1

0.1875T T

T

− = �






The system is spheres A and B and the ring D.

Initial velocities: ( )0cos30 sin30

Av= − ° − °v i j

( )0cos30 sin30

0

B

D

v= − ° − °

=

v i j

v

Locate the mass center.

( ) ( )0

0

4

sin30 cos30 sin30 cos30

1

4

A Bm m m

ml ml

l

= +

= − ° + ° + − ° − °

= −

r r r

i j i j

r i

Velocity of mass center.

( ) ( )0 0

0

4

cos30 sin30 cos30 sin30

1

4

A Bm m m

mv mv

v

= +

= − ° − ° + ° − °

= −

v v v

i j i j

v j

(a) Motion of mass center 0

t= +r r v

0

1 1

4 4l v t= − −r i j �

(b) / /G AG A B G B

m m= × + ×H r v r v

( )

( )

0

0

1cos30 cos30 sin 30

4

1 cos30 cos30 sin 30

4

l l mv

l l mv

= − + ° × − ° − °

+ − − ° × ° − °

i j i j

i j i j

0

7

4G

lmv=H k �

(c) 2 2 2

0

1 1

2 2A B

T mv mv mv= + =

2

0T mv= �






Let m be the mass of one ball.

Conservation of linear momentum: 0

( ) ( )m mΣ = Σv v

0 0 0( ) ( ) ( )

A B C A B Cm m m m m m+ + = + +v v v v v v

Dividing by m and applying numerical data,

(0.5 ft/s) [(3.75 ft/s) ( ) ] [( ) ( ) ] (6.5 ft/s) 0 0B y C x C yv v v+ + + + = + +i i j i j i

Components:

: 0.5 3.75 ( ) 6.5C x

x v+ = + ( ) 2.25 ft/sC xv = �

: ( ) ( ) 0B y C yy v v+ = (1)

Conservation of angular momentum about O:

0[ ( )] [ ( )]m mΣ × = Σ ×r v r v

where rA = 0, rB = 0, (1.5 ft)(cos30 sin30 )C

= ° + °r i j

( )( )1.5 cos30 + sin 30 [ ( ) ( ) ] 0C x C ym v m vi j i j° ° × + =

Since their cross product is zero, the two vectors are parallel.

( ) ( ) tan30 2.25 tan30 1.2990 ft/sC y C xv v= ° = ° =

From (1), ( ) 1.2990 ft/sB yv = −

( ) 1.299 ft/sB yv = − �

(2.25 ft/s) (1.299 ft/s)C

+=v i j�






Let m be the mass of one ball.


( ) ( )m mΣ = Σv v

0 0 0( ) ( ) ( )

A B C A B Cm m m m m m+ + = + +v v v v v v

Dividing by m and applying numerical data,

0 [(6 ft/s) ( ) ] [( ) ( ) ] (8 ft/s) 0 0B y C x C yv v v+ + + + = + +i j i j i

Components:

: 6 ( ) 8C x

x v+ = ( ) 2 ft/sC xv = �

: ( ) ( ) 0B y C yy v v+ = (1)


0[ ( )] [ ( )]m mΣ × = Σ ×r v r v

where rA = 0, rB = 0, (1.5 ft)(cos45 sin 45 )C

= ° + °r i j

(1.5)(cos45 sin 45 ) [ ( ) ( ) ] 0C x C ym v m v° + ° × + =i j i j

Since their cross product is zero, the two vectors are parallel.

( ) ( ) tan 45 2 tan 45 2 ft/sC y C xv v= ° = ° =

From (1), ( ) 2 ft/sB yv = −

( ) 2.00 ft/sB yv = − �

(2.00 ft/s) (2.00 ft/s)C

+=v i j�







A B A B

A B

W W W W

g g g g

+ = +

v v v

After multiplying by g, ( ) ( ) ( )7.2 5.76 1.44 4.8 2.4 2.4A B Bx yv v v+ = + +i j j i j

i: ( )41.472 2.4B xv= ( ) 17.28 ft/s

B xv =

j: ( )10.348 4.8 2.4A B yv v= − ( ) 4.32 2B Ay

v v= −

Speeds relative to the mass center: ( ) ( )( )1 13 8 8 ft/s

3 3A

u lω= = =

( ) ( )( )2 23 8 16 ft/s

3 3B

u lω= = =

Initial kinetic energy: ( ) ( )2 2 2 2

1 0 0

1 1 1

2 2 2

A B A BA Bx y

W W W WT v v u u

g g g g

= + + + +

( ) ( ) ( )2 22 2

1

1 7.2 1 4.8 1 2.45.76 1.44 8 16 18.2517 ft lb

2 32.2 2 32.2 2 32.2T

= + + + = ⋅

Final kinetic energy: ( ) ( )2 22

2

1 1 1

2 2 2

A B BA B Bx y

W W WT v v v

g g g= + +

( ) ( )2 22

2

1 4.8 1 2.4 1 2.417.28 4.32 2

2 32.2 2 32.2 2 32.2A A

T v v = + + −

20.2236 0.6440 11.8234

A Av v= − +

Conservation of energy: 1 2T T=

(a) 20.2236 0.6440 6.4283 0, 6.9919 ft/s

A A Av v v− − = = 6.99 ft/s

A=v �

( ) ( )( ) ( ) ( )4.32 2 6.9919 9.6638 ft/s 17.28 ft/s 9.6638 ft/sB Byv = − = − = −v i j

19.80 ft/sB

=v 29.2°�






( ) ( )( ) ( )

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

0 0 0 01

2

3 3

7.2 4.8 2.47.44 5.76 1.0 8 2.0 16 6.0047 ft lb s

32.2 32.2 32.2

A B B B

O A B G A Bx x

W W W l W lH y m m v H y v u u

g g g g

= − + + = − + + +

= − + + = − ⋅ ⋅

( ) ( )2

( )A BO A A B B A B yy

W WH m v a m v b v a v b

g g= + = +

4.8 2.4(6.9919) ( 9.6638)(24) 1.0423 17.2868

32.2 32.2a a

= + − = −

( ) ( )2 1

1.0423 17.2868 6.0047O O

H H a= − = −

(b) 10.82 fta = 10.82 fta = �







A B A B

A B

W W W W

g g g g

+ = +

v v v

( ) ( )0

9 6 37.68 10.8 6.72

32.2 32.2 32.2

= + −

v j i j

(a) ( ) ( )03.6 ft/s 2.88 ft/s= +v i j

04.61 ft/s=v 38.7°�

Let Al be the distance from G to A and

Bl be the distance from G to B.

or 2A B A

A B B A A

B

W W Wl l l l l

g g W= = =

Let ω be the spin rate.

Initial kinetic energy: ( ) ( )2 22

1 0

1 1 1

2 2 2

A B A B

A B

W W W WT v l l

g g g gω ω

= + + +

( ) ( ) ( )

( )

2 22 2

1

2

1 9 1 6 1 33.6 2.88 2

2 32.2 2 32.2 2 32.2

2.9703 0.27950

A A

A

T l l

l

ω ω

ω

= + + +

= +

Final kinetic energy: 2 2

2

1 1

2 2

A B

A B

W WT v v

g g= +

( ) ( )2 2 2

2

1 6 1 37.68 10.8 6.72 13.0324

2 32.2 2 32.2T

= + + =

Conservation of energy: 2 1

T T= .

( ) ( )213.0324 2.9703 0.27950 6.000 ft/s

A Al lω ω= + =


( ) ( ) ( ) ( ) ( )

( )( ) ( ) ( )( )

( ) ( )( )

0 0 0 01

2

2

9 6 30 7.5 3.6 2 2

32.2 32.2 32.2

7.5466 0.55901 7.5466 3.3540

A B A BO A A B By x

A A A

A A

W W W WH x v y v l l l l

g g g g

l l l

l l

ω ω

ω ω

ω

= + − + +

= − + +

= − + = − +

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

( ) ( )

2

2 1

6 37.68 5.58 6.72 21.6

32.2 32.2

5.5382 ft lb s

: 5.5382 7.5466 3.3540 0.600 ft

A BO A B y

O O A A

W WH v a v b

g g

H H l l

= + = + −

= − ⋅ ⋅

= − = − + =

(b) 2 1.200 ft B A A Bl l l l l= = = + 1.800 ftl = �

(c) 6.00

10.000.600

A

A

l

l

ωω = = = 10.00 rad/sω = �






Velocities in m/s. Lengths in meters. Assume masses are 1.0 for each ball.

Before impacts: ( ) ( ) ( )00 0 04 , 0

A B Cv= = = =v i i v v

After impacts: ( ) ( )1.92 , , A B B B C Cx yv v v= − = + =v j v i j v i

Conservation of linear momentum: 0 A B C

= + +v v v v

i: ( ) ( )4 0 4B C B Cx xv v v v= + + = −

j: ( ) ( )0 1.92 0 1.92B By yv v= − + + =


0

1 1 1 1

2 2 2 2A B C

v v v v= + +

( ) ( ) ( ) ( )22 2 2 21 1 1 1 14 1.92 1.92 4

2 2 2 2 2C Cv v= + + − +

24 3.6864 0

C Cv v− + =

( ) ( )( )24 4 4 3.6864

2 0.56 2.56 or 1.442

Cv

± −= = ± =

Conservation of angular momentum about :B′

( )( )( ) ( )( )

00.75 1.8

0.75 4 1.8 1.65 1.92 2.712

2.712

A C

C

C

v a v cv

cv

c

v

= − +

= − − =

=

If 1.44,Cv = 1.8833 off the table. Rejectc =

If 2.56,Cv = 1.059c =

Then, ( ) 4 2.56 1.44, 1.44 1.92B Bxv = − = = +v i j

Summary.

(a) 2.40 m/sB

=v 53.1°�

2.56 m/sC

=v �

(b) 1.059 mc = �






Velocities in m/s. Lengths in meters. Assume masses are 1.0 for each ball.

Before impacts: ( ) ( ) ( )00 0 05 , 0

A B Cv= = = =v i i v v

After impacts: ( ) ( ), , 3.2A A B B B Cx yv v v= − = + =v j v i j v i

Conservation of linear momentum: 0 A B C

= + +v v v v

i: ( ) ( )5 0 3.2 1.8B Bx xv v= + + =

j: ( ) ( )0 0 A B B Ay yv v v v= − + + =


0

1 1 1 1

2 2 2 2A B C

v v v v= + +

( ) ( ) ( ) ( ) ( )2 2 2 2 21 1 1 1 15 1.8 3.2

2 2 2 2 2A Av v= + + +

(a) 25.76 2.4

A Av v= = 2.40 m/s

A=v �

( ) 2.4B yv = 1.8 2.4

B= +v i j 3.00 m/s

B=v 53.1°�

Conservation of angular momentum about :B′

( )00.75 1.8

A Cv a v cv= − +

0

1.8 0.75A A C

av v cv v= + −

( )( ) ( )( ) ( )( )1.8 2.4 1.22 3.2 0.75 5 4.474= + − =

(b) 4.474 4.474

2.4A

a

v

= = 1.864 ma = �






Use a frame of reference that is translating with the mass center G of the system. Let

0v be its velocity.

0 0v=v i

The initial velocities in this system are ( ) ( )0 0,

A B′ ′v v and ( )

0,

C′v each having a magnitude of .lω They are

directed 120° apart. Thus,

( ) ( ) ( )0 0 0

0A B C′ ′ ′+ + =v v v

(a) Conservation of linear momentum:

( ) ( ) ( ) ( ) ( ) ( )0 0 00 0 0A B C A B Cm m m m m m′ ′ ′+ + = − + − + −v v v v v v v v v

( ) ( ) ( )0 0 00

A B Cv v v v v v= − + − − + −j i j i i i

Resolve into components.

i: ( )0 0

1 13 0 4.5

3 3C Cv v v v− = = =

01.500 m/s=v �

j: 0 2.6 m/sA B B Av v v v− = = =

Conservation of angular momentum about G:

( ) ( ) ( )2

0 0 03

G A A B B C Cml m rω= = × − + × − + × −H k r v v v v r v v

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

( )( ) ( )( )

2

0

2 2

3

10.260 2.6 0.150 4.5 0.45033 m /s

3

A B A C C A B C

A C A C

l v v v

a d v av dv

l

ω

ω

= − × + × − + +

= × + − × = +

= + =

k r r j r i r r r i

i v j j i k

Conservation of energy: ( )2 2 2 2

1

1 332 2

T ml mlω ω= =

0 0

0 0

0 0

2.6 1.5 3.00 m/s

2.6 1.5 3.00 m/s

4.5 1.5 3.00 m/s

A A

B B

C C

− = − − =

− = − − − =

− = − − =

v v j i v v

v v j i v v

v v i i v v

( ) ( ) ( )2 2 2

2 0 0 0

1 1 1

2 2 2A B C

T m m m= − + − + −v v v v v v

1 2T T=

( ) ( ) ( )2 2 22 23 1 1 13 3 3

2 2 2 2ml m m mω = + +

3 m/slω =





(b) 2 2

0.45033 m /s0.1501 m

3 m/s

ll

l

ωω

= = = 150.1 mml = �

(c) 3 m/s

0.1501

l

l

ωω = = 19.99 rad/sω = �






Use a frame of reference that is translating with the mass center G of the system. Let 0

v be its velocity.

0 0v=v i

The initial velocities in this system are ( ) ( )0 0, ,

A B′ ′v v and ( )

0,

C′v each having a magnitude of .lω They are

directed 120° apart. Thus,

( ) ( ) ( )0 0 0

0A B C′ ′ ′+ + =v v v

Conservation of linear momentum:

( ) ( ) ( ) ( ) ( ) ( )0 0 00 0 0A B C A B Cm m m m m m

′ ′ ′+ + = − + − + −v v v v v v v v v

( ) ( ) ( )0 0 00

A B Cv v v v v v= − + − − + −j i j i i i


i: 0

3 0Cv v− = ( )( )0

3 3 0.4 1.2 m/sCv v= = =

j: 0A Bv v− =

B Av v=

Initial kinetic energy: 2 2 2

1 0

1 13 3

2 2T mv ml ω = +

Final kinetic energy: 2 2 2 2 2

2

1 1 1 1

2 2 2 2A B C A C

T mv mv mv mv mv= + + = +

Conservation of energy: 2 1

T T= Solve for 2

Av .

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

0

3 3 1 3 3 10.4 0.75 1.2

2 2 2 2 2 2A Cv v l vω= + − = + −

(a) 2 20.36375 m /s ,= 0.6031 m/s

Av = 0.603 m/s

A=v �

0.603 m/sB

=v �

1.200 m/sC

=v �





Use a frame of reference moving with velocity 0.v

Conservation of angular momentum about G.

( ) ( ) ( )2

0 0 03

G A A B B C Cml m r m mω= = × − + × − + × −H k r v v v v r v v

( ) ( ) ( ) ( )( )2

03

A B A C C A B Cl v v vω = − × + × − + +k r r j r i r r r i

( ) ( ) ( ) ( ) ( )3A C A C

l l a d v av dvω = × + − × = +k i v j j i k

( )( )( ) ( )3 0.075 0.75 0.130 1.200Av d= +

(b) ( )( )0.1406 0.1083 0.603 0.0753 md = − = 75.3 mmd = �






Mass flow rate. As the fluid moves from section 1 to section 2 in time ∆t, the mass ∆m moved is

( )m A lρ∆ = ∆

Then, 1

( )dm m A lAv

dt t t

ρ ρ∆ ∆= = =∆ ∆

Data: 3 2 6 2

11000 kg/m , 500 mm 500 10 m , 25 m/sA vρ = = = × =

6(1000)(500 10 )(25) 12.5 kg/sdm

dt

−= × =


:

1( ) 0m v P t∆ − ∆ =

m dmP v v

t dt

∆= =∆

(12.5)(25)P =

312 NP = �






Consider velocities measured with respect to the plate, which is moving

with velocity V. The velocity of the stream relative to the plate is

1u v V= − (1)

Mass flow rate. As the fluid moves from section 1 to section 2 in time ∆t,

the mass ∆m moved is

( )m A lρ∆ = ∆

Then ( )dm m A l

Audt t t

ρ ρ∆ ∆= = =∆ ∆

(2)


( ) ( ) 0m u P t∆ − ∆ =

2m dmP u u Au

t dtρ∆= = =

∆

Pu

Aρ=

From (1), 1 1

PV v u v

Aρ= − = −

Data: 2 6 2400 N, 600 mm 600 10 mP A

−= = = ×

3

130 m/s, 1000 kg/mv ρ= =

6

40030

(1000)(600 10 )V −= −

× 4.18 m/sV = �






Let F be the force that the wedge exerts on the stream. Assume that the fluid speed is constant. 60 ft/s.v =

Volumetric flow rate: 3475 gal/min ft /sQ = = 1.0583

Mass flow rate: ( )3 362.4slug/ft 1.0584 ft /s 2.051 slug/s

32.2

dmQ

dtρ = = =

Velocity vectors: ( )1, cos30 sin 30v v= = ° + °v i v i j

( )2cos45 sin 45v= ° − °v i j

Impulse – momentum principle:

( ) ( ) 1 22 2

m mm t

∆ ∆∆ + ∆ = +v F v v

( ) ( )

( )( )( )( ) ( )

1 2

1 1

2 2

1 1cos30 sin 30 cos 45 sin 45

2 2

2.051 60 ft/s 0.21343 0.10355

26.26 lb 12.74 lb

m

t

dmv

dt

∆ = + − ∆

= ° + ° + ° − ° −

= − −

= − −

F v v v

i j i j i

i j

i j

Force that the stream exerts on the wedge:

( ) ( )26.26 lb 12.74 lb− = +F i j drag 26.3 lb= �

lift 12.74 lb= �






For a fixed observer, the upstream velocity is ( )48 ft/s .=v i

Volumetric flow rate: 3500 gal/min ft /sQ = = 1.1141


32.2

dmQ

dtρ = = =

Use a frame of reference that is moving with the wedge to the left at 12 ft/s. In this frame of reference the

upstream velocity vector is

( ) ( )48 12 60 ft/s .= − − =u i i i

For the moving frame of reference the mass flow rate is ( )602.1590 2.6987 slug/s.

48

dm u dm

dt v dt

′ = = =

Velocity vectors: ( )1, cos30 sin 30u u= = ° + °u i u i j

( )2cos45 sin 45u= ° − °u i j

Let F be the force that the wedge exerts on the stream.

Impulse-momentum principle:

( ) ( ) 1 22 2

m mm t

∆ ∆∆ + ∆ = +u F u u

( ) ( )

( )( )( )( ) ( )

1 2

1 1

2 2

1 1cos30 sin30 cos45 sin 45

2 2

2.6987 60 0.21343 0.10355

34.6 lb 16.76 lb

m

t

dmu

dt

∆ = + − ∆

′ = ° + ° + ° − ° −

= − −

= − −

F u u u

i j i j i

i j

i j

Force that the stream exerts on the wedge

( ) ( )34.6 lb 16.76 lb− = +F i j drag 34.6 lb= �

lift 16.76 lb= �






Let F be the force exerted on the chips. Apply the impulse-momentum principle to the chips. Assume that

the feed velocity is negligible.

( ) ( ) Ct m∆ = ∆F v

( )cos25 sin 25C

m dmv

t dt

∆ = = ° + ° ∆ F v i j

( )( )1060 cos25 sin 25

32.2

= ° + °

i j

( ) ( )16.89 lb 7.87 lb= +i j

0: 0x x x

F D FΣ = − =

16.89 lbx

D =

Force on truck hitch at D:

16.89 lbx

D− = − 16.89 lbx

D− = �






Initial momentum: ( ) 0.A

m∆ =v

Impulse – momentum principle.

( )( ) ( )x yF F t m+ ∆ = ∆i j v

( )o o

cos35 sin35x y

m dmF F v

t dt

∆ + = = + ∆ i j v i j

x component:

oEngine thrust cos35x

dmF v

dt= =

Data: 3 388 m /min m /s

60Q = = 31000 kg/mρ =

( ) 81000 133.333 kg/s

60

dmQ

dtρ = = =

( )( ) o

133.333 50 cos35 5461 Nx

F = =

5.46 kNx

F = �






Weight: (600)(9.81) 5886 NW mg= = =


Moments about F:

1 1 1 2

( ) (3 ) (2 ) ( ) ( ) ( ) 3( )m v a mv a m v a W t c R t L m v h∆ + ∆ + ∆ + ∆ − ∆ = ∆

1 2

16 3

m mR cW av hv

L t t

∆ ∆ = + + ∆ ∆

Data: 6 m, 4 m, 1.5 m, 0.8 m, 40 kg/sm dm

L c a ht dt

∆= = = = =∆

[ ]1(4)(5886) (6) (1.5) (3) (40) (3) (0.8) (4) (40) 4040 N

6R = + − =

4040 N=R �






Assume A B

u u u= =


Moments about O :

( ) ( ) ( )A c B

R m u x t R m u∆ + ∆ = ∆F kk r

( ) ( ) ( ) 0c B Ax t R m u u∆ = ∆ − =r F k

The line of action of F passes through point O.

Components : ( ) ( ) ( )sin cosA B

m u F t m uα θ∆ + ∆ = ∆

sin (1 cos )m

F ut

α θ∆= −∆

(1)

Components : ( ) ( )0 cos sinF t m uα θ+ ∆ = ∆

cos sinm

F ut

α θ∆=∆

(2)

Dividing (2) by (1),

22sin

1 cos 2tan tan

sin 22sin cos

2 2

θθ θα θ θθ

−= = =

.2

θα =

Thus point C lies at the midpoint of arc AB.






( ) ( )( )80013.333 L/s 1.000 kg/L 13.333 L/s 13.333 kg/s

60

dmQ Q

dtρ= = = = =

( ) ( )( )30 m/s 30 m/s sin 40 cos 40B C

= = ° + °v j v i j

Apply the impulse – momentum principle.

x components: ( ) ( )( )0 30sin 40x

A t m+ ∆ = ∆ °

( ) ( )( )30sin 40 13.333 30sin 40x

mA

t

∆= ° = °∆

257Nx

A =

y components: ( )( ) ( ) ( )( )30 30cos40y

m A t m∆ + ∆ = ∆ °

( ) ( )30cos40 30 13.333 30cos40 30y

mA

t

∆= ° − = ° −∆

93.6 N= − 93.6 Ny

A =

moments about :A ( )( )( ) ( )0.060 30

Am M t∆ + ∆

( )( )( ) ( )( )( )0.180 30cos40 0.300 30sin 40m m= ∆ ° − ∆ °

( ) ( ) ( )1.8 1.6484A

m M t m∆ = ∆ − ∆

( ) ( )( )3.4484 13.333 3.4484 46.0 N mA

mM

t

∆= − = − = − ⋅∆

46.0 N mA

= ⋅M �

274 N=A 20.0° �






Mass flow rate: 3 3

2

62.4 lb/ft 40 ft /min1.29193 lb s/ft

60 s/min32.2 ft/s

dmQ

dt g

γ= = = ⋅

75 ft/sA Bv v= =

Use impulse - momentum principle.

moments about :D ( )( ) ( )( ) ( )15 23 15

sin 60 cos6012 12 12

A Am v m v C t

− ∆ ° + ∆ ° − ∆

( ) 3

12B

m v

= ∆

( )( )( )15 15 23 3sin 60 cos60 1.29193 75 0.37420

12 12 12 12A

mC v

t

∆ = − ° + ° − = − ∆

29.006 lbC = − 0, 29.0 lbx y

C C= = − �

x component: ( ) ( ) ( )cos60A x B

m v D t m v∆ ° + ∆ = ∆

( ) ( )( )cos60 1.29193 75 75cos60x B A

mD v v

t

∆ = − ° = − ° ∆ 48.4 lb

xD = �

y component: ( ) ( ) ( )sin 60 0A ym v C t D t∆ ° + ∆ + ∆ =

( )( )sin 60 29.006 1.29193 75 sin 60y A

mD C v

t

∆= − − ° = + − °∆

54.9 lby

D = − �






Volumetric flow rate: 3300 gal/min 0 ft /sQ = = .6684


32.2

dmQ

dtρ = = =

90 ft/sA Bv v= =

Use the impulse – momentum principle.

Moments about C: ( ) ( ) ( ) cosA P B

m v a W t l m v bθ∆ − ∆ = ∆

(a) ( ) ( )( ) ( )( )( )( )

90 4 /12 90 1/12cos 1.2954 0.7287

40 1

A B

p

m v a v b

t W lθ

−∆ −= = =∆

43.23θ = ° 43.2θ = °�

x components: ( ) ( ) ( ) cosA x B

m v C t m v θ∆ + ∆ = ∆

( ) ( )( )cos 1.2954 90cos 90 31.63 lbx B A

mC v v

tθ θ∆= − = − = −

∆

y components: ( ) ( ) ( )0 siny p BC t W t m v θ+ ∆ − ∆ = − ∆

( ) ( )( )sin 40 1.2954 90 sin 39.84 lby p B

mC W v

tθ θ∆= − = − = −

∆

(b) [31.63 lb=C ] [39.84 lb+ ] 50.9 lb=C 51.6°�






Volumetric flow rate: 3300 gal/min 0 ft /sQ = = .6684


32.2

dmQ

dtρ = = =

, 45A Av v v θ= = = °

Use the impulse-momentum principle.

moments about C: ( ) ( ) ( ) cosA p Bm v a W t l m v bθ∆ − ∆ = ∆

(a) ( )( ) ( )( )( )

( ) cos 40 1 cos 45

87.338 ft/s4 1/

1.295412 12

pA B

W lv v

a b m t

θ °= = = =

− ∆ ∆ −

87.3 ft/sv = �

x components: ( ) ( ) ( ) cosA x B

m v C t m v θ∆ + ∆ = ∆

( ) ( )[ ]cos 1.2954 87.338cos45 87.338 33.137 lbx B A

mC v v

tθ∆= − = ° − = −

∆

y components: ( ) ( ) ( )0 siny p BC t W t m v θ+ ∆ − ∆ = − ∆

( )( )sin 40 1.2954 87.338 sin 45 40.0 lby p B

mC W v

tθ∆= − = − ° = −

∆

(b) [33.137 lb=C ] [40 lb+ ] 51.9 lb=C 50.4°�






Symbols: mass flow ratedm

dt=

exhaust relative to the airplaneu =

speed of airplanev =

drag forceD =


( ) ( ) ( )m v D t m u∆ + ∆ = ∆

m dm D

t dt u v

∆ = =∆ −

Data: 900 km/h = 250 m/sv =

600 m/su =

35 kN 35000ND = =

35000

600 250

dm

dt=

− 100 kg/sdm

dt= �






Let F be the force exerted on the slipstream of one engine. Then, the force exerted on the airplane is −2F as

shown.

Statics.

0B

MΣ =

( ) ( )( )0.9 4.8 2 0W F− =

( )( )

( )( )0.9 6000

2 4.8F =

562.5 lb=

Calculation of .

dm

dt

mass density volume density area length= × = × ×

( ) ( ) ( )B B

B B B

A v tm A l A v t

g

γρ ρ ∆∆ = ∆ = ∆ =

B B

m dm A v

t dt g

γ∆ = =∆

Force exerted on the slipstream: ( )B A

dmF v v

dt= −

Assume that Av , the speed far upstream, is negligible.

( ) 2 20

4

B B

B B

A vF v D v

g g

γ γ π = − =

( )( )( )

( ) ( )2 2 2

2 2

4 562.5 32.247058.9 ft /s

6.6 0.075B

Fgv

Dπ γ π= = =

84.0 ft/s B

=v �






Let F be the force exerted on the slipstream of one engine.

( )B A

dmF v v

dt= −

Calculation of .

dm

dt mass density volume density area length= × = × ×

( ) ( ) ( )B B

B B B

A v tm A l A v t

g

γρ ρ ∆∆ = ∆ = ∆ =

2 or

4

B B

B

m A v dmD v

t g dt g

γ γ∆ π = = ∆

Assume that ,Av the velocity far upstream, is negligible.

( ) ( ) ( )2 22 0.0750 6.6 60 286.87 lb

4 32.2 4B B

F D v vg

γ π π = − = =

The force exerted by two slipstreams on the airplane is 2 .F− 2 573.74 lbF− =

Statics.

0:B

MΣ =

( )0.9 4.8 2 9.3 0W F A− − − =

( )( ) ( )( )10.9 6000 4.8 573.74

9.3A = −

284.5 lb= 285 lb=A �

0: 2 0x x

F F B= − − =

2 573.74 lbx

B F= − =

( )0: 284.5 6000 0y y y

F A B W BΣ = + − = + − =

5715.5 lby

B =

[573.74 lb=B ] [5715.5 lb+ ] 5740 lb=B 84.3° �






Use a frame of reference moving with the plane. Apply the impulse-momentum principle. Let F be the force that

the plane exerts on the air.

x components: ( ) ( ) ( )A Bm u F t m u∆ + ∆ = ∆

( ) ( )B A B A

m dmF u u u u

t dt

∆= − = −∆

(1)

moments about B: ( ) ( ) 0A B

e m u M t− ∆ + ∆ =

B A

dmM e u

dt= (2)

Let d be the distance that the line of action is below B.

B

Fd M= B A

B A

M eud

F u u= =

− (3)

Data: 90 kg/s,dm

dt= 600 m/s,

Bu = 4 me =

(a) 480 km/h 133.333 m/sA

u = =

From (1), ( )( ) 390 600 133.333 42 10 NF = − = × 42.0 kNF = �

From (2), ( )( )

( )4 133.333

600 133.333d =

− 1.143 md = �

(b) 960 km/h 266.67 m/sA

u = =

From (1), ( )( ) 390 600 266.67 30 10 NF = − = × 30.0 kNF = �

From (2), ( )( )

( )4 266.67

600 266.67d =

− 3.20 md = �






The thrust on the fluid is ( )B A

dmF v v

dt= −

Calculation of dm

dt. mass density volume density area length= × = × ×

( ) ( )B B Bm A l A v tρ ρ∆ = ∆ = ∆

B B

m dmA v

t dtρ∆ = =

∆

whereB

A is the area of the slipstream well below the helicopter and Bv is the corresponding velocity in the

slipstream. Well above the blade, 0.Av ≈

Hence, 2

BF Avρ=

( ) ( ) ( )2 23

3 2

1.21 kg/m 9 m 24 m/s4

44.338 10 kg m/s

π =

= × ⋅

44.3 kNF =

The force on the helicopter is 44.3 kN .

Weight of helicopter: 15kNH

=W

Weight of payload: P P

W=W

Statics: 0y H PF F W WΣ = − − =

44.3 15 29.3 kN.P H

W F W= − = − = 29.3 kNW = �






Let dm

dt= mass flow rate, u = discharge velocity relative to the airliner, speed of airliner,v = and

thrustF = of the engines.

0F D− = ( ) 0dm

u v Ddt

− − =

Configuration before control surface malfunction:

1

720= 22.36 slug/s, 1860 ft/s, 560 mi/h 821.33 ft/s

32.2

dmu v

dt= = = =

( )( ) 1 122.36 1860 821.33 0 23225 lbD D− − = =

Drag force factor: ( )

2 2 21

1 1 1 1 2 2

1

23225 0.03443 lb s /ft

821.33

DD k v k

v= = = = ⋅

After control surface malfunction: 2 2

2 11.2 0.04131 lb s /ftk k= = ⋅

When the new cruising speed is attained,

( ) 2

2 2 20

dmu v k v

dt− − =

( )( ) 2

2 222.36 1860 0.04131 0v v− − =

Solving for v2, 2768.6 ft/sv =

2524 mi/hv = �






Apply the impulse - momentum principle to the moving air. Use a frame of reference that is moving with the

airplane. Let F be the force on the air.

270 km/h 75 m/s

600 m/s

v

u

= ==

( ) ( ) ( )2 sin 20

2

mm v F t u

∆− ∆ + ∆ = °

( ) ( )

( )( ) 3

sin 20 sin 20

120 75 600sin 20 33.6 10 N

m dmF v u v u

t dt

F

∆= + ° = + °∆

= + ° = ×

Force on airplane is .−F 33.6 kN=F �






Symbols: n = number engines operating

dm

dt= mass flow rate for one engine

u = discharge velocity relative to jetliner = 800 m/s

v = speed of jetliner

F = thrust force ( )dm

n u vdt

= −

D = drag force 2kv=

Force balance.

0F D− = 2( )dm

n u v kvdt

− =

2 1

( )

v dm

n u v k dt=

−

All 3 engines operating: 3, 900km/h = 250 m/sn v= =

2(250) 1

37.879 m/s3(800 250)

dm

k dt= =

−

(a) One engine lost: n = 2

2

37.8792(800 )

v

v

=−

275.758 60606 0v v+ − =

211.2m/sv = 760 km/hv = �

(b) 2 engines lost: n = 1

2

37.879800

v

v

=−

237.879 30303 0v v+ − =

156.16 m/sv =

562 km/hv = �






Let u be the velocity of the stream relative to the velocity of the blade. ( )u v V= −

Mass flow rate: A

mAv

tρ∆ =

∆


( ) ( ) ( ) cost

m u F t m u θ∆ − ∆ = ∆

( )1 cos ( ) (1 cos )t A A

mF u Av v V

tθ ρ θ∆= − = − −

∆

where Ft is the tangential force on the fluid.

The force Ft on the fluid is directed to the left as shown. By Newton’s law of action and reaction, the

tangential force on the blade is Ft to the right.

Output power: out( ) (1 cos )

t A AP FV Av v V Vρ θ= = − −

(a) V for maximum power output.

out ( 2 ) (1 cos ) 0

A

dPA v V

dVρ θ= − − =

1

2Av V= �

(b) Maximum power.

out max

1 1( ) (1 cos )

2 2A A A A

P Av v v vρ θ = − −

3

out max

1( ) (1 cos )

4A

P Avρ θ= − �

Input power = rate of supply of kinetic energy of the stream

2 2 3

in

1 1 1 1( )

2 2 2A A A

mP m v v Av

t tρ∆ = ∆ = = ∆ ∆





(c) Efficiency. out

in

p

pη =

3

( ) (1 cos )

1

2

A A

A

Av v V V

Av

ρ θηρ

− −=

2 1 (1 cos )A A

V V

v vη θ

= − −

�






Data: 240

slugs/s,32.2

dm

dt= = 7.4534 2200 ft/s,u = 570 mi/h 836 ft/sv = =

( ) ( )( )7.4534 2200 836 10166 lbdm

F u vdt

= − = − =

(a) Power used to propel airplane:

( )( ) 6

110166 836 8.499 10 ft lb/sP Fv= = = × ⋅

propulsion power 15450 hp= �

Power of kinetic energy of exhaust:

( ) ( )( )22

1

2P t m u v∆ = ∆ −

( ) ( )( )2 2 6

2

1 17.4534 2200 836 6.934 10 ft lb/s

2 2

dmP u v

dt= − = − = × ⋅

(b) Total power: 6

1 215.433 10 ft lb/sP P P= + = × ⋅

total power 28060 hp= �

(c) Mechanical efficiency: 6

1

6

8.499 100.551

15.433 10

P

P

×= =×

mechanical efficiency 0.551= �






Kinetic energy of fluid in slipstream passing in time .t∆

2 21 1 mass speed density volume speed

2 2T∆ = × = × ×

21

density area length speed2

= × × ×

( ) ( )2 21 1

2 2A l v Av t vρ ρ= ∆ = ∆

31

2

TAv

tρ∆ =

∆

Input power 31

2

dTAv

dtρ= =

Data: 31.2 kg/mρ =

( )22 26.5 33.183 m

4 4A d

π π= = =

30 km/h 8.333 m/sv = =

(a) ( )( )( )3 311.2 33.183 8.333 11.521 10 N m/s

2

dT

dt= = × ⋅

11.52 kJ/s�

Input power 11.521 kWdT

dt= =

(b) Output power ( )( )0.4 11.521 4.61 kW= =

output power 4.61 kW= �






Kinetic energy of fluid in slipstream passing in time .t∆

2 21 1 mass speed density volume speed

2 2T∆ = × = × ×

21

density area length speed2

= × × ×

( ) ( )2 21 1

2 2A l v Av t vρ ρ= ∆ = ∆

31

2

TAv

tρ∆ =

∆ (1)

Data: output power 3.5 kW 3500 W= =

3500input power 10000 W

0.35= =

input power 10,000 W, 36 km/h 10 m/sdT

vdt

= = = =

Using (1), ( )( )( )( )

2

3 3

2 10000216.667 m

1.2 10

dTA

dtvρ= = =

(a) ( )( )24 16.6674

4

Ad A d

ππ π

= = =

4.61 md = �

(b) From above, 10.00 kJ/sdT

dt= �






Mass flow rate:

mass density volume

density area length

= ×= × ×

( ) ( )m bd l bdv tρ ρ∆ = ∆ = ∆

dm m

bdvdt t

ρ∆= =∆

1 dm

Q bdvdtρ

= =

Continuity of flow: 1 2

Q Q Q= =

1 2

1 2

,

Q Qv v

bd bd=

Resultant pressure forces:

1 1 2 2 p d p dγ γ= =

2

1 1 1 1

1 1

2 2F p bd bdγ= =

2

2 2 2 2

1 1

2 2F p bd bdγ= =

Apply impulse-momentum principle to water between

sections 1 and 2.





( ) ( ) ( ) ( )1 1 2 2m v F t F t m v∆ + ∆ − ∆ = ∆

( )1 2 2 1

mv v F F

t

∆ − = −∆ ( )2 2

2 1

1 2

1

2

Q QQ b d d

bd bdρ γ

⋅ − = −

( ) ( )( )

2

2 1

1 2 2 1

1 2

1

2

Q d db d d d d

bd d

ργ

−= + −

Noting that ,gγ ρ=

( )1 2 1 2

1

2Q b gd d d d= + �






Mass flow rate:

mass density volume

density area length

= ×= × ×

( ) ( )m bd l bdv tρ ρ∆ = ∆ = ∆

dm m

bdvdt t

ρ∆= =∆

1 dm

Q bdvdtρ

= =

Continuity of flow: 1 2

Q Q Q= =

1 2

1 2

,

Q Qv v

bd bd=

Resultant pressure forces:

1 1 2 2 p d p dγ γ= =

2

1 1 1 1

1 1

2 2F p bd bdγ= =

2

2 2 2 2

1 1

2 2F p bd bdγ= =

Apply impulse-momentum principle to water between

sections 1 and 2.





( ) ( ) ( ) ( )1 1 2 2m v F t F t m v∆ + ∆ − ∆ = ∆

( )1 2 2 1

mv v F F

t

∆ − = −∆ ( )2 2

2 1

1 2

1

2

Q QQ b d d

bd bdρ γ

⋅ − = −

( ) ( )( )

2

2 1

1 2 2 1

1 2

1

2

Q d db d d d d

bd d

ργ

−= + −

Noting that ,gγ ρ=

( )1 2 1 2

1

2Q b gd d d d= +

Data: 1 2

9.81 m/s, 3 m, 1.25 m, 1.5 mg b d d= = = =

( )( )( )( ) 313 9.81 1.25 1.5 1.25 1.5 15.09 m /s

2Q = + =

315.09 m /sQ = �






From hydrostatics, the pressure at section 1 is 1

.p rh ghρ= =

The pressure at section 2 is 2

0.p =

Calculate the mass flow rate using section 2.


( ) ( )2 2A Am l v tρ ρ∆ = ∆ = ∆

2A

dm mv

dt tρ∆= =

∆

Apply the impulse-momentum principle to fluid between sections 1 and 2.

( ) ( ) ( )1 1 1m v p A t m v∆ + ∆ = ∆

1 1 1

dm dmv p A v

dt dt+ =

( ) ( )1 1 1 2 1

dmp A v v A v v v

dtρ= − = −

But 1v is negligible,

1,p ghρ= and 2v gh=

( )1 22ghA A ghρ ρ= or

1 22A A=

2 22

4 4D d

π π =

2

Dd = �






The flow through each arm is 1.25 gal/min.

3

3 3gal 1 ft 1 min5 11.1408 10 ft /s

min 7.48 gal 60 sQ − = = ×

( )3

3 3

2

3

62.4 lb/ftQ 11.1408 10 ft /s

32.2 ft/s

21.590 10 lb s/ft

dmQ

dt g

γρ

−

= = = ×

= × ⋅

Consider the moment about O exerted on the fluid stream of one arm.

Apply the impulse-momentum principle. Compute moments about O.

First, consider the geometry of triangle OAB. Using first the law of

cosines,

( ) ( ) ( ) ( )2 2 2 o6 4 2 6 4 cos120OA = + −

76 in. 0.72648 ftOA = =

Law of sines. o

sin sin120

4 76

β =

o

23.413 ,β =

o o

60 36.587α β= − =

Moments about O:

( )( )( ) ( ) ( )( ) ( )( )( )0 sinO O s

m v M t OA m v OA m OAα ω∆ + ∆ = ∆ − ∆

( ) ( )

( ) ( )( ) ( )( )

2

23

sin

21.590 10 0.72648 60 sin 36.587 0.72648

0.56093 0.011395 lb ft

O s

mM OA v OA

tα ω

ω

ω

−

∆ = − ∆ = × −

= − ⋅

Moment that the stream exerts on the arm is .

OM−

Friction couple for one arm:

( )10.275 0.06875 lb ft

4F

M = = ⋅

Balance of moments on one arm:

0 F O F O

M M M M− = =

0.06875 0.56093 0.011395ω= −

43.19 rad/s 412 rpmω = =

412 rpmω = �






Consider the conservation of the horizontal component of momentum of the railroad car of mass 0

m and the

sand mass .qt

( ) 0 0

0 0 0

0

m vm v m qt v v

m qt= + =

+ (1)

0 0

0

dx m vv

dt m qt= =

+

Integrating, using 0

0x = and x L= when ,L

t t=

( )0 0 0 0

0 00 0

0

0 0 0

0

ln ln

ln

L Lt t

L

L

m v m vL vdt dt m qt m

m qt q

m v m qt

q m

= = = + − ++=

∫ ∫

0

0 0 0

lnL

m qt qL

m m v

+ =

0 0/0

0

qL m vLm qte

m

+ =

(a) Final mass of railroad car and sand 0 0/

0 0

qL m vLm qt m e+ = �

(b) Using (1), 0 0/0 0 0 0

0 0

qL m vL

L

m v m vv e

m qt m

−= =+

0 0/

0

qL m vLv v e

−= �






Apply the impulse-momentum principle.

Moments about C : ( )( ) ( ) ( ) ( )( )0.9 3 1.8 1.65A B

m v D t W t m v− ∆ + ∆ − ∆ = − ∆

( )

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

1.8 10.9 1.65

3 3

1.8 4000 1100 0.9 4.5 1.65 4.5 2287.5 N

3 3

A B

mD W v v

t

∆= + −∆

= + − =

2.29 kN=D

�

x components: ( ) ( ) ( )A x Bm v C t m v∆ + ∆ = ∆

( ) ( )( )100 4.5 4.5 0x B A

mC v v

t

∆= − = − = ∆

y components: ( ) ( ) ( )0 0y

C t D t W t+ ∆ + ∆ − ∆ =

4000 2287.5 1712.5 Ny

C W D= − = − =

1.712 kN=C �






Let ρ be the mass per unit length of chain. Apply the impulse - momentum to the entire chain. Assume that the

reaction from the floor it equal to the weight of chain still in contact with the floor.

Calculate the floor reaction.

( )R g l yρ= − 1y

R mgl

= −

Apply the impulse-momentum principle.

( ) ( ) ( ) ( )yv P t R t gl t y y vρ ρ ρ+ ∆ + ∆ − ∆ = + ∆

( ) ( ) ( )P t y v gl t R tρ ρ∆ = ∆ + ∆ − ∆

(a) ( )yP v gl l y g

tρ ρ ρ∆= + − −

∆ Let

y dyv

t dt

∆ = =∆

2P v gyρ ρ= + ( )2mP v gy

l= + �

(b) From above, 1y

mgl

= −

R

�






(a) Let ρ be the mass per unit length of chain. The force P supports the weight of chain still off the floor.

P gyρ= mgy

Pl

= �

(b) Apply the impulse-momentum principle to the entire chain.

( ) ( ) ( ) ( )yv P t R t gl t g y y vρ ρ ρ− + ∆ + ∆ − ∆ = − + ∆

( ) ( ) ( ) ( )R t gl t P t g y vρ ρ∆ = ∆ − ∆ − ∆

y

R gl gy vt

ρ ρ ρ ∆= − −∆

Let 0.t∆ → Then, y dy

vt dt

∆ = = −∆

( ) 2R g l y vρ ρ= − + ( ) 2mg l y v

l = − + R

�






Let ρ be the mass per unit length of chain. Consider the impulse-momentum applied to the link being brought to

rest at point C.

Calculation of .m∆

( ) ( )m l v tρ ρ∆ = ∆ = ∆


( ) 0m v C t− ∆ + ∆ =

( ) 0v t v C tρ− ∆ + ∆ =

2C vρ=

Impulse – momentum applied to the moving portion of the chain. Consider only the changes in momentum and

forces contributing to moments about O in the diagram.

Moments about O:

( ) [ ] ( )m v gh C t m vρ∆ + − ∆ = ∆

C ghρ=

Equating the two expressions for C,

2v ghρ ρ=

2vh

g= �






Let ρ be the mass per unit length of chain. Assume that the weight of any chain above the hole is supported by

the floor. It and the corresponding upward reaction of the floor are not shown in the diagrams.

Case 1. Apply the impulse-momentum principle to the entire chain.

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

( )( )

yv gy t y y v v

yv y v y v y v

y vy vgy v y

t t t

ρ ρ ρ

ρ ρ ρ ρ

ρ ρ ρ ρ

+ ∆ = + ∆ + ∆

= + ∆ + ∆ + ∆ ∆

∆ ∆∆ ∆= + +∆ ∆ ∆

Let 0.t∆ → ( )dy dv dgy v y yv

dt dt dtρ ρ ρ ρ= + =

Multiply both sides by .yv ( )2 dgy v yv yv

dtρ ρ=

Let dy

vdt

= on left hand side. ( )2 dy dgy yv yv

dt dtρ ρ=

Integrate with respect to time. ( ) ( )2g y dy yv d yvρ ρ=∫ ∫

( )231 1

3 2gy yvρ ρ= or 2 2

3v gy= (1)

Differentiate with respect to time. 2 2

23 3

dv dyv g gvdt dt

= =





(a)

1

3

dva g

dt= = 0.333g=a �

(b) Set in (1)y l= 2 2

3v gl= 0.817 gl=v �

Case 2. Apply conservation of energy using the floor as the level for from which the potential energy is

measured.

1 10, 0T V= =

2

2 2

1,

2 2

yT mv V gyρ= = −

1 1 2 2T V T V+ = +

2 21 10

2 2mv gyρ= −

2 2

2 gy gyv

m l

ρ= = (1)

Differentiating with respect to y, 2

2dv gy

vdy l

=

(a) Acceleration. dv gy

a vdy l

= = gy

l=a �

(b) Setting in (1),y l= 2v gl= gl=v �

Note: The impulse-momentum principle may be used to obtain the force that the edge of the hole exerts

on the chain.






750 lb/s,dW

dt=

1 75023.292 lb s/ft

32.2

dm dW

dt g dt= = = ⋅

Thrust of one engine: ( )( ) 3 12500 23.292 291.15 10 lb

dmP u

dt= = = ×

For 3 engines, 33 873.4 10 lbP = ×

Thrust 873 kips= �






Thrust of one engine: 3

3 1200 10400 10 lb.

3P

×= = ×

But, dm

P udt

=

3400 10

32 lb s/ft12500

dm P

dt u

×= = = ⋅

( )( )32.2 32 1030 lb/sdW dm

gdt dt

= = = 1030 lb/sdW

dt= �






Thrust, .

dmP u

dt=

F P mg maΣ = − =

P u dm

a g gm m dt

= − = −

Data: 15 kg/sdm

dt=

As rocket is fired: 1500 kgm =

0

159.81 0.01 9.81

1500

ua u= − = − (1)

As all the fuel is consumed: 1

1500 1200 300 kg.m = − =

1

159.81 0.05 9.81

300

u

a u= − = − (2)

From the given data, 2

1 0220 m/sa a− = (3)

Using (1) and (2) for 1a and

0a and substituting into (3),

0.04 220u = 5500 m/su = �






Thrust: dm

P u uqdt

= =

Since u and dm

dt are constant, P is also constant.

:F maΣ =

P mg ma− =

( )P m a g= +

( )min maxm a g= +

( )( )1500 25 9.81= +

352.215 10 N= ×

(a) Fuel consumption rate.

3

52.215 10

450

Pq

u

×= =

116.0 kg/sq = �

(b) Mass of fuel consumed.

( )( )fuel116.0 15.6m qt= =

fuel

1810 kgm = �






Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time ∆t.

: ( )( ) ( )( )mv m m v v m v v v= − ∆ + ∆ + ∆ + ∆ −

( ) ( ) ( )( ) ( ) ( )( ) ( )mv m v m v m v m v m v m v= + ∆ − ∆ − ∆ ∆ + ∆ + ∆ ∆ − ∆

( ) ( ) 0m v u m∆ − ∆ =

( )dmm t

dt∆ = − ∆

v dv u dm

t dt m dt

∆ = = −∆

1 1 1

0 00

v t m

v m

u dm dmdv dt u

m dt m= − = −∫ ∫ ∫

1 0

1 0

0 1

ln lnm m

v v u u

m m

− = − =

0 1 0

1

expm v v

m u

− =

Data: 1 0

2430 m/s,v v− = 0

4200 m/s 5000 kgu m= =

1

5000 2430exp 1.7835

4200m

= =

1

2800 kgm =

fuel 0 1

5000 2800m m m= − = − fuel

2200 kgm = �






Data from Problem 14.97: 0

5000 kg, 4200 m/sm u= =

1 0 fuel

5000 1500 3500 kgm m m= − = − =

Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time ∆t.

: ( )( ) ( )( )mv m m v v m v v v= − ∆ + ∆ + ∆ + ∆ −

( ) ( ) ( )( ) ( ) ( )( ) ( )mv m v m v m v m v m v m v= + ∆ − ∆ − ∆ ∆ + ∆ + ∆ ∆ − ∆

( ) ( ) 0m v u m∆ − ∆ =

( )dmm t

dt∆ = − ∆

v dv u dm

t dt m dt

∆ = = −∆

1 1 1

0 00

v t m

v m

u dm dmdv dt u

m dt m= − = −∫ ∫ ∫

1 0

1 0

0 1

ln lnm m

v v u u

m m

− = − =

1 0

50004200 ln

3500v v v∆ = − = 1498 m/sv∆ = �






Apply conservation of momentum to the rocket plus the fuel.

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

mv m m v v m v v v

mv m v m v m v m v m v m v

= − ∆ + ∆ + ∆ + ∆ −

= + ∆ − ∆ − ∆ ∆ + ∆ + ∆ ∆ − ∆

( ) ( ) 0m v u m∆ − ∆ =

( )dmm t

dt∆ = − ∆

v dv u dm

t dt m dt

∆ = = −∆

1 1 1

0 00

v t m

v m

u dm dmdv dt u

m dt m= − = −∫ ∫ ∫

1 0 0

1 0

0 1 1

ln ln lnm m W

v v u u um m W

− = − = =

Data: 1 0

360 ft/sv v− =

( )0 111,600 lb, 11,600 1000 10,600 lbW W= = − =

11,600

360 ln10,600

u= 3990 ft/su = �






Apply conservation of momentum to the rocket plus the fuel.

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

mv m m v v m v v v

mv m v m v m v m v m v m v

= − ∆ + ∆ + ∆ + ∆ −

= + ∆ − ∆ − ∆ ∆ + ∆ + ∆ ∆ − ∆

( ) ( ) 0m v u m∆ − ∆ =

( )dmm t

dt∆ = − ∆

v dv u dm

t dt m dt

∆ = = −∆

1 1 1

0 00

v t m

v m

u dm dmdv dt u

m dt m= − = −∫ ∫ ∫

1 0

1 0

0 1

ln lnm m

v v u u

m m

− = − =

0 1 0 0

1 1

expm v v W

m u W

− = =

Data: 1 0

450 ft/s,v v− = 5400 ft/su =

0 1 fuel 1

1200 lbW W W W= + = +

1

1

1200 450exp 1.08690

5400

W

W

+ = =

1

12000.08690

W=

113810 lbW = �






See sample Problem 14.8 for derivation of

0 0

0 0

ln lnm m qt

v u gt u gtm qt m

−= − = − −−

(1)

Note that g is assumed to be constant.

Set dy

vdt

= in (1) and integrate with respect to time.

0

0 0 0

0

20

0

0

ln

1ln

2

h t t

t

mh dy vdt u gt dt

m qt

m qtu dt gt

m

= = = − −

−= − −

∫ ∫ ∫

∫

Let 0

0

m qtz

m

−= 0

qdz dt

m= − or 0

mdt dz

q= −

( )0 0

2 20 0

20 0 0 0 0

0 0 0 0

20 0

0 0

0 0 0

0 0

1 1ln ln

2 2

1ln 1 ln 1

2

11 ln 1 1

2

ln 1 1 ln 1

zz

z z

m u m uh z dz gt z z z gt

q q

m u m qt m qt m mgt

q m m m m

m u qt m qtgt

q m m

m u m qt m qtut

q m m

= − = + −

− −= − − − −

−= − − + −

− −= − + − −

∫

2

20 0

0

1

2

1ln

2

gt

m u m qtut ut gt

q m

−

−= + − −

20 0

0

1ln

2

m mh u t t gt

q m qt

= − − − −

(2)





Data: 20

0

7300226.7 lb s /ft

32.2

Wm

g= = = ⋅

260

8.0745 lb s/ft32.2

wq

g= = = ⋅

&

2fuel

fuel

4000124.22 lb s /ft

32.2

Wm

g= = = ⋅

fuel124.22

15.385 s8.0745

mt

q= = =

2

0226.7 124.22 102.48 lb s /ftm qt− = − = ⋅

1500 ft/su =

( )( )2226.7 226.7 11500 15.385 15.385 ln 32.2 15.385

8.0745 102.48 2h

= − − −

4150 fth = �






Thrust force dm

P u uqdt

= =

Mass of rocket plus unspent fuel 0

m m qt= −

Acceleration: 0

P uqa

m m qt= =

−

Integrating with respect to time to obtain the velocity,

0 00 0

0

t t qv v adt v u dt

m qt= + = +

−∫ ∫

( ) 0

0 0 0 0

0

ln ln lnm qt

v u m qt m v um

− = − − − = − (1)

Integrating again to obtain the displacement,

0

0 0 0

0

lnt m qt

s s v t u dtm

−= + − ∫

Let 0 0

0 0

or

m qt q mz dz dt dt dz

m m q

−= = − = −

( ) ]0 0

0 0

0 0

00 0 0 0

0 0

0 0 0 0

0 0

0 0

0

0 0 0

0 0

0 0

ln ln

ln 1 ln 1

1 ln 1 1

ln 1 1 ln 1

zz

z z

o

m u m us s v t zdz z z z

q q

mm u m qt m qt ms v t

q m m m m

m u qt m qts v t

q m m

m u m qt m qts v t ut

q m m

= + + = −

− −= + + − − −

−= + + − − +

− −= + + − + − −

∫

0 0

0 0

0

lnm u m qt

s v t ut utq m

−= + + + −

0 0

0 0

0

lnm m

s s v t u t tq m qt

= + + − − −

(2)





Data: 0

7500 ft/s 7500 360 7860 ft/sv v= = + =

( )( )fuel fuel

100060 s 0.5176 slug/s

32.2 60

m Wt q

t gt= = = = =

0

0

11600360.25 slugs

32.2

Wm

g= = =

0

11600 1000329.19 slugs

32.2 32.2m qt− = − =

0

0s =

From (1), 329.19

7860 7500 ln360.25

u= −

360.25

360 ln329.19

u= 3993 ft/su =

From (2), ( )( ) 360.25 360.250 7500 60 3993 60 60 ln

0.5176 329.19s

= + + − −

6460.6 10 ft= × 87.2 mis = �






See sample Problem 14.8 for derivation of

0 0

0 0

ln lnm m qt

v u gt u gtm qt m

−= − = − −−

(1)

Note that g is assumed to be constant.

Set dy

vdt

= in (1) and integrate with respect to time.

0

0 0 0

0

20

0

0

ln

1ln

2

h t t

t

mh dy vdt u gt dt

m qt

m qtu dt gt

m

= = = − −

−= − −

∫ ∫ ∫

∫

Let 0 0

0 0

or


m m q

−= = − = −

( ) ]0 0

2 20 0

20 0 0 0 0

0 0 0 0

20 0

0 0

0 0 0

0 0

1 1ln ln

2 2

1ln 1 ln 1

2

11 ln 1 1

2

1ln 1 1 ln 1

2

zz

z z

m u m uh zdz gt z z z gt

q q

m u m qt m qt m mgt

q m m m m

m u qt m qtgt

q m m

m u m qt m qtut

q m m

= − = − −

− −= − − − −

−= − − + −

− −= − + − − −

∫

2

20 0

0

1ln

2

gt

m u m qtut ut gt

q m

−= + − −

20 0

0

1ln

2

m mh u t t gt

q m qt

= − − − −

(2)

Data: 2

0960 kg, 10 kg/s, 3600 m/s, 9.81 m/sm q u g= = = =

fuel

fuel

800800 kg, 80 s

10

mm t

q= = = =





(a) ( )( )2 3960 960 13600 80 80 ln 9.81 80 153.4 10 m

10 960 800 2h

= − − − = × −

153.4 kmh = �

(b) From equation (1), ( )( )9603600ln 9.81 80

960 800v = −

−

5670 m/s=v �






Data from Problem 14.97: 0

5000 kg, 4200 m/sm u= =

0 0

0, 0v s= =

Thrust force dm

P u uqdt

= =

Mass of satellite plus unspent fuel 0

m m qt= −

Acceleration: 0

P uqa

m m qt= =

−

Integrating with respect to time to obtain the velocity,

0 00 0

0

t t qv v adt v u dt

m qt= + = +

−∫ ∫ (1)

( ) 0

0 0 0 0

0

ln ln lnm qt

v u m qt m v um

− = − − − = −

Integrating again to obtain the displacement,

0

0 0 0

0

lnt m qt

s s v t u dtm

−= + − ∫

Let 0 0

0 0

or


m m q

−= = − = −

( ) ]0 0

0 0

0 0

00 0 0 0

0 0

0 0 0 0

0 0

0 0

0

0 0 0

0 0

0 0

ln ln

ln 1 ln 1

1 ln 1 1

ln 1 1 ln 1

zz

z z

o

m u m us s v t zdz z z z

q q

mm u m qt m qt ms v t

q m m m m

m u qt m qts v t

q m m

m u m qt m qts v t ut

q m m

= + + = −

− −= + + − − −

−= + + − − +

− −= + + − + − −

∫

0 0

0 0

0

lnm u m qt

s v t ut utq m

−= + + + −

0 0

0 0

0

lnm m

s s v t u t tq m qt

= + + − − −

(2)

Using the data,

( )( )5000 5000

0 0 4200 80 80 ln18.75 5000 18.75 80

s

= + + − − −

356.367 10 m= × 56.4 kms = �






Let F be the thrust force, and dm

dt be the mass flow rate.

Absolute velocity of exhaust: ev u v= −

Thrust force: ( )dmF u v

dt= −

Power of thrust force: ( )1

dmP Fv u v v

dt= = −

Power associated with exhaust: ( ) ( ) ( )( )22

2

1 1

2 2e

P t m v m u v∆ = ∆ = ∆ −

( )22

1

2

dmP u v

dt= −

Total power supplied by engine: 1 2

P P P= +

( ) ( ) ( )2 2 21 1

2 2

dm dmP u v v u v u v

dt dt

= − + − = −

Mechanical efficiency: 1useful power

total power

P

Pη = =

( )2 2

2 u v v

u v

η−

=−

( )2v

u v

η =+

�

1η = when .u v= The exhaust, having zero velocity, carries no power away.






Let F be the thrust force and dm

dt be the mass flow rate.

Absolute velocity of exhaust: ev u v= −

Thrust force: dm

F udt

=

Power of thrust force: 1

dmP Fv uv

dt= =

Power associated with exhaust: ( ) ( ) ( )( )22

2

1 1

2 2e

P t m v m u v∆ = ∆ = ∆ −

( )22

1

2

dmP u v

dt= −

Total power supplied by engine: 1 2

P P P= +

( ) ( )2 2 21 1

2 2

dm dmP uv u v u v

dt dt

= + − = +

Mechanical efficiency: 1useful power

total power

P

Pη = =

( )2 2

2uv

u v

η =+

�

1η = when .u v= The exhaust, having zero velocity, carries no power away.






The weights are 30 lb, 40 lb, and 50 lb.A B C

W W W= = =

Initial velocities: ( )0

9 ft/sAv = , ( )

06 ft/s

Bv = , ( )

0and 0.

Cv =

There are no horizontal external forces acting during the impacts, and the baggage carrier is free to coast

between the impacts.

(a) Suitcase A is thrown first.

Let 1v be the common velocity of suitcase A and the carrier after the first impact and

2v be the common

velocity of the two suitcases and the carrier after the second impact.

Initial momenta: ( ) ( )0 0, , and 0.A B

A B

W Wv v

g g

Suitcase A impacts carrier. Conservation of momentum:

( ) ( ) ( )( )0

1 10

30 90 3.375 ft/s

80

A AA A C

A

A C

W vW W Wv v v

g g W W

++ = = = =+

Suitcase B impacts on suitcase A and carrier. Conservation of momentum:

( ) 1 20

B A C A B C

B

W W W W W Wv v v

g g g

+ + ++ =

( ) ( ) ( )( ) ( )( )10

2

40 6 80 3.3754.25 ft/s

120

B B A C

A B C

W v W W vv

W W W

+ + += = =

+ + 24.25 ft/s=v �

(b) Suitcase B is thrown first.

Let 3v be the common velocity of suitcase B and the carrier after the first impact and

4v be the common

velocity of all after the second impact.

Suitcase B impacts the carrier. Conservation of momentum:

( ) ( ) ( )( )0

3 30

40 60 2.6667 ft/s

90

B BB B C

B

B C

W vW W Wv v v

g g W W

++ = = = =+

Suitcase A impacts on suitcase B and carrier. Conservation of momentum:

( ) 3 40

A B C A B C

A

W W W W W Wv v v

g g g

+ + ++ =

( ) ( ) ( )( ) ( )( )30

4

30 9 90 2.66674.25 ft/s

120

A A B C

A B C

W v W W vv

W W W

+ + += = =

+ + 44.25 ft/s=v �






The weights are 30 lb, ?, and 50 lb.A B C

W W W= = =

Initial velocities: ( ) ( )0 0

7.2 ft/sA Bv v= = ( )

0, 0

Cv =

Final velocity: 3.6 ft/sfv =

(a) Conservation of momentum:

( ) ( )0 0

0A B A B C

A B f

W W W W Wv v v

g g g

+ ++ + = (1)

( )( ) ( ) ( )( )30 7.2 7.2 30 50 3.6B B

W W+ = + +

20.0 lbB

W = �

(b) Equation (1) shows that the final velocity is independent of the order in which the suitcases are thrown.

3.60 ft/sf =v �






Linear momentum of each particle expressed in kg m/s.⋅

12 6 6

8 6

8 16 8

A A

B B

C C

m

m

m

= + += += − + +

v i j k

v i j

v i j k

Position vectors, (meters): 3 , 1.2 2.4 3 , 3.6A B C

= = + + =r j r i j k r i

( )2Angular momentum about , kg m /s .O ⋅

( ) ( ) ( )

( ) ( ) ( )

0 3 0 1.2 2.4 3 3.6 0 0

12 6 6 8 6 0 8 16 8

18 36 18 24 12 28.8 57.6

0 4.8 9.6

O A A A B B B C C Cm m m= × + × + ×

= + +−

= − + − + − + − +

= − +

H r v r v r v

i j k i j k i j k

i k i j k j k

i j k

( ) ( )2 24.80 kg m /s 9.60 kg m /sO

= − ⋅ + ⋅H j k�






Position vectors, (meters): 3 , 1.2 2.4 3 , 3.6A B C

= = + + =r j r i j k r i

( ) Mass center:a ( )A B C A A B B C Cm m m m m m+ + = + +r r r r

( )( ) ( )( ) ( )( )9 3 3 2 1.2 2.4 3 4 3.6

1.86667 1.53333 0.66667

= + + + +

= + +

r j i j k i

r i j k

( ) ( ) ( )1.867 m 1.533m 0.667 m= + +r i j k�

Linear momentum of each particle, ( )2kg m /s .⋅

12 6 6

8 6

8 16 8

A A

B B

C C

m

m

m

= + += += − + +

v i j k

v i j

v i j k

(b) Linear momentum of the system, ( )kg m/s.⋅

12 28 14A A B B C C

m m m m= + + = + +v v v v i j k

( ) ( ) ( )12.00 kg m/s 28.0 kg m/s 14.00 kg m/sm = ⋅ + ⋅ + ⋅v i j k�

Position vectors relative to the mass center, (meters).

1.86667 1.46667 0.66667

0.66667 0.86667 2.33333

1.73333 1.53333 0.66667

A A

B B

C C

′ = − = − + −′ = − = − + +′ = − = − −

r r r i j k

r r r i j k

r r r i j k





(c) Angular momentum about G, ( )2kg m /s .⋅

( ) ( )( )

1.86667 1.46667 0.66667 0.66667 0.86667 2.33333

12 6 6 8 6 0

1.73333 1.53333 0.66667

8 16 8

12.8 3.2 28.8 14 18.6667 10.9333

1.6 8.5333 15.4667

2.8 13.3333

G A A A B B B C C Cm m m′ ′ ′= × + × + ×

= − − + −

+ − −−

= + − + − + −

+ − − +

= − + −

H r v r v r v

i j k i j k

i j k

i j k i j k

i j k

i j 24.2667k

( ) ( ) ( )2 2 22.80 kg m /s 13.33 kg m /s 24.3 kg m /sG

= − ⋅ + ⋅ − ⋅H i j k�

( ) ( ) ( )2 2 2

1.86667 1.53333 0.66667

12 28 14

2.8 kg m /s 18.1333 kg m /s 33.8667 kg m /s

m× =

= ⋅ − ⋅ + ⋅

i j k

r v

i j k

( ) ( )2 24.8 kg m /s 9.6 kg m /sG

m+ × = − ⋅ + ⋅H r v j k

Angular momentum about O.

( ) ( ) ( )

( ) ( ) ( )( ) ( )2 2

0 3 0 1.2 2.4 3 3.6 0 0

12 6 6 8 6 0 8 16 8

18 36 18 24 12 28.8 57.6

4.8 kg m /s 9.6 kg m /s

O A A A B B B C C Cm m m= × + × + ×

= + +−

= − + − + − + − +

= − ⋅ + ⋅

H r v r v r v

i j k i j k i j k

i k i j k j k

j k

Note that

O Gm= + ×H H r v






Choose x axis pointing east, y axis north, and z axis vertical.

Velocities before collision:

( )0

5.5 mi 5280 ft/miHelicoptor: 121 ft/s

4 min 60 s/minHv = ⋅ =

( )0

10 mi 5280 ft/miAirplane: 220 ft/s

4 min 60 s/minA xv = ⋅ =

( )0

7.5 mi 5280 ft/mi165 ft/s

4 min 60 s/minA yv

= − ⋅ = −

( ) ( ) ( ){ }( ) ( )

( ) ( )

0 0 0 0

Mass center:

6000 3000121 220 165

9000 9000

154 ft/s 55 ft/s

H AH A Ax y

A H A H

m m

v v v

m m m m

= + + + +

= + −

= −

v i i j

i i j

i j

No external forces act during impact. Assume that only gravity acts after the impact. Motion of mass center

after impact:

( )2 2

0 0

1154 55 3600 16.1

2t z gt t t t

= + − = − + −

r v k i j k

Time of fall. 2 3600

14.953 s16.1

t t= =

( )( ) ( )( ) ( ) ( )154 14.953 55 14.953 2302.8 ft 822.42 ft= − = −r i j i j

( ) ( ) ( ) ( ) ( )1 1 2 2H A H H H H A A

m m m m m+ = + +r r r r

( ) ( ) ( ) ( ) ( )

( )( )

( )( ) ( )( )

1 1 2 2

1

19000 2302.8 822.42

3000

2000 1500 300 4000 1800 1500

A H A H H H H

A

m m m m

m

= + − −

= −

− − − −

r r r r

i j

i j i j

( ) ( )3510 ft 267 ft= −i j

( )Coordinates of point : 3510 ft, 267 ftA − �






Choose x axis pointing east, y axis north, and z axis vertical.

Velocities before collision:

( )0

5.5 mi 5280 ft/miHelicoptor: 121 ft/s

4 min 60 s/minHv = ⋅ =

( )0

10 mi 5280 ft/miAirplane: 220 ft/s

4 min 60 s/minA xv = ⋅ =

( )0

7.5 mi 5280 ft/mi165 ft/s

4 min 60 s/minA yv

= − ⋅ = −

( ) ( ) ( ){ }( ) ( )

( ) ( )

0 0 0 0

Mass center:

6000 3000121 220 165

9000 9000

154 ft/s 55 ft/s

H AH A Ax y

A H A H

m m

v v v

m m m m

= + + + +

= + −

= −

v i i j

i i j

i j

No external forces act during impact. Assume that only gravity acts after the impact. Motion of mass center

after impact:

( )2 2

0 0

1154 55 3600 16.1

2t z gt t t t

= + − = − + −

r v k i j k

Time of fall. 2 3600

14.953 s16.1

t t= =

( )( ) ( )( ) ( ) ( )154 14.953 55 14.953 2302.8 ft 822.42 ft= − = −r i j i j

( ) ( ) ( ) ( ) ( )1 1 2 2H A H H H H A A

m m m m m+ = + +r r r r

( ) ( ) ( ) ( ) ( )

( )( )

( )( ) ( )( )

2 1 1

2

1

19000 2302.8 822.42

4000

3000 3600 240 2000 1200 600

H H A A A H H

H

r m m m m

m

= + − −

= −

− + − −

r r r

i j

i j i j

( ) ( )1881 ft 1730 ft= −i j ( )2

Coordinates of point : 1881 ft, 1730 ftH − �






Use a frame of reference moving with the mass center.


0A A B B

B

A B

A

m v m v

m

v v

m

′ ′= − +

′ ′=


( ) ( ) ( )

( ) ( )

( )

2

2 2 2

2

1 1 1 1

2 2 2 2

2

2

B

A A B B A B B B

A

B A B

B

A

A

B

B A B

mV m v m v m v m v

m

m m mv

m

m Vv

m m m

′ ′ ′ ′= + = +

+′=

′ =+

Data: 2 25 30.15528 lb s /ft, 0.09317 lb s /ft

32.2 32.2A B

m m= = ⋅ = = ⋅

90 ft lbV = ⋅

( )( )( )( )( )

2 0.15528 9034.75 34.75 ft/s

0.09317 0.24845B Bv v′ ′= = = 30°

( )0.0931734.75 20.85 20.85 ft/s

0.15528A Av v′ ′= = = 30°

Velocities of A and B:

[24 ft/sA

=v ] [20.85 ft/s+ ]30° 12.00 ft/sA

=v 60.3° �

[24 ft/sB

=v ] [34.75 ft/s+ ]30° 56.8 ft/sB

=v 17.8° �






Locate the mass center.

Let l be the distance between A and B.

( )B A B Am l m m l= +

2

7

5

7

B

A

A B

B

ml l l

m m

l l

= =+

=

(a) Linear momentum.

( )( )02.5 3.5 8.75

AL m v= = = 8.75 kg m/s= ⋅L �

Angular momentum about G:

( )( )( )0 0

2 20.210 2.5 3.5 0.525

7 7G A A A

H l m v lm v= = = =

20.525 kg m /s

G= ⋅H �

(b) There are no resultant external forces acting on the system;

therefore, L and G

H are conserved.

:L A A B B

m v m v L+ = 2.5 1.0 8.75A Bv v+ = (1)

:G

H B B B A A A Gl m v l m v H− =

( )( ) ( )( )5 20.210 1.0 0.210 2.5 0.525

7 7B Av v− =

0.15 0.15 0.525B Av v− = (2)

Solving (1) and (2) simultaneously, 1.5 m/s, 5 m/sA Bv v= =

1.500 m/sA

=v �

5.00 m/sB

=v �






For steady flow 1 2

Q Q Q+ = (1)

Assume that the fluid speed is constant.

Velocity vectors: ( ) 1 2sin cos , , v v vθ θ= − = − =v i j v i v i

Let Pj be the force that plate C exerts on the fluid.


( ) ( ) ( ) ( )1 21 2m P t m m∆ + ∆ = ∆ + ∆v j v v

( ) ( )

1 2

1 2

m m mP

t t t

∆ ∆ ∆= + −∆ ∆ ∆

j v v v

( ) ( ) ( )1 2

sin cosdm dm dm

v v v vdt dt dt

θ θ = − + − −

i i i j

( )1 2sin cosQ v Q v Qvρ ρ ρ θ θ= − + − −i i i j


i: 1 2 2 1

0 sin sinQ v Q v Qv Q Q Qρ ρ ρ θ θ= − + − = − (2)

j: cosP Qvρ θ= (3)

Data: 2 462.41.93789 lb s /ft , 90 ft/s

32.2v

g

γρ = = = ⋅ =

3 3 3 3

1 226 gal/min 57.932 10 ft /s, 130 gal/min 289.66 10 ft /sQ Q− −= = × = = ×





From (1), 3 3347.59 10 ft /sQ −= ×

(a) From (2), 2 1sin 0.66667 41.810

Q Q

Qθ θ−= = = °

41.8θ = °�

From (3), ( )( )( )31.93789 347.59 10 90 cos 41.810 45.2 lb.P

−= × ° =

(b) Force that stream exerts on plate C:

45.2 lbP− =j �






Calculation of or .

m dm

t dt

∆∆


( ) ( ) mm A l Av t Av

tρ ρ ρ∆∆ = ∆ = ∆ =

∆

3 2

2 2 2

62.4 lb/ft 1.5 in60 ft/s 1.21118 lb s/ft

32.2 ft/s 144 in /ft

dmAv

dtρ= = ⋅ ⋅ = ⋅

Apply the principle of impulse-momentum.

moments about D: ( ) ( ) ( ) ( )12 16 10 4

12 12 12 12A B

m v C t W t m v − ∆ − ∆ + ∆ = − ∆

( )16 10 4 12B A

mC W v v

t

∆= + −∆

( )( ) ( ) ( )( ) ( )( )10 10 1.21118 4 60 12 60 481.37 = + − = −

30.085 lbC = − 30.1 lb=C �

x components: ( ) ( ) ( )A x Bm v D t m v∆ + ∆ = ∆

( ) ( ) ( )( )1.21118 60 60 0x B A B A

m dmD v v v v

t dt

∆= − = − = − =∆

0x

D = �

y components: ( ) ( ) ( )0 0y

C t D t W t+ ∆ + ∆ − ∆ =

( )10 30.085 40.085 lby

D W C= − = − − = 40.1 lb=D �






Calculation of dm

dt at a section in the airstream:


( )m A l Av tρ ρ∆ = ∆ = ∆

m dm

Avt dt

ρ∆ = =∆

(a) ( )ThrustB A

dm

dt= −v v where

Bv is the velocity just downstream

of propeller and A

v is the velocity far upstream. Assume A

v is

negligible.

( ) 2 2Thrust

4Av v D v

πρ ρ = =

( ) ( )2 2 23600 1.21 2 3.801

4v v

π = =

30.774m/sv = 30.8 m/sv = �

(b) ( ) ( )2212 30.774

4 4

dmQ Av D v

dt

π πρ

= = = =

3

96.7 m /sQ = �

(c) Kinetic energy of mass :m∆

( ) ( ) ( )2 2 21 1 1

2 2 2T m v A l v Av t vρ ρ∆ = ∆ = ∆ = ∆

3 2 31 1

2 2 4

T dTAv D v

t dt

πρ ρ∆ = = = ∆

( ) ( ) ( )2 3 311.21 2 30.774 55.4 10 N m/s

2 4

π = = × ⋅

55.4 kWdT

dt= �






From Eq. (14.44) of the text book, the thrust is

( )( ) 3 2 310 kg/s 3600 m/s 36 10 kg m/s 36 10 Ndm

P udt

= = = × ⋅ = ×

F maΣ =

P mg ma− = P

a gm

= − (1)

(a) At the start of firing, 2

0960 kg, 9.81 m/s m m g= = =

From (1), 3

236 109.81 27.69 m/s

960a

×= − = 227.7 m/s=a �

(b) As the last particle of fuel is consumed,

960 800 160 kg,m = − = ( )29.81 m/s assumedg =

From (1), 3

236 109.81 215.19 m/s

160a

×= − = 2215 m/s=a �

solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion cap 14

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bcv v e v v

b b f fm v

b c fm v

b c bcm v

b fv v v0

b c bc b b c

b cv v vinitial velocities

b fv v vinitial values