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 Cm 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 Cm m v m v m v+ = +
( )( )6750 0 5400 2.2222BCv = + 1.77778 m/sBCv =
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 BCm 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= + +
( )( )0 0.028 550
1.7058 m/s
5 0.028 4
B
f
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
0.028 550
3.0628 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 ka 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 g
a
m
µ +
= =
2
6.1656 m/s=
Acceleration of block relative to cart.
( ) 2
/ 4.905 6.1656 11.0706 m/sAB 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
3.0628
0.424 m
2 2 11.0706
AB 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 3700
The masses are 124.2 slugs, 114.9 slugs, and 1366.5 slugs
32.2 32.2 32.2
A B Fm 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 Fm v m v m v+ + =
There are no horizontal external forces acting during the time period under consideration. Momentum is
conserved.
0 A A B B F Fm 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 W
m 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 Fm v m v m v+ + =
conserved.
0 A B F
A A B B F F A B F
W W W
m v m v m v v v v
g g g
= + + = + +
Solving for ,FW A A B B
F
F
W v W v
W
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 lb
1.02
FW
− + −
= − =
24.8 tonsFW =

( ) ( )
( )
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 Am 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 Cm 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 A
f
A B C
m v
v
m m m
×
= = =
+ + ×
4.00 km/hf =v

The masses are m for the bullet and Am and Bm for the blocks.
(a) The bullet passes through block A and embeds in block B. Momentum is conserved.
( ) ( )0 0Initial momentum: 0 0A Bmv m m mv+ + =
Final momentum: B A A B Bmv m v m v+ +
0Equating, B A A B Bmv mv m v m v= + +
( )( ) ( )( ) 3
0
3 3 2.5 5
43.434 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: 0Amv 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/s
43.434 10
A Amv m v
v
m
−
−
× −−
= = =
×
1 293 m/s=v

(a) Woman dives first.
Conservation of momentum:
( )1 1
120 300 180
16 0v v
g g
+
− − =
( )( )
1
120 16
3.20 ft/s
600
v = =
Man dives next. Conservation of momentum:
( )1 2 2
300 180 300 180
16v v v
g g g
+
− = − + −
( )( )1
2
480 180 16
9.20 ft/s
480
v
v
+
= = 2 9.20 ft/s=v
(b) Man dives first.
( )1 1
180 300 120
16 v v
g g
+
′ ′− −
( )( )
1
180 16
4.80 ft/s
600
v′ = =
Woman dives next. Conservation of momentum:
( )1 2 2
300 120 300 120
16v v v
g g g
+
′ ′ ′− = − + −
( )( )1
2
420 120 16
9.37 ft/s
420
v
v
′ +
′ = = 2 9.37 ft/s′ =v

(a) Woman dives first.
( )1 1
120 300 180
16 0v v
g g
+
− − + =
( )( )
1
120 16
3.20 ft/s
600
v = =
Man dives next. Conservation of momentum:
( )1 2 2
300 180 300 180
16v v v
g g g
+
= − + −
( )( )1
2
480 180 16
2.80 ft/s
480
− +
= =
v
v 2 2.80 ft/s=v
(b) Man dives first.
( )1 1
180 300 120
16 0v v
g g
+
′ ′− − =
( )( )
1
180 16
4.80 ft/s
600
v′ = =
Woman dives next. Conservation of momentum:
( )1 2 2
300 120 300 120
16v v v
g g g
+
′ ′ ′− = + −
( )( )1
2
420 120 16
0.229 ft/s
420
v
v
′− +
′ = = −
2 0.229 ft/s′ =v

The masses are 9 kg.A B Cm 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
2
In 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, OH is given as ( )2
1.8 kg m /s− ⋅ k
Equating the two expressions for OH 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 Ca v v v 1.333 m/sAv = ( )1.333 m/sA =v j
0.333 m/sBv = ( )0.333 m/sB =v i
1.000 m/sCv = ( )1.000 m/sC =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 r
r
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 2
1.800 kg m /s 0.900 kg m /s 3.60 kg m /sG = ⋅ + ⋅ − ⋅H i j k

The masses are 9 kg.A B Cm 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 r
r
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
( ) ( ) ( )
( )
0 A 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 2
λ λ= ⋅ = ⋅L r L L r r
( )
( )
2
2 2
2
45
50 , 50 N s/m
0.9
λ 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 Ca v v v
3.333 m/sAv = ( )3.33 m/sA =v j
1.6667 m/sBv = ( )1.667 m/sB =v i
3.333 m/sCv = ( )3.33 m/sC =v k
(b) Angular momentum about O expressed in 2
kg 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 2
9.00 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 1
42 63 168 252A AW
g g g
= + = +
v
i j i j
( ) ( )
4 1
42 63 168 252B BW
g g g
= − + = − +
v
i j i j
( ) ( )
28 1
9 6 252 168C CW
g g g
= − = − −
v
j k j k−−−−
Angular momentum of the system about O expressed in ft lb s.⋅ ⋅
( ) ( ) ( ){ }
( )
1
3 6 0 0 6 3 3 0 3
168 252 0 168 252 0 0 252 168
1
252 756 504 1008 756 504 756
1
0 0 0
A A B B C C
O A B C
W W W
g g g
g
g
g
= × + × + ×
 
 
= + + 
 − − − 
= − + − − + + + −
= + +
v v v
H 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 CW W W W= + + = + + =
( )( ) ( )( ) ( )( ){ }1
4 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 r
r
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
( )( ) ( )( ) ( )( )
1
4 42 63 4 42 63 28 9 6
g
 = + + − + + − − i j i j j k
( )
1
252 168
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.
( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )
1
0.333 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 v
H r r r
i j k i j k i j k
( ) ( ) ( ){ }
( )
1
672 448 700 84 56 112 308 56 84
1
896 448 672 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
1
3 6 0 0 6 3 3 0 3
168 252 0 168 252 0 0 252 168
252 756 504 1008 756 504 756
1
0 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 W
g
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
( )2
Angular 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 2
34 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 Cm 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, ( )2
kg 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 2
2.00 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 2
34 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 G m= + ×H H r v

The mass center moves as if the projectile had not exploded.
( ) ( )( ) ( )( )
( ) ( )
22
0
1 1
60 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
1
20 120 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
1
500 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 Cm m m m m m+ + = + +v v v v (1)
2 1 t− =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 3600 Bv t=j
( )
( )( )
432440
30.034
3600 3.6363
Bv = = 30.0 ft/sBv =

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 Cm m m m m m+ + = + +v v v v (1)
2 1 t− =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 za 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 0
0 165 1.5 247.5 mxx x v t= + = + =
( ) ( )( )22
0 0
1 1
15 0 9.81 1.5 3.9638 m
2 2
yy y v t gt= + − = + − =
( )0 0
0zz 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 15
247.5 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 2255 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 xa v x= = =
2
9.81 m/sya g= − = −
0 28 9.81y yv v a t t= + = −
2 2
0 0
1
60 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 BDefinition 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.
( ) ( )
( ) ( )
2.1
3 m/s, 3tan30 m/s
0.7
2.1
2.333 m/s, 2.3333tan m/s
0.9
C
C Cx y
C
D
D Dx y
D
x
v v
t
x
v 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 y
m v′=
Conservation of momentum of system:
( ) ( ) ( )0: 0A B A A C C D D xx
m v m m v m v m v′ ′ ′+ = + +
( ) ( ) ( )4.8 0 3 2.3333
2 2
A
m m
m mv′+ = + +
( )a 2.13 m/sA′ =v
( ) ( ) ( ) ( ) ( ): 0 0A B A A C C D Dy yy
m m m v m v m v′ ′ ′+ = + +
( ) ( )0 0 0 3tan30 2.3333tan
2 2
m m
θ+ = + ° −
( )b
3
tan tan30 0.7423
2.3333
θ = ° = 36.6θ = °
( ) ( ) ( ) ( )
22 2 2
3 3tan30C C Cx y
v v v= + = + o
3.46 m/sCv =
( ) ( ) ( ) ( )
22 2 2
2.3333 2.3333tan36.6D D Dx y
v v v= + = + o
2.91m/sDv =

( )
( )
( )
( )
0 0Velocity vectors: cos30 sin30
sin7.4 cos7.4
sin 49.3 cos49.3
cos45 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 Cm m m= =
( ) ( ) ( )
( )
4 cos30 sin30 sin7.4 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.
( ) ( )
( ) ( )
: sin7.4 sin 49.3 4cos30 2.1cos45
: cos7.4 cos49.3 4sin30 2.1sin 45
A B
A B
v v
v v
° + ° = − °
° − ° = − °
i
j
o
o
Solving simultaneously,
(a) 2.01 m/sAv =
(b) 2.27 m/sBv =

( ) ( )0
190 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
2
t gt= + −r r v j
0
0
1
2
gt
t
−
= +
r r
v j
where ( ) ( ) ( )1600 ft 2400 ft 400 ft= − +r i j k
0 0=r
( )( )
( ) ( ) ( )
0
1600 2400 400 1
32.2 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 10000
133.333 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 v
v
i j k
( ) 2
0 0
1
Vertical motion:
2
yy y v t gt= + −
( ) 2 21
0 45 3.6735 32.2 or 0.22817 2.7950 0
2
t t t t= + − − − =
Solving for , 1.7898 st t =
Horizontal motion: ,x zx 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
λλλλ
λλλλ
λλλλ
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 4A 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
1
Along , 18 9 18
27
1
Along , 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 4
60 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 60 108
19.5 27
810 54 52
27 22.5
32400 144 108 72
19.5 27 22.5
A B
B C
A B C
v v
v v
v v v
   
= − +   
   
   
− = −   
   
     
− = − − −     
     
Solving, 119.944
19.5
Av
= 2340 ft/sAv =
76.635
27
Bv
= 2070 ft/sBv =
95.160
22.5
Cv
= 2140 ft/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
1
Along , 18 9 18
27
1
Along , 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/s
0.010
A D
A
A
r
v
t
= = =
/ 27
1500 ft/s
0.018
B D
B
B
r
v
t
= = =
C/ 22.5
1875 ft/s
0.012
D
C
C
r
v
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 Cm m m m= + +v v v v
( ) ( ) ( )0 750 1800 1000 500 1000 1125 1500A B CW W W W
v
g 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 CW W W W= + + = (4)
Solving equations (1), (2), and (4) simultaneously,
(a) 2.88 lb, 2.16 lb, 0.96 lbA B CW 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, 0
n
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 AM 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 Am and Bm for the blocks.
The bullet passes through block A and embeds in block B. Momentum is conserved.
( ) ( )0 0Initial momentum: 0 0A Bmv m m mv+ + =
Final momentum: B A A B Bmv m v m v+ +
0Equating, B A A B Bmv mv m v m v= + +
( )( ) ( )( ) 3
0
3 3 2.5 5
43.434 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: 0Amv 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/s
43.434 10
A Amv m v
v
m
−
−
× −−
= = =
×
(a) Bullet passes through block A. Kinetic energies.
( )( )22 3
0 0
1 1
Before: 43.434 10 500 5429 J
2 2
T mv −
= = × =
( )( ) ( )( )2 22 2 3
1 1
1 1 1 1
After: 43.434 10 292.79 3 3 1875 J
2 2 2 2
A AT 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 1
Before: 43.434 10 292.79 1861.7 J
2 2
T mv −
= = × =
( ) ( )( )22
3
1 1
After: 2.54343 5 31.8 J
2 2
B BT 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 Bm m= = 5400 kgcm =
Initial velocities: 0 0( ) ( ) 0,A Bv v= = 0( ) 8 km/h = 2.2222 m/scv =
Velocities after first collision:
1( ) 0,Av = 1 1( ) ( ) 1.77778 m/sB cv v= =
Velocities after second collision
2.9630 m/s,Av = 1.18519 m/sB cv v= =
Initial kinetic energy: 2 2 2
0 0 0 0
1 1 1
( ) ( ) ( )
2 2 2
A A B B c BT m v m v m v= + +
2 3
0
1
0 0 (5400)(2.2222) 13.3333 10 J
2
T = + + = ×
Kinetic energy after the first collision:
( )22 2
1 1 11
1 1 1
( ) ( )
2 2 2
A A B B c cT m v m v m v= + +
2 2 31 1
0 (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 2
A A B B c cT 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 1 2.6667 10 JT T− = ×
2.67 kJ
Kinetic energies before and after second collision:
2 1 10.67 kJT T= =

( )( )22 20004000 3700
The masses are 124.2 slugs, 114.9 slugs, and 1366.5 slugs
32.2 32.2 32.2
A B Fm 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 Fm v m v m v+ + =
conserved.
0 A A B B F Fm 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 1
Initial kinetic energy: 0
2 2 2
A A B B CT m v m v v= + + =
2 2 2
2
1 1 1
Final kinetic energy:
2 2 2
A A B B C CT m v m v m v= + +
( )( ) ( )( ) ( )( )2 2 2
2
1 1 1
124.2 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 1 3243 ft lbT T= − = ⋅ 1U 2 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,AW = 2.16 lb,BW = 0.96 lbcW =
Speeds of fragments: 1950 ft/s,Av = 1500 ft/s,Bv = 1875 ft/scv =
Total kinetic energy before the explosion.
( )22 3
0 0
1 1 6
1464 199.69 10 ft lb
2 2 32.2
W
T 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 W
T v v v
g g g
= + +
( ) ( ) ( )2 2 21 2.88 1 2.16 1 0.96
1950 1500 1875
2 32.2 2 32.2 2 32.2
     
= + +     
     
3
297.92 10 ft lb= × ⋅
Increase in kinetic energy. 3
1 0 98.2 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 m
E m
m m
m m
E m
m m
− ⋅ −
′ ′= ⋅ =
+
− ⋅ −
′ ′= ⋅ =
+
v v v v
v v
v v v v
v v
( ) :a Ratio / /A B B AE E m m=
( ) 135 km/h 37.5 m/sAb = =v , 90 km/h 25 m/sB = =v
62.5 m/sA B− =v v
( )( ) ( )
( )( )
2 2
3
2
2400 1350 62.5
607.5 10 J
2 3750
AE = = × 608 kJAE =
( ) ( )( )
( )( )
2 2
6
2
2400 1350 62.5
1.08 10 J
2 3750
BE = = × 1080 kJBE =

( ): A B A A B BVelocity 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 m
E m
m m
m m
E m
m m
− ⋅ −
′ ′= ⋅ =
+
− ⋅ −
′ ′= ⋅ =
+
v v v v
v v
v v v v
v v
( ) ( )2 2
0 00 0
1 1
Energies from tests: ,
2 2
A A B BE 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
mE
S
E m m v
mE
S
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 0v
( )
( )
Thus, cos30 sin30
cos30 sin30
B B
C C
v
v
= ° + °
= ° − °
v i j
v i j
By symmetry, A Av=v i
0Conservation of momentum: A B Cm m m m= + +v v v v
0
component: 0 0 sin30 sin30
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 2
A B Cmv 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
= + −
− = − + = −
+ = − = − = −
= = =
00.200A v=v
00.693B v=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 Bv= ° + °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 1
cos30 sin30
2 2 2
1
2 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 1
cos30 , sin 30 ,
2 4
3
cos30 sin30
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 Cv= ° − °v i j
Conservation of momentum: A A Cm m m′ = +v v v

( ) ( )
( ) ( )
( ) ( )
( ) ( )
0
0
component: cos30 ,
1
cos30 cos30
4
component: sin30
3
sin30 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 3
cos30 sin30
2 16 16 2 4 4
1 1 1
cos30 cos 30
2 16 2
3 3
sin30 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 3
0 cos30 sin30 2
2 2
C Cv v v
 
= − ° + ° +  
 
( )
( )
0 0
0 0 0
0 0 0
1 3 3
cos30 sin30
4 4 4
1 3 1
cos30
4 4 8
3 3 3
sin30
4 4 8
C
A x
A y
v v v
v v v v
v v v v
 
= ° + ° =  
 
= − ° = −
= − + ° = −
00.250A v=v 60°
00.866B v=v 30°
00.433C v=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 ( )0 0A 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 v
T 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 v
m gh m v
m m
+ = +
+ =
+
2 2
2 0
0
1
2
A
A
B A B
m v
h m v
m g m m
 
= − 
+ 
2
0
2
A
A B
m v
h
m m g
=
+

Velocity vectors: ( )0 0 cos30 sin30v= ° − °v i j 0 15 ft/sv =
A Av= −v j
( )
( )
sin30 cos30
cos30 sin30
B B
C C
v
v
= ° − °
= ° + °
v i j
v i j
0 A B Cm m m m= + +v v v v
Divide by m and resolve into components.
i: 0 cos30 sin30 cos30B Cv v v° = ° + °
j: 0 sin30 cos30 sin30A B Cv v v v− ° = − − ° + °
Solving for and ,B Cv v
( ) ( )0 0
3 1
2 2
B A C Av v v v v v= − = +
0
1 1 1 1
2 2 2 2
Divide by 1
2
m 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
1
7.5 ft/s
2
Av v= = 7.50 ft/sAv =
( )
3
15 7.5 6.4952 ft/s
2
Bv = − = 6.50 ft/sBv =
( )
1
15 7.5 11.25 ft/s
2
Cv = + = 11.25 ft/sCv =

Velocity vectors: ( )0 0 cos45 sin 45v= ° + °v i j 0 15 ft/sv =
A Av=v j
( )
( )
sin60 cos60
cos60 sin60
B B
C C
v
v
= ° − °
= ° + °
v i j
v i j
0 A B Cm m m m= + +v v v v
i: 0 cos45 sin 60 cos60B Cv v v° = ° + °
j: 0 sin 45 cos60 sin 60A B Cv v v v° = − ° + °
Solving for and ,B Cv v
0 00.25882 0.5 0.96593 0.86603B A C Av v v v v v= − = −
0
1 1 1 1
2 2 2 2
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
= + + + −
= + +
00.70711 10.607 ft/sAv v= = 10.61 ft/sAv =
00.61237 9.1856 ft/sBv v= = 9.19 ft/sBv =
00.35356 5.303 ft/sCv v= = 5.30 ft/sCv =

1
sin ,
3
θ =
8
cos ,
3
θ = 19.471θ = °
Velocity vectors 0 0v= −v j
( )cos sinA Av θ θ= −v i j
( )/ sin cosB A Bu θ θ= − −v i j
/B A B A= +v v v
C Cv=v j
Conservation of momentum: 0 /2A B C A B A Cm 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 Cv v u vθ θ= + + −
Solving for and ,A Bv u ( ) ( )0 0
1
0.94281
6
A C B Cv v v u v v= + = +
0
1 1 1 1
2 2 2 2
( )2 2 2 21 1 1
2 2 2
A A B Cmv m v u mv= + + +
Divide by 1
2
m and substitute for and .A Bv u
( ) ( ) ( )
2
2 222 2
0 0 0
1
2 0.94281
6
C C Cv v v v v v
 
= + + + + 
 
( )22 2
0 0 00.94445 0.02857C C Cv v v v v v− = + = 00.0286C v=v
[0 00.17143 0.17143A Av v v= =v ]19.471° , 00.1714A v=v 19.5°
[0 / 00.96975 0.96975B B Au v v= v ]19.471°
/B A B A= +v v v 0.985B =v 80.1°

1 8
sin , cos , 19.471
3 3
θ θ θ= = = °
C strikes B.
0 0orB C B Cm m m v v v′ ′= + = −v v v
( )22 2
0
1 1 1
2 2 2
B Cmv m v mv′= +
( )22 2
0 0 C Cv v v v= − +
0Cv =
0Bv v′ =
Cord becomes taut.
Velocity vectors:
A Av=v θ
/B A Bu=v θ
Conservation of momentum: /B A A B Am m m m′ = + +v v v v
+ :θ sin 2B Av vθ′ = 0
1 1
sin
2 6
A Bv v vθ′= =

(a) 0
6
A
v
=v 19.5°
+ :θ cosB Bv uθ′ = 0
cos 8
3
B B
v
u v θ′= =
0
1
19.471
6
B v
 
= ° 
 
v [ 00.94281v+ ]19.471°
[ 00.95743B v=v ]80.8° 00.957B v=v 80.5°
0C =v
Initial kinetic energy: 2
1 0
1
2
T mv=
Final kinetic energy: 2 2 2
2
1 1 1
2 2 2
A B CT mv mv mv= + +
( ) ( )
2
22 2
0 0
1 1 1
.95743 0 0.94444
2 6 2
mv mv
  
= + + =  
   
(b) Fraction lost: 1 2
1
1 0.94444
0.05555
1
T T
T
− −
= =
Fraction of energy lost = 0.0556

(a) Use part (a) of sample Problem 14.4 with A Bm m m.= =
0 0
1
2
m
v v v
m m
= =
+
0
1
2
A Bv v v= =
(b) Consider initial position and position when 0θ = again.
Conservation of momentum
0 A A B Bmv m v m v= +
0B Av v v+ = (1)
Conservation of energy
2 2 2
0
1 1 1
2 2 2
A Bmv mv mv= + (2)
Substituting (1) into (2),
2 2 21 1 1
( )
2 2 2
A B A Bm v v mv mv+ = +
0A Bmv v =
Either 0Av = with 0Bv v=
or 0Bv = with 0Av v=
(c) Consider positions when maxθ θ= and min.θ θ=
Since / 0B Av = at these states
B Av v=
Conservation of momentum
0 A Bmv mv mv= +
0
1
2
B Av v v= =

Conservation of energy would show that the elevation of B is the same for maxθ 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 AN m g m a− °= − ° (1)
:F ma∑ = 1 /sin30 cos30B B A B AN m a m a°= °− (2)
Wedge:
:F ma∑ = 1 sin30 A AN m a°= (3)
Rearranging (1), (2), and (3) and applying numerical data,
1 (6sin30 ) (6)(9.81)cos30AN a+ ° = ° (1)
1 /(sin30 ) 6 (6cos30 ) 0A B AN a a° + − ° = (2)
1(sin30 ) 10 0AN a° − = (3)
1 44.325N,N = 2
2.2163 m/s ,Aa = 2
/ 6.8243 m/sB 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 s
6.8243
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 0.9A C C A B Bm v m v m m v= + +
( )
( )( ) ( )0
0.90.9
0.5 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/sCv =
Moments about C : ( )00.9 0.9 1.8A A B B D Dm v m m v m v= + +
( )
( )( ) ( )( )0 0.90.9
0.25 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/sDv =
(b) Initial kinetic energy:
( )22
1 0
1 1
7.5 12 540 N m
2 2
AT m v= = = ⋅
( ) ( ) ( )
2 2 2
2
2 2 2
1 1 1
( )
2 2 2
1 1 1
15 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
0.786
540
= =
( )1 2
1
0.786
T T
T
−
=

There are no external forces. Momentum is conserved.
(a) Moments about D : 00.9 1.8 0.9A C C B Bm v m v m v= +
( )( ) ( )( )0
0.9 0.9
0.5 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 : 00.9 1.8 0.9A D D B Bm v m v m v= +
( )( ) ( )( )0
0.9 0.9
0.25 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 1
7.5 12 540 N m
2 2
AT m v= = = ⋅
( ) ( ) ( )
2 2 2
2
2 2 2
1 1 1
2 2 2
1 1 1
7.5 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
0.727
540
= =
( )1 2
1
0.727
T T
T
−
=

(a) Linear and angular momentum.
0 0A A B Bm m mv= + = +L v v i
0mv=L i
0
2
( ) ( ) 0
3 3
G
l l
mv= × + − ×H j i j i
0
2
3
G lmv= −H k
Motion of mass center G. Since there is no external force,
0 constantA A B Bm m mv= + = =L v v i
03m mv=v i 0
1
3
v=v i
Motion about mass center.
( ) ( )G G i i i A A A B B Bm m m′ ′ ′ ′ ′ ′ ′= = Σ × = × + ×H H r v r v r v
where
2 1
3 3
A Bl , l′ ′ ′ ′= = −r j r j
2 1
,
3 3
A Bl lθ θ′ ′ ′ ′= = −v i v i& &
Thus,
2 2 1 1 1
2
3 3 3 3 3
G l ml l m lθ θ
       
′ ′ ′ ′= × + − × ⋅       
       
H j i j i& &
22
3
ml θ= − k&
But HG is constant.
2 0
0
2 2
3 3
v
ml lmv
l
θ θ− = − =k k &
0
2 2
3 3
Av l vθ′ = =&
0
1 1
3 3
Bv l vθ′ = =&

(b) After 180º rotation.
0 0
1 2
3 3
A A v v′= + = −v v v i i
0
1
3
A v= −v i
0 0
1 1
3 3
B B v v′= + = +v v v i i
0
2
3
B v=v i

Masses: 2125
3.882 lb s /ft, 2 , 3 .
32.2
A B A C Am 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
0 0A 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 Cm × + × + × =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 2
Am mv
 
= − −  
 
v i j
( ) 01
3 1
2 2
Bm mv
 
= −  
 
v i j
( ) ( )1 01 1A Bm m mv= + = −L v v j
1
0
1
2 2
v
m
= = −
L
v 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 2
T mv mv mv= + =
State 2. The cord is taut. Conservation of linear momentum:
(a) 0
1
2
D v= = −v v j 00.500Dv v=
Let ( )2
andA 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 Almu lmu lmu= + =H k k k

(b) Conservation of angular momentum:
( ) ( )2 1G G=H H
0
3
2
2
Almu lmv=k k 0
3
4
A Bu u v= =
00.750u v=
( ) 2 2 2
2
2 2
0 0
1 1 1
2
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
0.1875
T T
T
−
=

The system is spheres A and B and the ring D.
Initial velocities: ( )0 cos30 sin30A v= − ° − °v i j
( )0 cos30 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 4
l v t= − −r i j
(b) / /G A G A B G Bm m= × + ×H r v r v
( )
( )
0
0
1
cos30 cos30 sin30
4
1
cos30 cos30 sin30
4
l l mv
l l mv
 
= − + ° × − ° − ° 
 
 
+ − − ° × ° − ° 
 
i j i j
i j i j
0
7
4
G lmv=H k
(c) 2 2 2
0
1 1
2 2
A BT 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 xx 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 2.25tan30 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.
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 [(6 ft/s) ( ) ] [( ) ( ) ] (8 ft/s) 0 0B y C x C yv v v+ + + + = + +i j i j i
Components:
: 6 ( ) 8C xx 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 2tan 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

Conservation of linear momentum: 0
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 y
v v v+ = + +i j j i j
i: ( )41.472 2.4 B x
v= ( ) 17.28 ft/sB x
v =
j: ( )10.348 4.8 2.4A B y
v v= − ( ) 4.32 2B Ay
v v= −
Speeds relative to the mass center: ( ) ( )( )
1 1
3 8 8 ft/s
3 3
Au lω= = =
( ) ( )( )
2 2
3 8 16 ft/s
3 3
Bu lω= = =
Initial kinetic energy: ( ) ( )2 2 2 2
1 0 0
1 1 1
2 2 2
A B A B
A Bx y
W W W W
T v v u u
g g g g
  = + + + +    
( ) ( ) ( )2 22 2
1
1 7.2 1 4.8 1 2.4
5.76 1.44 8 16 18.2517 ft lb
2 32.2 2 32.2 2 32.2
T
     
= + + + = ⋅     
     
Final kinetic energy: ( ) ( )2 22
2
1 1 1
2 2 2
A B B
A B Bx y
W W W
T v v v
g g g
= + +
( ) ( )2 22
2
1 4.8 1 2.4 1 2.4
17.28 4.32 2
2 32.2 2 32.2 2 32.2
A AT v v
     
= + + −     
     
2
0.2236 0.6440 11.8234A Av v= − +
Conservation of energy: 1 2T T=
(a) 2
0.2236 0.6440 6.4283 0, 6.9919 ft/sA A Av v v− − = = 6.99 ft/sA =v
( ) ( )( ) ( ) ( )4.32 2 6.9919 9.6638 ft/s 17.28 ft/s 9.6638 ft/sB By
v = − = − = −v i j
19.80 ft/sB =v 29.2°

( ) ( )( ) ( )
( ) ( ) ( )( ) ( )( )
0 0 0 01
2
3 3
7.2 4.8 2.4
7.44 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 l
H y m m v H y v u u
g g g g
 
= − + + = − + + + 
 
     
= − + + = − ⋅ ⋅     
     
( ) ( )2
( )A B
O A A B B A B yy
W W
H 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 32.2
a a
   
= + − = −   
   
( ) ( )2 1
1.0423 17.2868 6.0047O OH H a= − = −
(b) 10.82 fta = 10.82 fta =

Conservation of linear momentum: 0
A B A B
A B
W W W W
g g g g
 
+ = + 
 
v v v
( ) ( )0
9 6 3
7.68 10.8 6.72
32.2 32.2 32.2
   
= + −   
   
v j i j
(a) ( ) ( )0 3.6 ft/s 2.88 ft/s= +v i j 0 4.61 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 W
l 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 W
T v l l
g g g g
ω ω
 
= + + + 
 
( ) ( ) ( )
( )
2 22 2
1
2
1 9 1 6 1 3
3.6 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 W
T v v
g g
= +
( ) ( )2 2 2
2
1 6 1 3
7.68 10.8 6.72 13.0324
2 32.2 2 32.2
T
   
= + + =   
   
Conservation of energy: 2 1T T= .
( ) ( )2
13.0324 2.9703 0.27950 6.000 ft/sA Al lω ω= + =
( ) ( ) ( ) ( ) ( )
( )( ) ( ) ( )( )
( ) ( )( )
0 0 0 01
2
2
9 6 3
0 7.5 3.6 2 2
32.2 32.2 32.2
7.5466 0.55901 7.5466 3.3540
A B A B
O A A B By x
A A A
A A
W W W W
H 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 3
7.68 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 B
O A B y
O O A A
W W
H v a v b
g g
H H l l
   
= + = + −   
   
= − ⋅ ⋅
= − = − + =
(b) 2 1.200 ftB A A Bl l l l l= = = + 1.800 ftl =
(c)
6.00
10.00
0.600
A
A
l
l
ω
ω = = = 10.00 rad/sω =

Velocities in m/s. Lengths in meters. Assume masses are 1.0 for each ball.
Before impacts: ( ) ( ) ( )00 0 0
4 , 0A B Cv= = = =v i i v v
After impacts: ( ) ( )1.92 , ,A B B B C Cx y
v 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 x
v v v v= + + = −
j: ( ) ( )0 1.92 0 1.92B By y
v v= − + + =
0
1 1 1 1
2 2 2 2
A B Cv v v v= + +
( ) ( ) ( ) ( )22 2 2 21 1 1 1 1
4 1.92 1.92 4
2 2 2 2 2
C Cv v= + + − +
2
4 3.6864 0C Cv v− + =
( ) ( )( )2
4 4 4 3.6864
2 0.56 2.56 or 1.44
2
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 Bx
v = − = = +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 0
5 , 0A B Cv= = = =v i i v v
After impacts: ( ) ( ), , 3.2A A B B B Cx y
v 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 x
v v= + + =
j: ( ) ( )0 0A B B Ay y
v v v v= − + + =
0
1 1 1 1
2 2 2 2
A B Cv v v v= + +
( ) ( ) ( ) ( ) ( )2 2 2 2 21 1 1 1 1
5 1.8 3.2
2 2 2 2 2
A Av v= + + +
(a) 2
5.76 2.4A Av v= = 2.40 m/sA =v
( ) 2.4B y
v = 1.8 2.4B = +v i j 3.00 m/sB =v 53.1°
Conservation of angular momentum about :B′
( )00.75 1.8 A Cv a v cv= − +
01.8 0.75A A Cav 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 1
3 0 4.5
3 3
C Cv v v v− = = = 0 1.500 m/s=v
j: 0 2.6 m/sA B B Av v v v− = = =
Conservation of angular momentum about G:
( ) ( ) ( )2
0 0 03G A A B B C Cml m rω= = × − + × − + × −H k r v v v v r v v
( ) ( ) ( ) ( )( )
( ) ( ) ( ) ( )
( )( ) ( )( )
2
0
2 2
3
1
0.260 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 3
3
2 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 2
A B CT m m m= − + − + −v v v v v v
1 2T T=
( ) ( ) ( )2 2 22 23 1 1 1
3 3 3
2 2 2 2
ml m m mω = + +
3 m/slω =

(b)
2 2
0.45033 m /s
0.1501 m
3 m/s
l
l
l
ω
ω
= = = 150.1 mml =
(c)
3 m/s
0.1501
l
l
ω
ω = = 19.99 rad/sω =

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
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: 03 0Cv v− = ( )( )03 3 0.4 1.2 m/sCv v= = =
j: 0A Bv v− = B Av v=
Initial kinetic energy: 2 2 2
1 0
1 1
3 3
2 2
T mv ml ω
   
= +   
   
Final kinetic energy: 2 2 2 2 2
2
1 1 1 1
2 2 2 2
A B C A CT mv mv mv mv mv= + + = +
Conservation of energy: 2 1T T= Solve for 2
Av .
( ) ( ) ( ) ( )2 2 2 22 2 2
0
3 3 1 3 3 1
0.4 0.75 1.2
2 2 2 2 2 2
A Cv v l vω= + − = + −
(a) 2 2
0.36375 m /s ,= 0.6031 m/sAv = 0.603 m/sA =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 03G 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
( ) ( ) ( ) ( ) ( )3 A C A Cl 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 l
Av
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/s
dm
dt
−
= × =
: 1( ) 0m v P t∆ − ∆ =
m dm
P 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
Au
dt t t
ρ
ρ
∆ ∆
= = =
∆ ∆
(2)
( ) ( ) 0m u P t∆ − ∆ =
2m dm
P u u Au
t dt
ρ
∆
= = =
∆
P
u
Aρ
=
From (1), 1 1
P
V v u v
Aρ
= − = −
Data: 2 6 2
400 N, 600 mm 600 10 mP A −
= = = ×
3
1 30 m/s, 1000 kg/mv ρ= =
6
400
30
(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: 3
475 gal/min ft /sQ = = 1.0583
Mass flow rate: ( )3 362.4
slug/ft 1.0584 ft /s 2.051 slug/s
32.2
dm
Q
dt
ρ
 
= = = 
 
Velocity vectors: ( )1, cos30 sin30v v= = ° + °v i v i j
( )2 cos45 sin 45v= ° − °v i j
Impulse – momentum principle:
( ) ( ) 1 2
2 2
m m
m t
∆ ∆
∆ + ∆ = +v F v v
( ) ( )
( )( )( )
( ) ( )
1 2
1 1
2 2
1 1
cos30 sin30 cos45 sin 45
2 2
2.051 60 ft/s 0.21343 0.10355
26.26 lb 12.74 lb
m
t
dm
v
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
500 gal/min ft /sQ = = 1.1141
32.2
dm
Q
dt
ρ
 
= = = 
 
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 ( )
60
2.1590 2.6987 slug/s.
48
dm u dm
dt v dt
′
= = =
Velocity vectors: ( )1, cos30 sin30u u= = ° + °u i u i j
( )2 cos45 sin 45u= ° − °u i j
Let F be the force that the wedge exerts on the stream.
Impulse-momentum principle:
( ) ( ) 1 2
2 2
m m
m t
∆ ∆
∆ + ∆ = +u F u u
( ) ( )
( )( )( )
( ) ( )
1 2
1 1
2 2
1 1
cos30 sin30 cos45 sin 45
2 2
2.6987 60 0.21343 0.10355
34.6 lb 16.76 lb
m
t
dm
u
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 dm
v
t dt
∆  
= = ° + ° 
∆  
F v i j
( )( )
10
60 cos25 sin 25
32.2
 
= ° + ° 
 
i j
( ) ( )16.89 lb 7.87 lb= +i j
0: 0x x xF D FΣ = − =
16.89 lbxD =
Force on truck hitch at D:
16.89 lbxD− = − 16.89 lbxD− =

Initial momentum: ( ) 0.Am∆ =v
Impulse – momentum principle.
( )( ) ( )x yF F t m+ ∆ = ∆i j v
( )o o
cos35 sin35x y
m dm
F F v
t dt
∆  
+ = = + 
∆  
i j v i j
x component:
o
Engine thrust cos35x
dm
F v
dt
= =
Data: 3 38
8 m /min m /s
60
Q = = 3
1000 kg/mρ =
( )
8
1000 133.333 kg/s
60
dm
Q
dt
ρ
 
= = = 
 
( )( ) o
133.333 50 cos35 5461 NxF = =
5.46 kNxF =

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
1
6 3
m m
R cW av hv
L t t
∆ ∆ 
= + + ∆ ∆ 
Data: 6 m, 4 m, 1.5 m, 0.8 m, 40 kg/s
m dm
L c a h
t dt
∆
= = = = =
∆
[ ]
1
(4)(5886) (6)(1.5)(3)(40) (3)(0.8)(4)(40) 4040 N
6
R = + − =
4040 N=R

Assume A Bu u u= =
Moments about O :
( ) ( ) ( )A c BR 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 Bm u F t m uα θ∆ + ∆ = ∆
sin (1 cos )
m
F u
t
α θ
∆
= −
∆
(1)
Components : ( ) ( )0 cos sinF t m uα θ+ ∆ = ∆
cos sin
m
F u
t
α θ
∆
=
∆
(2)
Dividing (2) by (1),
2
2sin1 cos 2tan tan
sin 22sin cos
2 2
θ
θ θ
α
θ θθ
−
= = =
.
2
θ
α =
Thus point C lies at the midpoint of arc AB.

( )
( )( )
800
13.333 L/s 1.000 kg/L 13.333 L/s 13.333 kg/s
60
dm
Q Q
dt
ρ= = = = =
( ) ( )( )30 m/s 30 m/s sin 40 cos40B C= = ° + °v j v i j
Apply the impulse – momentum principle.
x components: ( ) ( )( )0 30sin 40xA t m+ ∆ = ∆ °
( ) ( )( )30sin 40 13.333 30sin 40x
m
A
t
∆
= ° = °
∆
257 NxA =
y components: ( )( ) ( ) ( )( )30 30cos40ym A t m∆ + ∆ = ∆ °
( ) ( )30cos40 30 13.333 30cos40 30y
m
A
t
∆
= ° − = ° −
∆
93.6 N= − 93.6 NyA =
moments about :A ( )( )( ) ( )0.060 30 Am M t∆ + ∆
( )( )( ) ( )( )( )0.180 30cos40 0.300 30sin 40m m= ∆ ° − ∆ °
( ) ( ) ( )1.8 1.6484Am M t m∆ = ∆ − ∆
( ) ( )( )3.4484 13.333 3.4484 46.0 N mA
m
M
t
∆
= − = − = − ⋅
∆
46.0 N mA = ⋅M
274 N=A 20.0°

Mass flow rate:
3 3
2
62.4 lb/ft 40 ft /min
1.29193 lb s/ft
60 s/min32.2 ft/s
dm
Q
dt g
γ
= = = ⋅
75 ft/sA Bv v= =
Use impulse - momentum principle.
moments about :D ( )( ) ( )( ) ( )
15 23 15
sin60 cos60
12 12 12
A Am v m v C t
     
− ∆ ° + ∆ ° − ∆     
     
( )
3
12
Bm v
 
= ∆  
 
( )( )( )
15 15 23 3
sin60 cos60 1.29193 75 0.37420
12 12 12 12
A
m
C v
t
∆   
= − ° + ° − = −   ∆   
29.006 lbC = − 0, 29.0 lbx yC C= = −
x component: ( ) ( ) ( )cos60A x Bm v D t m v∆ ° + ∆ = ∆
( ) ( )( )cos60 1.29193 75 75cos60x B A
m
D v v
t
∆ 
= − ° = − ° 
∆ 
48.4 lbxD =
y component: ( ) ( ) ( )sin60 0A ym v C t D t∆ ° + ∆ + ∆ =
( )( )sin60 29.006 1.29193 75 sin60y A
m
D C v
t
∆
= − − ° = + − °
∆
54.9 lbyD = −

300 gal/min 0 ft /sQ = = .6684
32.2
dm
Q
dt
ρ
 
= = = 
 
90 ft/sA Bv v= =
Use the impulse – momentum principle.
Moments about C: ( ) ( ) ( )cosA P Bm v a W t l m v bθ∆ − ∆ = ∆
(a) ( )
( )( ) ( )( )
( )( )
90 4/12 90 1/12
cos 1.2954 0.7287
40 1
A B
p
m v a v b
t W l
θ
−∆ −
= = =
∆
43.23θ = ° 43.2θ = °
x components: ( ) ( ) ( ) cosA x Bm v C t m v θ∆ + ∆ = ∆
( ) ( )( )cos 1.2954 90cos 90 31.63 lbx B A
m
C v v
t
θ θ
∆
= − = − = −
∆
y components: ( ) ( ) ( )0 siny p BC t W t m v θ+ ∆ − ∆ = − ∆
( ) ( )( )sin 40 1.2954 90 sin 39.84 lby p B
m
C W v
t
θ θ
∆
= − = − = −
∆
(b) [31.63 lb=C ] [39.84 lb+ ] 50.9 lb=C 51.6°

300 gal/min 0 ft /sQ = = .6684
32.2
dm
Q
dt
ρ
 
= = = 
 
, 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 cos45
87.338 ft/s
4 1/
1.2954
12 12
p
A B
W l
v v
a b m t
θ °
= = = =
 − ∆ ∆
− 
 
87.3 ft/sv =
x components: ( ) ( ) ( ) cosA x Bm v C t m v θ∆ + ∆ = ∆
( ) ( )[ ]cos 1.2954 87.338cos45 87.338 33.137 lbx B A
m
C v v
t
θ
∆
= − = ° − = −
∆
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
m
C W v
t
θ
∆
= − = − ° = −
∆
(b) [33.137 lb=C ] [40 lb+ ] 51.9 lb=C 50.4°

Symbols: mass flow rate
dm
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/s
dm
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.
0BMΣ =
( ) ( )( )0.9 4.8 2 0W F− =
( )( )
( )( )
0.9 6000
2 4.8
F =
562.5 lb=
Calculation of .
dm
dt
mass density volume density area length= × = × ×
( ) ( )
( )B B
B B B
A v t
m A l A v t
g
γ
ρ ρ
∆
∆ = ∆ = ∆ =
B Bm dm A v
t dt g
γ∆
= =
∆
Force exerted on the slipstream: ( )B A
dm
F v v
dt
= −
Assume that Av , the speed far upstream, is negligible.
( ) 2 2
0
4
B B
B B
A v
F v D v
g g
γ γ π 
= − =  
 
( )( )( )
( ) ( )
2 2 2
2 2
4 562.5 32.24
7058.9 ft /s
6.6 0.075
B
Fg
v
Dπ γ π
= = =
84.0 ft/sB =v

Let F be the force exerted on the slipstream of one engine.
( )B A
dm
F v v
dt
= −
Calculation of .
dm
dt
mass density volume density area length= × = × ×
( ) ( )
( )B B
B B B
A v t
m A l A v t
g
γ
ρ ρ
∆
∆ = ∆ = ∆ =
2
or
4
B B
B
m A v dm
D v
t g dt g
γ γ∆ π 
= =  
∆  
Assume that ,Av the velocity far upstream, is negligible.
( ) ( ) ( )2 22 0.075
0 6.6 60 286.87 lb
4 32.2 4
B BF D v v
g
γ π π    
= − = =    
    
The force exerted by two slipstreams on the airplane is 2 .F− 2 573.74 lbF− =
Statics.
0:BMΣ =
( )0.9 4.8 2 9.3 0W F A− − − =
( )( ) ( )( )
1
0.9 6000 4.8 573.74
9.3
A  = − 
284.5 lb= 285 lb=A
0: 2 0x xF F B= − − =
2 573.74 lbxB F= − =
( )0: 284.5 6000 0y y yF A B W BΣ = + − = + − =
5715.5 lbyB =
[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 dm
F u u u u
t dt
∆
= − = −
∆
(1)
moments about B: ( ) ( ) 0A Be m u M t− ∆ + ∆ =
B A
dm
M e u
dt
= (2)
Let d be the distance that the line of action is below B.
BFd M= B A
B A
M eu
d
F u u
= =
−
(3)
Data: 90 kg/s,
dm
dt
= 600 m/s,Bu = 4 me =
(a) 480 km/h 133.333 m/sAu = =
From (1), ( )( ) 3
90 600 133.333 42 10 NF = − = × 42.0 kNF =
From (2),
( )( )
( )
4 133.333
600 133.333
d =
−
1.143 md =
(b) 960 km/h 266.67 m/sAu = =
From (1), ( )( ) 3
90 600 266.67 30 10 NF = − = × 30.0 kNF =
From (2),
( )( )
( )
4 266.67
600 266.67
d =
−
3.20 md =

The thrust on the fluid is ( )B A
dm
F 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 dm
A v
t dt
ρ
∆
= =
∆
where BA 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/s
4
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 PW=W
Statics: 0y H PF F W WΣ = − − =
44.3 15 29.3 kN.P HW 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− = ( ) 0
dm
u v D
dt
− − =
Configuration before control surface malfunction:
1
720
= 22.36 slug/s, 1860 ft/s, 560 mi/h 821.33 ft/s
32.2
dm
u 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
D
D 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 2 0
dm
u v k v
dt
− − =
( )( ) 2
2 222.36 1860 0.04131 0v v− − =
Solving for v2, 2 768.6 ft/sv = 2 524 mi/hv =

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

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

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