PROBLEM 13.1
A 400-kg satellite is placed in a circular orbit 6394 km above the surface of the earth. At this elevation the
acceleration of gravity is 4.09 m/s2. Knowing that its orbital speed is 20 000 km/h, determine the kinetic
energy of the satellite.
SOLUTION
Given: Mass of satellite,
m  400 kg
Speed of satellite,
v  20.0  103 km/h
Find: Kinetic energy, T


 1h 
v  20.0  103 km/h 
 1000 m/km 
 3600 s 
 5555 m/s
T 
2
1 2 1
mv   400 kg  5.555  103 m/s
2
2


T  6.17  109 N  m
Note: Acceleration of gravity has no effect on the mass of the satellite.
T  6.17 GJ 
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PROBLEM 13.2
A 0.5-kg stone is dropped down the “bottomless pit” at
Carlsbad Caverns and strikes the ground with a speed of
30 m/s. Neglecting air resistance, determine (a) the
kinetic energy of the stone as it strikes the ground and the
height h from which it was dropped, (b) Solve Part a
assuming that the same stone is dropped down a hole on
the moon. (Acceleration of gravity on the moon =
1.63 m/s2.)
SOLUTION
Mass of stone:
m = 0.5 kg
Initial kinetic energy:
T1 = 0
(a)
(rest)
Kinetic energy at ground strike:
T2 =
1 2 1
mv2 = (0.5)(30)2 = 225 N ⋅ m
2
2
T2 = 225 N ⋅ m W
Use work and energy:
where
T1 + U1→2 = T2
U1→2 = wh = mgh
0 + mgh =
h=
(b)
On the moon:
1 2
mv2
2
v22
(30) 2
=
2 g 2(9.81)
h = 45.9 m W
g = 1.63 m/s2
T2 = 225 N ⋅ m W
T1 and T2 will be the same, hence
h=
v22
(30)2
=
2 g 2(1.63)
h = 276 m W
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4
PROBLEM 13.3
A baseball player hits a 160 g baseball with an initial velocity of 40 m/s
at an angle of 40° with the horizontal as shown. Determine (a) the kinetic
energy of the ball immediately after it is hit, (b) the kinetic energy of the
ball when it reaches its maximum height, (c) the maximum height above
the ground reached by the ball.
SOLUTION
m = 160 g = 0.16 kg
Mass of baseball:
(a)
Kinetic energy immediately after hit.
v = v0 = 40 m/s
T1 =
(b)
1 2 1
mv = (0.16)(40) 2
2
2
T1 = 128 N ⋅ m W
Kinetic energy at maximum height:
v = v0 cos 40° = 40cos 40° = 30.642 m/s
T2 =
Principle of work and energy:
1 2 1
mv = (0.16)(30.642)2
2
2
T2 = 75.1 N ⋅ m W
T1 + U1→2 = T2
U1→2 = T2 − T1 = −52.887 N ⋅ m
Work of weight:
U1→2 = −mgd
Maximum height above impact point.
d=
(c)
T2 − T1
−52.887
=
= 33.69 m
−W
−(0.16)(9.81)
33.7 m W
Maximum height above ground:
h = 33.7 + 0.6
h = 34.3 m W
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5
PROBLEM 13.4
A 500-kg communications satellite is in a circular geosynchronous orbit and completes one revolution about
the earth in 23 h and 56 min at an altitude of 35800 km above the surface of the earth. Knowing that the radius
of the earth is 6370 km, determine the kinetic energy of the satellite.
SOLUTION
Radius of earth:
R  6370 km
Radius of orbit:
r  R  h  6370  35800  42170 km  42.170  106 m
Time one revolution:
t  23 h  56 min
t  (23 h)(3600 s/h)  (56 min)(60 s/min)  86.160  103 s
Speed:
v
2 r 2 (42.170  106 )

 3075.2 m/s
t
86.160  103
Kinetic energy:
T
1 2
mv
2
T
1
(500 kg)(3075.2 m/s)2  2.3643  109 J
2
T  2.36 GJ 
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PROBLEM 13.5
In an ore-mixing operation, a bucket full of ore is suspended
from a traveling crane which moves along a stationary bridge.
The bucket is to swing no more than 4 m horizontally when
the crane is brought to a sudden stop. Determine the maximum
allowable speed v of the crane.
SOLUTION
v1 = v
v2 = 0
1
T1 = mv 2
2
T2 = 0
U1- 2 = - mgh
d=4m
2
AB = (10 m)2 = d 2 + y 2 = ( 4 m)2 + y 2
y 2 = 100 - 16 = 84
y = 84
h = 10 - y = 10 - 84 = 0.8349 m
U1- 2 = - m(9.81)(0.8349) = - 0.8190 m
T1 + U1- 2 = T2
1 2
mv - 0.8190 m = 0
2
v 2 = (2)(0.8190) = 16.38
v = 4.05 m/s b
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7
PROBLEM 13.6
In an ore-mixing operation, a bucket full of ore is suspended from
a traveling crane which moves along a stationary bridge. The crane
is traveling at a speed of 3 m/s when it is brought to a sudden stop.
Determine the maximum horizontal distance through which the
bucket will swing.
SOLUTION
Refer to free body diagram in Problem 13.13.
v1 = v = 3 m/s
T1 =
1 2 1
mv = m(3 m)2 = 4.5 m
2
2
T2 = 0
U1- 2 = - mgh
T1 + U1- 2 = T2
4.5 m - mgh = 0
h=
4.5
= 0.4587
9.81
2
AB = (10)2 = d 2 + y 2 = d 2 + (10 - 0.4587)2
100 = d 2 + 91.04
d 2 = 8.96
d = 2.99 m b
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8
PROBLEM 13.7
Determine the maximum theoretical speed that may be achieved over a distance of 110 m by a car starting
from rest assuming there is no slipping. The coefficient of static friction between the tires and pavement is 0.75,
and 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels.
Assume (a) front-wheel drive, (b) rear-wheel drive.
SOLUTION
Let W be the weight and m the mass.
(a)
W  mg
N  0.60W  0.60 mg
Front wheel drive:
s  0.75
Maximum friction force without slipping:
F   s N  (0.75)(0.60W )  0.45mg
U12  Fd  0.45 mgd
T1  0,
Principle of work and energy:
T2 
1 2
mv2
2
T1  U12  T2
1 2
mv2
2
v22  (2)(0.45 gd )  (2)(0.45)(9.81 m/s 2 )(110 m)  971.19 m2 /s 2
0  0.45mgd 
v2  31.164 m/s
(b)
v2  112.2 km/h 
N  0.40W  0.40mg
Rear wheel drive:
s  0.75
Maximum friction force without slipping:
F   s N  (0.75)(0.40W )  0.30mg
U12  Fd  0.30mgd
T1  0,
Principle of work and energy:
T2 
1 2
mv2
2
T1  U12  T2
1
mv22
2
2
v2  (2)(0.30) gd  (2)(0.30)(9.81 m/s 2 )(110 m)  647.46 m2 /s2
0  0.30 mgd 
v2  25.445 m/s
v2  91.6 km/h 
Note: The car is treated as a particle in this problem. The weight distribution is assumed to be the same for
static and dynamic conditions. Compare with sample Problem 16.1 where the vehicle is treated as a rigid
body.
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PROBLEM 13.8
A 2000-kg
kg automobile starts from rest at point A on a 6o
incline and coasts through a distance of 150 m to point B. The
brakes are then applied, causing the automobile to come to a
stop at point C, 20 m from B.. Knowing that slipping is
impending during the braking period and neglecting air
resistance and rolling resistance, determine (a)) the speed of the
automobile at point B, (b)) the coefficient of static friction
between the tires and the road.
SOLUTION
Given:: Automobile Weight W = mg = (2000 kg) (9.81)
W  19, 620 N
Initial Velocity A,
vA  0 m/s
Incline Angle,
  6
Vehicle brakes at impending slip for 20 m from B to C
vC  0
Find;; speed of automobile at point B, vB
Coefficient of static friction, 
(a)
U A  B  WhA  B  (19620 N) (150 m)sin 6
 307.63  103 N  m
U A  B  TB  TA 
307.63  103 N  m 
1 2
mv  0
2
1
(2000
(2000kg)
kg) vB2  0
2
vB  17.54 m/s 
(b)
U A C  WhA C  Fd B C  TC  TA  0
d B C  20 m
F  N
Where   coefficient of static friction
U AC  (19620 N)(sin
N) (sin 6) (170 m)  F (20 m)
F   (19620 N) cos 6
(19620 N) (sin 6) (170 m)   (19620 N) (cos 6) (20 m)  0

170
tan 6 0.893
20
  0.893 
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PROBLEM 13.9
A package is projected up a 15 incline at A with an initial
velocity of 8 m/s. Knowing that the coefficient of kinetic
friction between the package and the incline is 0.12, determine
(a) the maximum distance d that the package will move up the
incline, (b)) the velocity of the package as it returns to its
original position.
SOLUTION
(a)
Up the plane from A to B
B:
1 2 1W
W
mv A 
(8 m/s)2  32
TB  0
2
2 g
g
U AB  (W sin15  F )d
F  k N  0.12 N
TA 
F  0 N  W cos15  0 N  W cos15
U AB  W (sin15  0.12 cos15)d  Wd (0.3747)
TA  U AB  TB : 32
W
 Wd (0.3743)  0
g
d
(b)
32
(9.81)(0.3747)
d  8.70 m 
Down the plane from B to A: (F reverses direction)
1W 2
vA
TB  0
2 g
U BA  (W sin15  F )d
TA 
d  8.71 m/s
 W (sin15  0.12cos15)(8.70 m/s)
U BA  1.245W
TB  U BA  TA
0  1.245W 
1W 2
vA
2 g
v A2  (2)(9.81)(1.245)
 24.43
v A  4.94 m/s
vA  4.94 m/s
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15 
PROBLEM 13.10
A 1.4 kg model rocket is launched vertically from rest with a constant thrust of 25 N until the rocket reaches
an altitude of 15 m and the thrust ends. Neglecting air resistance, determine (a) the speed of the rocket when
the thrust ends, (b) the maximum height reached by the rocket, (c) the speed of the rocket when it returns to
the ground.
SOLUTION
Weight:
W  mg  (1.4)(9.81)  13.734 N
(a)
T1  0
First stage:
U12  (25  13.734)(15)  169.0 N  m
T1  U12  T2
T2 
1 2
mv  U12  169.0 N  m
2
2U1 2

m
v2 
(b)
(2) (169.0)
1.4
v2  15.54 m/s 
Unpowered flight to maximum height h:
T2  169.0 N  m
T3  0
U 23  W (h  15)
T2  U 23  T3
W (h  15)  T2
h  15 
(c)
T2 169.0


W 13.734
h  27.3 m 
Falling from maximum height:
T3  0
T4 
U 34  Wh  mgh
T3  U 34  T4 :
1 2
mv4
2
0  mgh 
1 2
mv4
2
v42  2 gh  (2)(9.81 m/s2 )(27.3 m)  535.6 m2 /s 2
v4  23.1 m/s 
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PROBLEM 13.11
Packages are thrown down an incline at A with a velocity of 1
m/s. The packages slide along the surface ABC to a conveyor
belt which moves with a velocity of 2 m/s. Knowing that
k  0.25 between the packages and the surface
ABC,determine the distance d if the packages are to arrive at C
with a velocity of 2 m/s.
SOLUTION
On incline AB:
N AB  mg cos 30
FAB  k N AB  0.25 mg cos 30
U A B  mgd sin 30  FAB d
 mgd (sin 30   k cos30)
On level surface BC:
N BC  mg
xBC  7 m
FBC  k mg
U B C    k mg xBC
At A,
TA 
1 2
mvA and vA  1 m/s
2
At C,
TC 
1 2
mvC
2
and vC  2 m/s
Assume that no energy is lost at the corner B.
TA  U AB  U B C  TC
Work and energy.
1 2
1
mvA  mgd (sin 30  k cos30)  k mg xBC  mv02
2
2
Dividing by m and solving for dd,
 vC2 /2 g   k xBC  v A2 /2 g 

d
(sin 30  k cos 30)

(2) 2/(2)(9.81)  (0.25)(7)  (1)2/(2)(9.81)
sin 30  0.25 cos 30
d  6.71 m 
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PROBLEM 13.12
Packages are thrown down an incline at A with a velocity of
1 m/s. The packages slide along the surface ABC to a conveyor
belt which moves with a velocity of 2 m/s. Knowing that
d  7.5 m and k  0.25 between the packages and all
surfaces, determine (a)the
)the speed of the package at C, (b)the
distance a package will slide on the conveyor belt before it
comes to rest relative to the belt.
SOLUTION
(a)
On incline AB:
N AB  mg cos 30
FAB  k N AB  0.25 mg cos 30
UA B  mgd sin 30  FAB d
 mgd (sin 30  k cos 30)
On level surface BC:
N BC  mg
xBC  7 m
FBC  k mg
U B C   k mg xBC
At A,
1
TA  mvA2 and vA  1 m/s
2
At C,
1
TC  mvC2
2
and vC  2 m/s
Assume that no energy is lost at the corner B.
Work and energy.
TA  U AB  U B C  TC
1 2
1
mvA  mgd (sin 30  k cos30)  k mg xBC  mv02
2
2
Solving for vC2 ,
vC2  vA2  2 gd (sin 30  k cos30)  2 k g xBC
 (1)2  (2)(9.81)(7.5)(sin 30  0.25cos30)  (2)(0.2
(2)(0.25)(9.81)(7)
5)(9.81)(7)
 8.3811 m2/s2
(b)
vC  2.90 m/s 
Box on belt: Let xbelt be the distance moves by a package as it slides on the belt.
Fy  ma y
N  mg  0
N  mg
Fx  k N  k mg
At the end of sliding,
v  vbelt  2 m/s
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PROBLEM 13.12 (Continued)
Principle of work and energy:
1 2
1 2
mvC  k mg xbelt  mvbelt
2
2
2
v 2  vbelt
xbelt  C
2 k g

8.3811  (2)2
(2)(0.25)(9.81)
xbelt  0.893 m 
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PROBLEM 13.13
Boxes are transported by a conveyor belt with a velocity v0 to a
fixed incline at A, where they slide and eventually fall off at B.
Knowing that m k = 0.40, determine the velocity of the conveyor belt
if the boxes leave the incline at B with a velocity of 2.5 m/s.
SOLUTION
1
TA = mv02
2
1
1
TB = mv B2 = m (2.5 m/s)2
2
2
TB = 3.125 m
U A- B = (W sin 15∞ - m k N )(6 m)
 Â F = 0 N - W cos 15∞ = 0
N = W cos 15∞
U A- B = W (sin 15∞ - 0.40 cos15∞)(6 m)
U A- B = - 0.76531 mg
TA + U A- B = TB
1 2
mv0 - 0.76531 mg = 3.125 m
2
v02 = (2)(3.125 + 0.76531 g)
v02 = 21.265 m2 / s2
v 0 = 4.61 m/s
15∞ b
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16
PROBLEM 13.14
Boxes are transported by a conveyor belt with a velocity v0 to a
fixed incline at A, where they slide and eventually fall off at B.
Knowing that m k = 0.40, determine the velocity of the conveyor belt
if the boxes are to have zero velocity at B.
SOLUTION
1 2
mv0 TB = 0
2
U A- B = (W sin 15∞ - m k N )(6 m)
TA =
 Â F = 0 N - cos 15∞ = 0
N = W cos 15∞
U A- B = W (sin 15∞ - 0.40 cos 15∞)(6 m)
U A- B = - 0.76531 mg
TA + U A- B = TB
1 2
mv0 - 0.76531 mg = 0
2
v02 = 2(0.76531)(9.81)
v02 = 15.0154 m2 / s2
v 0 = 3.87 m/s
15° b
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17
PROBLEM 13.15
A 1200
1200-kg trailer is hitched to a 1400-kg
kg car. The car and trailer are
traveling at 72 km/h when the driver applies the brakes on both the car
and the trailer. Knowing that the braking forces exerted on the car and
the trailer are 5000 N and 4000 N, respectively, determine (a)
( the
distance traveled by the car and trailer before they come to a stop,
(b)) the horizontal component of the force exerted by the trailer hitch
on the car.
SOLUTION
Let position 1 be the initial state at velocity v1  72 km/h  20 m/s and position 2 be at the end of braking
(v2  0). The braking forces and FC  5000 N for the car and 4000 N for the trailer.
(a)
Car and trailer system.
( d  braking distance)
1
(mC  mT )v12
2
U12  ( FC  FT )d
T1 
T2  0
T1  U12  T2
1
(mC  mT )v12  ( FC  FT )d  0
2
d
(b)
(mC  mT )v12 (2600)(20)2

 57.778
2( FC  FT )
(2)(9000)
d  57.8 m 
Car considered separately.
Let H be the horizontal pushing force that the trailer exerts on the car through the hitch.
1
mC v12
2
U12  ( H  FC ) d
T1 
T2  0
T1  U12  T2
1
mC v12  (H  FC )d  0
2
H  FC 
mC v12
(1400)(20) 2
 5000 
2d
(2)(57.778)
H  154 N
Trailer hitch force on car:
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
PROBLEM 13.16
A trailer truck enters a 2 percent uphill grade traveling at 72 km/h and reaches a speed of 108 km/h in 300 m.
The cab has a mass of 1800 kg and the trailer 5400 kg. Determine (a)the average force at the wheels of the
cab, (b)the average force in the coupling between the cab and the trailer.
SOLUTION
Initial speed:
v1  72 km/h  20 m/s
Final speed:
v2  108 km/h  30 m/s
Vertical rise:
h  (0.02)(300)  6.00 m
Distance traveled:
d  300 m
(a)
Traction force. Use cab and trailer as a free body.
m  1800  5400  7200 kg
T1  U12  T2
Work and energy:
Ft 
W  mg  (7200)(9.81)  70.632 103 N
1 2
1
mv1  Wh  Ft d  mv22
2
2
1 1 2
1 1

1

mv1  Wh  mv12  
(7200)(30) 2  (70.632  103 )(6.00)  (7200)(20)2 

d  2
300
2
2



 7.4126  103 N
(b)
Ft  7.41 kN 
Coupling force Fc . Use the trailer alone as a free body.
W  mg  (5400)(9.81)  52.974 103 N
m  5400 kg
Assume that the tangential force at the trailer wheels is zero.
Work and energy:
T1  U12  T2
1 2
1
mv1  Wh  Fc d  mv22
2
2
The plus sign before Fc means that we have assumed that the coupling is in tension.
Fc 
1 1 2
1
1 1

1

mv2  Wh  mv12  
(5400)(30)2  (52.974  103 )(6.00)  (5400)(20)2 

d  2
2
300
2
2



 5.5595  103 N
Fc  5.56 kN (tension) 
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PROBLEM 13.17
The subway train shown is traveling at a speed of 50 km/h
when the brakes are fully applied on the wheels of cars B and
C, causing them to slide on the track, but are not applied on
the wheels of car A. Knowing that the coefficient of kinetic
friction is 0.35 between the wheels and the track, determine
(a) the distance required to bring the train to a stop, (b) the
force in each coupling.
SOLUTION
m k = 0.35 FB = (0.35)(50, 000)(9.81) = 171.675 kN
FC = (0.35)( 40, 000)(9.81) = 137.34 kN
125
m/s ¨
9
T1 + U1- 2 = T2
v1 = 50 km/h =
(a)
Entire train:
v2 = 0 T2 = 0
2
1
Ê 125 ˆ
- (171675 + 137340) x = 0
( 40, 000 + 50, 000 + 40, 000) Á
Ë 9 ˜¯
2
x = 40.576 m
(b)
x = 40.576 m b
Force in each coupling: Recall that x = 40.576 m
Car A: Assume FAB to be in tension.
T1 + V1- 2 = T2
2
1
Ê 125 ˆ
- FAB ( 40.576) = 0
( 40, 000) Á
Ë 9 ˜¯
2
FAB = + 95081.4 N
FAB = 95.1 kN (tension) b
Car C:
T1 + U1- 2 = T2
2
1
Ê 125 ˆ
+ ( FBC - 137340)( 40.576) = 0
( 40, 000) Á
Ë 9 ˜¯
2
FBC - 137340 = - 95081.4
FBC = + 42258.6 N
FBC = 42.3 kN (tension ) b
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20
PROBLEM 13.18
Solve Problem 13.17, assuming that the brakes are applied only
on the wheels of car A.
PROBLEM 13.17 The subway train shown is traveling at
a speed of 50 km/h when the brakes are fully applied on the
wheels of cars B and C, causing them to slide on the track,
but are not applied on the wheels of car A. Knowing that the
coefficient of kinetic friction is 0.35 between the wheels and
the track, determine (a) the distance required to bring the train
to a stop, (b) the force in each coupling.
SOLUTION
(a)
Entire train:
FA = m N A = (0.35)( 40, 000)(9.81) = 137, 340 N
125
v1 = 50 km/h =
m/s
9
T1 + v1- 2 = T2
v2 = 0 T2 = 0
2
1
Ê 125 ˆ
- (137, 340) x = 0
( 40, 000 + 50, 000 + 40, 000) Á
Ë 9 ˜¯
2
x = 91.296 m
(b)
x = 91.3 m b
Force in each coupling:
Car A: Assume FAB to be in tension.
T1 + v1- 2 = T2
2
1
Ê 125 ˆ
- (137340 + FAB )(91.296) = 0
( 40, 000) Á
Ë 9 ˜¯
2
137340 + FAB = 42, 258.4
FAB = - 95, 081.6 N
Car C:
FAB = 95.1 kN (compression) b
T1 + v1- 2 = T2
2
1
Ê 125 ˆ
+ FBC (91.296) = 0
40, 000 Á
Ë 9 ˜¯
2
FBC = - 42, 258.4 N
FBC = 42.3 kN (compression) b
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21
PROBLEM 13.19
Blocks A and B have masses of 11 kg and 5 kg, respectively, and they are both
at a height h = 2 m above the ground when the system is released from rest. Just
before hitting the ground, Block A is moving at a speed of 3 m/s. Determine
(a) the amount of energy dissipated in friction by the pulley, (b) the tension in
each portion of the cord during the motion.
SOLUTION
Energy dissipated.
v1 = 0 T1 = 0
v2 = v A = 3 m / s = v B
1
T2 = ( mA + mB ) v22
2
Ê 16 ˆ
T2 = Á kg˜ (3 m/s)2 = 72 J
¯
Ë 2
(a)
U1- 2 = mA g (2) - mB g (2) - E p
U1- 2 = (6 kg)(9.81 m/s2 )(2 m) - E p
U1- 2 = 117.72 - E p
T1 + U1- 2 = T2
0 + 117.72 - E p = 72
E p = 117.72 - 72
(b)
E p = 45.7 J b
Block A:
T1 = 0 T2 =
1
Ê 11 ˆ
mA v22 = Á kg˜ (3 m/s)2 = 49.5 J
Ë2 ¯
2
U1- 2 = ( mA g - TA)(2) = [(11 kg)(9.81 m/s2 ) - TA ] [2 m]
U1- 2 = 215.82 - 2TA
T1 + U1- 2 = T2
0 + 215.82 - 2TA = 49.5
Block B:
T1 = 0 T2 =
T1 + U1- 2 = T2
TA = 83.2 N b
1
Ê5 ˆ
mB v22 = Á kg˜ (3 m/s)2 = 22.5 J
Ë2 ¯
2
U1- 2 = - mB g (2) + TB (2) = - (5 kg) (9.81 m/s2 ) (2 m) + 2TB
U1- 2 = - 98.1 + 2TB
0 - 98.1 + 2TB = 22.5
TB = 60.3 N b
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22
PROBLEM 13.20
The system shown is at rest when a constant 150-N force is applied
to collar B. (a) If the force acts through the entire motion, determine
the speed of collar B as it strikes the support at C. (b) After what
distance d should the 150-N force be removed if the collar is to
reach support C with zero velocity?
SOLUTION
Kinematics:
mA = 3 kg
(a)
v A = 2v B
W A = mA g = (3 kg) (9.81 m/s2 )
W A = 29.43 N
150–N force acts through entire 0.6 m motion of B.
U1- 2 = (150 N )(0.6 m) - (29.43 N ) (1.2 m) = 54.68 J
1
1
T1 = 0 T2 = mA ( v A )22 + mB ( v B )22
2
2
1
1
= (3 kg) ÈÎ2( v B )22 ˘˚ + (8 kg) ( v B )22 = 10( v B )22
2
2
T1 + U1- 2 = T2 :
0 + 54.68 = 10( v B )22
( v B )2 = 2.338 m/s
(b)
vB = 2.34 m/s
b
Initial and final velocities are zero.
T1 = 0
T2 = 0
Remove 150-N force after B moves distance d.
U1- 2 = (150 N )d - (29.43 N ) (1.2 m)
= 150d - 35.316 J
T1 + U1- 2 = T2 : 0 + 150d - 31.316 = 0
d = 0.2354 m
d = 235 mm b
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23
PROBLEM 13.21
Car B is towing car A at a constant speed of 10 m/s on an uphill
grade when the brakes of car A are fully applied causing all four
wheels to skid. The driver of car B does not change the throttle
setting or change gears. The masses of the cars A and B are
1400 kg and 1200 kg, respectively, and the coefficient of
kinetic friction is 0.8. Neglecting air resistance and rolling
resistance, determine (a) the distance traveled by the cars
before they come to a stop, (b) the tension in the cable.
SOLUTION
Given: Car B tows car A at 10 m/s uphill.
k  0.8
Car A brakes so 4 wheels skid.
Car B continues in same gear and throttle setting.
Find: (a) Distance d, traveled to stop
(b) Tension in cable
(a)
F1  traction force (from equilibrium)
F1  (1400 g )sin 5  (1200 g )sin 5
F  0.8N A
 2600(9.81)sin 5
For system: A  B
U1 2  [( F1  1400 g sin 5  1200 g sin 5)  F ]d
 T2  T1  0 
Since
1
1
mA  Bv2   (2600)(10)2
2
2
( F1  1400 g sin 5  1200g sin 5)  0
 Fd  0.8[1400(9.81) cos 5]d  130, 000 N  m
d  11.88 m 
(b)
Cable tension, T
U1 2  [T  0.8N A ](11.88)  T2  T1
(T  0.8(1400)(9.81) cos5)11.88  
1400
(10)2
2
(T  10945)  5892
 5.053 kN
T  5.05 kN 
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PROBLEM 13.22
The system shown is at rest when a constant 250-N force is applied to
block A. Neglecting the masses of the pulleys and the effect of friction
in the pulleys and between block A and the horizontal surface,
determine (a) the velocity of block B after block A has moved 2 m, (b)
the tension in the cable.
SOLUTION
Constraint of cable:
x A  3 yB  constant
x A  3yB  0
v A  3vB  0
Let F be the tension in the cable.
Block A:
mA  30 kg, P  250 N, (T1 ) A  0
(T1 ) A  (U12 ) A  (T2 ) A
1
mA v A2
2
1
0  (250  F )(2)  (30)(3vB )2
2
0  ( P  F )(x A ) 
500  2F  135vB2
Block B:
(1)
mB  25 kg, WB  mB g  245.25 N
(T1 ) B  (U12 ) B  (T2 ) B
1
mB vB2
2
2 1
(3F )  245.25)    (25) vB2
3 2
0  (3F  WB )(yB ) 
2F  163.5  12.5 vB2
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(2)
PROBLEM 13.22 (Continued)
Add Eqs. (1) and (2) to eliminate F.
500  163.5  147.5vB2
vB2  2.2814 m 2 /s2
(a)
Velocity of B.
(b)
Tension in the cable.
From Eq. (2),
v B  1.510 m/s
2 F  163.5  (12.5)(2.2814)
F  96.0 N 
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
PROBLEM 13.23
The system shown is at rest when a constant 250
250-N
N force is applied to
block A.. Neglecting the masses of the pulleys and the effect of friction
in the pulleys and assuming that the coefficients of friction between
block A and the horizontal surface are  s  0.25 and k  0.20,
determine ((a) the velocity of block B after block A has moved 2 m, (b)
(
the tension in the cable
cable.
SOLUTION
Check the equilibrium position to see if the blocks move. Let F be the tension in the cable.
Block B:
3 F  mB g  0
F
Block A:
mB g (25)(9.81)

 81.75 N
3
3
Fy  0:
N A  mA g  0
N A  mA g  (30)(9.81)  294.3 N
Fx  0: 250  FA  F  0
FA  250  81.75  168.25 N
s N A  (0.25)(294.3)  73.57 N
Available static friction force:
Since FA  s N A , the blocks move.
The friction force, FA, during sliding is
FA  k N A  (0.20)(294.3)  58.86 N
Constraint of cable:
x A  3 yB  constant
 x A  3yB  0
v A  3vB  0
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PROBLEM 13.23 (Continued)
Block A:
m A  30 kg, P  250 N, (T1 ) A  0.
(T1 ) A  (U1 2 ) A  (T2 ) A
1
mA v A2
2
1
0  (250  58.86  F )(2)  (30)(3vB ) 2
2
0  ( P  FA  F )( x A ) 
382.28  2F  135vB2
(1)
M B  25 kg, WB  mB g  245.25 N
Block B:
(T1 ) B  (U12 ) B  (T2 ) B
1
mB vB2
2
2 1
(3F  245.25)    (25)vB2
3 2
0  (3F  WB )(yB ) 
2F  163.5  12.5vB2
(2)
Add Eqs. (1) and (2) to eliminate F.
382.28  163.5  147.5vB2
vB2  1.48325 m 2 /s2
v B  1.218 m/s
(a)
Velocity of B:
(b)
Tension in the cable:
From Eq. (2),

F  91.0 N 
2 F  163.5  (12.5)(1.48325)
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PROBLEM 13.24
Two blocks A and B,, of mass 4 kg and 5 kg, respectively, are connected
by a cord which passes over pulleys as shown. A 3 kg collar C is placed
on block A and the system is released from rest. After the blocks have
moved 0.9 m, collar C is removed and blocks A and B continue to move.
Determine the speed of block A just before it strikes the ground.
SOLUTION
Position  to Position .
v1  0
T1  0
At  before C is removed from the system
1
1
( mA  mB  mC )v22  (12 kg)v22  6v22
2
2
U1 2  (m A  mC  mB ) g (0.9 m)
T2 
U1 2  (4  3  5)( g )(0.9 m)  (2 kg)(9.81 m/s 2 )(0.9 m)
U1 2  17.658 J
T1  U1 2  T2 :
0  17.658  6v22
v22  2.943
At Position , collar C is removed from the system.
Position  to Position .
T2 
1
9 
(mA  mB )v22   kg  (2.943)  13.244 J
2
2 
1
9
T3  (mA  mB )(v3 )2  v32
2
2
U 23  ( mA  mB )( g )(0.7 m)  ( 1 kg)(9.81 m/s 2 )(0.7 m)  6.867 J
T2  U 2 3  T3
13.244  6.867  4.5v32
v32  1.417
vA  v3  1.190 m/s
vA  1.190 m/s 
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PROBLEM 13.25
Four 3-kg packages are held in place by friction on a conveyor
which is disengaged from its drive motor. When the system is
released from rest, package 1 leaves the belt at A just as
package 4 comes onto the inclined portion of the belt at B.
Determine (a) the velocity of package 2 as it leaves the belt at
A, (b) the velocity of package 3 as it leaves the belt at A.
Neglect the mass of the belt and rollers.
SOLUTION
Given: Conveyor is disengaged, packages held by friction and system is released from rest. Neglect mass of
belt and rollers. Package 1 leaves the belt as package 4 comes onto the belt.
Find:(a)Velocity of package 2 as it leaves the belt at A.
(b) Velocity of package 3 as it leaves the belt at A.
(a) Package 1 falls off the belt, and 2, 3, 4 move down.
2.4
 0.8 m
3
1

T2  3  mv22 
2


T2 
3
 3 kg  v22
2
T2  4.5v22
U1 2   3 W  0.8    3 3 kg   9.81 m/s 2  0.8 
U1 2  70.632 J
T1  U1 2  T2
0  70.632  4.5v22
v22  15.696
v2  3.9618
v2  3.96 m/s 
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PROBLEM 13.25 (Continued)
(b) Package 2 falls off the belt and its energy is lost to the system, and 3 and 4 move down 2 ft.
1

T2   2   mv22 
2


T2   3 kg 15.696 
T2  47.088 J 
 
1

T3   2   mv32    3 kg  v32
2

T3  3v32


U 2  3  2 W  0.8    2  3 kg  9.81 m/s 2  0.8 m 
U 2  3  47.088 J
T2  U 2  3  T3  47.088  47.088  3v32
v32  31.392
v3  5.6029
v3  5.60 m/s 
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PROBLEM 13.26
A 3-kg
kg block rests on top of a 22-kg
kg block supported by, but not attached to, a spring of
constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum
speed reached by the 22-kg block, (b)) the maximum height reached by the 2-kg
2
block.
SOLUTION
Call blocks A and B.
(a)
mA  2 kg, mB  3 kg
Position 1: Block B has just been removed.
Spring force:
FS  (mA  mB ) g  k x
Spring stretch:
x1  
(mA  mB ) g
(5 kg)(9.81 m/s 2 )
 1.22625 m

k
40 N/m
Let position 2 be a later position while the spring still contacts block A.
Work of the force exerted by the spring:
(U12 )e  
x2
 k x dx
x1
x
2
1
1
1
  k x 2  k x12  k x22
2
2
2
x1

Work of the gravitational force:
1
1
(40)( 1.22625) 2  (40) x22  30.074  20 x22
2
2
(U1 2 ) g   m A g ( x2  x1 )
  (2)(9.81)( x2  1.22625)  19.62 x2  24.059
Total work:
Kinetic energies:
U12  20 x22  19.62 x2  6.015
T1  0
T2 
Principle of work and energy:
1
1
m A v22  (2)v22  v22
2
2
T1  U12  T2
0  20 x22  19.62 x2  6.015  v22
Speed squared:
v22  20x22  19.62x2  6.015
At maximum speed,
dv2
0
dx2
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(1)
PROBLEM 13.26 (Continued)
Differentiating Eq. (1), and setting equal to zero,
2v2
Substituting into Eq. (1),
dv2
 40 x2  19.62  0
dx
19.62
x2  
 0.4905 m
40
v22  (20)(0.4905)2  (19.62)(0.4905)  6.015  10.827 m2 /s2
v2  3.29 m/s 
Maximum speed:
(b)
Position 3: Block A reaches maximum height. Assume that the block has separated from the spring.
Spring force is zero at separation.
Work of the force exerted by the spring:
(U13 )e  
0
1
1
 kxdx  2 kx  2 (40)(1.22625)  30.074 J
x1
2
1
2
Work of the gravitational force:
(U13 ) g  mA gh  (2)(9.81)h  19.62 h
Total work:
U13  30.074  19.62 h
At maximum height,
v3  0, T3  0
Principle of work and energy:
T1  U13  T3
0  30.074  19.62 h  0
h  1.533 m 
Maximum height:
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PROBLEM 13.27
Solve Problem 13.26, assuming that the 2-kg block is attached to the spring.
PROBLEM 13.26 A 3-kg block rests on top of a 2-kg block supported by, but not
attached to, a spring of constant 40 N/m. The upper block is suddenly removed.
Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height
reached by the 2-kg block.
SOLUTION
Call blocks A and B.
(a)
mA  2 kg, mB  3 kg
Position 1: Block B has just been removed.
Spring force:
FS  (mA  mB ) g  kx1
Spring stretch:
x1  
(m A  mB ) g
(5 kg)(9.81 m/s 2 )

 1.22625 m
k
40 N/m
Let position 2 be a later position. Note that the spring remains attached to block A.
Work of the force exerted by the spring:
(U12 )e  
x2
 kxdx
x1
x
2
1
1
1
  kx 2  kx12  kx22
2
2
2
x1

Work of the gravitational force:
1
1
(40)(1.22625) 2  (40) x22  30.074  20 x22
2
2
(U1 2 ) g   m A g ( x2  x1 )
 (2)(9.81)( x2  1.22625)  19.62 x2  24.059
Total work:
Kinetic energies:
U12  20 x22  19.62 x2  6.015
T1  0
T2 
Principle of work and energy:
1
1
m A v22  (2)v22  v22
2
2
T1  U12  T2
0  20 x22  19.62 x2  6.015  v22
Speed squared:
v22  20x22  19.62x2  6.015
At maximum speed,
dv2
0
dx2
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(1)
PROBLEM 13.27 (Continued)
Differentiating Eq. (1) and setting equal to zero,
2v2
dv2
 40 x2  19.62  0
dx2
x2  
Substituting into Eq. (1),
19.62
 0.4905 m
40
v22  (20)(0.4905)2  (19.62)(0.4905)  6.015  10.827 m1 /s2
v2  3.29 m/s 
Maximum speed:
(b)
Maximum height occurs when v2 0.
Substituting into Eq. (1),
0  20 x22  19.62 x2  6.015
Solving the quadratic equation
x2  1.22625 m and 0.24525 m
Using the larger value,
x2  0.24525 m
Maximum height:
h  x2  x1  0.24525  1.22625
h  1.472 m 
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35
PROBLEM 13.28
A 4 kg collar C slides on a horizontal rod between springs A and B. If
the collar is pushed to the right until spring B is compressed 50 mm
and released, determine the distance through which the collar will
travel, assuming (a) no friction between the collar and the rod, (b) a
coefficient of friction m k = 0.35 .
SOLUTION
k B = 2400 N/m
k A = 3600 N/m
(a)
Since the collar C leaves the spring at B and there is no friction, it must engage the spring at A.
TA = 0
U A- B = Ú
TB = 0
0.05
0
y
k B xdx - Ú k A ¥ dx
0
Ê 2400 ˆ
Ê 3600 ˆ 2
y
(0.05)2 - Á
U A- B = Á
˜
Ë 2 ¯
Ë 2 ˜¯
TA + U A- B = TB
0 + 3 - 1800 y 2 = 0
1
y2 =
fi y = 0.04082 m = 40.82 mm
600
Total distance
d = 50 + 400 + (150 - 40.82) mm
d = 559 mm b
(b)
Assume that C does not reach the spring at B because of friction.
N =W = 4 g
F f = (0.35)( 4 ¥ 9.81) = 13.734 N
TA = TD = 0
U A- D = Ú
0.05
0
2400 xdx - F f ( y ) = 3 - 13.734 y
TA + U A- D = TD 0 + 3 - 13.734 y = 0
y = 0.2184 m = 218.4 mm
The collar must travel 400 - 150 + 50 = 300 mm before it engages the spring at B. Since y = 218.4 mm
it stops before engaging the spring at B.
d = 218 mm b
Total distance
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36
PROBLEM 13.29
A 3.4 kg collar is released from rest in the position shown, slides down
the inclined rod, and compresses the spring. The direction of motion is
reversed and the collar slides up the rod. Knowing that the maximum
deflection of the spring is 12.7 cm, determine (a) the coefficient of kinetic
friction between the collar and the rod, (b) the maximum speed of the
collar.
SOLUTION
Position 1, initial condition
Position 2, spring deflected 5 inches
Position 3, initial contact of spring with collar
2
 45.7 + 12.7  1
 45.7 + 12.7 
 12.7 
U1 → 2 = − F 
 sin 30°
 + 33.35 
 − (876) 

100
2
100
100
(Friction)
(Spring)
(Gravity)
T1 = T2 = 0, ∴ U1 − 2 = 0
2
 58.4  1 (
 12.7 
 58.4 
− 876) 
+ 33.35 
0 = − µ (33.35) (0.866) 
(0.5)


 100  2
 100 
 100 
µ = 0.1590 W
(a)
(b) Max speed occurs just before contact with the spring
1
 45.7 
 45.7 
2
+ 33.35 
U1 → 2 = − µ (33.35) (0.866) 
(0.5) = T3 = (3.4) vmax
 100 
 100 
2
vmax = 1.8 m/s W
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PROBLEM 13.30
A 10
10-kg block is attached to spring A and connected to spring B by a cord and
pulley. The block is held in the position shown with both springs unstretched when
the support is removed and the block is released with no initial velocity. Knowing
that the constant of each spring is 2 kN/m, determine ((a)) the velocity of the
t block
after it has moved down 50 mm, ((b)) the maximum velocity achieved by the block.
SOLUTION
(a)
W weight of the block  10 (9.81)  98.1 N
xB 
1
xA
2
1
1
k A ( x A ) 2  k B ( xB ) 2
2
2
(Gravity) (Spring A) (Spring B)
U1 2  W ( x A ) 
U1  2  (98.1 N)(0.05 m) 

1
(2000 N/m)(0.05 m)2
2
1
(2000 N/m)(0.025 m)2
2
U1 2 
1
1
(m)v 2  (10 kg)v2
2
2
4.905  2.5  0.625 
1
(10)v
(10) 2
2
v  0.597 m/s 
(b)
Let x  distance moved down by the 10 kg block
U1  2  W ( x ) 
2
1
1  x
1
k A ( x) 2  k B    (m)v 2
2
2 2
2
d 1
k

(m)v 2   0  W  k A ( x)  B (2 x)
dx  2
8

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PROBLEM 13.30 (Continued)
0  98.1  2000 ( x) 
2000
(2x)  98.1  (2000  250) x
8
x  0.0436 m (43.6 mm)
For
x  0.0436, U  4.2772  1.9010  0.4752 
1
(10)v 2
2
vmax  0.6166 m/s
vmax  0.617 m/s 
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PROBLEM 13.31
A 5-kg collar A is at rest on top of, but not attached to, a spring
with stiffness k1 400 N/m; when a constant 150-N force is
applied to the cable. Knowing A has a speed of 1 m/s when the
upper spring is compressed 75 mm, determine the spring
stiffness k2. Ignore friction and the mass of the pulley.
SOLUTION
Use the method of work and energy applied to the collar A.
T1  U12  T2
Since collar is initially at rest,
T1  0.
In position 2, where the upper spring is compressed 75 mm and v2  1.00 m/s, the kinetic energy is
T2 
1 2 1
mv2  (5 kg)(1.00 m/s)2  2.5 J
2
2
As the collar is raised from level A to level B, the work of the weight force is
(U12 ) g  mgh
where m  5 kg, g  9.81 m/s2 and h  450 mm  0.450 m
Thus, (U12 ) g  (5)(9.81)(0.450)  22.0725 J
In position 1, the force exerted by the lower spring is equal to the weight of collar A.
F1  mg  (5 kg)(9.81 m/s)  49.05 N
As the collar moves up a distance x1, the spring force is
F  F1  k1 x2
until the collar separates from the spring at
xf 
F1 49.05 N

 0.122625 m  122.625 mm
k1 400 N/m
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PROBLEM 13.31 (Continued)
Work of the force exerted by the lower spring:
(U12 )1 
xf
 ( F  k x)dx
0
1
 F1 x f 

1
1 2
1
1
kx f  k1 x 2f  k1 x 2f  k1 x 2f
2
2
2
1
(400 N/m)(0.122625) 2  3.0074 J
2
In position 2, the upper spring is compressed by y  75 mm  0.075 m. The work of the force exerted
by this spring is
1
1
(U12 )2   k2 y 2   k2 (0.075)2  0.0028125 k 2
2
2
Finally, we must calculate the work of the 150 N force applied to the cable. In position 1, the length AB is
(l AB )1  (450)2  (400)2  602.08 mm
In position 2, the length AB is (l AB )2  400 mm.
The displacement d of the 150 N force is
d  (l AB )1  (l AB )2  202.08 mm  0.20208 m
The work of the 150 N force P is
(U12 ) P  Pd  (150 N)(0.20208 m)  30.312 J
Total work:
U1 2  22.0725  3.0074  0.0028125k 2  30.312
 11.247  0.0028125k2
Principle of work and energy:
T1  U12  T2
0  11.247  0.0028125k2  2.5
k2  3110 N/m
k2  3110 N/m 
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PROBLEM 13.32
A piston of mass m and cross-sectional
sectional area A is equilibrium
under the pressure p at the center of a cylinder closed at both
ends. Assuming that the piston is moved to the left a distance
a/2
/2 and released, and knowing that the pressure on each side of
the piston varies inversely with the volume, determine the
velocity of the piston as it again reaches the center of the
cylinder. Neglect friction between the piston and the cylinder
and express your answer in terms of m,, a, p, and A.
SOLUTION
Pressures vary inversely as the volume
pL Aa

P
Ax
pR
Aa

P
A(2a  x)
Initially at ,
v0
x
T1  0
At ,
pR 
pa
x
pa
(2a  x)
a
2
x  a, T2 
U1 2 
pL 
1 2
mv
2
a
a
1
1

 ( p  p ) Adx   paA  x  2a  x  dx
a/2
L
R
a/2
U1 2  paA[ln x  ln (2a  x)]aa/2

a
 3a  
U1 2  paA ln a  ln a  ln    ln   
2
 2 


3a 2 
4
U1 2  paA ln a 2  ln
  paA ln  
4 
3

4 1
T1  U12  T2 0  paA ln    mv 2
3 2
v2 
2 paA ln  43 
m
 0.5754
paA
m
v  0.759
paA

m
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PROBLEM 13.33
An uncontrolled automobile traveling at 100 km/h strikes squarely a highway crash cushion of the type shown
in which the automobile is brought to rest by successively crushing steel barrels. The magnitude F of the force
required to crush the barrels is shown as a function of the distance x the automobile has moved into the cushion.
Knowing that the weight of the automobile is 1000 kg and neglecting the effect of friction, determine (a) the
distance the automobile will move into the cushion before it comes to rest, (b) the maximum deceleration of
the automobile.
SOLUTION
(a)
250
m/s
9
Assume auto stops in 1.5 m d 4 m.
100 km/h =
v1 =
250
m/s
9
1 2 1
Ê 250 ˆ
mv1 = (1000) Á
Ë 9 ˜¯
2
2
T1 = 385, 802.5 J
= 385.8025 kJ
v2 = 0
T2 = 0
2
T1 =
U1- 2 = - {(80 kN )(1.5 m) + (120 kN )( d - 1.5)}
= - (120 + 120 d - 180)
= - (120d - 60) kN ◊ m
T1 + U1- 2 = T2
385.8025 = 120d - 60
d = 3.715 m
d = 3.72 m b
Assumption that d 4 m is ok.
(b)
Maximum deceleration occurs when F is largest. For d = 3.715 m, F = 120 kN. Thus, F = maD
120, 000 = 1000 aD
aD = 120 m/s2 b
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43
PROBLEM 13.34
A 300-g brass (nonmagnetic) block A and a 200-g steel magnet B are in equilibrium
in a brass tube under the magnetic repelling force of another steel magnet C located
at a distance x = 4 mm from B. The force is inversely proportional to the square of
the distance between B and C. If block A is suddenly removed, determine (a) the
maximum velocity of B, (b) the maximum acceleration of B. Assume that air
resistance and friction are negligible.
SOLUTION
(a)
Calculating K.
 F = ( mA + mB ) g - k / ( 4 ¥ 10 -3 m)2
k = ( 4 ¥ 10 -3 m)2 (0.5 kg)( g m/s2 )
k = 8 ¥ 10 -6 g N ◊ m
v1 = 0 T1 = 0
v2 = v
1
T2 = mB v 2 = 0.1v 2 N ◊ m
2
U1-2 = Ú

2

1
U1-2 = Ú
( F - mB g ) dx
Ê 8 ¥ 10 -6 g
ˆ
- 0.2 g ˜ dx
2
4 ¥ 10 Á
x
Ë
¯
x
T1 + U1-2 = T2
-3
0+Ú
Ê 8 ¥ 10 -6 g
ˆ
- 0.2 g ˜ dx = 0.1v 2
Á
2
4 ¥ 10 Ë
x
¯
x
-3
d (0.1v 2 )
=0
dx
For maximum v,
Thus,
8 ¥ 10 -6 g
x2
- 0.2 g = 0
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44
PROBLEM 13.34 (continued)
x = 0.00632 m
At vmax ,
0+Ú
0.00632 Ê 8 ¥ 10
0.004
Á
Ë
x
-6
g
2
ˆ
2
- 0.2 g ˜ dx = 0.1vmax
¯
0.00632 m
È - (8 ¥ 10 -6 )
˘
2
(9.81) - 0.2(9.81)k ˙
0+Í
= 0.1vmax
x
Î
˚0.004 m
vmax = 0.1628 m/s
vm = 162.8 mm/s b
(b)
Maximum acceleration at x = 0.004 m where SF is maximum.
 F = k /x 2 - WB = mB a
(8 ¥ 10 -6 )(9.81)/(0.004)2 - (0.2)(9.881) = (0.2)am
a m = 14.72 m/s2 ≠ b
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45
PROBLEM 13.35
Nonlinear springs are classified as hard or soft, depending upon the curvature of their force-deflection
force
curve
(see figure). If a delicate instrument having a mass of 5 kg is placed on a spring of length l so that its base is
just touching the undeformed sspring
pring and then inadvertently released from that position, determine the
maximum deflection xm of the spring and the maximum force Fm exerted by the spring, assuming (a)
( a linear
spring of constant k 3 kN/m, ((b) a hard, nonlinear spring, for which F  (3 kN/m)( x  160x 2 ) .
SOLUTION
W  mg  (5 kg) g
W  49.05 N
T1  T2  0, T1  U1 2  T2 yields U12  0
Since
U12  Wxm 
(a)
xm
xm
 Fdx  49.05x   Fdx  0
m
0
For F  kx  (300 N/m) x
Eq. (1):
49.05 xm 
xm
 3000x dx  0
0
49.05 xm  1500 xm2  0
xm  32.7  103 m  32.7 mm 
Fm  3000 xm  3000(32.7 103 )
(b)
(1)
0
For
Eq. (1)
Fm  98.1 N 
F  (3000 N/m) x(1  160 x2 )
49.05xm 
xm
 3000( x  160x )dx  0
3
0
1

49.05 xm  3000  xm2  40 xm4   0
2


Solve by trial:
(2)
xm  30.44  103 m
xm  30.4 mm 
Fm  (3000)(30.44  103 )[1  160(30.44  103 )2 ]
Fm  104.9 N 
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PROBLEM 13.36
A meteor starts from rest at a very great distance from the earth. Knowing that the radius of the earth is 6370
km and neglecting all forces except the gravitational attraction of the earth, determine the speed of the meteor
(a) when it enters the ionosphere at an altitude of 1000 km, (b) when it enters the stratosphere at an altitude of
50 km, (c) when it strikes the earth’s surface.
SOLUTION
R  6370 km
Given:
U1 2  T2  T1

GMm GMm

r2
r1
Since r1 is very large,
GMm
0
r1
thus
T1  0
1
GMm
mv 2 
2
r2
v2
R2
 g
2
r2
v2 
2 gR 2
r2



2 9.81 m/s 2  6,370, 000 m 
(a) For
2
r2
r2  6,370,000 m  1,000,000 m  7,370,000 m
v  10.3933 km/s
(b) For
v  10.39 km/s 
r2  6,370,000 m  50,000 m  6, 420,000 m
v  11.1358 km/s
(c) For
v  11.14 km/s 
r2  6,370,000 m
v  11.1794 km/s
v  11.18 km/s 
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PROBLEM 13.37
Express the acceleration of gravity g h at an altitude h above the surface of the earth in terms of the
acceleration of gravity g 0 at the surface of the earth, the altitude h and the radius R of the earth. Determine
the percent error if the weight that an object has on the surface of earth is used as its weight at an altitude of
(a) 1 km, (b) 1000 km.
SOLUTION
F=
At earth’s surface, (h = 0)
GM E m
(h + R)
GM E m / R 2
( Rh + 1)
2
mg h
GM E m
= mg 0
R2
GM E
= g0
R2
gh =
Thus,
=
2
gh =
g0
⎛h ⎞
⎜ +1⎟
⎝R ⎠
GM E
R2
⎛h ⎞
⎜ +1⎟
⎝R ⎠
2
2
R = 6370 km
At altitude h, “true” weight
F = mg h = WT
Assumed weight
W0 = mg0
Error = E =
gh =
W0 − WT mg 0 − mg h g0 − g h
=
=
W0
mg 0
g0
g0
( Rh + 1)
g0 −
2
E=
g0
(1 + ) = ⎡⎢1 −
h
R
2
g0
⎣⎢
(
1
1 + Rh
⎤
2⎥
) ⎦⎥
(a)
h = 1 km:
1
⎡
⎤
P = 100E = 100 ⎢1 −
2⎥
1
⎢⎣ (1 + 6370 ) ⎥⎦
P = 0.0314% W
(b)
h = 1000 km:
1
⎡
⎤
P = 100E = 100 ⎢1 −
2⎥
1000
⎢⎣ (1 + 6370 ) ⎥⎦
P = 25.3% W
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48
PROBLEM 13.38
A golf ball struck on earth rises to a maximum height of 60 m
and hits the ground 230 m away. How high will the same golf
ball travel on the moon if the magnitude and direction of its
velocity are the same as they were on earth immediately after
the ball was hit? Assume that the ball is hit and lands at the
same elevation in both cases and that the effect of the
atmosphere on the earth is neglected, so that the trajectory in
both cases is a parabola. The acceleration of gravity on the
moon is 0.165 times that on earth.
SOLUTION
Solve for hm .
At maximum height, the total velocity is the horizontal component of the velocity, which is constant and the
same in both cases.
1 2
mv
2
U1 2  mge he
Earth
U1 2  mg m hm
Moon
T1 
Earth
1 2
1
mv  mge he  mvH2
2
2
Moon
1 2
1
mv  mgm hm  mvH2
2
2
1 2
mvH
2
hm ge

he g m
 g e he  g m hm  0
Subtracting
T2 
 ge 
hm  (60 m) 

 0.165 ge 
hm  364 m 
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PROBLEM 13.39
The sphere at A is given a downward velocity v 0 of magnitude
5 m/s and swings in a vertical plane at the end of a rope of
length l  2 m attached to a support at O.
O Determine the angle 
at which the rope will break, knowing that it can withstand a
maximum tension
ension equal to twice the weight of the sphere.
SOLUTION
1 2 1
mv0  m (5)2
2
2
T1  12.5 m
T1 
1 2
mv
2
U1 2  mg (l )sin 
T2 
T1  U1 2  T2
12.5m  2mg sin  
1 2
mv
2
25  4 g sin   v2
(1)
Newton’s law at.
v2
v2
m

2
v2  4 g  2 g sin 
2mg  mg sin   m
(2)
Substitute for v2 from Eq. (2) into Eq. (1)
25  4 g sin   4 g  2 g sin 
sin  
(4)(9.81)  25
 0.2419
(6)(9.81)
  14.00 
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PROBLEM 13.40
The sphere at A is given a downward velocity v 0 and swings
in a vertical circle of radius l and center O.. Determine the
smallest velocity v0 for which the sphere will reach Point B as
it swings about Point O (a) if AO is a rope, (b)) if AO is a
slender rod of negligible mass.
SOLUTION
1 2
mv0
2
1
T2  mv 2
2
U1 2  mgl
T1 
1 2
1
mv0  mgl  mv 2
2
2
2
2
v0  v  2 gl
T1  U1 2  T2
Newton’s law at
(a)
For minimum v, tension in the cord must be zero.
Thus, v2  gl
v02  v2  2gl  3gl
(b)
v0  3 gl 
Force in the rod can support the weight so that v can be zero.
v02  0  2 gl
Thus,
v0  2 gl 
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51
PROBLEM 13.41
A bag is gently pushed off the top of a wall at A and swings in a vertical
plane at the end of a rope of length l Determine the angle for which the rope
will break, knowing that it can withstand a maximum tension
tens
equal to twice
the weight of the bag.
SOLUTION
Use work - energy: position 1 is at A, position 2 is at B.
T1  U12  T2
(1)
Where
T1  0; U1 2  mg l sin  ; T2 
1 2
mvB
2
Substitute
0  mg l sin  
1 2
mvB
2
vB2  2g l sin 
(2)
For T = 2 W use Newton
Newton’s 2nd law.
Fn  man  2W  W sin  
mvB2
l
(3)
Substitute (2) into (3)
2 mg  mg sin   2 mg
l sin 
l
2  3sin 
or sin 
2
   41.81
3
  41.8 
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PROBLEM 13.42
A roller coaster starts from rest at A, rolls down the track to B,
describes a circular loop of 12 m diameter, and moves up and
down past Point E. Knowing that h = 18 m and assuming no
energy loss due to friction, determine (a) the force exerted by his
seat on a 80 kg rider at B and D, (b) the minimum value of the
radius of curvature at E if the roller coaster is not to leave the
track at that point.
SOLUTION
Let yp be the vertical distance from Point A to any Point P on the track. Let position 1 be at A and position
2 be at P. Apply the principle of work and energy.
T1 = 0
T2 =
1 2
mvP
2
U1→2 = mgyP
T1 + U1→ 2 = T2 : 0 + mgyP =
1 2
mvP
2
vP2 = 2 gyP
Magnitude of normal acceleration at P:
( aP ) n =
(a)
vP2
ρP
=
2 gyP
ρP
Rider at Point B.
yB = h = 18 m
ρB = r = 6 m
an =
(2 g )(18)
= 6g
6
ΣF = ma:
N B − mg = m(6 g )
N B = 7 mg = 7(80)(9.81) = 5493.6 N
N B = 5490 N W
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53
PROBLEM 13.42 (Continued)
Rider at Point D.
yD = h − 2r = 18 − 12 = 6 m
ρD = 6 m
an =
(2 g )(6)
= 2g
6
ΣF = ma:
N D + mg = m(2 g )
N D = mg = (80)(9.81) = 784.8 N
N D = 785 N W
(b)
Car at Point E.
yE = h − r = 18 − 6 = 12 m
NE = 0
ΣF = man :
mg = m ⋅
2 gyE
ρE
ρ E = 2 yE
ρ = 24 m W
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54
PROBLEM 13.43
In Problem 13.42, determine the range of values of h for which
the roller coaster will not leave the track at D or E, knowing
that the radius of curvature at E is ρ = 22.5 m. Assume no
energy loss due to friction.
PROBLEM 13.42 A roller coaster starts from rest at A, rolls
down the track to B, describes a circular loop of 12-m
diameter, and moves up and down past Point E. Knowing that
h = 18 m and assuming no energy loss due to friction,
determine (a) the force exerted by his seat on a 80 kg rider at B
and D, (b) the minimum value of the radius of curvature at E if
the roller coaster is not to leave the track at that point.
SOLUTION
Let yp be the vertical distance from Point A to any Point P on the track. Let position 1 be at A and position
2 be at P. Apply the principle of work and energy.
T1 = 0
T2 =
1 2
mvP
2
U1→2 = mg yP
T1 + U1→ 2 = T2 : 0 + mgy p =
1 2
mvP
2
vP2 = 2 g yP
Magnitude of normal acceleration of P:
( aP ) n =
vP2
ρP
=
2 g yP
ρP
The condition of loss of contact with the track at P is that the curvature of the path is equal to ρp and the
normal contact force N P = 0.
Car at Point D.
ρD = r = 6 m
y D = h − 2r
( aD ) n =
2 g (h − 2r )
r
ΣF = ma
2 g ( h − 2r )
r
2 h − 5r
N D = mg
r
N D + mg = m
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55
PROBLEM 13.43 (Continued)
For N D > 0
2 h − 5r > 0
5
h > r = 15 m
2
Car at Point E.
ρE = ρ = 22.5 m
yE = h − r = h − 6 m
( aE ) n =
2 g (h − 6)
22.5
ΣF = ma
2mg (h − 6)
22.5
⎛ 34.5 − 2h ⎞
N E = mg ⎜
⎝ 22.5 ⎟⎠
N E − mg = −
N E > 0,
For
34.5 − 2h > 0
h < 17.25 m
Range of values for h:
15 m ≤ h ≤ 17.25 m W
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56
PROBLEM 13.44
A small block slides at a speed v on a horizontal surface. Knowing that
h 0.9 m, determine the required speed of the block if it is to leave the
cylindrical surface BCD when  30°.
SOLUTION
At Point C where the block leaves the surface BCD the contact force is reduced to zero. Apply Newton’s
second law at Point C.
n-direction:
N  mg cos   man  
mvC2
h
vc2  gh cos
With N 0, we get
Apply the work-energy principle to the block sliding over the path BC. Let position 1 correspond to Point B
and position 2 to C.
T1 
1 2
mvB
2
T2 
1
1
mvC2  mgh cos
2
2
U1 2  weight  change in vertical distance
 mgh (1  cos  )
1 2
1
mvB  mg (1  cos  )  mgh cos 
2
2
2
vB  gh cos   2 gh(1  cos  )  gh(3cos   2)
T1  U12  T2 :
Data:
g  9.81 m/s 2 , h  0.9 m,   30.
vB2  (9.81)(0.9)(3cos 30  2)  5.2804 m 2 /s 2
vB  2.30 m/s 
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PROBLEM 13.45
A small block slides at a speed v = 2.5 m/s on a horizontal surface at a
height h = 1m above the ground. Determine (a) the angle q at which it
will leave the cylindrical surface BCD, (b) the distance x at which it will
hit the ground. Neglect friction and air resistance.
SOLutiOn
Block leaves surface at C when the normal force N = 0.
mg cos q = man
+
g cos q =
vC2
n
vC2 = gh cosq = gy
(1)
Work-energy principle.
TB =
(a)
1 2 1
mv = m(2.5)2 = 3.125 m
2
2
1 2
mvC
2
TB + U B -C = TC
TC =
Using Eq. (1)
3.125 m + mg ( h - y ) =
1 2
mvC
2
3.125 + g ( h - yC ) =
1
g yC
2
U B -C = W ( h - y ) = mg ( h - yC )
(2)
3
g yC
2
(3.125 + gh)
yC =
3
2g
3.125 + gh =
( )
yC =
(3.125 + 9.81 ¥ 1)
3
2 (9.81)
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58
PROBLEM 13.45 (continued)
yC = 0.87903 m
yC = h cos q
(b)
cos q =
yC 0.87903
=
= 0.87903
h
1
(3)
q = -28.5∞ b
From (1) and (3)
vC = gy
vC = (9.81)(0.87903)
vC = 2.9365 m/s
At C:
( vC ) x = vC cos q = (2.9365)(cos 28.5∞) = 2.5806 m/s
( vC ) y = - vC sin q = -(2.9365) sin(28.5∞) = -1.40118 m/s
y = yC + ( vC ) y t -
At E:
1 2
gt = 0.87903 - 1.40118 t - 4.905 t 2
2
yE = 0 4.905t 2 + 1.40118t - 0.87903 = 0
neglecting the –ve root: t = 0.3039 s
At E:
x = h(sin q ) + ( vC ) x t = (1) sin(28.5∞) + (2.5806)(0.3039)
x = 0.47716 + 0.78424 = 1.2614 m
x = 1.261 m b
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59
PROBLEM 13.46
A chair-lift is designed to transport 1000 skiers per hour from the base A to
the summit B. The average mass of a skier is 70 kg and the average speed of
the lift is 75 m/min. Determine (a) the average power required, (b) the
required capacity of the motor if the mechanical efficiency is85 percent and
if a 300 percent overload is to be allowed.
SOLUTION
Note: Solution is independent of speed.
(a)
Average power 
U (1000)(70 kg)(9.81 m/s 2 )(300 m)
Nm

 57, 225
t
3600 s
s
Average power  57.2 kW 
(b)
Maximum power required with 300% over load

100  300
(57.225 kW)  229 kW
100
Required motor capacity (85% efficient)
Motor capacity 
229 kW
 269 kW 
0.85
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PROBLEM 13.47
It takes 15 s to raise a 1200-kg car and the supporting 300-kg hydraulic
car-lift platform to a height of 2.8 m. Determine (a) the average output
power delivered by the hydraulic pump to lift the system, (b) the average
power electric required, knowing that the overall conversion efficiency
from electric to mechanical power for the system is 82 percent.
SOLUTION
(a)
( PP ) A  ( F )(vA )  (mC  mL )( g )(v A )
vA  s/t  (2.8 m)/(15 s)  0.18667 m/s
( PP ) A  [(1200 kg)  (300 kg)](9.81 m/s2 )(0.18667 m/s)3
(b)
( PP ) A  2.747 kJ/s
( PP ) A  2.75 kW 
( PE ) A  ( PP )/  (2.75 kW)/(0.82)
( PE ) A  3.35 kW 
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PROBLEM 13.48
The velocity of the lift of Problem 13.47 increases uniformly from zero to
its maximum value at mid-height
height 7.5 s and then decreases uniformly to
zero in 7.5 s. Knowing that the peak power output of the hydraulic pump is
6 kW when the velocity is maximum, determine the maximum life force
provided by the pump.
PROBLEM 13.47 It takes 15 s to raise a 1200-kg
kg car and the supporting
300-kg hydraulic car-lift
lift platform to a height of 2.8 m. Determine
(a)) the average output power delivered by the hydraulic pump to lift the
system, (b)) the average power electric required, knowing that the overall
conversion efficiency from electric to mechanical power for the
th system is
82 percent.
SOLUTION
Newton’s law
Mg  (M C  M L ) g  (1200  300) g
Mg  1500 g
F  F  1500 g  1500a
(1)
Since motion is uniformly accelerated, a constant
Thus, from (1), F is constant and peak power occurs when the velocity is a maximum at 7.5 s.
a
vmax
7.5 s
P  (6000 W)  ( F )(vmax )
vmax  (6000)/F
a  (6000)/(7.5)( F )
Thus,
(2)
Substitute (2) into (1)
F  1500 g  (1500)(6000)/(7.5)( F )
F 2  (1500 kg)(9.81 m/s 2 ) F 
(1500 kg)(6000 N  m/s)
0
(7.5 s)
F 2  14, 715F  1.2  106  0
F  14,800 N
F  14.8 kN 
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PROBLEM 13.49
(a) A 120-lb woman rides a 15-lb
lb bicycle up a 33-percent
percent slope at a constant speed of 5 ft/s. How much power
must be developed by the woman? (b)) A 180
180-lb man on an 18-lb
lb bicycle starts down the same slope and
maintains a constant speed of 20 ft/s by braking. How much ppower
ower is dissipated by the brakes? Ignore air
resistance and rolling resistance.
SOLUTION
tan  
3
100
  1.718
W  WB  WW  15  120
(a)
W  135 lb
PW  W  v  (W sin  ) (v)
PW  (135)(sin 1.718)(5)
(a)
PW  20.24 ft  lb/s
PW  20.2 ft  lb/s 
W  WB  Wm  18  180
(b)
W  198 lb
Brakes must dissipate the power generated by the bike and the man going down the slope at 20 ft/s.
PB  W  v  (W sin  )(v)
PB  (198)(sin 1.718)(20)
PB  118.7 ft  lb/s 




(b)
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PROBLEM 13.50
A power specification formula is to be
derived for electric motors which drive
conveyor belts moving solid material at
different rates to different heights and
distances. Denoting the efficiency of the
motors by h and neglecting the power
needed to drive the belt itself, derive a
formula (a) in the SI system of units for
the power P in kW, in terms of the mass
flow rate m in kg/h, the height b and
horizontal distance l in meters.
SOLutiOn
(a)
Material is lifted to a height b at a rate, ( m kg/h )( g m/s2 ) = [mg ( N/h)]
Thus,
DU [mg ( N/h)][b( m)] Ê mgb ˆ
N ◊ m/s
=
=Á
Ë 3600 ˜¯
(3600 s/h)
Dt
1000 N ◊ m/s = 1 kw
Thus, including motor efficiency, h,
P ( kw ) =
mgb ( N ◊ m/s)
Ê 1000 N ◊ m/s ˆ
(3600) Á
˜¯ (h)
Ë
kw
P ( kw ) = 0.278 ¥ 10 -6
mgb
b
h
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64
PROBLEM 13.51
A 1400-kg automobile starts from rest and travels 400 m during a
performance test. The motion of the automobile is defined by the relation
x  4000ln(cosh 0.03t ), where x and t are expressed in meters and seconds,
respectively. The magnitude of the aerodynamic drag is D  0.35v 2 , where
D and v are expressed in newtons and m/s, respectively. Determine the
power dissipated by the aerodynamic drag when (a) t 10 s, (b) t 15 s.
SOLUTION
Motion is determined as a function of time as
x  4000ln  cosh 0.03t 
Velocity
v
dx
1


 4000 
 sinh 0.03t  0.03

dt
 cosh 0.03t 
v
Power dissipated
120sinh 0.03t
cosh 0.03t


P  Dv  0.35v 2 v  0.35v 3
3  sinh 0.03t 
P  0.35 120  

 cosh 0.03t 
3
0.03t
 e0.03t 
3 e
 604.8 10   0.03t

 e0.03t 
e
3
3
(a) t  10 s,
P  14.95 10  W  14.95 kW 
(b) t  15 s,
P  45.4 10  W  45.4 kW 
3
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65
PROBLEM 13.52
A 1400-kg automobile starts from rest and travels 400 m during a
performance test. The motion of the automobile is defined by the relation
2
a  3.6e0.0005x , where a and x are expressed in m/s and meters,
respectively. The magnitude of the aerodynamic drag is D  0.35v 2 ,
where D and v are expressed in newtons and m/s, respectively. Determine
the power dissipated by the aerodynamic drag when (a) x  200 m,
(b) x  400 m.
SOLUTION
Motion is defined by the following function:
a  3.6e0.0005x
v
v
dv
dx
x 0.0005 x
0 vdv  3.60 e

dx
3.6 0.0005 x u
e du

0.0005 0
v2
 7200 e 0.0005 x  1
2




v 2  14400 1  e 0.0005 x


1
v  120 1  e 0.0005 x 2
Power dissipated  P  Dv  0.35v3
3
3
P  604.8 10  1  e0.0005 x  2
3
x  200 m,
P  17.75 10  W  17.75 kW 
(b) x  400 m,
P  46.7 10  W  46.7 kW 
(a)
3
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PROBLEM 13.53
The fluid transmission of a 15-Mg truck allows the engine to deliver an essentially constant power of 50 kW
to the driving wheels. Determine the time required and the distance traveled as the speed of the truck is
increased (a) from 36 km/h to 54 km/h, (b) from 54 km/h to 72 km/h.
SOLUTION
For constant power, P:
P  Fv
 mav
m
dv
v
dt
Separate variables

t
dt 
0
v1 m dv
0
 P v
 t
m  v12 v02 
  
P  2
2
(1)
Distance
P  mv
dv dx
dv
 mv 2
dx dt
dx
Separate variables

m v1 2
x
dx 
v dv
0
v0
P

 x
m  v13 v03 
  
P  3
3
(2)
with numbers
(a) v0  36 km/h  10 m/s; v1  54 km/h  15 m/s, so,
t

x
15  103 kg
15 m/s 2  10 m/s 2 

2 50 × 10 W 

3
15  103
3  50  10 
3

 t  18.75 s 
153  103   237.5 m 


 x  238 m 
(b) v0  54 km/h  15 m/s; v1  72 km/h  20 m/s
t

x
15  103
 20 2  152   26.25 s

3 

2
50
10
 

 t  26.2 s 

15  103
 3  50  10 
3
 203  153   462.5 m 


 x  462 m 
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67
PROBLEM 13.54
The elevator E has a weight of 6600 lbs when fully loaded and is connected as
shown to a counterweight W of weight of 2200 lb. Determine the power in hp
delivered by the motor (a) when the elevator is moving down at a constant speed
of 1 ft/s, (b) when it has an upward velocity of 1 ft/s and a deceleration of
0.18 ft/s2 .
SOLUTION
(a) Acceleration  0
Counterweight
Elevator
Motor
Fy  0: TW  WW  0
F  0: 2TC  TW  6600  0
TW  2200 lb
Kinematics:
TC  2200 lb
2 xE  xC , 2 xE  xC , vC  2vE  2 ft/s
P  TC  vC  (2200 lb)(2 ft/s)  4400 lb  ft/s  8.00 hp
P  8.00 hp 
aE  0.18 ft/s2 , vE  1 ft/s
(b)
Counterweight
Elevator
Counterweight:
F  Ma : TW  W 
W
(aW )
g
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PROBLEM 13.54 (Continued)
TW  (2200 lb) 
(2200 lb)(0.18 ft/s 2 )
(32.2 ft/s 2 )
TW  2212 lb
Elevator
F  ma
2TC  TW  WE 
2TC  (2212 lb)  (6600 lb) 
WE
(a E )
g
(6600 lb)(0.18 ft/s 2 )
(32.2 ft/s 2 )
2TC  4351 lb
TC  2175.6 lb
vC  2 ft/s (see part(a))
P  TC  vC  (2175.6 lb)(2 ft/s)  4351.2 lb  ft/s
 7.911 hp



P  7.91 hp 
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69
PROBLEM 13.55
A force P is slowly applied to a plate that is attached to two springs and causes a deflection x0 . In each of the
two cases shown, derive an expression for the constant ke , in terms of k1 and k2 , of the single spring
equivalent to the given system, that is, of the single spring which will undergo the same deflection x0 when
subjected to the same force P.
SOLUTION
System is in equilibrium in deflected x0 position.
Case (a)
Force in both springs is the same  P
x0  x1  x2
Thus,
x0 
P
ke
x1 
P
k1
x2 =
P
k2
P P P
 
ke k1 k2
1
1 1
 
ke k1 k2
Case (b)
ke 
k1k2

k1  k2
Deflection in both springs is the same  x0
P  k1 x0  k2 x0
P  (k1  k2 ) x0
P  ke x0
Equating the two expressions for
P  (k1  k2 ) x0  ke x0
ke  k1  k2 
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70
PROBLEM 13.56
A loaded railroad car of mass m is rolling at a
constant velocity v0 when it couples with a
massless bumper system. Determine the
maximum deflection of the bumper assuming the
two springs are (a) in series (as shown),(b) in
parallel.
SOLUTION
Let position A be at the beginning of contact and position B be at maximum deflection.
1 2
mv0
2
VA  0
(zero force in springs)
TB  0
(v  0 at maximum deflection)
TA 
1 2 1
k1 x1  k2 x22
2
2
where x1is deflection of spring k1 and x2 is that of spring k2.
VB 
Conservation of energy:
TA  VA  TB  VB
1 2
1
1
mv0  0  0  k1 x12  k2 x22
2
2
2
k1x12  k2 x22  mv02
(a)
(1)
Springs are in series.
Let F be the force carried by the two springs.
F
k1
Then,
x1 
Eq. (1) becomes
1
1 
F 2     mv02
k
k
2 
 1
so that
The maximum deflection is
and x2 
F
k2
1
1 
F  v0 m /   
 k1 k 2 
1
1 
  x1  x2     F
 k1 k2 
1
1
1 
1 
    v0 m /   
 k1 k2 
 k1 k2 
1
1 
 v0 m   
k
k
2 
 1
  v0 m(k1  k2 )/k1k2 
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PROBLEM 13.56 (Continued)
(b)
Springs are in parallel.
x1  x2  
Eq. (1) becomes
(k1  k2 ) 2  mv02
  v0
m

k1  k2
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72
PROBLEM 13.57
A 750-g collar can slide along the horizontal rod shown. It is
attached to an elastic cord with an undeformed length of 300 mm
and a spring constant of 150 N/m. Knowing that the collar is
released from rest at A and neglecting friction, determine the
speed of the collar (a) at B, (b) at E.
SOLUTION
T1  V1  T2  V2
Use conservation of energy
(1)
(a) Position 1 is at A and position 2 is at B
T1  0;
V1 
1 2
kx1
2
where
x1     0
1
  5002  4002  350 2  2  729.726 mm
So    0  429.726 mm
V1 
At B
1
1500 N/m  0.429726 m 2  13.8498 J
2
1
  350 2  400 2  2  531.507 mm     0  231.507 mm
1 2 1
mv2   0.75 v22  0.375 v22
2
2
1
2
V2  150  0.231507   4.01966 J
2
T2 
Substituting into (1)
0  13.8498  0.375 v22  4.01966
v2  vB  5.12 m/s 
(b) At E
At E
T1  0; V1  13.8498  same as before 
1
  3502  500 2  2  610.328 mm     0  310.328 mm
1 2
mvE  0.375 vE2
2
1
1
2
vE  kx 2  150  0.310328  7.223 J
2
2
T3 
0  13.8498  0.375 vE2  7.223  vE  4.20 m/s 
Substituting into (1)
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73
PROBLEM 13.58
A 1.8 kg collar can slide without friction along a horizontal rod and is in
equilibrium at A when it is pushed 25 mm to the right and released from
rest. The springs are undeformed when the collar is at A and the constant
of each spring is 490 kN/m. Determine the maximum velocity of the
collar.
SOLUTION
A1 =
(0.15) 2 + (0.225) 2 = 0.2704 m
A0 =
(.15)2 + (.2 )2 = .25 m
Stretch = 0.2704 – 0.25 = 0.0204 m
A 2 = (0.175) 2 + (0.15) 2 = 0.2305 m
Stretch = 0.2305 – 0.25 = –0.01951 m
T1 = 0, V2 = 0
T2 =
V1 =
( )
1 2 1
mv2 =
1.8 v22
2
2
(
1
( 490 × 103 N/m ) S12 + S22
2
)
V1 = (245000) (.000797) = 195.22 N.m
0
0
T1 + V1 = T2 + V2
v22 = 216.91
v2 = 14.73 ms W
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74
PROBLEM 13.59
A 18-kg collar can slide without friction along a horizontal rod and
is released from rest at A.. The undeformed lengths of springs BA and
CA are 250 mm and 225 mm,, respectively, and the constant of each
spring is 490 kN/m.. Determine the velocity of the collar when
whe it has
moved 25 mm to the right.
SOLUTION
k  490,000 N/m
T1  0
Initially spring AB underformed,
undeformed, spring
springAC stretched
stretch by 0.025 m
1
1
2
2
k  l1    490,000.025 m 
2
2
 153.125 N  m
V1 
1  0.152  0.2252  0.2704 m
S1  Stretch  0.2704  0.25  0.02042 m
 2  0.152  0.1752  0.23049 m
S2  Stretch  0.23049  0.225  0.00549 m
T2 
1 2 1
mv2  (1.8)V22  0.9V22
2
2
V2 
1
 490,000 S12  S22
2


2
2
 245,000  0.02042   0.00549  


 109.544 N  m
T1  V1  T2  V2
153.125  0.9 v22  109.544
v2  6.96 m/s 
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75
PROBLEM 13.60
A 500-g
g collar can slide without friction on the curved rod BC in a horizontal
plane. Knowing that the undeformed length of the spring is 80 mm and that
k  400 kN/m, determine (a)) the velocity that the collar should be given at A
to reach B with zero velocity, (b)) the velocity of the collar when it eventually
reaches C.
SOLUTION
(a)
Velocity at A:
TA 
1 2  0.5

mv A  
kg  v A2
2
 2

TA  (0.25)v A2
LA  0.150 m  0.080 m
LA  0.070 m
1
k ( L A ) 2
2
1
VA  (400  103 N/m)(0.070 m) 2
2
VA  980 J
VA 
vB  0
TB  0
LB  0.200 m  0.080 m  0.120 m
1
1
k (LB ) 2  (400  103 N/m)(0.120 m) 2
2
2
VB  2880 J
VB 
Substitute into conservation of energy.
TA  VA  TB  VB
v A2 
0.25v A2  980  0  2880
(2880  980)
(0.25)
v A2  7600 m 2 /s 2
vA  87.2 m/s 

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76
PROBLEM 13.60 (Continued)
(b)
Velocity at C:
Since slope at B is positive, the component of the spring force FP , parallel to the rod, causes the block
to move back toward A.
TB  0, VB  2880 J [from part (a)]
1 2 (0.5 kg) 2
mvC 
vC  0.25vC2
2
2
LC  0.100 m  0.080 m  0.020 m
TC 
VC 
1
1
k ( LC ) 2  (400  103 N/m)(0.020 m) 2  80.0 J
2
2
Substitute into conservation of energy.
TB  VB  TC  VC
0  2880  0.25vC2  80.0
vC2  11, 200 m2 /s2
vC  105.8 m/s 
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77
PROBLEM 13.61
For the adapted shuffleboard device in Prob 13.28, you decide to utilize an elastic
cord instead of a compression spring to propel the puck forward. When the cord is
stretched directly between points A and B,, the tension is 20 N. The 425 gram puck
is pla
placed
ced in the center and pulled back through a distance of 400
40 mm; a force of
100 N is required to hold it at this location. Knowing that the coefficient of friction
is 0.3, determine how far the puck will travel.
SOLUTION
Given:
Sketch:
 k  0.3
m  0.425 kg
FC,2  20 N, Force in cord at positon 2
li = length of cord at positon i
si = stretch of cord in position i
lO  unstretched length of cord
At Position 1:


l1 = 2 0.32  0.42  l1  1 m
 40 
 =tan 1    53.13
 30 
Find Force in cord:
FBD:
For Equilibrium:
F  0
x
2 FC ,1 sin   100  0
FC ,1  62.5 N
At Position 2:
l2 = 0.6 m
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PROBLEM 13.61 (Continued)
Find Stiffness of cord, k:
FC ,1  ks1 , FC ,2  k s2
FC ,1  FC ,2  k (s1  s2 )
62.5  20  k [(1.0  l0 )  (0.6  l0 )]
k  106.25 N/m
Find stretch in cord at positions 1, 2 and 3:
s1 
FC ,1
s2  s3 
k
62.5 N

106.25 N/m
FC ,2
k
20 N

106.25 N/m
 0.5882 m
 0.1882 m
Work Energy from position 1 to 3:
T1  Vg1  Ve1  U1NC
3  T3  Vg3  Ve3
where:
1
Ve1  ks12
2

U 1NC
3

x1  d
0
k mgdx
1
Ve3  ks32
2
Therefore:
1 2
1
ks1  k mg  x1  d   ks32
2
2
Substitute known values:
1
1
106.25 0.58822  0.3  0.425 9.81 0.4  d   106.25 0.1882 2
2
2
d  12.79 m 
Solve for d:
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79
PROBLEM 13.62
An elastic cable is to be designed for bungee jumping from a tower
40 m high. The specifications call for the cable to be 25 m long
when unstretched, and to stretch to a total length of 30 m when
a 300 kg weight is attached to it and dropped from the tower.
Determine (a) the required spring constant k of the cable, (b) how
close to the ground a 90 kg man will come if he uses this cable to
jump from the tower.
SOLutiOn
(a)
Conservation of energy.
V1 = 0 T1 = 0 V1 = mg (30)
Datum at ➁:
V1 = (30 m)(300 m)(9.81)
= 8.829 ¥ 10 4 J
V2, g = 0 T2 = 0
1
V2 = Vg + Ve = 0 + k (5 m)2
2
T1 + V1 = T2 + V2
0 + 8.829 ¥ 10 4 = 0 + 12.5k
k = 7060 N/m b
(b)
From (a),
k = 7063.2 N/m
T1 = 0
W = 90 g = 90 ¥ 9.81 N
V1 = 90 ¥ 9.81 ¥ ( 40 - d )
T2 = 0
Datum:
1
V2 = Vg + Ve = 0 + (7063.2)( 40 - 25 - d )2
2
V2 = 3531.6(15 - d )2
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80
PROBLEM 13.62 (continued)
d = distance from the ground
T1 + V1 = T2 + V2
0 + (90)(9.81)( 40 - d ) = 0 + 3531.6(15 - d )2
3531.6d 2 - 104615.1d + 759294 = 0
d = 16.903 m,12.7198 m
Discard 16.903 m (assumes cord acts in compression when rebound occurs).
d = 12.72 m b
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81
PROBLEM 13.63
It is shown in mechanics of materials that when an elastic
beam AB supports a block of weight W at a given Point B,
the deflection yst (called the static deflection) is proportional
to W. Show that if the same block is dropped from a height h
onto the end B of a cantilever beam AB and does not bounce
off, the maximum deflection ym in the ensuing motion can be
expressed as ym = yst (1 + 1 + 2h/yst ). Note that this formula is
approximate, since it is based on the assumption that there is no
energy dissipated in the impact and that the weight of the beam is
small compared to the weight of the block.
SOLutiOn
Denote by k an equivalent spring constant.
Static deflection of beam is then
yst =
W
k
(1)
Drop W from height h.
1
T1 = 0 V1 = Wh T2 = 0 V2 = -Wym + k ym2
2
1
T1 + V1 = T2 + V2 0 + Wh = 0 - Wym + k ym2
2
From Equation (1),
W = k yst
k yst( h + ym ) =
1
k ym2
2
Ê
ym = yst Á1 +
Ë
ym2 - 2 yst ym - 2 yst h = 0
2h ˆ
b
yst ˜¯
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82
PROBLEM 13.64
A 2-kg collar is attached to a spring and slides without friction in
a vertical plane along the curved rod ABC. The spring is
undeformed when the collar is at C and its constant is 600 N/m. If
the collar is released at A with no initial velocity, determine its
velocity (a) as it passes through B, (b) as it reaches C.
SOLUTION
Spring elongations:
At A,
x A  250 mm  150 mm  100 mm  0.100 m
At B,
xB  200 mm  150 mm  50 mm  0.050 m
At C ,
xC  0
Potential energies for springs.
1 2 1
kx A  (600)(0.100)2  3.00 J
2
2
1
1
(VB )e  kxB2  (600)(0.050)2  0.75 J
2
2
(VC )e  0
(VA )e 
Gravitational potential energies: Choose the datum at level AOC.
(VA ) g  (VC ) g  0
(VB ) g  mg y  (2)(9.81)(0.200)  3.924 J
Kinetic energies:
TA  0
1 2
mvB  1.00 vB2
2
1
TC  mvC2  1.00 vC2
2
TB 
(a)
Velocity as the collar passes through B.
Conservation of energy:
TA  VA  TB  VB
0  3.00  0  1.00 vB2  0.75  3.924
vB2  6.174 m2 /s2
v B  2.48 m/s
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
PROBLEM 13.64 (C
(Continued)
(b)
Velocity as the collar reaches C.
Conservation of energy:
TA  VA  TC  VC
0  3.00  0  1.00vC2  0  0
vC2  3.00 m2 /s2
vC  1.732 m/s 
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PROBLEM 13.65
A 500-g collar can slide without friction along the semicircular rod BCD.
The spring is of constant 320 N/m and its undeformed length is 200 mm.
Knowing that the collar is released form rest at B, determine (a) the speed
of the collar as it passes through C, (b) the force exerted by the rod on the
collar at C.
SOLUTION
Given:
m  500 g
Sketch:
k  320 N/m
Undeformed length of spring  200 mm
vB  0
Finding Speed at C:
 300 2  1502   752  343.69318 mm
LAB 
k  320 N/m
At B
TB  0
VB  VB e  VB  g
LAB  343.69318 mm  200 mm
LAB  143.69318 mm  0.14369318 m
VB e 
1
1
2
2
k  LAB    320 N/m  0.1436932 m 
2
2
VB e  3.303637 J
VB  g  Wr   0.5 kg   9.81 m/s 2   0.15 m   0.73575 J
VB  VB e  VB  g  3.303637 J  0.73575 J  4.03939 J
At C
TC 
1 2 1
mvC   0.5 kg  vC2
2
2
 
TC  0.25vC2
VC e 
1
2
k  LAC 
2
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PROBLEM 13.65 (Continued)
LAC  309.23 mm  200 mm  109.23 mm  0.10923 m
1
2
VC e   320 N/m  0.10923 m 2  1.90909 J
TB  VB  TC  VC
0  4.0394  0.25vC2  1.90909
vC2 
(a)
Find force of rod on collar AC
4.0394  1.90909
 8.5212 m2 /s 2
0.25
Free Body Diagram
vC  2.92 m/s 
Fz  0 (no friction)
F  Fxi  Fy j
  tan 1
75
 14.04
300
Fe   k LAC   cos i  sin k 
Fe   320 0.10923 cos14.04i  sin14.04k 
Fe  33.909i  8.4797k (N)


F   Fx  33.909  i  Fy  4.905 j  8.4797k 
mv 2
j  mgk
r
8.5212 m /s 
F  4.905 N   0.5 
2
Fx  33.909 N  0
y
2
0.15 m
Fx  33.909 N
Fy  33.309 N
(b)
F  33.9 N i  33.3 N j 
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86
PRoBlEM 13.66
A thin circular rod is supported in a vertical plane by a bracket at A.
Attached to the bracket and loosely wound around the rod is a spring
of constant k = 50 N/m and undeformed length equal to the arc of
circle AB. A 250 gm collar C, not attached to the spring, can slide
without friction along the rod. Knowing that the collar is released
from rest at an angle q with the vertical, determine (a) the smallest
value of q for which the collar will pass through D and reach Point A,
(b) the velocity of the collar as it reaches Point A.
Solution
(a)
Smallest angle q occurs when the velocity at D is close to zero.
vC = 0
vD = 0
TC = 0
TD = 0
V = Ve + Vg
Point C.
DLBC = (0.3 m) = (0.3q ) m
1
(VC )e = k ( DLBC )2
2
1
(VC )e = (50)(0.3q )2 = 2.25q 2
2
R = 300 mm = 0.3 m
(VC ) g = WR(1 - cos q )
(VC ) g = (0.25)(9.81)(0.3)(1 - cosq )
(VC ) g = 0.73575(1 - cos q )
VC = (VC )e + (VC ) g = 2.25q 2 + 0.7375(1 - cos q )
Point D.
(VD )e = 0 (spring is unattached)
(VD ) g = W (2 R) = (0.25)(9.81)(0.6) = 1.4715
TC + VC = TD + VD
2.25q 2 - 0.7375(1 - cos q ) = 1.4715
2.25q 2 - 0.7375 cos q = 0.734
By trial,
q = 0.7548 rad
q = 43.2∞ b
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87
PRoBlEM 13.66 (continued)
(b)
Velocity at A.
Point D.
VD = 0 TD = 0 VD = 1.4715
Point A.
1 2 1
mv A = (0.25)v 2A
2
2
2
TA = 0.125v A
VA = (VA ) g = W ( R) = (0.25)(9.81)((0.3) = 0.73575
TA =
TA + VA = TD + VD
2
0.125v A + 0.73575 = 1.4715
v 2A = 5.886 m2 /s2
vA = 2.43 m/s Ø b
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88
PROBLEM 13.67
Cornhole is a game that requires you to toss beanbags
through a hole in a wooden board. People with limited
arm mobility often have difficulty enjoying this favorite
tailgating activity. An adapted launching device attaches
to a wheelchair so that points O and A are fixed. The
device mimics an underhand throw by utilizing an elastic
band to power the arm OC, which rotates about pin O.
The elastic cord has an unstretched length of 300 mm
and is attached to fixed point A and to point B on the
arm. The combined
bined weight of the beanbag and holder at
C is 20 N,, and you can neglect the weight of the rod OB.
Knowing that the starting position is 30 degrees from the
horizontal as shown in the figure, determine the spring
constant if the velocity of the bean bag is 10 m/s when
the bag is released at an angle of = 45 degrees.
SOLUTION
Given:
lO  0.3 m
Sketch:
20 N
= 2.0.39
2.039 kg
kg
9.81 m/s 2
v2  10 m/s
m
2  45
li = length of cord at positon i
si = stretch of cord in position i
hi = height of beanbag at position i
At Position 1:
From law of cosines:
l12 = 0.62  0.62  2(0.6)(0.6)cos 30
l1  1.1591 m
Stretch in cord:
s1  l1  l0
s1  0.8591 m
Height of beanbag:
h1  0.9 sin 30
h1  0.45 m
At Position 2:
From law of cosines:
l22 = 0.62  0.62  2(0.6)(0.6)cos 45
l2  0.4592 m
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PROBLEM 13.67 (Continued)
Stretch in cord:
s2  l2  l0
s2  0.1592 m
Height of beanbag:
h2  0.9sin 45
h2  0.6364 m
Work Energy from position 1 to 2:
T 1  Vg1  Ve1  U1NC
2  T2  Vg2  Ve2
Vg1  mgh1 , Ve1 
where:
1 2
ks1
2
1 2
mv2 , Vg2  mgh2
2
1
Ve2  ks22
2
T2 
1
1
1
mgh1  ks12  mv22  mgh2  ks22
2
2
2
Therefore:
Substitute known values:
1
1
1
 20 0.45  2 k  0.85912  2  2.039 102   20 0.6364  2 k 0.15922
 9  0.3690k  101.95  12.728  0.01267k
Solve for k:
k  275.6 N/m
k  276 N/m 
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90
PROBLEM 13.68
A spring is used to stop a 50-kg package which is moving down a 20º
incline. The spring has a constant k  30 kN/m and is held by cables so
that it is initially compressed 50 mm. Knowing that the velocity of the
package is 2 m/s when it is 8 m from the spring and neglecting friction,
determine the maximum additional deformation of the spring in
bringing the package to rest.
SOLUTION
Let position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by the
cable. Let position 2 be the position of maximum compression. Let x be the additional compression of the
spring. Use the principle of conservation of energy. T1  V1  T2  V2 .
Position 1:
1 2 1
mv1  (50)(2)2  100 J
2
2
V1g  mgh1  (50)(9.81)(8 sin 20)  1342.09 J
T1 
V1e 
Position 2:
1 2 1
ke1  (30  103 )(0.050) 2  37.5 J
2
2
1 2
mv2  0 since v2  0.
2
V2 g  mgh2  (50)(9.81)( x sin 20)  167.76 x
T2 
V2e 
1 2 1
ke2  (30  103 )(0.05  x)2  37.5  1500 x  15000 x 2
2
2
Principle of conservation of energy:
100  1342.09  37.5  167.61x  37.5  1500 x  15000 x 2
15,000 x 2  1332.24 x  1442.09  0
Solving for x,
x  0.26882 and 0.357 64
x  0.269 m 
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91
PROBLEM 13.69
Solve Problem 13.68 assuming the kinetic coefficient of friction between
the package and the incline is 0.2.
PROBLEM 13.68 A spring is used to stop a 50-kg package which is
moving down a 20
20 incline. The spring has a constant k  30 kN/m
and is held by cables so that it is initially compressed 50 mm. Knowing
that the velocity of the package is 2 m/s when it is 8 m from the spring
and neglecting friction, determine the maximum additional deformation
of the spring in bringing the package to rest.
SOLUTION
Let position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by the
cable. Let position 2 be the position of maximum compression. Let x be the additional compression of the
spring. Use the principle of work and energy. T1  V1  U12  T2  V2
Position 1.
1 2 1
mv1  (50)(2)2  100 J
2
2
V1g  mgh1  (50)(9.81)(8sin 20)  1342.09 J
T1 
V1e 
Position 2.
1 2 1
ke1  (30  103 )(0.05)2  37.5 J
2
2
1 2
mv2  0 since v2  0.
2
V2 g  mgh2  (50)(9.81)( x sin 20)  167.76 x
T2 
V2e 
1 2 1
ke2  (30  103 )(0.05  x )2  37.5  1500 x  15,000 x 2
2
2
Work of the friction force.
Fn  0
N  mg cos 20  0
N  mg cos 20
 (50)(9.81) cos 20
 460.92 N
F f  k N
 (0.2)(460.92)
 92.184
U12   F f d
 92.184(8  x)
 737.47  92.184 x
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92
PROBLEM 13.69 (Continued)
Principle of work and energy:
T1  V1  U12  T2  V2
100         x
 167.76 x  37.5  1500 x  15, 000 x 2
15,000 x 2  1424.42 x  704.62  0
Solving for x,
x  0.17440 and 0.26936
x  0.1744 m 
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93
PROBLEM 13.70
A section of track for a roller coaster consists of two circular
arcs AB and CD joined by a straight portion BC. The radius
of AB is 27 m and the radius of CD is 72 m. The car and its
occupants, of total mass 250 kg, reach Point A with
practically no velocity and then drop freely along the track.
Determine the normal force exerted by the track on the car as
the car reaches point B. Ignore air resistance and rolling
resistance.
SOLUTION
Calculate the speed of the car as it reaches Point B using the principle of conservation of energy as the car
travels from position A to position B.
Position A:
vA  0,
TA 
Position B:
VB  mgh
1 2
mvA  0, VA  0 (datum)
2
where h is the decrease in elevation between A and B.
TB 
Conservation of energy:
1 2
mvB
2
TA  VA  TB  VB :
1 2
mvB  mgh
2
vB2  2 gh
00
 (2)(9.81 m/s 2 )(27 m)(1  cos 40)
 123.94 m 2 /s 2
Normal acceleration at B:
( aB ) n 
vB2 123.94 m 2 /s 2

 4.59 m/s 2

27 m
(aB )n  4.59 m/s 2
50°
Apply Newton’s second law to the car at B.
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PROBLEM 13.70 (Continued)
50Fn  man : N  mg cos 40  man
N  mg cos 40  man  m( g cos 40  an )
cos 40  4.59 m/s 2 ]
 (250 kg)[(9.81 m/s2 ))cos
 1878.7  1147.5
N  731 N 
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95
PROBLEM 13.71
A section of track for a roller coaster consists of two circular
arcs AB and CD joined by a straight portion BC. The radius of
AB is 27 m and the radius of CD is 72 m. The car and its
occupants, of total mass 250 kg, reach Point A with
practically no velocity and then drop freely along the track.
Determine the maximum and minimum values of the normal
force exerted by the track on the car as the car travels from A
to D. Ignore air resistance and rolling resistance.
SOLUTION
Calculate the speed of the car as it reaches Point P, any point on the roller coaster track. Apply the principle of
conservation of energy.
Position A:
vA  0,
TA 
Position P:
VP  mgh
1 2
mvA  0, VA  0 (datum)
2
where h is the decrease in elevation along the track.
TP 
Conservation of energy:
1
mv 2
2
TA  VA  TP  VP
00
1 2
mv  mgh v2  2 gh
2
(1)
Calculate the normal force using Newton’s second law. Let  be the slope angle of the track.
Fn  man : N  mg cos  man
N  mg cos   man
Over portion AB of the track,
and
(2)
h   (1  cos  )
an  
mv 2

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PROBLEM 13.71 (Continued)
where  is the radius of curvature. (   27 m)
N  mg cos  
2mg  (1  cos  )
 mg (3cos   2)

At Point A (  0)
N A  mg  (250)(9.81)  2452.5 N
At Point B (  40)
N B  (2452.5)(3cos 40  2)
N B  731 N
Over portion BC,
  40,
an  0
(straight track)
N BC  mg cos 40  2452.5cos 40
N BC  1879 N
Over portion CD,
h  hmax  r (1  cos )
and
an 
mv 2
r
where r is the radius of curvature. (r  72 m)
2mgh
r
h

 mg cos   2mg  max  1  cos  
 r

N  mg cos  
2h 

 mg  3cos   2  max 
r 

which is maximum at Point D, where
 2h 
N D  mg 1  max 
r 

hmax  27  18  45 m,
Data:
r  72 m
 (2) (45) 
 5520 N
N D  (2452.5) 1 
72 

Summary:

minimum (just above B ):
maximum (at D ):
731 N 
5520 N 
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PROBLEM 13.72
A 500 g collar is attached to a spring and slides without friction along
a circular rod in a vertical plane. The spring has an undeformed
length of 125 mm and a constant k = 150 N/m . Knowing that
the collar is released from being held at A, determine the speed
of the collar and the normal force between the collar and the rod as
the collar passes through B.
SOLutiOn
For the collar,
m = 0.5 kg
For the spring,
k = 150 N/m  O = 0.125 m
At A:
 A = 0.175 + 0.25 = 0.425 m
 A -  O = 0.425 m - 0.125 m = 0.3 m
At B:
 B = (0.175 + 0.125)2 + (0.125)2 = 0.325 m
 B -  O = 0.325 - 0.125 = 0.2 m
Velocity of the collar at B.
Use the principle of conservation of energy.
TA + VA = TB + VB
where
TA =
1 2
mv A = 0
2
1
VA = k ( A -  O )2 + W (0)
2
1
= (150)(0.3)2 + 0 = 6.75 J
2
1 2 1
TB = mv B = (0.5)v B2 = 0.25v B2
2
2
1
VB = k ( B -  O )2 + Wh
2
1
= (150)(0.2)2 + (0.5)(9.81)( - 0.125)
2
= 2.38688 J
0 + 6.75 = 0.25v B2 = 2.38688
v B2 = 17.4525
m2
s2
vB = 4.18 m/s ¨ b
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98
PROBLEM 13.72 (continued)
Forces at B.
Fs = k ( B -  O ) = (150)(0.2) = 30 N
5
13
r = 0.125 m
sina =
mv B2
r
0.5(17.4525)
=
0.125
= 69.81 N
+
man =
SFy = ma y : Fs sin b - W + N = man
N = man + W - Fs sin a
Ê 5ˆ
= 69.81 + (0.5)(9.81) - 30 Á ˜
Ë 13 ¯
N = 63.2 N ≠ b
N = 63.1765N
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99
PROBLEM 13.73
A 4.5 kg collar is attached to a spring and slides without friction along a
fixed rod in a vertical plane. The spring has an undeformed length of
356 cm and a constant k = 700 N/m. Knowing that the collar is released from
rest in the position shown, determine the speed of the collar at (a) point A,
(b) point B.
SOLUTION
Spring length = .356 2 + .712 2
= .796 m
Stretch
= .796 m − .356 m
= .44 m
T0 + V0 = 0 + 0 +
(a) At A:
67.76 =
1
( 700 N/m )( .44 m )2 = 67.76 N.m
2
1
1
(4.5 kg ) vA2 + 2 ( 700 N/m ) (.356 2 − .356 )2
2
v A = 5.17 m/s W
(b) At B:
67.76 =
1
1
2
(4.5 ) vB2 + 2 ( 700 )(.356) − 44 (.356)
2
vB = 4.17 m/s W
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100
PROBLEM 13.74
A 200-g package is projected upward with a velocity v0
by a spring at A; it moves around a frictionless loop
and is deposited at C. For each of the two loops shown,
determine (a) the smallest velocity v0 for which the
package will reach C, (b) the corresponding force
exerted by the package on the loop just before the
package leaves the loop at C.
SOLutiOn
(a) The smallest velocity at B will occur when the force exerted by the tube on the package is zero.
+
(1)
mv B2
r
v B2 = gr = (9.81 m/s2 )(0.5 m)
SF = 0 + mg =
v B2 = 4.905 m2/s2
At A:
TA =
1 2
mv0 VA = 0
2
At B:
TB =
1 2 1
mv B = m ( 4.905) = 2.453 m
2
2
VB = mg (2.5 + 0.5) = 3 mg
TA + VA = TB + VB
1 2
mv0 + 0 = 2.453 m + 3 mg
2
v02 = 2[(2.453) + 3(9.81) J = 63.77
v0 = 7.99 m/s b
1
TC = mvC2 VC = mg (2.5 m)
2
TA + VA = TC + VC
At C:
1 2
1
mv0 + 0 = mvC2 + 2.5 mg
2
2
2
vC = [63.77 - (5.0)(9.81)]
vC2 = 14.72 m2 /s2
(b)
+
SF = N C =
mvC2 (0.2 kg)(14.72 m2/s2 )
=
r
(0.5 m)
Package on tube,
NC = 5.89 N ¨ b
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101
PROBLEM 13.74 (continued)
(2) (a)
The velocity at B can be nearly equal to zero since the weight of the package is supported by the
tube.
v B = 0, TB = 0
Thus,
VB = mg (2.5 m + 0.5 m)
VB = 3 mg
1
TA = mv02 VA = 0
2
TB + VB = TA + VA
1
0 + 3 mg = mv02 + 0
2
v02 = 6 g
v0 = 7.67 m/s b
1
TC = mvC2 VC = mg (2.5 m)
2
1 2
1
TA + VA = TC + VC
mv0 + 0 = mvC2 + 2.5 mg
2
2
2
2 2
vC = 6 g - 5 g = 9.81 m /s
(b)
+
SF = N C =
mvC2
r
N C = (0.2 kg)(9.81 m/s2 )/(0.5 m)
Package on tube,
NC = 3.92 N ¨ b
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102
PROBLEM 13.75
If the package of Problem 13.74 is not to hit the
horizontal surface at C with a speed greater than
3.5 m/s, (a) show that this requirement can be
satisfied only by the second loop, (b) determine
the largest allowable initial velocity v0 when the
second loop is used.
SOLutiOn
Loop (1). The smallest allowable velocity at B will occur when the force exerted by the tube on the
package is zero
+
(a)
SF = 0 + mg = mv B2 /r
v B2 = gr = (9.81 m/s2 )(0.5 m) = 4.905 m2 /s2
v B = 2.215 m/s
The velocity at B cannot be less than 2.215 m/s if the package is to maintain contact with the tube.
For vC to be as small as possible, v B must be as small as possible; that is, v B = 2.215 m/s.
At B:
TB =
1 2 1
mv B = m (2.215)2
2
2
TB = 2.453 m
VB = mg (2.5 + 0.5) = 3 mg
At C:
1
TC = mvC2
2
VC = 2.5 mg
TB + VB = TC + VC
1
2.453 m + 3 mg = mvC2 + 2.5 mg
2
2
vC = 2[2.453 + 0.5(9.81 m/s2 )]
vC2 = 14.72 m2/s2
vC = 3.836 m/s 3.5 m/s
b
Thus, Loop (1) cannot meet the requirement.
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103
PROBLEM 13.75 (continued)
(b)
Loop (2).
1 2
mv0
2
VA = 0
At A:
At C:
TA =
vC = 3.5 m/s
1
m (3.5)2
2
TC = 6.125 m
TC =
vC = 2.5 mg
TA +
1
= TC + VC
A
1 2
mv0 + 0 = 6.125 m + 2.5 mg
2
v02 = 2 (6.125 + 2.5 g ) = 61.3 m2/s2
v0 = 7.83 m/s b
Note: A larger velocity at A would result in a velocity at C greater than 3.5 m/s.
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104
PROBLEM 13.76
A small package of weight W is projected into a
vertical return loop at A with a velocity v0. The
package travels without friction along a circle
of radius r and is deposited on a horizontal
surface at C.. For each of the two loops shown,
determine (a)) the smallest velocity v0 for which
the package will reach the horizontal surface at
C, (b) the corresponding
orresponding force exerted by the
loop on the package as it passes Point B.
SOLUTION
Loop 1:
(a)
Newton’s second law at position C:
F  ma:
v2
mg  m c
r
vc2  gr
Conservation of energy between position A and B.
1 2
mv0
2
VA  0
TA 
1 2
1
mvC  mgr
2
2
VC  mg (2r )  2mgr
TC 
TA  VA  TC  VC :
1 2
1
mv0  0  mgr  2mgr
2
2
v02  5gr
Smallest velocity v 0:
v0 
5 gr

(b) Conservation of energy between positions A and B.
(b) TB 
1 2
mvB ;
2
VB  mg (r )
TA  VA  TB  VB:
1 2
1
mvA  0  mvB2  mgr
2
2
1
1
m(5 gr )  0  mvB2  mgr 
2
2
vB2  3gr

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105
PROBLEM 13.76 (C
(Continued)
Newton’s second law at position B.
v2
3gr
man  m B  m
 3 mg
r
r
F  Feff :
N B  3 mg
N B  3W
Force exerted by the loop:

Loop 2:
(aa)
At point C, vc  0
Conservation of energy between positions A and C..
1 2
mvC  0
2
VC  mg (2r )  2mgr
TC 
TA  VA  TC  VC :
1 2
mv0  0  0  2mgr
2
v02  4 gr
Smallest velocity v 0:
(bb)
v0 
4 gr

Conservation of energy between positions A and B.
B
TA  VA  TB  VB:
1 2
1
mv0  0  mvB2  mgr
2
2
1
1
m(4gr )  mvB2  mgr  vB2  2 gr
2
2
Newton’s second law at position B.
v2
2 gr
man  m B  m
 2 mg
r
r
F  eff : N  2 mg
Force exerted by loop:
N  2W
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
PROBLEM 13.77
The 1 kg ball at A is suspended by an inextensible cord and given an initial
horizontal velocity of 5 m/s. If l  0.6 m and xB  0, determine yB so that the
ball will enter the basket.
SOLUTION
Let position 1 be at A.
v1  v0
Let position 2 be the point described by the angle where the path of the ball changes from circular to
parabolic. At position 2, the tension Q in the cord is zero.
Relationship between v2 and  based on Q  0. Draw the free body diagram.
F  0: Q  mg sin   man 
With
Q  0, v22  gl sin 
mv22
l
(1)
or v2  gl sin 
Relationship among v0 , v2 and  based on conservation of energy.
T1  V1  T2  V2
1 2
1
mv0  mgl  mv22  mgl sin 
2
2
v02  v22  2 gl (1  sin  )
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(2)
PROBLEM 13.77 (Continued)
Eliminating v2 from Eqs. (1) and (2),
v02  gl sin   2 gl (1  sin  )
sin  
 1  (5)2

1  v02
 2   0.74912
  2  
3  gl

 3  (9.81)(0.6)
  48.514
From Eq. (1),
v22  (9.81)(0.6)sin 48.514  4.4093 m 2/s 2
v2  2.0998 m/s
x and y coordinates at position 2.
x2  l cos   0.6cos 48.514  0.39746 m
y2  l sin   0.6sin 48.514  0.44947 m
Let t2 be the time when the ball is a position 2.
Motion on the parabolic path. The horizontal motion is
x  v2 sin   2.0998 sin 48.514
 1.5730 m/s
x  x2  1.5730(t  t2 )
At Point B,
xB  0
0  0.39746  1.5730(t B  t2 ) tB  t2  0.25267 s
The vertical motion is
y  y2  v2 cos  (t  t2 ) 
At Point B,
1
g (t  t2 )2
2
yB  y2  v2 cos  (t B  t2 ) 
1
g (tB  t2 )2
2
yB  0.44947  (2.0998 cos 48.514)(0.25267)
1
 (9.81)(0.25267) 2
2
 0.48779 m
yB  0.448 m 
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108
PROBLEM 13.78
The pendulum shown is released from rest at A and swings through 90°
before the cord touches the fixed peg B.. Determine the smallest value of
a for which the pendulum bob will describe a circle about the peg.
SOLUTION
Use conservation of energy from the point of release ((A) and the top of the circle.
T1  V1  T2  V2
(1) (datum at lowest point)
where
T1  0;
V1  mg 
1
mv 2 ; V2  mgz  mg  2    a 
2
At 2
T2 
Substituting into (1)
0  mg  
1
m v2  2mg    a 
2
(2)
We need another equation – use Newton’s 2nd law at the top.  Tension, T0  0 at top
 Fn  man  m g 
m v2

v2  g   g    a 
Substituting into (2)
mg  
1
mg    a   2 mg    a 
2
2    a  4  4a
5 a  3
a
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3

5
PROBLEM 13.79*
Prove that a force F(x, y, z) is conservative if, and only if, the following relations are satisfied:
 Fx  Fy

y
x
 Fy
z

 Fz
y
 Fz  Fx

x
z
SOLUTION
For a conservative force, Equation (13.22) must be satisfied.
Fx  
We now write
Since
V
x
 Fx
 2V

y
 x y
Fy  
 Fy
x
V
y
Fz  

 2V
 y x
V
z
 Fx  Fy


y
x
 2V
 2V

:
 x y  y x
We obtain in a similar way
 Fy
z

 Fz
y
 Fz  Fx


x
z
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PROBLEM 13.80
The force F  ( yzi  zxj  xyk )/xyz acts on the particle P( x, y , z ) which moves in space. (a) Using the
relation derived in Problem 13.79, show that this force is a conservative force. (b) Determine the potential
function associated with F.
SOLUTION
Fx 
(a)
yz
xyz
Fy 
zx
xyz
Fz 
xy
xyz

1
1
 Fy  y
 Fx   x 

0

0
y
y
x
x
Thus,
 Fx  Fy

y
x
The other two equations derived in Problem 13.79 are checked in a similar way.
(b)
Recall that
Fx  
V
,
x
Fy  
V
,
y
Fz  
V
z
Fx 
1
V

x
x
V   ln x  f ( y , z )
(1)
Fy 
1
V

y
y
V   ln y  g ( z , x)
(2)
Fz 
1
V

z
z
V   ln z  h( x, y )
(3)
Equating (1) and (2)
 ln x  f ( y, z )   ln y  g ( z , x)
Thus,
f ( y, z )   ln y  k ( z )
(4)
g ( z , x)   ln x  k ( z )
(5)
Equating (2) and (3)
 ln z  h( x, y )   ln y  g ( z , x)
g ( z , x )   ln z  l ( x)
From (5),
g ( z , x)   ln x  k ( z )
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PROBLEM 13.80 (Continued)
Thus,
k ( z )   ln z
l ( x)   ln x
From (4),
f ( y, z )   ln y  ln z
Substitute for f ( y, z ) in (1)
V   ln x  ln y  ln z
V   ln xyz 

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112
PROBLEM 13.81*
A force F acts on a particle P(x, y) which moves in the xy plane.
Determine whether F is a conservative force and compute thework of
F when P describes the path ABCA knowing that
(a) F   kx  y  i   kx  y  j, (b) F   kx  y  i   x  ky  j.
SOLUTION
(a)
a
U AB   0 kxdx  k
a2
2
Fx  Fy ,  F isnormal to BC, U BC  0
U CA   0   a  u du 
a
U ABCA   k  1
(b) From Problem 13.79,
a 2
2
a2
, not conservative 
2
Fy
Fx
1
y
x
Conservative, U ABCA  0 
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PROBLEM 13.82*
The potential function associated with a force P in space is known
to be V ( x, y, z)  ( x 2  y 2  z 2 )1/2. (a) Determine the x, y, and z
components of P.(b) Calculate the work done by P from O to D by
integrating along the path OABD, and show that it is equal to the
negative of the change in potential from O to D.
SOLUTION
(a)
(b)
Px  
V
 [ ( x 2  y 2  z 2 )1/ 2 ]

 x ( x 2  y 2  z 2 ) 1/ 2
x
x

Py  
V
 [ ( x 2  y 2  z 2 )1/ 2 ]

 y ( x 2  y 2  z 2 ) 1/ 2
y
y

Pz  
V
 [ ( x 2  y 2  z 2 )1/ 2 ]

 z ( x 2  y 2  z 2 ) 1/ 2
z
z

U OABD  U OA  U AB  U BD
O–A: Py and Px are perpendicular to O–A and do no work.
x  y  0 and Pz  1
Also, on O–A
UO A 
Thus,

a
0
Pz dz 
a
 dz  a
0
A–B: Pz and Py are perpendicular to A–B and do no work.
y  0, z  a and Px 
Also, on A–B
Thus,
U A B 
x
(x  a 2 )1/2
2
xdx
a
 (x  a )
0
2
2 1/ 2
 a ( 2  1)
B–D: Px and Pz are perpendicular to B–D and do no work.
On
BD,
ka
za
Py 
y
( y  2a 2 )1/2
2
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PROBLEM 13.82* (Continued)
Thus,
U BD 

a
y
dy  ( y 2  2a 2 )1/ 2
2
1/
2
0
0 ( y  2a )
a
2
U BD  (a 2  2a 2 )1/ 2  (2a 2 )1/ 2  a ( 3  2)
U OABD  U O  A  U A B  U B  D
 a  a ( 2  1)  a ( 3  2)
UOABD  a 3 
VOD  V ( a, a, a )  V (0, 0, 0)
 (a 2  a 2  a 2 )1/ 2  0
Thus,
VOD   a 3 
U OABD  VOD
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PROBLEM 13.83*
((a) Calculate the work done from D to O by the force P of
Problem 13.82 by integrating along the diagonal of the cube.
((b) Using the result obtained and the answer to part b of Problem
13.82, verify that the work done by a conservative force around
the closed path OABDO is zero.
PROBLEM 13.82 The potential function associated with a force
1/ 2
.
P in space is known to be V(x, y, z)  ( x2  y 2  z 2 )1/2
((a) Determine the x, y, and z components of P.(b) Calculate the
work done by P from O to D by integrating along the path OABD,
and show that it is equal to the negative of the change in potential
from O to D.
SOLUTION
From solution to (a) of Problem 13.82
P
(a)
U OD 
xi  yj  zk
( x  y 2  z 2 )1/ 2
2
D
 P  dr
O
r  xi  yj  zk
dr  dxi  dyj  dzk
P
xi  yj  zk
( x  y 2  z 2 )1/2
2
Along the diagonal.
Thus,
xyz
P  dr 
3x
 3
(3 x 2 )1/ 2
UOD 

a
0
UOD  3a 
3 dx  3a
U OABDO  U OABD  U DO
(b)
From Problem 13.82
UOABD  3a
at left
The work done from D to O along the diagonal is the negative of the work done from O to D.
D
Thus,
U DO  U OD   3a
[see part (a)]
U OABDO  3a  3a  0
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
PROBLEM 13.84*
The force F  ( xi  yj  zk )/(x2  y 2  z 2 )3/2 acts on the particle P( x, y, z ) which moves in space. (a) Using
the relations derived in Problem 13.79, prove that F is a conservative force. (b) Determine the potential
function V(x, y, z) associated with F.
SOLUTION
x
Fx 
(a)
2
2
( x  y  z 2 )3/2
y
Fy  2
2
( x  y  z 2 )1/2
x   32  (2 y )
 Fx
 2
 y ( x  y 2  z 2 )5/2
 Fy
x
Thus,

y   32  2 y
( x 2  y 2  z 2 )5/2
 Fx  Fy

y
x
The other two equations derived in Problem 13.79 are checked in a similar fashion.
(b)
Recalling that
V
V
V
, Fy  
, Fz  
x
y
z
V
x
Fx  
V 
dx
2
2
x
( x  y  z 2 )3/2
Fx  

V  ( x 2  y 2  z 2 ) 1/2  f ( y , z )
Similarly integrating  V/ y and  V/ z shows that the unknown function f ( x, y ) is a constant.
V

1
2
2
( x  y  z 2 )1/2
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
PROBLEM 13.85
(a) Determine the kinetic energy per unit mass which a missile must have after being fired from the surface of
the earth if it is to reach an infinite distance from the earth. (b) What is the initial velocity of the missile
(called the escape velocity)? Give your answers in SI units and show that the answer to part b is independent
of the firing angle.
SOLUTION
g  9.81 m/s2
At the surface of the earth,
r1  R  6370 km  6.37 106 m
Centric force at the surface of the earth,
GMm
R2
GM  gR 2  (9.81)(6.37  106 )2  398.06  1012 m3 /s2
F  mg 
Let position 1 be on the surface of the earth ( r1  R ) and position 2 be at r2  OD. Apply the conservation of
energy principle.
T1  V1  T2  V2
1 2 GMm 1 2 GMm
mv1 
 mv2 
2
r1
2
r2
GMm GMm


R
T1 T2 GM T2
 
  gR
m m
R
m
T1  T2 
For the escape condition set
T2
0
m
T1
 gR  (9.81 m/s 2 )(6.37  106 m)  62.49  106 m2 /s 2
m
T1
 62.5 MJ/kg 
m
(a)
1 2
mvesc  mgr
2
vesc  2 gR
(b)
vesc  (2)(9.81)(6.37  106 )  11.18  103 m/s
vesc  11.18 km/s 
Note that the escape condition depends only on the speed in position 1 and is independent of the direction of
the velocity (firing angle).
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PROBLEM 13.86
A satellite describes an elliptic orbit of minimum altitude 606 km
above the surface of the earth. The semimajor and semiminor axes
are 17,440 km and 13,950 km, respectively. Knowing that the speed
of the satellite at Point C is 4.78 km/s, determine (a) the speed at
Point A, the perigee, (b) the speed at Point B, the apogee.
SOLUTION
rA  6370  606  6976 km  6.976  106 m
rC  (17440  6976)2  (13950) 2  17438.4 km  17.4384  106 m
rB  (2)(17440)  6976  27904 km  27.904  106 m
For earth,
R  6370 km  6.37 106 m
GM  gR2  (9.81 m/s2 )(6.370  106 )2  398.06 1012 m3 /s 2
vC  4.78 km/s  4780 m/s
(a)
Speed at Point A: Use conservation of energy.
TA  VA  TC  VC
1 2 GMm 1 2 GMm
mvA 
 mvC 
2
rA
2
rC
 1
1 
v A2  vC2  2GM   
 rA rC 
1
1


 (4780) 2  (2)(398.06  1012 ) 

6
6
6.976
10
17.4384
10




 91.318  106 m2 /s 2
VA  9.556  103 m/s
vA  9.56 km/s 
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PROBLEM 13.86 (Continued)
(b)
Speed at Point B: Use conservation of energy.
TB  VB  TC  VC
1 2 GMm 1 2 GMm
 mvC 
mvB 
2
rB
2
rC
1
1 
vB2  vC2  2GM   
 rB rC 
1
1


 (4780) 2  (2)(398.06  1012 ) 

6
6
 27.904  10 17.4384  10 
 5.7258  106 m 2 /s 2
vB  2.39  103 m/s
vB  2.39 km/s 
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PROBLEM 13.87
While describing a circular orbit 300 km above the earth, a space vehicle launches a 3600-kg communications
satellite. Determine (a) the additional energy required to place the satellite in a geosynchronous orbit at an
altitude of 35,770 km above the surface of the earth, (b) the energy required to place the satellite in the same
orbit by launching it from the surface of the earth, excluding the energy needed to overcome air resistance.
(A geosynchronous orbit is a circular orbit in which the satellite appears stationary with respect to the ground.)
SOLutiOn
For any circular orbit of radius r, the total energy
E = T +V =
1 2 GMm
mv 2
r
M = mass of the earth
m = 3600 kg = satellite mass
Newton’s second law
F = man :
GMm
r
2
=
mv 2
r
v2 =
GM
r
GM
GMm
1 2
mv = m
V =2
2r
r
1 GMm GMm
1 GMm
E = T +V =
=2 r
2 r
r
2
1 gRE m
GM = gRE2
E=2 r
2
1 (9.81 m/s )(6370 ¥ 103 m)2 (3600 kg)
E=2
r
716.15 ¥ 1015
( N ◊ m)
E=r
T=
For a geosynchronous orbit.
( r2 = 42.140 ¥ 106 m)
EGS = -
716 ¥ 1015
42.140 ¥ 106
= - 17.003 ¥ 109 J
= - 17.003 GJ
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121
PROBLEM 13.87 (continued)
(a)
At 300 km ( r1 = 6.67 ¥ 106 m).
E300 = -
716 ¥ 1015
6.67 ¥ 106
= - 107.42 ¥ 109 J
= - 107.42 GJ
Additional energy
DE300 = EGS - E300
DE300 = - 17.003 + 107.42
(b)
DE300 = 90.4 GJ b
Launch from the earth ( RE = 6370 km).
At launch pad,
EE = V = -
gR2 m
GMm
=- E
RE
RE
E E = - (9.81 m/s2 )(6370 ¥ 103 m)(3600 kg)
E E = - 224.96 ¥ 109 J = - 224.96 GJ
Additional energy
DE E = EGS - ES
DE E = - 17.003 + 224.96
DE E = 208 GJ b
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122
PROBLEM 13.88
How much energy per kilogram should be imparted to a satellite in order to place it in a circular orbit at an
altitude of (a) 600 km, (b) 6000 km?
SOLUTION
r1 = R = 6.37 ¥ 106 m ; v1 = 0
Before launching:
E1 = T1 + V1 = 0 In circular orbit of radius r2 .
GMm
gR2 m
== - mgR
R
R
[cf. Eq. 12.30]
Newton’s second law
F = man :
v2
GMm
=m 2
2
r2
r2
GM gR2
=
r2
r2
1
GMm
E2 = T2 + V2 = m v22 2
r2
v22 =
E2 =
Energy imparted is
1 gR2 gR2 m
1 gR2 m
=m
2 r2
2
r2
r2
DE = E2 - E1
=-
1 gR2 m
- ( - mgR)
2 r2
Ê
Rˆ
= Rmg Á1 2r2 ˜¯
Ë
Energy per kg is
(a)
(b)
D
Ê
E
Rˆ
= Rg Á1 m
Ë 2r2 ˜¯
D
Ê
E
(6370) ˆ
= (6.37 ¥ 106 )(9.81) Á1 m
Ë
2(6970) ˜¯
D
E
= 33.9 MJ/kg b
m
D
Ê
E
6370 ˆ
= (6.37 ¥ 106 )(9.81) Á1 m
Ë
2(12, 370) ˜¯
D
E
= 46.4 MJ/kg b
m
r2 = 6370 + 600 = 6970 km
r2 = 6370 + 6000 = 12,370 km
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123
PROBLEM 13.89
Knowing that the velocity of an experimental space probe fired
from the earth has a magnitude vA  32.5 Mm/h at Point A,
determine the speed of the probe as it passes through Point B.
SOLUTION
rA  R  hA  6370  4300  10740 km  10.670 106 m
rB  6370  12700  19070 km  19.070 106 m
GM  gR2  (9.81)(6.370  106 )2  398.06  1012 m3 /s2
vA  32.5 Mm/h  9.0278  103 m/s
Use conservation of energy.
TB  VB  TA  VA
1 2 GMm 1 2 GMm
mvB 
 mv A 
2
rB
2
rA
1 1
vB2  v A2  2GM   
 rB rA 
1
1


 (9.0278  103 ) 2  (2)(398.06  1012 ) 

6
6
19.070  10 10.670  10 
 48.635  106 m 2 /s 2
vB  6.97  103 m/s
vB  25.1 Mm/h 
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PROBLEM 13.90
A spacecraft is describing a circular orbit at an altitude of 1500 km above
the surface of the earth. As it passes through Point A, its speed is reduced by
40 percent and it enters an elliptic crash trajectory with the apogee at Point
A. Neglecting air resistance, determine the speed of the spacecraft when it
reaches the earth’s surface at Point B.
SOLUTION
Circular orbit velocity
vC2
GM
 2 , GM  gR 2
r
r
vC2 
GM
gR 2
(9.81 m/s 2 )(6.370  106 m)2


r
r
(6.370  106 m  1.500  106 m)
vC2  50.579  106 m2/s2
vC  7112 m/s
Velocity reduced to 60% of vC gives vA  4267 m/s.
Conservation of energy:
TA  VA  TB  VB
1
GM m
1
GM m
m v A2 
 m vB2 
2
rA
2
rB
1
9.81(6.370  106 ) 2
vB2 9.81(6.370  106 ) 2


(4.267  103 ) 2 
2
2
(7.870  106 )
(6.370  106 )
vB  6.48  103 m/s
vB  6.48 km/s 
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PROBLEM 13.91
Observations show that a celestial body traveling at 2 ¥ 106 km/h appears to be describing about Point B a circle
of radius equal to 60 light years. Point B is suspected of being a very dense concentration of mass called a black
hole. Determine the ratio MB/MS of the mass at B to the mass of the sun. (The mass of the sun is 330,000 times
the mass of the earth, and a light year is the distance traveled by light in one year at a speed of 3 ¥ 105 km/s.)
SOLutiOn
One light year is the distance traveled by light in one year.
Speed of light = 3 ¥ 108 m/s
r = (60 ¥ 365 ¥ 24 ¥ 3600) s ¥ (3 ¥ 108 m/s)
r = 5.67648 ¥ 1017 m
Newton’s second law
F=
GM B m
r2
=m
v2
r
rv 2
G
2
GM earth = gRearth
MB =
= (9.81 m/s2 )(6.37 ¥ 106 m)2
= 3.98059 ¥ 1014 ( m3 /s2 )
M sun = 330, 000 M E : GM sun = 330, 000 GM earth
GM sun = (330, 000)(3.98059 ¥ 1014 )
= 1.313596 ¥ 1020 m3 /s2
G=
1.31359 ¥ 1020
M sun
MB =
rv 2 M sun
rv 2
=
G 1.31359 ¥ 1020
(5.67648 ¥ 1017 )(5 ¥ 106 )2
MB
=
M sun
(81) ¥ 1.31359 ¥ 1020
MB
= 1.334 ¥ 109 b
M sun
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126
PROBLEM 13.92
(a) Show that, by setting r  R  y in the right-hand member of Eq. (13.17) and expanding that member in a
power series in y/R, the expression in Eq. (13.16) for the potential energy Vg due to gravity is a first-order
approximation for the expression given in Eq. (13.17). (b) Using the same expansion, derive a second-order
approximation for Vg.
SOLUTION
Vg  
WR 2
WR 2
WR
setting r  R  y : Vg  

r
R y
1  Ry
y

Vg  WR 1  
R


1
 (1) y ( 1)(2)  y 2

 WR 1 

 


1 R
1 2  R 


We add the constant WR, which is equivalent to changing the datum from r   to r  R :
 y  y 2

Vg  WR      
 R  R 

(a)
First order approximation:
 y
Vg  WR    Wy 
R
[Equation 13.16]
(b)
Second order approximation:
 y  y 2 
Vg  WR     
 R  R  
Vg  Wy 
Wy 2

R
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PROBLEM 13.93
Collar A has a mass of 3 kg and is attached to a spring of constant
1200 N/m and of undeformed length equal to 0.5 m. The system is set
in motion with r  0.3 m, v 2 m/s, and vr  0. Neglecting the mass
of the rod and the effect of friction, determine the radial and
transverse components of the velocity of the collar when r  0.6 m.
SOLUTION
Let position 1 be the initial position.
r1  0.3 m
(vr )1  0,
(v )1  2 m/s,
v1  2 m/s
x1  r1  l0  (0.3  0.5)  0.2 m
Let position 2 be when r  0.6 m.
r2  0.6 m
(vr ) 2  ?,
(v )2  ?,
v2  ?
x2  r2  l0  (0.6  0.5)  0.1 m
Conservation of angular momentum:
r1m(v )1  v2 m(v )2
r (v )
(0.3)(2)
 1.000 m/s
(v ) 2  1  1 
r2
0.6
Conservation of energy:
T1  V1  T2  V2
1 2 1 2 1 2 1 2
mv1  kx1  mv2  kx2
2
2
2
2
k 2
v22  v12 
x1  x22
m
1200 
 (2) 2 
(0.2) 2  (0.1) 2   16 m 2 /s 2

3
(vr )22  v22  (v ) 2  16  1  15 m 2 /s 2


vr  3.87 m/s
vr  3.87 m/s 
v  1.000 m/s 
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PROBLEM 13.94
Collar A has a mass of 3 kg and is attached to a spring of constant
1200 N/m and of undeformed length equal to 0.5 m. The system is set
in motion with r  0.3 m, v 2 m/s, and vr 0. Neglecting the mass
of the rod and the effect of friction, determine (a) the maximum
distance between the origin and the collar, (b) the corresponding
speed. (Hint: Solve the equation obtained for r by trial and error.)
SOLUTION
Let position 1 be the initial position.
r1  0.3 m
(vr )1  0,
(v )1  2 m/s,
v1  2 m/s
x1  r1  l0  0.3  0.5  0.2 m
1 2 1
mv  (3)(2) 2  6 J
2
2
1
1
V1  kx12  (1200)(0.2)2  24 J
2
2
T1 
Let position 2 be when r is maximum. (vr )2  0
r2  rm
x2  (rm  0.5)
1 2 1
mv2  (3)(v ) 22  1.5(v ) 22
2
2
1 2 1
V2  kx2  (1200)(rm  0.5) 2
2
2
T2 
 600(rm  0.5)2
Conservation of angular momentum:
r1m(v )1  v2 m(v )2
r
(0.3)
0.6
(v )2  1 (v ), 
(2) 
r2
rm
rm
Conservation of energy:
T1  V1  T2  V2
6  24  1.5(v ) 22  600(rm  0.5)2
2
 0.6 
2
30  (1.5) 
  600(rm  0.5)
r
 m 
0.54
f (rm )  2  600(rm  0.5) 2  30  0
rm
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PROBLEM 13.94 (Continued)
Solve for rm by trial and error.
rm (m)
0.5
1.0
0.8
0.7
0.72
0.71
f ( rm )
–27.8
120.5
24.8
–4.9
0.080
–2.469
rm  0.72 
(a)
Maximum distance.
(b)
Corresponding speed.
(0.01)(0.08)
 0.7197 m
2.467  0.08
rm  0.720 m 
(v )2 
0.6
 0.8337 m/s
0.7197
(vr )2  0
v2  0.834 m/s 
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PROBLEM 13.95
A 1.8-kg collar A and a 0.7-kg collar B can slide without friction
on a frame, consisting of the horizontal rod OE and the vertical
rod CD, which is free to rotate about CD. The two collars are
connected by a cord running over a pulley that is attached to
the frame at O. At the instant shown, the velocity vA of collar
A has a magnitude of 2.1 m/s and a stop prevents collar B from
moving. If the stop is suddenly removed, determine (a) the
velocity of collar A when it is 0.2 m from O, (b) the velocity of
collar A when collar B comes to rest. (Assume that collar B does
not hit O, that collar A does not come off rod OE, and that the
mass of the frame is negligible.)
SOLutiOn
(a)
Conservation of angular momentum about D, C.
(0.1 m)( mA )( v A ) = (0.2 m)( mA )( v A¢ )T
Ê 0.1ˆ
( v A¢ )T = Á
(2.1 m/s)
Ë 0.2 ˜¯
= 1.05 m/s
Conservation of energy.
1 v A = 2.1 m/s T1 = 1 (1.8 kg)(2.1 m/s)2 = 3.969 J
2
vB = 0
Choose datum for B at its initial position and note that the potential energy of A does not change. Thus,
we take V1 = 0 .
2 ( v A¢ )T = 1.050 m/s ( v A¢ ) R = v B¢
(kinematics)
1
1
mA ÈÎ( v A¢ )T2 + ( v A¢ )2R ˘˚ + mB ( v B¢ )2
2
2
1
1
T2 = (1.8 kg) ÈÎ(1.050 m//s)2 + ( v A¢ )2R ˘˚ + (0.7 kg)( v A¢ )2R
2
2
2
T2 = 0.9923 + 1.25( v A¢ ) R
T2 =
V2 = mB g (0.1 m) = (0.7 kg)(9.81 m/s2 )(0.1 m) = 0.6867 J
T1 + V1 = T2 + V2
3.969 + 0 = 0.9923 + 1.25( v A¢ )2R + 0.6867
( v A¢ )2R = 1.832 m2 /s2 ;
( v A¢ ) R = 1.354 m/s
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131
PROBLEM 13.95 (continued)
v A¢ = ( v A¢ )T2 + ( v A¢ )2R
= [(1.05)2 + (1.354)2 ]1/2
= 1.713 m/s
(v ¢ )
q = tan -1 A T
( v A¢ ) R
1.05
= tan -1
1.354
= 37.8∞
(b)
v A¢ = 1.713 m/s
37.8∞ b
When B comes to rest, the distance x moved by A and B is unknown.
Conservation of angular momentum about D, C.
(0.1 m)( mA )( v A ) = (0.1 m + x m)( mA )( v A¢ )T
( v A¢ ) R = ( v B¢ ) = 0 ,
Kinematics:
Thus,
( v A¢ )T = v A¢
v A = 2.1 m/s
(0.1)(2.1) = (0.1 + x )v A¢
0.21
- 0.1
x=
v A¢
Conservation of energy.
At 1 ,
v A = 21 m/s
vB = 0
1
T1 = mA v 2A
2
1
= (1.8 kg)(2.1)2
2
T1 = 3.969 J
V1 = 0
At 2 ,
v B¢ = 0, ( v A¢ ) R = 0
1
mA v A¢ 2 = 0.9v A¢ 2
2
V2 = mB gx = (0.7 kg)(9.81 m/s2 ) x
V2 = 6.867 x
T2 =
T1 + V1 = T2 + V2
3.969 + 0 = 0.9v A¢ 2 + 6.867 x
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132
PROBLEM 13.96
A 0.7-kg ball that can slide on a horizontal frictionless surface is attached
to a fixed point O by means of an elastic cord of constant k = 150 N/m
and undeformed length 600 mm. The ball is placed at point A, 800 mm
from O, and given an initial velocity v 0 perpendicular to OA. Determine
(a) the smallest allowable value of the initial speed v0 if the cord is not to
become slack, (b) the closest distance d that the ball will come to point O
if it is given half the initial speed found in part a.
SOLUTION
The cord will not go slack if v2 is perpendicular to the undeformed cord
length, L0 , at N
Conservation of angular momentum
0.8v1 = 0.6v2
v2 =
0.8
v1 = 1.333v0
0.6
T1 =
1 2
mv0 = 0.35v02
2
Conservation of energy
v1 = v0
Point M
V1 =
1
1
2
2
k ( L − L0 ) = (150 N/m )(0.8 m − 0.6 m ) 2
2
V1 = 3J
T2 =
Point N
∆L = 0
V =0
1 2
mv2 = 0.35v22
2
T1 + V1 = T2 + V2 : 0.35vB2 + 3 = 0.35v22 + 0
From conservation of angular momentum
v2 = 1.3158vB
2
0.35v02 (1.3158 ) − 1 = 3


v02 =
(3J )
= 11.72 m 2 /s 2
0.35
kg
0.7313
(
)(
)
v0 = 3.42 m/s W
continued
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PROBLEM 13.96 CONTINUED
The ball travels in a straight line after the cord goes slack.
Conservation of angular momentum
(0.8)(1.71) = dv
d =
1.368
v
Conservation of energy
v1 = 1.71 m/s
Point M
T1 =
1 2 1
2
mv1 = (0.7 kg )(1.71 m/s ) = 1.0234 J
2
2
V1 =
1
1
2
2
k ( L − L0 ) = (150 N/m )(0.8 m − 0.6 m ) = 3J
2
2
T3 =
Point O
1 2
mv3 = 0.35v 2
2
V3 = 0
T1 + V1 = T3 + V3 : 1.0234 + 3 = 0.35v 2 + 0
v = 3.39 m/s
From conservation of momentum
d =
1.368 1.368
=
= 404 mm
v
3.39
d = 404 mm W
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PROBLEM 13.97
A 0.7-kg ball that can slide on a horizontal frictionless surface is attached
to a fixed point O by means of an elastic cord of constant k = 150 N/m
and undeformed length 600 mm. The ball is placed at point A, 800 mm
from O, and given an initial velocity v 0 perpendicular to OA, allowing
the ball to come within a distance d = 270 mm of point O after the cord
has become slack. Determine (a) the initial speed v0 of the ball, (b) its
maximum speed.
SOLUTION
(a) Conservation of angular momentum: About O
0.8v0 = 0.27v
v = 2.963v0
Conservation of energy
v1 = v0
Point M
V1 =
T1 =
1 2
mv0 = 0.35v02
2
1
1
2
2
k ( L1 − L0 ) = (150 N/m )(0.8 m − 0.6 m ) 2
2
V1 = 3 J
v2 = v
Point N
T2 =
1 2
mv = 0.35v 2
2
V2 = 0 ( cord is slack )
T1 + V1 = T2 + V2 : 0.35v02 + 3 = 0.35v 2 + 0
From conservation of angular momentum,
v = 3.125v0
2
0.35v02 (3.125 ) − 1 = 3


v02 =
(3J )
(0.35 kg )(8.7656 )
v02 = 0.9779 m 2 /s 2
v0 = 0.989 m/s W
(b)
Maximum velocity occurs when the ball is at its minimum distance
from O, (when d = 0.27 m)
vm = 3.125v0 = (3.125 )( 0.9889 ) = 3.09 m/s
vm = 3.09 m/s W
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PROBLEM 13.98
Using the principles of conservation of energy and conservation of angular momentum, solve part a of Sample
Problem 12.9.
SOLUTION
R = 6370 km
r0 = 500 km + 6370 km
r0 = 6870 km = 6.87 × 106 m
v0 = 36,900 km/h
=
36.9 × 106 m
3.6 × 103 s
= 10.25 × 103 m/s
Conservation of angular momentum:
r0 mv0 = r1mv A,
r0 = rmin , r1 = rmax
⎛ 6.870 × 106 ⎞
⎛r ⎞
3
v A′ = ⎜ 0 ⎟ v0 = ⎜⎜
⎟⎟ (10.25 × 10 )
r
r
1
⎝ 1⎠
⎝
⎠
v A′ =
70.418 × 109
r1
(1)
Conservation of energy:
Point A:
v0 = 10.25 × 103 m/s
1 2 1
mv0 = m(10.25 × 103 ) 2
2
2
TA = (m)(52.53 × 106 )(J)
TA =
VA = −
GMm
r0
GM = gR 2 = (9.81 m/s 2 )(6.37 × 106 m) 2
GM = 398 × 1012 m3 /s 2
r0 = 6.87 × 106 m
VA = −
(398 × 1012 m3 /s 2 )m
(6.87 × 106 m)
= −57.93 × 106 m (J)
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136
PROBLEM 13.98 (Continued)
Point A′:
1 2
mv A′
2
GMm
VA′ = −
r1
TA′ =
=−
398 × 1012 m
(J)
r1
TA + VA = TA′ + VA′
52.53 × 106 m − 57.93 × 106 m =
1
398 × 1012 m
m v A2 ′ −
r1
2
Substituting for vA′ from (1)
−5.402 × 106 =
(70.418 × 109 )2 398 × 1012
−
r1
(2)(r1 ) 2
−5.402 × 106 =
(2.4793 × 1021 ) 398 × 1012
−
r1
r12
(5.402 × 106 )r12 − (398 × 1012 )r1 + 2.4793 × 1021 = 0
r1 = 66.7 × 106 m, 6.87 × 106 m
rmax = 66,700 km W
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137
PROBLEM 13.99
Solve sample Problem 13.8, assuming that the elastic cord is
replaced by a central force F of magnitude (80/r2) N directed
toward O.
PROBLEM 13.8 Skid marks on a drag racetrack indicate
that the rear (drive) wheels of a car slip for the first 20 m of
the 400-m track.(a) Knowing that the coefficient of kinetic
friction is 0.60, determine the speed of the car at the end of
the first 20-m portion of the track if it starts from rest and the
front wheels are just off the ground. (b) What is the maximum
theoretical speed for the car at the finish line if, after skidding
for 20 m, it is driven without the wheels slipping for the
remainder of the race? Assume that while the car is rolling
without slipping, 60 percent of the weight of the car is on the
rear wheels and the coefficient of static friction is 0.75.
Ignore air resistance and rolling resistance.
SOLUTION
(a)
The force exerted on the sphere passes through O. Angular momentum about O is conserved.
Minimum velocity is at B, where the distance from O is maximum.
Maximum velocity is at C, where distance from O is minimum.
rA mvA sin 60° = rm mvm
(0.5 m)(0.6 kg)(20 m/s)sin 60° = rm (0.6 kg)vm
vm =
8.66
rm
(1)
Conservation of energy:
At Point A,
1 2 1
mv A = (0.6 kg)(20 m/s)2 = 120 J
2
2
−80
80
V = Fdr = 2 dr =
,
r
r
−80
VA =
= −160 J
0.5
TA =
∫
At Point B,
TB =
∫
1 2 1
mvm = (0.6 kg)vm2 = 0.3vm2
2
2
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138
PROBLEM 13.99 (Continued)
and Point C :
VB =
−80
rm
TA + VA = TB + VB
120 − 160 = 0.3vm2 −
80
rm
(2)
Substitute (1) into (2)
2
⎛ 8.66 ⎞ 80
−40 = (0.3) ⎜
⎟ −
rm
⎝ rm ⎠
rm2 − 2 rm + 0.5625 = 0
rm′ = 0.339 m and rm = 1.661 m
rmax = 1.661 m W
rmin = 0.339 m W
(b)
Substitute rm′ and rm from results of part (a) into (1) to get corresponding maximum and minimum
values of the speed.
vm′ =
8.66
= 25.6 m/s
0.339
vmax = 25.6 m/s W
vm =
8.66
= 5.21 m/s
1.661
vmin = 5.21 m/s W
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139
PROBLEM 13.100
A spacecraft is describing an elliptic orbit of minimum altitude
hA  2400 km and maximum altitude hB  9600 km above
the surface of the earth. Determine the speed of the spacecraft
at A.
SOLUTION
rA  6370 km  2400 km
rA  8770 km
rB  6370 km  9600 km
 15,970 km
rA mvA  rB mvB
Conservation of momentum:
r
8770
vB  A v A 
v A  0.5492v A
rB
15,970
Conservation of energy:
TA 
1 2
mv A
2
VA 
GMm
rA
TB 
1 2
mvB
2
(1)
VB 
GMm
rB
GM  gR 2  (9.81 m/s 2 )(6370  103 m) 2  398.1  1012 m3/s 2
(398.1  1012 ) m
 45.39  106 m
3
8770  10
(398.1  1012 ) m
VB 
 24.93 m
(15,970  103 )
VA 
TA  VA  TB  VB :
1
1
m v A2  45.39  106 m  m vB2  24.93  106 m
2
2
(2)
Substituting for vB in (2) from (1)
v A2 [1  (0.5492)2 ]  40.92  106
v 2A  58.59  106 m2/s2
vA  7.65  103 m/s
vA  27.6  103 km/h 
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PROBLEM 13.101
While describing a circular orbit, 300 km above the surface of the
earth, a space shuttle ejects at Point A an inertial upper stage (IUS)
carrying a communication satellite to be placed in a geosynchronous
orbit (see Problem 13.87) at an altitude of 36000 km above the
surface of the earth. Determine (a) the velocity of the IUS relative
to the shuttle after its engine has been fired at A, (b) the increase in
velocity required at B to plae the satellite in its final orbit.
SOLUTION
For earth,
R = 6370 km = 6370 × 103 m
g = 9.81 m/s 2
GM = gR 2 = (9.81)(6370 × 103 )2 = 3.98059 × 1014 m3 /s 2
Speed on a circular orbits of radius r, rA, and rB.
B
F = man
GMm mv 2
=
r
r2
GM
v2 =
r
v=
GM
r
rA = (6370 + 300) km = 6670 × 103 m
(v A )circ =
3.98059 × 1014
= 7725.2 m/s
6670 × 103
rB = 6370 + 36000 = 42370 km = 42370 × 103 m
(vB )circ =
3.98059 × 1014
= 3065.1 m/s
42370 × 103
Calculate speeds at A and B for path AB.
Conservation of angular momentum: mrAvA sin φ A = mrB vB sin φ A
vB =
rAv A sin 90° 6670 × 103 v A
=
= 0.15742 v A
rB sin 90°
42370 × 103
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141
PROBLEM 13.101 (Continued)
Conservation of energy:
TA + VA = TB + VB
1 2 GMm 1 2 GMm
= mvB −
mv A +
rA
rB
2
2
⎛ 1 1 ⎞ 2GM (rB − rA )
v A2 − vB2 = 2GM ⎜ − ⎟ =
rA rB
⎝ rA rB ⎠
v A2 − (0.15742v A ) 2 =
(2)(3.98059 × 1014 )(35700 × 103 )
(6670 × 103 )(42370 × 103 )
0.97522v A2 = 1.0057 × 108
v A = 10.155 × 103 m/s
vB = (0.15742)(10.155 × 103 ) = 1.5986 × 103 m/s
(a)
(b)
Increase in speed at A:
Δv A = 10.155 × 103 − 7.7252 × 103 = 2.4298 × 103 m/s
Δv A = 2430 m/s W
ΔvB = 3.0651 × 103 − 1.5986 × 103 = 1.4665 × 103 m/s
ΔvB = 1470 m/s W
Increase in speed at B:
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142
PROBLEM 13.102
A spacecraft approaching the planet Saturn reaches Point A with a
velocity vA of magnitude 20 ¥ 103 m/s. It is to be placed in an elliptic orbit
about Saturn so that it will be able to periodically examine Tethys, one
of Saturn’s moons. Tethys is in a circular orbit of radius 300 ¥ 103 about
the center of Saturn, traveling at a speed of 11.1 ¥ 103. Determine (a) the
decrease in speed required by the spacecraft at A to achieve the desired
orbit, (b) the speed of the spacecraft when it reaches the orbit of Tethys at B.
SOLutiOn
(a)
rA = 185 ¥ 106 m
rB = 300 ¥ 106 m
v A¢ = speed of spacecraft in the elliptical orbit after its speed has been
decreased
Elliptical orbit between A and B.
Conservation of energy.
1
mv A¢ 2
2
-GM sat m
VA =
rA
Point A:
TA =
M sa = mass of Saturn; determine GM sa from the speed of Tethys in its
circular orbit.
vcirc =
(Eq. 12.44)
GM sat
r
2
GM sat = rB vcirc
GM sat = (300 ¥ 106 )(11.1 ¥ 103 m/s)2
= 3.6963 ¥ 1016 m3/s2
VA = -
(3.6963 ¥ 1016 m3/s2 ) m
(185 ¥ 106 m)
= -199.8 ¥ 106 m
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143
PROBLEM 13.102 (continued)
Point B:
TB =
-GM sat m
(3.6963 ¥ 1016 ) m
1 2
mv B VB =
=rB
2
(300 ¥ 106 )
VB = -123.21 ¥ 106 m
TA + VA = TB + VB ;
1
1
m v A¢ 2 - 199.8 ¥ 106 m = m v B2 - 123.21 ¥ 106 m
2
2
v A¢ 2 - v B2 = 76.59 ¥ 106
Conservation of angular momentum.
rA mv A¢ = rB mv B
r
185 ¥ 106
v B = A v A¢ =
v A¢ = 0.616667v A¢
rB
300 ¥ 106
v A¢ 2 [1 - (0.616667)2 ] = 76.59 ¥ 106
v A¢ 2 = 123.5877 ¥ 106
v A¢ = 11,117 m/s
(a)
Dv A = v A - v A¢ = (20, 000 - 11,117)
(b)
r
Ê 37 ˆ
v B = A v A¢ = Á ˜ (11,117)
Ë 60 ¯
rB
Dv A = 8, 880 m/s b
v B = 6860 m/s b
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144
PROBLEM 13.103
A spacecraft traveling along a parabolic path toward the planet Jupiter is
expected to reach Point A with a velocity vA of magnitude 26.9 km/s. Its
engines will then be fired to slow it down, placing it into an elliptic orbit
which will bring it to within 100  103 km of Jupiter. Determine the
decrease in speed v at Point A which will place the spacecraft into the
required orbit. The mass of Jupiter is 319 times the mass of the earth.
SOLUTION
Conservation of energy.
Point A:
1
m(v A  v A ) 2
2
GM J m
VA 
rA
TA 
GM J  319GM E  319 gRE2
RE  6.37  106 m
GM J  (319)(9.81 m/s 2 )(6.37  106 m) 2
GM J  126.98  1015 m3 /s 2
rA  350  106 m
VA 
(126.98  1015 m3 /s 2 )m
(350  106 m)
VA  (362.8  106 ) m
Point B:
1 2
mvB
2
GM J m (126.98  1015 m3 /s 2 )m
)m

VB 
rB
(100  106 m)
TB 
VB  (1269.8  106 )m
TA  VA  TB  VB
1
1
m(v A  v A ) 2  362.8  106 m  mvB2  1269.8  106 m
2
2
(vA  vA )2  vB2  1814 106
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(1)
PROBLEM 13.103 (Continued)
Conservation of angular momentum.
rA  350  106 m
rB  100  106 m
rA m(v A  v A )  rB mvB
r 
vB   A  (v A  v A )
 rB 
 350 

 (v A  v A )
 100 
(2)
Substitute vB in (2) into (1)
(v A  v A ) 2 [1  (3.5)2 ]  1814  106
(v A  v A ) 2  161.24  106
(v A  v A )  12.698  103 m/s
(Take positive root; negative root reverses flight direction.)
vA  26.9  103 m/s
(given)
vA  (26.9  103 m/s  12.698  103 m/s)
v A  14.20 km/s 
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146
PROBLEM 13.104
As a first approximation to the analysis of a space flight from the
earth to Mars, it is assumed that the orbits of the earth and Mars are
circular and coplanar. The mean distances from the sun to the
earth and to Mars are 149.6  106 km and 227.8  106 km,
respectively. To place the spacecraft into an elliptical transfer orbit
at Point A, itsspeed is increased over a short interval of time to vA
which is fasterthan the earth’s orbital speed. When the spacecraft
reaches Point B on the elliptical transfer orbit, its speed vB is
increased to the orbital speed of Mars. Knowing that the mass of
the sun is 332.8  103 times the mass of the earth, determine the
increase in velocity required (a) at A, (b) at B.
SOLUTION
M  mass of the sun
GM  332.8(10)3 (9.81 m/s2 )(6.37  106 m)2  1.3247(10)20 m3 /s2
Circular orbits
Earth vE 
GM
 29.758 m/s
149.6(10)9
Mars vM 
GM
 24.115 m/s
227.8(10)9
Conservation of angular momentum
Elliptical orbit
v A (149.6)  vB (227.8)
Conservation of energy
1 2
GM
1
GM
vA 
 vB2 
9
2
2
149.6(10)
227.8(10)9
v A  vB
(227.8)
 1.52273 vB
(149.6)
1
1.3247(10) 20
1 2 1.3247(10) 20
(1.52273) 2 vB2 

vB 
2
2
149.6(10)9
227.8(10)9
0.65935vB2  3.0398(10)8
vB2  4.6102(10)8
vB  21, 471 m/s, vA  32,695 m/s
(a)
Increase at A,
vA  vE  32.695  29.758  2.94 km/s 
(b)
Increase at B,
vB  vM  24.115  21.471  2.64 km/s 
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PROBLEM 13.105
The optimal way of transferring a space vehicle from an inner circular
orbit to an outer coplanar circular orbit is to fire its engines as it passes
through A to increase its speed and place it in an elliptic transfer orbit.
Another increase in speed as it passes through B will place it in the
desired circular orbit. For a vehicle in a circular orbit about the earth at
an altitude h1  200 mi, which is to be transferred to a circular orbit at
an altitude h2  500 mi, determine (a) the required increases in speed
at A and at B, (b) the total energy per unit mass required to execute the
transfer.
SOLUTION
Elliptical orbit between A and B
Conservation of angular momentum
mrAvA  mrBvB
vA 
rB
7.170
vB 
vB
rA
6.690
rA  6370 km  320 km  6690 km,
rA  6.690  106 m
vA  1.0718vB
rB  6370 km  800 km  7170 km,
(1)
rB  7.170  106 m
R  (6370 km)  6.37  106 m
Conservation of energy
GM  gR2  (9.81 m/s2 )(6.37  106 m)2  398.060  1012 m3 /s2
Point A:
TA 
1 2
mv A
2
VA  
GMm
(398.060  1012 )m

rA
(6.690  106 )
VA  59.501  106 m
Point B:
TB 
1 2
mvB
2
VB  
GMm
(398.060  1012 )m

rB
(7.170  106 )
VB  55.5  106 m
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PROBLEM 13.105 (Continued)
TA  VA  TB  VB
1 2
1
mv A  59.501  106 m  mvB2  55.5  106 m
2
2
v2A  vB2  8.002  106
From (1)
vB2 [(1.0718)2  1]  8.002  106
vA  1.0718vB
vB2  53.79  106 m2/s2 ,
vB  7334 m/s
vA  (1.0718)(7334 m/s)  7861 m/s
Circular orbit at A and B
(Equation 12.44)
(v A )C 
GM

rA
398.060  1012
 7714 m/s
6.690  106
(vB )C 
GM

rB
398.060  1012
 7451 m/s
7.170  106
(a) Increases in speed at A and B
vA  v A  (vA )C  7861  7714  147 m/s 
vB  (vB )C  vB  7451  7334  117 m/s 
(b) Total energy per unit mass
E/m 
E/m 
1
[(vA )2  (vA )C2  (vB )C2  (vB )2 ]
2
1
[(7861)2  (7714)2  (7451)2  (7334)2 ]
2
E/m  2.01  106 J/kg 
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149
PROBLEM 13.106
During a flyby of the earth, the velocity of a spacecraft is 10.4 km/s
as it reaches its minimum altitude of 990 km above the surface at
Point A. At Point B the spacecraft is observed to have an altitude of
8350 km. Determine (a) the magnitude of the velocity at Point B,
(b) the angle B .
SOLUTION
At A:
hA  vr  [1.04(10)4 m/s][6.37(10)6 m  0.990(10)6 m]
hA  76.544(10)9 m2 /s
1
1
GM
(TA  VA )  v 2 
m
2
r

1
(9.81)[6.37(10)6 ]2
[1.04(10)4 ]2 
0
2
[6.37(10)6  0.990(10)6 ]
(Parabolic orbit)
At B:
1
1
GM
(TB  VB )  vB2 
0
m
2
rB
1 2
(9.81)[6.37(106 )]2
vB 
2
[6.37(10)6  8.35(10)6 ]
vB2  54.084(10)6
vB  7.35 km/s 
(a)
hB  vB sin BrB  76.544(10)9
sin B 
76.544(10)9
7.35(106 )[6.37(10)6  8.35(10)6 ]
 0.707483
B  45.0 
(b)
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150
PROBLEM 13.107
A space platform is in a circular orbit about the earth at an
altitude of 300 km. As the platform passes through A, a rocket
carrying a communications satellite is launched from the
platform with a relative velocity of magnitude 3.44 km/s in a
direction tangent
ngent to the orbit of the platform. This was intended
to place the rocket in an elliptic transfer orbit bringing it to
Point B,, where the rocket would again be fired to place the
satellite in a geosynchronous orbit of radius 42,140 km. After
launching, it was discovered that the relative velocity imparted
to the rocket was too large. Determine the angle  at which the
rocket will cross the intended orbit at Point C.
SOLUTION
R  6370 km
rA  6370 km  300 km
rA  6.67  106 m
rC  42.14  106 m
GM  gR 2
GM  (9.81 m/s)(6.37  106 m) 2
GM  398.1  1012 m3 /s 2
For any circular orbit:
Fn  man 
2
mv circ
r
2
mv
GMm
 m circ
2
r
r
GM
vcirc 
r
Fn 
Velocity at A:
(v A )circ 
GM
(398.1  1012 m3/s3 )

 7.726 103 m/s
rA
(6.67  106 m)
v A  (v A ) circ  (v A ) R  7.726  103  3.44  103  11.165  103 m/s
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151
PROBLEM 13.107 (Continued)
Velocity at C:
Conservation of energy:
TA  VA  TC  VC
1
GM m 1
GM m
m v A2 
 m vC2 
2
rA
2
rC
 1 1
vC2  v 2A  2GM   
 rC rA 
1
1


 (11.165  103 ) 2  2(398.1  1012 ) 

6
6 
6.67  10 
 42.14  10
vC2  124.67  106  100.48  106
 24.19  106 m 2 /s 2
vC  4.919  103 m/s
Conservation of angular momentum:
rA mv A  rC mvC cos 
rv
cos   A A
rC vC
(6.67  106 )(11.165  103 )
(42.14  106 )(4.919  103 )
cos   0.35926

  68.9 
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152
PROBLEM 13.108
A satellite is projected into space with a velocity v0 at a distance r0 from the
center of the earth by the last stage of its launching rocket. The velocity v0
was designed to send the satellite into a circular orbit of radius r0.
However, owing to a malfunction of control, the satellite is not projected
horizontally but at an angle  with the horizontal and, as a result, is
propelled into an elliptic orbit. Determine
ermine the maximum and minimum
values of the distance from the center of the earth to the satellite.
SOLUTION
For circular orbit of radius r0
F  man
v02 
GM
r0
v02
GMm

m
r0
r02
But v0 forms an angle  with the intended circular path.
For elliptic orbit.
Conservation of angular momentum:
r0 mv0 cos   rA mvA
r

v A   0 cos   v0
 rA

Conservation of energy:
1 2 GMm 1 2 GMm
mv0 
 mv A 
2
r0
2
rA
r0 
2GM 
v02  v A2 
1  
r0  rA 
Substitute for v A from (1)
  r 2
 2GM 
r0 
v02 1   0  cos 2   
1  
r
r
r
  A

0
A 



But
v02 
GM
,
r0
2
thus
r 

r 
1   0  cos 2   2 1  0 
r
r
A 
 A

2
r 
r 
cos 2   0   2  0   1  0
r
 A
 rA 
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(1)
PROBLEM 13.108 (Continued)
Solving for
r0
rA
r0  2  4  44cos
cos 2 
1  sin 


2
rA
2cos 
1  sin 2 
rA 
(1  sin  )(1  sin  )
r0  (1  sin  )r0
1  sin 
also valid for Point A
Thus,
rmax  (1  sin  )r0
rmin  (1  sin  )r0 
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154
PROBLEM 13.109
A space vehicle is to rendezvous with an orbiting laboratory which circles
the earth at a constant altitude of 360 km. The shuttle has reached an
altitude of 60 km when its engine is shut off, and its velocity v0 forms an
angle 0 = 50° with the vertical OB at that time. What magnitude should v0
have if the shuttle’s trajectory is to be tangent at A to the orbit of the
laboratory?
SOLUTION
Conservation of energy
1 2 GMm 1 2 GMm
mv0 
 mv A 
2
rB
2
rA
v 2A  v02 
So
(1)
2GM 
rB 
1  
rB 
rA 
Given
R  6370 km = 6.37  106 m

GM  gR 2   9.81 6.37  106
  398  10
2
12
rA  6370  360  6730 km  6.73  106 m
Substitute into (1)
rB  6370  60  6430 km  6.43  106 m
vA2  v02 


2 398  1012 
6.43  106 
1



6.43  106 
6.73  106 
v2A  v02  5.518  106
(2)
We need another equation  conservation of angular momentum
rB mv sin   rA mvA
r v sin   6.43  106 
 
vA  B 0
6
 v0 sin 50 
rA
 6.73  10 
Substitute into (2)
vA  0.7319 v0
2
 0.7319 v0   v02  5.518  106
0.46433 v02  5.518  106
v0  3477 m/s
v0  3450 m/s 
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155
PRoBlEM 13.110
A space vehicle is in a circular orbit at an altitude of 360 km above the
earth. To return to the earth, it decreases its speed as it passes through A
by firing its engine for a short interval of time in a direction opposite to
the direction of its motion. Knowing that the velocity of the space vehicle
should form an angle f B = 60o with the vertical as it reaches Point B at
an altitude of 60 km, determine (a) the required speed of the vehicle as it
leaves its circular orbit at A, (b) its speed at Point B.
Solution
rA = 6370 + 360 = 6730 km = 6.73 ¥ 106 m
rB = 6370 + 60 = 6430 km
(a)
rB = 6.43 ¥ 106 m
R = 6.37 ¥ 106 m
GM = gR2 = (9.81) ¥ (6.37 ¥ 106 m)2
GM = 3.9806 ¥ 1014 m3/s2
Conservation of energy.
1 2
mv A
2
-GMm
VA =
rA
TA =
-3.9806 ¥ 1014
6.73 ¥ 106
= -59.147 ¥ 106 m
=
1 2
mv B
2
-GMm
VB =
rB
TB =
-3.9806 ¥ 1014
6.43 ¥ 106
= -61.907 ¥ 106 m
TA + VA = TB + VB
1
1 2
mv A - 59.147 ¥ 106 m = mv B2 - 61.907 ¥ 106 m
2
2
=
v 2A = v B2 - 2.76 ¥ 106
(1)
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156
PRoBlEM 13.110 (continued)
Conservation of angular momentum.
rA mv A = rB mv B sin f B
vB =
( rA )v A
6730 Ê 1 ˆ
=
˜ vA
Á
( rB )(sin f B ) 6430 Ë sin 60∞ ¯
v B = 1.2086 v A
(2)
Substituting v B from (2) in (1)
v 2A = (1.2086v A )2 - 2.76 ¥ 106
v 2A [(1.2086)2 - 1] = 2.76 ¥ 106
v 2A = 5.9907 ¥ 106 m2 /s2
VA = 2447.6 m/s
v A = 2450 m/s b
(a)
(b)
From (2)
v B = 1.2086v A
= 1.2086(2447.6)
= 2958.16 m/s
v B = 2960 m/s b
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157
PRoBlEM 13.111*
In Problem 13.110, the speed of the space vehicle was decreased as it
passed through A by firing its engine in a direction opposite to the
direction of motion. An alternative strategy for taking the space vehicle
out of its circular orbit would be to turn it around so that its engine would
point away from the earth and then give it an incremental velocity Dv A
toward the center O of the earth. This would likely require a smaller
expenditure of energy when firing the engine at A, but might result in too
fast a descent at B. Assuming this strategy is used with only 50 percent of
the energy expenditure used in Problem 13.110, determine the resulting
values of f B and u B .
Solution
rA = 6370 + 360 = 6730 km = 6.73 ¥ 106 m
rB = 6370 + 60 = 6430 km
rB = 6.43 ¥ 106 m
R = 6.37 ¥ 106 m
GM = gR2 = (9.81) ¥ (6.37 ¥ 106 m)2
GM = 3.9806 ¥ 1014 m3/s2
Velocity in circular orbit at 360 km altitude.
Newton’s second law
GMm m( v A )2circ
=
rA
rA2
F = man :
( v A )circ =
GM
rA
3.9806 ¥ 1014
6.73 ¥ 106
= 7690.72 m/s
=
Energy expenditure.
From Problem 13.109,
v A = 2447.6 m/s
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673
158
PRoBlEM 13.111* (continued)
1
1
m( v A )2circ - mv 2A
2
2
1
1
DE109 = m(7690.72)2 - m(24476)2
2
2
6
DE109 = (26.5782 ¥ 10 ) m J
DE109 =
Energy,
Ê 26.5782 ¥ 106 ˆ
DE110 = (0.50) DE109 = Á
˜mJ
2
Ë
¯
Thus, additional kinetic energy at A is
1
m( Dv A )2 = DE110 = (13.2891 ¥ 106 ) m J
2
(1)
Conservation of energy between A and B.
TA =
1
m[( v A )2circ + ( Dv A )2 ]
2
TB =
1 2
mv B
2
VB =
VA =
-GMm
rA
-GMm
rA
TA + VA = TB + VB
1
1
m (7690.72)2 + (13.2891 ¥ 106 ) m - 59.147 ¥ 106 m = m v B2 - 61.907 ¥ 106 m
2
2
v B2 = 91.2454 ¥ 106 m2 /s2
v B = 9552.2 m/s
v B = 9560 m/s b
Conservation of angular momentum between A and B.
rA m( v A )circ = rB mv B sin f B
Ê r ˆ (v )
(6730) (7690.72)
sin f B = Á A ˜ A circ =
= 0.8427
(6430) (9552.2)
Ë rB ¯ ( v B )
f B = 57.4∞ b
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674
159
PROBLEM 13.112
Show that the values vA and vP of the speed of an earth satellite at the
apogee A and the perigee P of an elliptic orbit are defined by the
relations
v 2A 
2GM rP
rA  rP rA
vP2 
2GM rA
rA  rP rP
where M is the mass of the earth, and rA and rP represent,
respectively, the maximum and minimum distances of the orbit to the
center of the earth.
SOLUTION
Conservation of angular momentum:
rA mvA  rP mvP
r
v A  P vP
rA
(1)
Conservation of energy:
1 2 GMm 1 2 GMm
mvP 
 mv A 
2
rP
2
rA
(2)
Substituting for v A from (1) into (2)
2
vP2 
2GM  rP  2 2GM
   vP 
rP
rA
 rA 
  r 2 


 1   P   vP2  2GM  1  1 
  rA  
 rP rA 


rA2  rP2
rA2
r r
vP2  2GM A P
rA rP
rA2  rP2  (rA  rP )(rA  rP )
with
vP2 
2GM  rA 
  (3) 
rA  rP  rP 
Exchanging subscripts P and A
v 2A 
2GM  rP 
 
rA  rP  rA 
Q.E.D.
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
PROBLEM 13.113
Show that the total energy E of an earth satellite of mass m
describing an elliptic orbit is E  GMm/(rA  rP ), where M is the
mass of the earth, and rA and rP represent, respectively, the
maximum and minimum distances of the orbit to the center of the
earth. (Recall that the gravitational potential energy of a satellite was
defined as being zero at an infinite distance from the earth.)
SOLUTION
See solution to Problem 13.112 (above) for derivation of Equation (3).
vP2 
2GM rA
(rA  rP ) rP
Total energy at Point P is
E  TP  VP 

1 2 GMm
mvP 
2
rP
1  2GMm rA  GMm


2  (rA  r0 ) rP 
rP

rA
1
 GMm 
 
 rP (rA  rP ) rP 
(r  r  r )
 GMm A A P
rP (rA  rP )
E
GMm

rA  rP
Note: Recall that gravitational potential of a satellite is defined as being zero at an infinite distance from the
earth.
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PROBLEM 13.114*
A space probe describes a circular orbit of radius nR with a velocity v0 about a planet of radius R and center O.
Show that (a)) in order for the probe to leave its orbit and hit the planet at an angle  with the vertical, its
velocity must be reduced to  v0, where
  sin 
2(n  1)
n  sin 2 
(b) the probe will not hit the planet if  is larger than
2 /(1  n).
2
SOLUTION
(a)
Conservation of energy:
1
m ( v0 ) 2
2
GMm
VA  
nR
At A:
TA 
1
mv 2
2
GMm
VB  
R
At B:
TB 
M mass of planet
m mass of probe
TA  VA  TB  VB
1
GMm 1
GMm
m ( v0 )2 
 mv 2 
2
nR
2
R
(1)
Conservation of angular momentum:
nR m v0  Rmv sin 
v
n v0
sin 
(2)
Replacing v in (1) by (2)
( v0 ) 2 
2
2GM  n v0 
2GM



nR
R
 sin  
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(3)
PROBLEM 13.114* (Continued)
For any circular orbit.
an 
v2
r
Newton’s second law
2
GMm m (v ) circ

r
r2
GM
vcirc 
r
v0  vcirc 
For r  nR,
GM
nR
Substitute for v0 in (3)
2
GM 2GM
n 2 2  GM  2GM




nR
nR
R
sin 2   nR 

n2 
 2 1  2   2(1  n)
 sin  
2(1  n)(sin 2  ) 2(n  1)sin 2 
2 
 2
(sin 2   n2 )
( n  sin 2  )
  sin 
(b)
2(n  1)
n  sin 2 
2
Q.E.D.

2
n 1

Probe will just miss the planet if   90,
  sin 90
Note:
2(n  1)

n 2  sin 2 90
n2  1  (n  1)(n  1) 
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163
PROBLEM 13.115
A missile is fired from the ground with an initial velocity v 0 forming an angle  0 with the vertical. If the
missile is to reach a maximum altitude equal to  R , where R is the radius of the earth, (a)) show that the
required angle  0 is defined by the relation
sin 0  (1   ) 1 
  vesc 


1    v0 
2
where vesc is the escape velocity, (b)) determine the range of allowable values of v0 .
SOLUTION
rA  R
(a)
Conservation of angular momentum:
Rmv0 sin 0  rB mvB
rB  R   R  (1   ) R
vB 
Rv0 sin 0 v0 sin 0

(1   ) R
(1   )
(1)
Conservation of energy:
1 2 GMm 1 2
GMm
mv0 
 mvB 
2
R
2
(1   ) R
TA  VA  TB  VB
v02  vB2 
2 GMm 
1  2 GMm   
1

R  1   
R  1   
Substitute for vB from (1)

sin 2 0  2 GMm   
v02 1 



 (1   )2 
R 1 


From Equation (12.43):
2
vesc

2GM
R

sin 2 0  2   
v02 1 
v
2 

 esc  1   
 (1   ) 
sin 2 0
2
v  
 1   esc 
2
(1   )
 v0  1  
sin 0  (1   ) 1 
  vesc 


1    v0 
(2)
2
Q.E.D.
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PROBLEM 13.115 (Continued)
(b)
Allowable values of v0 (for which maximum altitude  R)
0  sin 2 0  1
For sin 0  0, from (2)
2
v  
0  1   esc 
 v0  1  
v0  vesc

1 
For sin 0  1, from (2)
2
v  
1
 1   esc 
2
(1   )
 v0  1  
2
 vesc 
1
1  1  2   2  1 2  



  1   

1   
 (1   )
1
 v0 
v0  vesc
1
2 
vesc

1 
 v0  vesc

1 
2 
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PROBLEM 13.116
A spacecraft of mass m describes a circular orbit of radius r1 around
the earth. ((a) Show that the additional energy  E which must be
imparted to the spacecraft to transfer it to a circular orbit of larger
radius r2 is
E 
GMm(r2  r1 )
2r1r2
where M is the mass of the earth. (b)) Further show that if the transfer
from one circular orbit to the other is executed by placing the
spacecraft on a transitional semielliptic path AB,, the amounts of
energy  EA and  EB which must be imparted at A and B are,
respectively, proportional to r2 and r1 :
 EA 
r2
r1  r2
E
 EB 
r1
r1  r2
E
SOLUTION
(a)
For a circular orbit of radius r
F  man :
GMm
v2

m
r
r2
GM
v2 
r
1
GMm
1 GMm
E  T  V  mv2 

2
r
2 r
(1)
Thus  E required to pass from circular orbit of radius r1 to circular orbit of radius r2 is
1 GMm 1 GMm

2 r1
2 r2
GMm(r2  r1 )
Q.E.D.
E 
2r1r2
 E  E1  E2  
(b)
For an elliptic orbit, we recall Equation (3) derived in
Problem 13.113 (with vP  v1 )
v12 
2Gm r2
(r1  r2 ) r1
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(2)
PROBLEM 13.116 (Continued)
At Point A: Initially spacecraft is in a circular orbit of radius r1.
GM
2
vcirc

r1
1 2
1 GM
Tcirc  mvcirc  m
2
2
r1
After the spacecraft engines are fired and it is placed on a semi-elliptic path AB, we recall
and
v12 
2GM r2

(r1  r2 ) r1
T1 
2GMr2
1 2 1
mv1  m
2
2 r1 (r1  r2 )
At Point A, the increase in energy is
E A  T1  Tcirc 
Recall Equation (2):
2GMr2
1
1 GM
m
 m
2 r1 (r1  r2 ) 2
r1
E A 
GMm(2r2  r1  r2 ) GMm(r2  r1 )

2r1 ( r1  r2 )
2r1 ( r1  r2 )
E A 
r2  GMm(r2  r1 ) 


r1  r2 
2r1r2

E A 
r2
E
(r1  r2 )
Q.E.D.
r1
E
(r1  r2 )
Q.E.D.
A similar derivation at Point B yields,
EB 
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167
PROBLEM 13.117*
Using the answers obtained in Problem 13.108, show that the intended circular orbit and the resulting elliptic
orbit intersect at the ends of the minor axis of the elliptic orbit.
PROBLEM 13.108 A satellite is projected into space with a velocity v0 at a distance r0 from the center of the
earth by the last stage of its launching rocket. The velocity v0 was designed to send the satellite into a circular
orbit of radius r0. However, owing to a malfunction of control, the satellite is not projected horizontally but at
an angle  with the horizontal and, as a result, is propelled into an elliptic orbit. Determine the maximum and
minimum values of the distance from the center of the earth to the satellite.
SOLUTION
If the point of intersection P0 of the circular and elliptic orbits is at an end of
the minor axis, then v0 is parallel to the major axis. This will be the case
only if   90   0 , that is if cos0   sin  . We must therefore prove that
cos0   sin 
(1)
We recall from Equation (12.39):
1 GM
 2  C cos 
r
h
r  rmin
When   0,
and rmin  r0 (1  sin  )
1
GM
 2 C
r0 (1  sin  )
h

(2)
(3)
r  rmax  r0 (1  sin  )
For   180,
1
GM
 2 C
r0 (1  sin  )
h
(4)
Adding (3) and (4) and dividing by 2:
GM
1 
1
1





2


2
r
1
sin

1
sin

h

0 
1

r0 cos 2 
Subtracting (4) from (3) and dividing by 2:
C
1 
1
1
  1  2sin 




2r0  1  sin  1  sin    2r0  1  sin 2 
sin 
C
r0 cos 2 
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
PROBLEM 13.117* (Continued)
Substituting for GM2 and C into Equation (2)
h
1
1

(1  sin  cos  )
r r0 cos 2 
Letting r  r0 and   0 in Equation (5), we have
cos 2   1  sin  cos  0
cos 2   1
sin 
sin 2 

sin 
  sin 
cos 0 
This proves the validity of Equation (1) and thus P0 is an end of the minor axis of the elliptic orbit.
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169
(5)
PROBLEM 13.118*
(a) Express in terms of rmin and vmax the angular momentum per unit mass, h,, and the total energy per unit
mass, E/m,, of a space vehicle moving under the gravitational attraction of a planet of mass M (Figure 13.15).
(b) Eliminating vmax between the equations obtained, derive the formula
1
rmin

2
GM 
2 E  h  


1
1
m  GM  
h2 


(c) Show that the eccentricity  of the trajectory of the vehicle can be expressed as
  1
2E  h 
m  GM 
2
(d ) Further show that the trajectory of the vehicle is a hyperbola, an ellipse, or a parabola, depending on
whether E is positive, negative, or zero.
SOLUTION
(a)
Point A:
Angular momentum per unit mass.
H0
m
r mv
 min max
m
h
h  rmin vmax
(1) 
Energy per unit mass
E 1
 (T  V )
m m
E 1 1 2
GMm  1 2
GM
  mvmax 
  vmax 
m m2
rmin  2
rmin
(b)
From Eq. (1): vmax  h/rmin substituting into (2)
E 1 h 2 GM


2
m 2 rmin
rmin
E
2
2 
 1 
2GM 1
m
  2  0

  2 
r
r
h
h
min
 min 
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(2) 
PROBLEM 13.118* (Continued)
1
Solving the quadratic:
rmin

2
2  mE 
GM
 GM 


 2 
h2
h2
 h 

GM 
2E  h 
1 1
m  GM 
h2 

Rearranging
1
rmin
(c)
2
(3) 
Eccentricity of the trajectory:
Eq. (12.39)
1 GM
 2 (1   cos  )
r
h
When   0,
cos  1 and r  rmin
Thus,
1
rmin
Comparing (3) and (4),
(d )

GM
(1   )
h2
  1
(4)
2E  h 
m  GM 
2
(5)
Recalling discussion in section 12.12 and in view of Eq. (5)

1. Hyperbola if   1, that is, if E  0

2. Parabola if   1, that is, if E  0

3. Ellipse if   1, that is, if E  0

Note: For circular orbit   0 and
2
1
2E  h 
0
m  GM 
GM
r
2
 GM  m
E  
,

 h  2
or
but for circular orbit
v2 
thus
1 (GM ) 2
1 GMm

E m

2
GMr
2 r
and
h2  v2 r 2  GMr ,
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PROBLEM 13.119
A 35,000 Mg ocean liner has an initial velocity of 4 km/h. Neglecting the frictional resistance of the water,
determine the time required to bring the liner to rest by using a single tugboat which exerts a constant force of
150 kN.
SOLUTION
m  35,000 Mg  35 106 kg
F  150  103 N
v1  4 km/hr  1.1111 m/s
mv1  Ft  0
6
(35  10 kg)(1.1111 m/s)  (150  103 N)t  0
t  259.26 s
t  4 min19 s 
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PROBLEM 13.120
A 1200-kg automobile is moving at a speed of 90 km/h when the brakes are fully applied, causing all four
wheels to skid. Determine the time required to stop the automobile (a) on dry pavement (mk = 0.75), (b) on an
icy road (mk = 0.10).
SOLutiOn
mv1 - m kWt = 0
t=
(a)
For m k = 0.75
t=
(b)
mv1
mv1
v
=
= 1
m kW m k mg m k g
25 m/s
(0.75)(9.81 m/s2 )
t = 3.40 s b
For m k = 0.10
t=
25 m/s
(0.10)(9.81 m/s2 )
t = 25.5 s b
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173
PROBLEM 13.121
A sailboat of mass 500 kg with its occupants is running
downwind at 12 km/h when its spinnaker is raised to increase
its speed. Determine the net force provided by the spinnaker
over the 10-s interval that it takes for the boat to reach a speed
of 18 km/h.
SOLutiOn
10
m/s t1- 2 = 10 sec
3
v2 = 18 km/h = 5 m/s
v1 = 12 km/h =
m ◊ v1 + imp1- 2 = mv2
Ê 10 ˆ
m Á ˜ + Fn (10 s) = m(5)
Ë 3¯
Ê 10 ˆ
(500) Á 5 - ˜
Ë
3 ¯ 250
Fn =
=
= 83.33 N
10
3
Fn = 83.3 N b
Note: Fn is the net force provided by the sails. The force on the sails is actually greater and includes the force
needed to overcome the water resistance on the hull.
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174
PROBLEM 13.122
A truck is hauling a 300-kg log out of a ditch
using a winch attached to the back of the truck.
Knowing the winch applies a constant force of
2500 N and the coefficient of kinetic friction
between the ground and the log is 0.45,
determine the time for the log to reach a speed
of 0.5m/s.
SOLUTION
Apply the principle of impulse and momentum to the log.
mv1  Imp1 2  mv2
Components in y-direction:
0  Nt  mgt cos 20  0
N  mg cos 20
Components in x-direction:
0  Tt  mgt sin 20  k Nt  mv2
(T  mg sin 20  k mg cos 20)t  mv2
[T  mg (sin 20  k cos 20)]t  mv2
Data:
T  2500 N, m  300 kg,
2
g  9.81 m/s ,
 k  0.45,
v2  0.5 m/s
[2500  (300)(9.81)(sin 20  0.45 cos 20 )] t  (300)(0.5)
248.95 t  150
t  0.603 s 
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PROBLEM 13.123
The coefficients of friction between the load and the flatbed trailer shown
are µs = 0.40 and µk = 0.35. Knowing that the speed of the rig is
88 km/h, determine the shortest time in which the rig can be brought to a
stop if the load is not to shift.
SOLUTION
v1 = 88 km/h = 24.4 m/s
Use impulse momentum
x-Direction
m v1 − µs m gt = 0
t=
v1
24.4 m/s
=
µs g ( 0.4 ) 9.81 m/s 2
(
)
t = 6.229 s W
Since this is the shortest time the load can be brought to rest and the load does not slide it is also the shortest
time the rig can be brought to rest.
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PROBLEM 13.124
Steep safety ramps are built beside mountain highways to enable
vehicles with defective brakes to stop. A 10,000 kg truck enters a 15°
ramp at a high speed v0 = 30 m/s and travels for 6 s before its speed is
reduced to 10 m/s. Assuming constant deceleration, determine (a) the
magnitude of the braking force, (b) the additional time required for the
truck to stop. Neglect air resistance and rolling resistance.
SOLUTION
m = 10,000 kg
Momentum in the x direction
x : mv0 − ( F + mg sin15°)t = mv1
(10,000)(30) − ( F + mg sin15°)6 = (10,000)(10)
F + mg sin15° = 3.333 × 104
(a)
F = 3.333 × 104 − (10,000)(9.81)(sin15°) = 7939.9 N
F = 7940 N W
(b)
mv0 − ( F + mg sin15°)t = 0
t = total time
(10,000)(30) − (3.333 × 104 )t = 0;
t = 9.00 s
Additional time = 9 – 6
t = 3.00 s W
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177
PROBLEM 13.125
Baggage on the floor of the baggage car of a high-speed train is not prevented from moving other than by
friction. Determine the smallest allowable value of the coefficient of static friction between a trunk and the
floor of the car if the trunk is not to slide when the train decreases its speed at a constant rate from 200 km/h
to 90 km/h in a time interval of 12 s.
SOLutiOn
v1 = 200 km/h = 55.56 m/s
v2 = 90 km/h = 25.0 m/s
+
m v1 - m s m gt1- 2 = m v2
(55.56 m/s) - m s (9.81 m/s2 )(12 s) = 25 m/s
ms =
(55.56 - 25.0)
= 0.2596
(9.81)(12)
m s = 0.260 b
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178
PROBLEM 13.126
The 18000-kg F-35B
35B uses thrust vectoring to allow it
to take off vertically. In one maneuver, the pilot
reaches the top of her static hover at 200 m. The
combined thrust and lift force on the airplane applied
at the end of the static hover can be expressed as F=
(44t + 2500t2)i + (250t2 + t + 176580)j,
176580) where F and
t are expressed in newtons
ewtons and seconds, respectively.
Determine (a) how long it will take the airplane to
reach a cruising speed of 1000 km/hr (cruising speed
is defined to be in the x--direction only), (b) the
altitude of the plane at this time.
SOLUTION
Given:
y1  200 m
Impulse-Momentum Diagram:
m  18000 kg
Fx  44t  2500t 2
Fy  250t 2  t  176580
vx ,2  1000 km/hr  277.78 m/s

Impulse-Momentum in the x-direction:
direction:
0  Fx dt  mvx ,2
  44t  2500t dt  18000 * 277.78
t
2
0
2500 3
t  22t 2  5, 000, 000
3
33 2
t3 
t  6000  0
1250
(a) Solve for t:
Impulse-Momentum in the y-direction:
direction:
t  18.162 s 

0  Fy dt  mvy,2
  250t  t  176580 dt  mv
t
2
0
y,2
250 3 1 2
t  t  176580t  mvy,2
3
2
Solve for velocity as function of time:
vy,2 
1  250 3 1 2

t  t  176580t 
m  3
2

From kinematics:
vy,2 
dy
dt
 vy,2 dt  dy
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PROBLEM 13.126 (Continued)
Integrating:
t
y2
 v dt   dy
0
y,2
200
y2
1 18.162  250 3 1 2

t  t  176580t  dt 
dy

200
m 0
2
 3



1  125 4 1 3

t  t  88290t 2   y2  200

18000  6
6

(b) Find y2 for t=18.162 s:
y2  1944 m
y2  1.944 km 
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180
PROBLEM 13.127
A truck is traveling down a road with a 4-percent grade at a speed of 80 km/h when its brakes are applied to
slow it down to 30 km/h. An antiskid braking system limits the braking force to a value at which the wheels of
the truck are just about to slide. Knowing that the coefficient of static friction between the road and the wheels
is 0.60, determine the shortest time needed for the truck to slow down.
SOLutiOn
q = tan -1
4
= 2.29∞
100
+
mv1 + S imp1- 2 = mv 2
mv1 + mg sinθt - Ft = mv2
200
m/s
N = W cosq W = mg
9
25
v2 = 30 km/h =
m/s F = m s N = m s mg cosq
3
v1 = 80 km/h =
Ê 200 ˆ
Ê 25 ˆ
+ ( m )(9.81)(sin 2.29∞)(t ) - (0.60)( m )(9.81)(cos 2.29∞)(t ) = ( m ) Á ˜
( m )Á
Ë 9 ˜¯
Ë 3¯
25 200
3
9
t=
(9.81sin 2.29∞ - 0.6 ¥ 9.81 ¥ cos 2.29∞)
t = 2.53 s b
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181
PROBLEM 13.128
In anticipation of a long 6 upgrade, a bus driver accelerates at a constant rate from 80 km/h to 100 km/h in
8 s while still on a level section of the highway. Knowing that the speed of the bus is 100 km/h as it begins to
climb the grade at time t 0 and that the driver does not cha
change
nge the setting of the throttle or shift gears,
determine (a) the speed of the bus when t 10 s, (b) the time when the speed is 60 km/h.
SOLUTION
Given:
v2  100 km h  27.778 m s , v1  80 km h  22.222 m s , t  8 s
Impulse Momentum Diagram for acceleration on level ground:
Impulse-Momentum in the x-direction
direction while on level ground:
mv1 
8
 Fdt  mv
2
0
F  8   m  v2  v1   F  0.6944m
Impulse Momentum Diagram for acceleration on incline:
Impulse-Momentum in the x-direction
direction while on incline:
mv2 
10
  F  mg sin 6  dt  mv
3
0
m  0.6944   9.81 sin 6  10   m  v3  27.778 
v3  24.47 m/s
(a) Solve for v3:
mv2 
v3  88.08 km h 
t
  F  mg sin 6  dt  mv
0
3
m  0.6944   9.81 sin 6   t   m 16.667  27.778 
(b)
Solve for t:
t  33.57 s 
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182
PRoBlEM 13.129
A light train made of two cars travels at 72 km/h. Car A weighs
18000 kg, and car B weighs 13000 kg. When the brakes are
applied, a constant braking force of 21.5 kN is applied to each
car. Determine (a) the time required for the train to stop after
the brakes are applied, (b) the force in the coupling between the
cars while the train is slowing down.
Solution
(a)
v1 = 72 km/h = 20 m/s
Entire train:
mA + mB = (18, 000 + 13, 000) = 31, 000 kg
:0 = -(21500 + 21500)t1- 2 + (31, 000)(20)
+
(31, 000 kg)(20 m/s)
( 43000 N )
= 14.41865 s
t1-2 =
(b)
Car A:
t1- 2 = 14.42 s b
mA = 18000 kg, t1- 2 = 14.4186 s
: 0 = -( FB + FC )t1- 2 + mA v1
+
0 = -(21500 + FC )14.4186 + (18, 000)(20)
FC = 3467.7 N
FC = 3470N (tension) b
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183
PRoBlEM 13.130
Solve Problem 13.129, assuming that a constant braking force
of 21.5 kN is applied to car B, but the brakes on car A are not
applied.
PROBLEM 13.129 A light train made of two cars travels at
72 km/h. Car A weighs 18000 kg, and car B weighs 13000 kg.
When the brakes are applied, a constant braking force of
21.5 kN is applied to each car. Determine (a) the time required
for the train to stop after the brakes are applied, (b) the force in
the coupling between the cars while the train is slowing down.
Solution
(a)
Entire train:
v1 = 72 km/h = 20 m/s
mA + mB = 31, 000 kg
+ :0 = -( 21500 N )t
1- 2 + (31, 000 kg)( 20 m/s)
t1- 2 = 28.837 s
(b)
t1- 2 = 28.8 s b
Car A:
+ :0 = - F (t
C
FC =
1- 2 ) + mA v1
t1- 2 = 28.8375 s
(18, 000)(20)
= 12, 483.9 N
(28.837)
FC = 12, 480 N (tension) b
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184
PROBLEM 13.131
A tractor-trailer rig with a 2000-kg
kg tractor, a 4500-kg
4500
trailer,
and a3600-kg
kg trailer is traveling on a level road at 90 km/h. The
brakes on the rear trailer fail and the antiskid system of the
tractor and front trailer provide the largest possible force which
will not cause the wheels to slide. Knowing that the coefficient
of static friction is 0.75, determine (a)) the shortest time for the
rig to a come to a stop, (b) the force in the coupling between
the two trailers during that time. Assume that the force exerted
by the coupling on each of the two trailers is horizontal.
SOLUTION
v1  90 km h  25 m s , v2  0
Given:
W1   6500 9.81  63765 N; W2   3600  9.81  35316 N
Impulse Momentum diagram for combined cab and trailer:
From Equilibrium:
N1  W1; N2  W2
Impulse-Momentum in the x-direction:
direction:
mv1 
F  0.75 N1
t
 0.75N dt  mv
0
1
2
0.75 N1t  m  v2  v1 
t
t
(a)

m v1  v2

0.75N1
10,100  25 
 5.2798 s
 0.75 63765
t  5.28 s 
Impulse Momentum diagram for the trailer alone:
Impulse-Momentum in the x-direction:
direction:
mv1 

5.2798
0
Cdt  mv2
5.2798C  m  v1  0 
C

3600 25  v2
5.2798
  C  17046 N
(b)
C  17.05 kN Compression
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PROBLEM 13.132
The system shown is at rest when a constant 150-N
150 force
is applied to collar B.. Neglecting the effect of friction,
determine (a)) the time at which the velocity of collar B
will be 2.5 m/s to the left, (b)) the corresponding tension
in the cable.
SOLUTION
Constraint of cord. When the collar B moves 1 unit to the left, the weight A moves up 2 units. Thus
vA  2vB
Masses and weights:
1
vB  vA
2
m A  3 kg
WA  29.43 N
mB  8 kg
Let T be the tension in the cable.
Principle of impulse and momentum applied to collar B.
: 0  150t  2Tt  mB (vB )2
For (vB )2  2.5 m/s
150t  2Tt  (8 kg)(2.5 m/s)
150t  2Tt  20
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(1)
PROBLEM 13.132 (Continued)
Principle of impulse and momentum applied to weight A.
: 0  Tt  WAt  mA (vA )2
Tt  WAt  mA (2VB 2 )
Tt  29.43t  (3 kg)(2)(2.5 m/s)
Tt  29.43t  15
(2)
To eliminate T multiply Eq. (2) by 2 and add to Eq. (1).
(a)
(b)
Time:
91.14t  50
From Eq. (2),
T
t  0.549 s 
15
 29.43
t
T  56.8 N 
Tension in the cable.
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PROBLEM 13.133
An 8-kg
kg cylinder C rests on a 4-kg platform A supported by a cord which
passes over the pulleys D and E and is attached to a 4-kg
kg block B. Knowing
that the system is released from rest, determine ((a)) the velocity of block B
after 0.8 s, ((b) the force exerted by the cylinder on the platform.


SOLUTION
(a)
Blocks A and C:
[(mA  mC )v]1  T (t12 )  (mA  mC ) gt12  [(mA  mC )v]2
(1)
0  (12 g  T )(0.8)  12v
Block B:
[mB v]1  (T )t12  mB gt12  (mB v)2

(2)
0  (T  4 g )(0.8)  4v

Adding (1) and (2), (eliminating T)
(12 g  4 g )(0.8)  (12  4)

v
(b)
(8 kg)(9.81 m/s 2 )(0.8 s)
16 kg
v  3.92 m/s 
Collar A:
(mAv)1  0 0  ( FC  mA g )
(3)
From Eq. (2) with v  3.92 m/s
4v
T
 4g
0.8
(4 kg)(3.92 m/s)
 (4 kg)(9.81 m/s 2 )
T
(0.8 s)
T  58.84 N


Solving for FC in (3)
FC 

(4 kg)(3.92 m/s)
 (4 kg)(9.81 m/s 2 )  58.84 N
(0.8 s)
FC  39.2 N 
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PROBLEM 13.134
An estimate of the expected load on overthe-shoulder seat belts is to be made before
designing prototype belts that will be evaluated
in automobile crash tests. Assuming that an
automobile traveling at 72 km/h is brought to
a stop in 110 ms, determine (a) the average
impulsive force exerted by a 100 kg man on
the belt, (b) the maximum force Fm exerted on
the belt if the force-time diagram has the shape
shown.
SOLutiOn
(a)
Force on the belt is opposite to the direction shown.
v1 = 72 km/h = 20 m/s
W = 100 ¥ 9.81 = 981 N
mv1 - Ú Fdt = mv 2
Ú Fdt = Fave Dt
Dt = 110 ¥ 10 -3 = 0.110 s
(100)(20) - Fave (0.110 s) = 0
Fave =
(b)
(100)(20)
= 18,181.8N
(0.110)
Impulse = area under F - t diagram =
Fave = 18.18 kN b
1
Fm (0.110 s)
2
From (a), impulse = Fave Dt = 2000
1
Fm (0.110) = 2000
2
Fm = 36363.6 N
Fm = 36.4 kN b
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189
PROBLEM 13.135
A 60-g
g model rocket is fired vertically. The engine
applies a thrust P which varies in magnitude as shown.
Neglecting air resistance and the change in mass of the
rocket, determine (a) the maximum speed of the rocket
as it goes up, (b)) the time for the rocket to reach its
maximum elevation.
SOLUTION
Mass:
m  0.060 kg
Weight:
mg  (0.060)(9.81)  0.5886 N
Forces acting on the model rocket:
Thrust:
P (t )(given function of t )
Weight:
W (constant)
Support:
S (acts until P  W )
Over
0  t  0.2 s:
13
t  65t
0.2
W  0.5886 N
P
S  W  P  0.5886  65t
Before the rocket lifts off,
S become zero when t  t1.
0  0.5886  65t1
t1  0.009055 s.
Impulse due to S: (t  t1 )
t
t1
0
0
 Sdt   Sdt
1
mgt1
2
 (0.5)(0.5886)(0.009055)
 0.00266 N  s

The maximum speed occurs when
dv
 a  0.
dt
At this time, W  P  0, which occurs at t2  0.8 s.
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PROBLEM 13.135 (Continued)
(a)
Maximum speed (upward motion):
Apply the principle of impulse-momentum to the rocket over 0  t  t2 .
0.8
 Pdt  area under the given thrust-time plot.
0
1
1
(0.2)(13)  (0.1)(13  5)  (0.8  0.3)(5)
2
2
 4.7 N  s

0.8
 Wdt  (0.5886)(0.8)  0.47088 N  S
0
m1v1 

0.8
0
Pdt 

0.8
0
Sdt 
0.8
 Wdt  mv
2
0
0  4.7  0.00266  0.47088  0.060 v2
v2  70.5 m/s 
(b)
Time t3 to reach maximum height: (v3  0)
mv1 

t3
0
Pdt 
t3
 Sdt  Wt  mv
0
3
0  4.7  0.00266  0.5886t3  0
3
t3  7.99 s 
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PROBLEM 13.136
A simplified model consisting of a single straight line is to be obtained for the
variation of pressure inside the 10
10-mm-diameter
diameter barrel of a rifle as a 20-g
20 bullet
is fired. Knowing that it takes 1.6 ms for the bullet to travel the length of the
barrel and
d that the velocity of the bullet upon exit is 700 m/s, determine the value
of p0.
SOLUTION
At
p  p0  c1  c2t
t  0,
c1  p0
At
t  1.6  10 3 s,
p0
0  c1  c2 (1.6  103 s)
c2 
p0
1.6  103 s
m  20  103 kg
0 A

1.6  103 s
0
pdt  mv2
 (10  103 ) 2
4
A  78.54  10 6 m 2
A
0 A

1.6 103 s
0
(c1  c2 t ) dt 
20  10 3
g

(c )(1.6  103 s)2 
3
(78.54  106 m 2 ) (c1 )(1.6  103 s)  2
  (20  10 kg)(700 m/s)
2


1.6  10 3 c1  1.280  10 6 c2  178.25  103
(1.6  10 3 m 2  s) p0 
(1.280  106 m 2  s 2 )
p0  178.25  103 kg  m/s
3
(1.6  10 s)
p0  222.8  106 N/m 2
p0  223 MPa 
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PROBLEM 13.137
A crash test is performed between an SUV A and a 1200-kg compact car B.
The compact car is stationary before the impact and has its brakes applied. A
transducer measures the force during the impact, and the force P varies as
shown. Knowing that the coefficients of friction between the tires and road
are s= 0.9 and k= 0.7,
7, determine (a) the time at which the compact car will
start moving, (b) the maximum speed of the car, (c) the time at which the car
will come to a stop.
SOLUTION
mB  1200 kg
Impulse-Momentum
Momentum Diagram:
s  0.9, k  0.7
0  t  0.1  P  1.5  106 t N
Given:
0.1  t  0.2  P  1.5  106 (0.2  t )N
vB,1  0
  P  F  dt  m v
Impulse-Momentum in the x-direction:
direction:
0
From Equilibrium:
N  mB g
(a) Compact car starts moving when:
Ff  s N  s mB g
f
(1)
B B,2
  P   m g  dt  0 or P   m g  0
Substitute into (1):
0
Solving for P:
P  10594.8 N when car starts to move
For 0  t  0.1, P  1.5  106 t
10594.8  1.5  106 t
s
B
s
(b) Maximum Velocity will occur near the end of the impulse when:
For
B
ts  0.00706 s 
P  Ff  k N  k mB g
0  t  0.2, P  1.5  106 (0.2  t )
0.7 1200 9.81  1.5  106  0.2  t 
tm  0.194506 s
Use (1) to find maximum velocity:
m
0  t 1.5  106 tdt  0.1
1.5  106  0.2  t  dt  t m k mB gdt  mBvB,max
0.1
t
t
s
vB,max 
s
t
1 
t
6 2 0.1
5
6 2 m
0.75

10
t

3

10
t

0.75

10
t
 k g tm


t
0.1
s


s
mB


vB,max = 12.45 m/s
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PROBLEM 13.137 (Continued)
(c) Use impulse momentum from t m until car stops at tstp:
mB vB ,max 
tstp
  P  F  dt  0
tm
f
mBvB,max  t 1.5  10  0.2  t  dt  t stp k mB gdt  0
0.2
t
m
m

mBVB,max  3  105 t  0.75  106 t 2
 t  k mB g tt  0
0.2
m
stp
m
tstp = 2.01 s
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PROBLEM 13.138
A crash test is performed between a 2000 kg SUV A and a compact
car B. A transducer measures the force during the impact, and the
force P varies as sh own. Knowing that the SUV is travelling
45 km/h when it hits the car, determine the speed of the SUV
immediately after the impact.
SOLUTION
m  2000 kg
Given:
0  t  0.1  P  1.5  106 t N
Impulse-Momentum
Momentum Diagram of SUV:
0.1  t  0.2  P  1.5  106 (0.2  t )N
v1  45 km/h  12.5 m/s

Impulse-Momentum in the x-direction:
direction:
mv1  Pdt  mv2
Put in load function P(t):
mv1  0 1.5  106t  0.1 1.5  106  0.2  t   mv2
Integrate:
mv1  0.75  106 t 2 0  3  105t  0.75  106 t 2
Solve for v2:
v2  v1 
0.1
0.2
0.1

 0.1  mv2
0.2
1
15,000
m
v2  5 m/s
v2 = 18 km/h
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195
PRoBlEM 13.139
A baseball player catching a ball can soften the impact by
pulling his hand back. Assuming that a 150 g ball reaches
his glove at 140 km/h and that the player pulls his hand back
during the impact at an average speed of 9 m/s over a distance
of 150 mm, bringing the ball to a stop, determine the average
impulsive force exerted on the player’s hand.
Solution
v = 140 km/h =
350
m/s
9
m = 0.15 kg
d
0.15
t=
=
= 1 s
60
vAV
9
( )
+ Æ 0 - FAV t + mv
FAV =
=
mv
t
Ê 350 ˆ
(0.15) Á
Ë 9 ˜¯
( 601 )
= 350 N
b
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196
PRoBlEM 13.140
A 46-g golf ball is hit with a golf club and leaves it with a velocity of 50 m/s. We assume that for 0 t t0,
where t0 is the duration of the impact, the magnitude F of the force exerted on the ball can be expressed as
F = Fm sin (p t /t0 ). Knowing that t0 = 0.5 ms, determine the maximum value Fm of the force exerted on the ball.
Solution
m = 46 g = 0.046 kg
t0 = 0.5 ms = 0.5 ¥ 10 -3 s
The impulse applied to the ball is
t0
Ú0
t0
F dt = Ú Fm sin
0
=-
t0
F t
pt
pt
dt = - m 0 cos
t0
p
t0 0
2F t
Fmt0
(cos p - cos 0) = m 0
p
p
Principle of impulse and momentum.
t0
mv1 + Ú F dt = mv 2
0
With v1 = 0,
Solving for Fm,
0+
2 Fmt0
= mv2
p
Fm =
p mv2 p ( 46 ¥ 10 -3 )(50)
=
= 7.23 ¥ 103 N
2t0
(2)(0.5 ¥ 10 -3 )
Fm = 7.23 kN b
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197
PROBLEM 13.141
The triple jump is a track-and-field
field event in which an athlete
gets a running start and tries to leap as far as he can with a hop,
step, and jump. Shown in the figure is the initial hop of the
athlete. Assuming that he approaches the takeoff line from the
left with a horizontal velocity of 10 m/s, remains in contact
with the ground for 0.18 s, and takes off at a 50° angle with a
velocity of 12 m/s, determine the vertical component of the
average impulsive force exerted by the ground on his foot. Give
your answer in terms of the weight W of the athlete.
SOLUTION
mv1  (P  W)t  mv 2
t  0.18 s
Vertical components
W
(12)(sin 50)
g
(12)(sin 50)
Pv  W 
W
(9.81)(0.18)
0  ( Pv  W )(0.18) 
Pv  6.21W 
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PROBLEM 13.142
The last segment of the triple jump track-and-field
track
event is the
jump, in which the athlete makes a final leap, landing in a sandsand
filled pit. Assuming that the velocity of a 80-kg
80
athlete just
before landing is 9 m/s at an angle of 35° with the horizontal
and that the athlete comes to a complete stop in 0.22 s after
landing, determine the horizontal component of the average
impulsive force exerted
ed on his feet during landing.
SOLUTION
m  80 kg
t  0.22 s
mv1  (P  W)t  mv 2
Horizontal components
m(9)(cos 35)  PH (0.22)  0
PH 
(80 kg)(9 m/s)(cos 35)
 2.6809 kN
(0.22 s)
PH  2.68 kN 
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PROBLEM 13.143
The design for a new cementless hip implant is to be studied using an
instrumented implant and a fixed simulated femur. Assuming the punch applies
an average force of 2 kN over a time of 2 ms to the 200 g implant determine
(a) the velocity of the implant immediately after impact, (b) the average
resistance of the implant to penetration if the implant moves1 mm before coming
to rest.
SOLUTION
m  200 g  0.200 kg
Fave  2 kN  2000 N
t  2 ms  0.002 s
(a)
Velocity immediately after impact:
Use principle of impulse and momentum:
v1  0
v2  ?
Imp1 2  Fave (t )
mv1  Imp1 2  mv 2
0  Fave (t )  mv2
v2 
(b)
Fave (t ) (2000)(0.002)

m
0.200
v2  20.0 m/s 
Average resistance to penetration:
x  1 mm  0.001 m
v2  20.0 ft/s
v3  0
Use principle of work and energy.
T2  U 23  T3
Rave 
or
1 2
mv2  Rave (x)  0
2
mv22
(0.200)(20.0) 2

 40  103 N
2( x)
(2)(0.001)
Rave  40.0 kN 
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200
PROBLEM 13.144
A 28-g steel-jacketed
jacketed bullet is fired with a velocity of 650 m/s toward
a steel plate and ricochets along path CD with a velocity 500 m/s.
Knowing that the bullet leaves a 50-mm
mm scratch on the surface of the
plate and assuming that it has an average speed of 600 m/s while in
contact with the plate, determine the magnitude and direction of the
impulsive force exerted by the plate on the bullet.
SOLUTION
Given:
v1  650 m/s,
m  0.028 kg,
v2  500 m/s
Impulse-momentum
momentum diagram of the bullet:
Impulse momentum in the x-dir
mv1 cos 20 Fx t  mv2 cos10
So,
Fx t  mv1 cos 20 mv2 cos10
 0.028  650 cos 20 0.028  500 cos10 3.3151 N s
y -dir
 mv1 sin 20 Fy t  mv2 sin10
So,
Fy t  mv2 sin10 mv1 sin 20
 0.028  500  sin10 0.028  650 sin 20 8.6558 N s
We need t.. The average velocity is 600 m/s
x  vave t; t 
x
0.05 m

 83.33  106 s
vave 600 m/s
So
3.3151

 39.78 kN 
83.33  106


8.6558


Fy 
103.87
kN

83.33  106
Fx 
F  111.2 kN
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
PRoBlEM 13.145
A 20-Mg railroad car moving at 4 km/h is to
be coupled to a 40-Mg car which is at rest
with locked wheels ( m k = 0.30). Determine
(a) the velocity of both cars after the coupling
is completed, (b) the time it takes for both cars
to come to rest.
Solution
(a)
The momentum of the system consisting of the two cars is conserved immediately before and after
coupling.
Before coupling
After coupling
+
Smv = Smv ¢
0 + (20 Mg)( 4 km/h) = (20 Mg + 40 Mg)( v ¢ )
v¢ =
(b)
(20)( 4)
(20 + 40)
v ¢ = 1.333 km/h ¨ b
After coupling
The friction force acts only on the 40 Mg car since its wheels are locked. Thus,
F f = m k N 40 = (0.30)( 40 ¥ 103 kg)(9.81 m/s2 )
F f = 117.72 ¥ 103 N
v1 = v ¢ = 1.333 km/h = 0.3704 m/s
From (a),
Impulse momentum
t
Smv1 + Ú F f dt = Smv2
0
t
(60 ¥ 103 kg)(0.3704 m/s) - Ú (117.72 ¥ 103 N )dt = 0
0
t=
(60 ¥ 103 )(0.3704)
(117.72 ¥ 103 )
t = 0.1888 s b
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202
PROBLEM 13.146
At an intersection car B was traveling south and car A was
traveling 30° north of east when they slammed into each other.
Upon investigation it was found that after the crash the two cars
got stuck and skidded off at an angle of 10° north of east. Each
driver claimed that he was going at the speed limit of 50 km/h and
that he tried to slow down but couldn’t avoid the crash because the
other driver was going a lot faster. Knowing that the masses of
cars A and B were 1500 kg and 1200 kg, respectively, determine
((a) which car was going faster, (b) the speed of thee faster of the
two cars if the slower car was traveling at the speed limit.
SOLUTION
(a)
Total momentum of the two cars is conserved.
 mv, x :
mAvA cos 30  (mA  mB )v cos 10
(1)
mv, y :
mAvA sin 30  mB vB  (mA  mB )v sin 10
(2)
Dividing (1) into (2),
mB v B
sin 30
sin 10


cos 30 m Av A cos 30 cos 10
vB (tan 30  tan 10)(mA cos 30)

vA
mB
vB
(1500)
cos 30
 (0.4010)
vA
(1200)
vB
 0.434
vA
v A  2.30 vB
A was going faster. 
Thus,
(b)
Since vB was the slower car.
vB  50 km/h
vA  (2.30)(50)
vA  115.2 km/h 
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203
PROBLEM 13.147
The 650-kg hammer of a drop-hammer
hammer pile driver falls from a height of
1.2 m onto the top of a 140-kg
kg pile, driving it 110 mm into the ground.
Assuming perfectly plastic impact (e  0), determine the average
resistance of the ground to penetration.
SOLUTION
Velocity of the hammer at impact:
Conservation of energy.
T1  0
VH  mg (1.2 m)
VH  (650 kg)(9.81 m/s 2 )(1.2 m)
V1  7652 J
1
m
2
650 2
VH2 
v  325 vH2
2
V2  0
T2 
T1  V1  T2  V2
0  7652  325 v 2
v 2  23.54 m 2 /s 2
v  4.852 m/s
Velocity of pile after impact:
Since the impact is plastic (e  0), the velocity of the pile and hammer are the same after impact.
Conservation of momentum:
The ground reaction and the weights are non
non-impulsive.
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PROBLEM 13.147 (C
(Continued)
Thus,
mH vH  (mH  m p )v 
v 
Work and energy:
mH vH
(650)

(4.852 m/s)  3.992 m/s
(mH  m p ) (650  140)
d  0.110 m
T2  U 2 3  T3
1
(mH  mH )(v ) 2
2
T3  0
T2 
1
(650  140)(3.992) 2
2
T2  6.295  103 J
T2 
U 2 3  (mH  m p ) gd  FAV d
 (650  140)(9.81)(0.110)  FAV (0.110)
U 2 3  852.49  (0.110) FAV
T2  U 2 3  T3
6.295 103  852.49  (0.110) FAV  0
FAV  (7147.5)/(0.110)  64.98  103 N
FAV  65.0 kN 
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PROBLEM 13.148
A small rivet connecting two pieces of sheet metal is being clinched
by hammering. Determine the impulse exerted on the rivet and the
energy absorbed by the rivet under each blow, knowing that the head
of the hammer has a mass of 750 g and that it strikes the rivet with a
velocity of 6 m/s. Assume that the hammer does not rebound and that
the anvil is supported by springs and (a) has an infinite mass (rigid
support), (b) has a mass of 4.5 kg.
SOLUTION
Weight and mass:
Hammer: mH = 0.75 kg
Anvil: Part a: WA = ∞
mA = ∞
Part b: mA = 4.5 kg
Kinetic energy before impact:
1
1
mH vH2 = (0.75)(6)2 = 13.5 N ⋅ m
2
2
Let v2 be the velocity common to the hammer and anvil immediately after impact. Apply the principle of
conservation of momentum to the hammer and anvil over the duration of the impact.
T1 =
: Σmv1 = Σ mv2
mH vH = (mH + mA )v2
v2 =
mH vH
mH + mA
TA =
1
1 mH2 vH2
(mH + mA )v22 =
2
2 mH + mA
T2 =
mH
T1
mH + m A
(1)
Kinetic energy after impact:
(2)
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206
PROBLEM 13.148 (Continued)
Impulse exerted on the hammer:
: mH vH − F (Δt ) = mH v2
F Δt = mH (vH − v2 )
(a)
(3)
WA = ∞ :
By Eq. (1),
v2 = 0
By Eq. (2),
T2 = 0
T1 − T2 = 13.50 N ⋅ m W
Energy absorbed:
By Eq. (3),
F ( Δt ) = (0.75)(6 − 0) = 4.5 N ⋅ S
The impulse exerted on the rivet the same magnitude but opposite to direction.
F Δt = 4.5 N ⋅ S W
(b)
WA = 9 lb:
By Eq. (1),
v2 =
(0.75)(6)
= 0.8511 m/s
5.25
By Eq. (2),
T2 =
(0.75)(13.5)
= 1.9286 N ⋅ m
(5.25)
T1 − T2 = 11.57 N ⋅ m W
Energy absorbed:
By Eq. (3),
F ( Δt ) = (0.75)(6 − 0.8571)
F ( Δt ) = 3.86 N ⋅ m W
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207
PRoBlEM 13.149
Bullet B has a mass 15 g and blocks A and C both
have a mass 2 kg. The coefficient of friction between
the blocks and the plane is m k = 0.25. Initially, the
bullet is moving at v0, and blocks A and C are at rest
(Fig. 1). After the bullet passes through A, it becomes
embedded in block C, and all three objects come to
stop in the positions shown (Fig. 2). Determine the
initial speed of the bullet v0.
Solution
Masses:
Bullet:
mB = 15 ¥ 10 -3 kg = 0.015kg
Blocks A and C:
mA = mC = 2 kg
Block C + bullet:
mC + mB = 2.015 kg
Normal forces for sliding blocks from N - mg = 0
Block A:
N A = mA g = (2)(9.81) = 19.62 N
Block C + bullet:
N C = ( mC + mB ) g = (2.015)(9.81) = 19.767N
Let v0 be the initial speed of the bullet;
v1 be the speed of the bullet after it passes through block A;
vA be the speed of block A immediately after the bullet passes through it;
vC be the speed of block C immediately after the bullet becomes embedded in it.
Four separate processes and their governing equations are described below.
1.
The bullet hits block A and passes through it. Use the principle of conservation of momentum.
( v A )0 = 0
mB v0 + mA ( v A )0 = mB v1 + mA v A
v0 = v1 +
2.
mA v A
mB
(1)
The bullet hits block C and becomes embedded in it. Use the principle of conservation of momentum.
( vC )0 = 0
mB v1 + mC ( vC )0 = ( mB + mC )vC
v1 =
( mB + mC )vC
mB
(2)
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731
208
PRoBlEM 13.149 (continued)
3.
Block A slides on the plane. Use principle of work and energy.
T1 + U1Æ2 = T2
2mk N Ad A
1
mA v 2A - m k N A d A = 0 or v A =
mA
2
d A = 0.15 m
4.
(3)
Block C with embedded bullet slides on the plane. Use principle of work and energy.
dC = 100 mm = 0.1 m
T1 + U1Æ2 = T2
2 m k N C dC
1
( mC + mB )vC2 - m k N C dC = 0 or vC =
mC + mB
2
(4)
Applying the numerical data:
(2)(0.25)(19.767)(0.1)
2.015
= 0.70035 m/s
From Eq. (4),
vC =
From Eq. (3),
vA =
(2)(0.25)(19.62)(0.15)
( 2)
= 0.85776 m/s
(2.015)(0.70035)
(0.015)
= 94.080 m/s
From Eq. (2),
v1 =
From Eq. (1),
v0 = 94.080 +
(2)(0.85776)
(0.015)
v0 = 208 m/s b
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209
PRoBlEM 13.150
A 90-kg man and a 60-kg woman stand at opposite ends of a
150-kg boat, ready to dive, each with a 5-m/s velocity relative
to the boat. Determine the velocity of the boat after they have
both dived, if (a) the woman dives first, (b) the man dives first.
Solution
(a)
Woman dives first.
Conservation of momentum:
-60(5 - v1 ) + 240v1 = 0
v1 =
300
= 1 m/s
300
Man dives next. Conservation of momentum:
240v1 = -150v2 + 90(5 - v2 )
v2 =
(b)
450 + 240 v1
= 0.875 m/s
240
v 2 = 0.875 m/s
b
Man dives first.
Conservation of momentum:
90(5 - v1¢) - 210 v1¢ = 0
v1¢ =
450
= 1.5 m/s
300
Woman dives next. Conservation of momentum:
-210 v1¢ = 150 v2¢ - 60(5 - v2¢ )
v2¢ = -210v1¢ = 210v2¢ - 300
v2¢ =
-210v1¢ + 300 -210 ¥ 1.5 + 300
=
= -0.0714 m/s
210
210
v 2¢ = -0.0714 m/s
b
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210
PROBLEM 13.151
A 75-g ball is projected from a height of 1.6 m with a horizontal
velocity of 2 m/s and bounces from a 400-g smooth plate
supported by springs. Knowing that the height of the rebound is
0.6 m, determine (a) the velocity of the plate immediately after
the impact, (b) the energy lost due to the impact.
SOLUTION
Just before impact
Just after impact
vy 
vy 
2 g (1.6)  5.603 m/s
2 g (0.6)  3.431 m/s
(a) Conservation of momentum: ( y )

mballvy  0  mballvy  mplatevplate
(0.075)(5.603)  0  0.075(3.431)  0.4vplate
vplate  1.694 m/s 
(b)
Energy loss
Initial energy
Final energy
(T  V )1 
(T  V )2 
1
(0.075)(2)2  0.075g (1.6)
2
1
1
(0.075)(2)2  0.075g (0.6)  (0.4)(1.694)2
2
2
Energy lost  (1.3272  1.1653) J  0.1619 J 
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211
PROBLEM 13.152
A ballistic pendulum is used to measure the speed of high-speed
high
projectiles. A 6-g bullet A is fired into a 1-kg
1
wood block B
suspended by a cord of length l=
= 2.2 m. The block then swings
through a maximum angle of = 60o. Determine (a)
( the initial speed
of the bullet vo, (b) the impulse imparted by the bullet on the block,
(c) the force on the cord immediately after the impact
SOLUTION
Given:
m A  6 g  0.006 kg
Impulse-Momentum Diagram in x--direction:
mB  1 kg
l  2.2 m
 =60
m A v0   m A  mB  v1
Impulse-Momentum in the x-direction
direction during impact:
(1)
Apply Work-Energy
Energy between the impact location and the maximum swing angle (Datum at the pivot, O):
T1  Vg1  Ve1  U1NC
 3  T2  Vg 2  Ve2
Hence:
where:
1
 mA  mB  v12   mA  mB  g    mA  mB  gl cos 
2
Solving for v1:
1
 mA  mB  v12
2
Vg1    mA  mB  gl
T1 
Vg1    mA  mB  gl cos 
v1  2 gl 1  cos  
 4.646 m/s
(a) Substitute into (1) and solve for v0:
v0 
 mA  mB 
mA
v1
v0  778.92 m/s 
(b) Write the Impulse Moment Equation for the block during impact:

0  Fdt  mB v1
 Fdt  4.646 J 
(c) FBD of Block just after impact:
 F  ma
n
n
v2
T   mA  mB  g   mA  mB  1
l
2
v

T   mA  mB   1  g 
 l

T  19.74 N 
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212
PROBLEM 13.153
A 25-g bullet is traveling with a velocity of 425 m/s when it impacts and
becomes embedded in a 2.5 kg wooden block. The block can move vertically
without friction. Determine (a) the velocity of the bullet and block immediately
after the impact, (b) the horizontal and vertical components of the impulse
exerted by the block on the bullet.
SOLUTION
Bullet: m = 0.025 kg.
Weight and mass.
Block: M = 2.5 kg.
(a)
Use the principle of impulse and momentum applied to the bullet and the block together.
Σm v1 + ΣImp1→ 2 = mv 2
mv0 cos30° + 0 = (m + M )v′
Components :
mv0 cos30°
(0.025)(425)cos 30°
=
m+M
2.525
v ′ = 3.644 m/s
v′ =
v ′ = 3.64 m/s
W
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736
213
PROBLEM 13.153 (Continued)
(b)
Use the principle of impulse and momentum applied to the bullet alone.
x-components:
−mv0 sin 30° + Rx Δt = 0
Rx Δt = mv0 sin 30° = (0.025)(425)sin 30°
= 5.3125 N ⋅S
y-components:
Rx Δt = 5.31 N ⋅ S W
−mv0 cos 30° + Ry Δt = −mv′
Ry Δt = m(v0 cos 30° − v′)
= 0.025(425 cos 30° − 3.644)
Ry Δt = 9.11 N ⋅ S W
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214
PRoBlEM 13.154
In order to test the resistance of a chain to impact, the chain is suspended from
a 120-kg rigid beam supported by two columns. A rod attached to the last link
is then hit by a 30-kg block dropped from a 1.5 m height. Determine the initial
impulse exerted on the chain and the energy absorbed by the chain, assuming
that the block does not rebound from the rod and that the columns supporting
the beam are (a) perfectly rigid, (b) equivalent to two perfectly elastic springs.
Solution
Velocity of the block just before impact.
T1 = 0 V1 = Wh = (30)(9.81)(1.5) = 441.45 J
1 2
mv V2 = 0
2
T1 + V1 = T2 + V2
1
0 + 441.45 = (30)v 2
2
2( 441.45)
v=
30
= 5.4249 m/s
T2 =
(a)
Rigid columns.
+
- mv + F Dt = 0
(30)(5.4249) = F Dt
F Dt = 162.747 N ◊ s ≠ on the block.
F Dt = 162.7 N ◊ s b
All of the kinetic energy of the block is absorbed by the chain.
1
T = (30)(5.4249)2
2
= 441.45 J
E = 441 J b
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738
215
PRoBlEM 13.154 (continued)
(b)
Elastic columns.
Momentum of system of block and beam is conserved.
mv = ( M + m)v ¢
30
m
(5.4249 m/s)
v¢ =
v=
(m + M )
150
v ¢ = 1.085 m/s
Referring to figure in Part (a),
- mv + F Dt = - mv ¢
F Dt = m( v - v ¢ )
= (30)(5.4249 - 1.085)
= 130.197 N ◊ s
F Dt = 130.2 N ◊ s b
1 2 1
1
mv - mv ¢ 2 - mv ¢ 2
2
2
2
1
1
1
= (30)(5.4249)2 - (30)(1.085)2 - (120)(1.085)2
2
2
2
= 353.151J
E=
E = 353J b
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739
216
PROBLEM 13.155
The coefficient of restitution between the two collars is known to be
0.70. Determine ((a) their velocities after impact, (b)) the energy loss
during impact.
SOLUTION
Impulse-momentum principle (collars A and B):
mv1  Imp12  mv 2
Horizontal components
Using data,
: mAvA  mB vB  mAvA  mB vB
(5)(1)  (3)(1.5)  5vA  3vB
5vA  3vB  0.5
or
(1)
Apply coefficient of restitution.
vB  vA  e(v A  vB )
vB  vA  0.70[1  (0.5)]
vB  vA  1.75
(a)
(2)
Solving Eqs. (1) and (2) simultaneously for the velocities,
vA  0.59375 m/s
v A  0.594 m/s

vB  1.15625 m/s
v B  1.156 m/s

1
1
1
1
mA v A2  mB vB2  (5)(1)2  (3)(1.5)2  5.875 J
2
2
2
2
1
1
1
1
T2  mA (vA )2  mB (vB ) 2  (5)(0.59375)2  (3)(1.15625) 2  2.8867 J
2
2
2
2
Kinetic energies: T1 
(b)
T1  T2  2.99 J 
Energy loss:
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217
PROBLEM 13.156
Collars A and B, of the same mass m,, are moving toward each other
with identical speeds as shown. Knowing that the coefficient of
restitution between the collars is e,, determine the energy lost in the
impact as a function of m, e and v.
SOLUTION
Impulse-momentum principle (collars A and B):
mv1  Imp12  mv 2
Horizontal components
: mAv A  mB vB  mAvA  mB vB
mv  m(v)  mvA  mvB
Using data,
vA  vB  0
or
(1)
Apply coefficient of restitution.
vB  vA  e(v A  vB )
vB  vA  e [v  (v)]
vB  vA  2ev
Subtracting Eq. (1) from Eq. (2),
2vA  2ev
vA  ev
Adding Eqs. (1) and (2),
(2)
v A  ev
2vB  2ev
v B  ev
vB  ev
1
1
1
1
m Av A2  mB vB2  mv 2  m(v )2  mv 2
2
2
2
2
1
1
1
1
2
2
T2  m A (vA )  mB (vB )  m(ev) 2  m(ev) 2  e 2 mv 2
2
2
2
2
Kinetic energies: T1 
T1  T2  (1  e2 ) mv2 
Energy loss:
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PROBLEM 13.157
One of the requirements for tennis balls to be used in official competition is that, when dropped onto a rigid
surface from a height of 2.5 m the height of the first bounce of the ball must be in the range 1.325 m £ h £ 1.45 m.
Determine the range of the coefficients of restitution of the tennis balls satisfying this requirement.
SOLutiOn
Uniform accelerated motion.
v = 2 gh
v ¢ = 2 gh¢
v¢
v
h¢
e=
h
Coefficient of restitution.
e=
Height of drop
h = 2.5 m
Height of bounce
1.325 m h¢ 1.45 m
Thus,
1.325
1.45
e
2.5
2.5
0.728 e 0.762 b
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219
PROBLEM 13.158
Two disks sliding on a frictionless horizontal plane with opposite velocities of the
same magnitude v0 hit each other squarely. Disk A is known to have a mass of 3 kg
and is observed to have zero velocity after impact. Determine (a) the mass of disk B,
knowing that the coefficient of restitution between the two disks is 0.5, (b) the range
of possible values of the mass of disk B if the coefficient of restitution between the
two disks is unknown.
SOLutiOn
Total momentum conserved.
+
mA v A + mB v B = mA v A¢ + mB v B¢
( mA )v0 + mB ( - v0 ) = 0 + mB v B¢
Êm
ˆ
v ¢ = Á A - 1˜ v0
Ë mB ¯
(1)
Relative velocities.
v B¢ - v A¢ = e ( v A - v B )
v ¢ = 2ev0
(2)
Subtracting Eq. (2) from Eq. (1) and dividing by v0 ,
mA
- 1 - 2e = 0
mB
mA
= 1 + 2e
mB
(a)
(b)
mB =
mA
1 + 2e
mB =
3
1 + (2)(0.5)
mB =
3
1 + (2)(1)
mB = 1.000 kg
mB =
3
1 + (2)(0)
mB = 3.00 kg
(3)
With mA = 3 kg and e = 0.5,
mB = 1.500 kg b
With mA = 3 kg and e = 1,
With mA = 3 kg and e = 0,
Range:
1.000 kg mB 3.00 kg b
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220
PROBLEM 13.159
To apply shock loading to an artillery shell, a 20-kg pendulum A is
released from a known height and strikes impactor B at a known
velocity v0. Impactor B then strikes the 1-kg artillery shell C.
Knowing the coefficient of restitution between all objects is e,
determine the mass of B to maximize the impulse applied to the
artillery shell C.
SOLUTION
mA  20 kg, mB  ?
First impact: A impacts B.
mv  Imp12  mv 2
Impulse-momentum:
mAv0  mAvA  mB vB
Components directed left:
20v0  20vA  mB vB
(1)
vB  vA  e(v A  vB )
Coefficient of restitution:
vB  vA  ev0
vA  vB  ev0
(2)
Substituting Eq. (2) into Eq. (1) yields
20v0  20(vB  ev0 )  mB vB
20v0 (1  e)  (mB )vB
vB 
Second impact: B impacts C.
Impulse-momentum:
20v0 (1  e)
20  mB
mB  ?, mC  1 kg
mv 2  Imp23  mv3
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(3)
PROBLEM 13.159 (Continued)
Components directed left:
mB vB  mB vB  mC vC
mB vB  mB vB  vC
(4)
vC  vB  e(vB  vC )
Coefficient of restitution:
vC  vB  evB
vB  vC  evC
(5)
Substituting Eq. (4) into Eq. (5) yields
mB vB  mB (vC  evB )  mC vC
mB vB (1  e)  (1  mB )vC
vC 
mB vB (1  e)
1  mB
vC 
20mB v0 (1  e) 2
(20  mB )(1  mB )
mC vC 
(1)(20) mB v0 (1  e)2
(20  mB )(1  mB )
(6)
Substituting Eq. (3) for vB in Eq. (6) yields
The impulse applied to the shell C is
To maximize this impulse choose mB such that
Z
mB
(20  mB )(1  mB )
is maximum. Set dZ /dmB equal to zero.
(20  mB )(1  mB )  mB [(20  mB )  (1  mB )]
dZ

0
dmB
(20  mB ) 2 (1  mB ) 2
20  21mB  mB2  mB (21  2mB )  0
20  mB2  0
mB  4.47 kg 
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PROBLEM 13.160
Packages in an automobile parts supply house are transported to
the loading dock by pushing them along on a roller track with
very little friction. At the instant shown, packages B and C are
at rest and package A has a velocity of 2 m/s. Knowing that the
coefficient
oefficient of restitution between the packages is 0.3, determine
(a) the velocity of package C after A hits B and B hits C, (b) the
velocity of A after it hits B for the second time.
SOLUTION
(a)
Packages A and B.
Totalmomentumconserved.
m Av A  mB vB  mA vA  mB vB
(8 kg)(2 m/s)  0  (8 kg)vA  (4 kg)vB
4  2vA  vB
(1)
(v A  vB )e  (vB  v A )
(2)(0.3)  vB  vA
(2)
Relative velocities.
Solving Equations (1) and (2) simultaneously,
vA  1.133 m/s 
vB  1.733 m/s 
Packages B and C.
mB vB  mC vC  mB vB  mC vC
(4 kg)(1.733 m/s)  0  4vB  6vc
6.932  4vB  6vC
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(3)
PROBLEM 13.160 (Continued)
Relativevelocities.
(vB  vC )e  vC  vB
(1.733)(0.3)  0.5199  vC  vB
(4)
Solving equations (3) and (4) simultaneously,
vC  0.901 m/s  
(b)
Packages A and B (second time),
Total momentum conserved
conserved.
(8)(1.133)  (4)(0.381)  8vA  4vB
10.588  8vA  4vB
(5)
Relative velocities.
(vA  vB )e  vB  vA
(1.133  0.381)(0.3)  0.2256  vB  vA
(6)
Solving (5) and (6) simultaneously,
vA  0.807 m/s
vA  0.807 m/s  
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PROBLEM 13.161
Three steel spheres of equal mass are suspended from the ceiling by cords of
equal length which are spaced at a distance slightly greater than the diameter of
the spheres. After being pulled back and released, sphere A hits sphere B, which
then hits sphere C. Denoting by e the coefficient of restitution between the
spheres and by v0 the velocity of A just before it hits B, determine (a) the
velocities of A and B immediately after the first collision, (b) the velocities of B
and C immediately after the second collision. (c) Assuming now that n spheres are
suspended from the ceiling and that the first sphere is pulled back and released as
described above, determine the velocity of the last sphere after it is hit for the first
time.(d) Use the result of part c to obtain the velocity of the last sphere when
n  8 and e  0.9.
SOLUTION
(a) First collision(between A and B)
The total momentum is conserved
mv A  mvB  mvA  mvB
v0  vA  vB
(1)
Relative velocities
 vA  vB  e   vB  vA 
v0e  vB  vA
(2)
Solving equations (1) and (2) simultaneously



(b) Second collision (Between B and C)
vA 
v0 1  e 

2
vB 
v0 1  e 

2
The total momentum is conserved.
mvB  mvC  mvB  mvC
Using the result from (a) for vB
v0 1  e 
 0  vB  vC
2
(3)
Relative velocities
 vB  0 e  vC  vB
Substituting again for vB from (a)
v0
1  e  e  v  v
  C B
2
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(4)
PROBLEM 13.161 (Continued)
Solving equations (3) and (4) simultaneously
vC 
e 
1  v0 1  e 
 v0 1  e 


2
2
2 
2
v 1  e 
vC  0

4
2
vB 
 
v0 1  e 

4
(c) For n spheres
n Balls
n  1th collision
We note from the answer to part (b), with n  3
vn  v3  vC 
v3 
or
v0 1  e 
4
v0 1  e 
2
2
 3 1
3 1
Thus for n balls
vn 
 n 1
v0 1  e 
2
n 1

(d) For n  8, e  0.90
From the answer to part (c) with n  8
vB 
 8 1
v0 1  0.9 
2
8 1

v0 1.9 
7
 2 7
v8  0.698v0 
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PROBLEM 13.162
At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg,
and 35 kg respectively. Car A is moving to the right with a velocity vA  2 m/s and car C has a velocity vB
 1.5 m/s to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8.
Determine the final velocity of each car, after all impacts, assuming (a) cars A and C hit car B at the same
time, (b) car A hits car B before car C does.
SOLUTION
Assume that each car with its rider may be treated as a particle. The masses are:
m A  200  40  240 kg,
mB  200  60  260 kg,
mC  200  35  235 kg.
Assume velocities are positive to the right. The initial velocities are:
v A  2 m/s vB  0 vC  1.5 m/s
Let vA , vB , and vC be the final velocities.
(a)
Cars A and C hit B at the same time. Conservation of momentum for all three cars.
m A v A  mB vB  mC vC  m Av A  mB vB  mC vC
(240)(2)  0  (235)(1.5)  240vA  260vB  235vC
(1)
Coefficient of restition for cars A and B.
vB  vA  e(vA  vB )  (0.8)(2  0)  1.6
(2)
Coefficient of restitution for cars B and C.
vC  vB  e(vB  vC )  (0.8)[0  (1.5)]  1.2
(3)
Solving Eqs. (1), (2), and (3) simultaneously,
vA  1.288 m/s vB  0.312 m/s vC  1.512 m/s
vA  1.288 m/s

vB  0.312 m/s

vC  1.512 m/s

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PROBLEM 13.162 (Continued)
(b)
Car A hits car B before C does.
First impact. Car A hits car B. Let vA and vB be the velocities after this impact. Conservation of
momentum for cars A and B.
m Av A  mB vB  m A vA  mB vB
(240)(2)  0  240vA  260vB
(4)
Coefficient of restitution for cars A and B.
vB  vA  e(vA  vB )  (0.8)(2  0)  1.6
(5)
Solving Eqs. (4) and (5) simultaneously,
vA  0.128 m/s, vB  1.728 m/s
vA  0.128 m/s
vB  1.728 m/s
Second impact. Cars B and C hit. Let vB and vC be the velocities after this impact. Conservation of
momentum for cars B and C.
mB vB  mC vC  mB vB  mC vC
(260)(1.728)  (235)(1.5)  260vB  235vC
(6)
Coefficient of restitution for cars B and C.
vC  vB  e(vB  vC)  (0.8)[1.728  (1.5)]  2.5824
(7)
Solving Eqs. (6) and (7) simultaneously,
vB  1.03047 m/s vC  1.55193 m/s
vB  1.03047 m/s
vC  1.55193 m/s
Third impact. Cars A and B hit again. Let vA and vB be the velocities after this impact. Conservation of
momentum for cars A and B.
mA vA  mB vB  mA vA  mB vB
(240)(0.128)  (260)( 1.03047)  240vA  260vB
(8)
Coefficient of restitution for cars A and B.
vB  vA  e(vA  vB )  (0.8)[0.128  (1.03047)]  0.926776
Solving Eqs. (8) and (9) simultaneously,
vA  0.95633 m/s
vB  0.02955 m/s
vA  0.95633 m/s
vB  0.02955 m/s
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PROBLEM 13.162 (Continued)
There are no more impacts. The final velocities are:
vA  0.956 m/s

vB  0.0296 m/s

vC  1.552 m/s

We may check our results by considering conservation of momentum of all three cars over all three
impacts.
mAv A  mB vB  mC vC  (240)(2)  0  (235)(1.5)
 127.5 kg  m/s
mA vA  mB vB  mC vC  (240)(0.95633)  (260)(0.02955)  (235)(1.55193)
 127.50 kg  m/s.
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PROBLEM 13.163
At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg,
and 35 kg respectively. Car A is moving to the right with a velocity vA  2 m/s when it hits stationary car B.
The coefficient of restitution between each car is 0.8. Determine the velocity of car C so that after car B
collides with car C the velocity of car B is zero.
SOLUTION
Assume that each car with its rider may be treated as a particle. The masses are:
m A  200  40  240 kg
mB  200  60  260 kg
mC  200  35  235 kg
Assume velocities are positive to the right. The initial velocities are:
vA  2 m/s, vB  0, vC  ?
First impact. Car A hits car B. Let vA and vB be the velocities after this impact. Conservation of momentum
for cars A and B.
m Av A  mB vB  m A vA  mB vB
(240)(2)  0  240vA  260vB
(1)
Coefficient of restitution for cars A and B.
vB  vA  e(vA  vB )  (0.8)(2  0)  1.6
(2)
Solving Eqs. (1) and (2) simultaneously,
vA  0.128 m/s
vB  1.728 m/s
vA  0.128 m/s
vB  1.728 m/s
Second impact. Cars B and C hit. Let vB and vC be the velocities after this impact. vB  0. Coefficient of
restitution for cars B and C.
vC  vB  e(vB  vC )  (0.8)(1.728  vC )
vC  1.3824  0.8vC
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PROBLEM 13.163 (Continued)
Conservation of momentum for cars B and C.
mB vB  mC vC  mB vB  mC vC
(260)(1.728)  235vC  (260)(0)  (235)(1.3824  0.8vC )
(235)(1.8)vC  (235)(1.3824)  (260)(1.728)
vC  0.294 m/s
vC  0.294 m/s
Note: There will be another impact between cars A and B.
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
PROBLEM 13.164
Two identical billiard balls can move freely on a horizontal
table. Ball A has a velocity v0 as shown and hits ball B, which is
at rest, at a Point C defined by  45°. Knowing that the
coefficient of restitution between the two balls is e 0.8 and
assuming no friction,, determine the velocity of each ball after
impact.
SOLUTION
Ball A: t-dir
mv0 sin   mvAt  vAt  v0 sin 
Ball B: t-dir
0  mB vBt  vBt  0
Balls A + B: n-dir
mv0 cos  0  m vAn  m vBn
(1)
Coefficient of restitution
vBn
  vAn  e (vAn  vBn )
vBn  vAn  e (v0 cos  0)
Solve (1) and (2)
1  e

1  e 
vAn  v0 
cos  ; vBn  v0 
 cos
2


 2 
With numbers
e  0.8;   45
vAt  v0 sin 45  0.707 v0
 1  0.8

vAn  v0 
cos 45   0.0707 v0
 2

vBt  0
 1  0.8 
vBn  v0 
 cos 45 0.6364 v0
 2 
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(2)
PROBLEM 13.164 ((Continued)
(A)
1
v" A  [(0.707 v0 ) 2  (0.0707v0 ) 2 ] 2  0.711v0

 0.711v0 
 0.0707 
  tan 1 
  5.7106
 0.707 
So
  45  5.7106  39.3
(B)
vA  0.711v0
39.3°
vB  0.636 v0
45°
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PROBLEM 13.165
Two identical pool balls of 57.15-mm diameter, may move freely on a
pool table. Ball B is at rest and ball A has an initial velocity v = v0 i.
(a) Knowing that b = 50 mm and e = 0.7, determine the velocity of
each ball after impact. (b) Show that if e = 1, the final velocities of the
balls from a right angle for all values of b.
SOLUTION
Geometry at instant of impact:
b
50
=
d 57.15
θ = 61.032°
sin θ =
Directions n and t are shown in the figure.
Principle of impulse and momentum:
Ball B:
Ball A:
Ball A, t-direction:
mv0 sin θ + 0 = m(vA )t
Ball B, t-direction:
0 + 0 = m(vB )t
Balls A and B, n-direction:
Coefficient of restitution:
(a)
(vA )t = v0 sin θ
(1)
(vB )t = 0
(2)
mv0 cos θ + 0 + m(vA )n + m(vB )n
(vA )n + (vB )n = v0 cosθ
(3)
(vB )n − (vA )n = e[v0 cos θ ]
(4)
e = 0.7. From Eqs. (1) and (2),
(v A )t = 0.87489v0
(1)′
(vB )t = 0
(2)′
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234
PROBLEM 13.165 (Continued)
From Eqs. (3) and (4),
(vA )n + (vB )n = 0.48432v0
(3)′
(vB )n − (v A )n = (0.7)(0.48432v0 )
(4)′
Solving Eqs. (5) and (6) simultaneously,
(v A )n = 0.072648v0
(vB ) n = 0.41167v0
v A = (v A ) 2n + (v A )t2
= (0.072648v0 ) 2 + (0.87489v02
= 0.87790v0
tan β =
(v A ) n 0.072648v0
=
= 0.083037
(v A )t
0.87489v0
β = 4.7468°
ϕ = 90° − θ − β
= 90° − 61.032° − 4.7468°
= 24.221°
(b)
v A = 0.878v0
24.2° W
v B = 0.412v0
61.0° W
e = 1. Eqs. (3) and (4) become
(vA )n + (vB )n = v0 cos θ
(3)′′
(vB )n − (vA )n = v0 cosθ
(4)′′
Solving Eqs. (3)′′ and (4)′′ simultaneously,
(vA )n = 0, (vB )t = v0 cosθ
But
(v A )t = v0 sin θ , and (vB )t = 0
vA is in the t-direction and vB is in the n-direction; therefore, the velocity vectors form a right angle.
B
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235
PROBLEM 13.166
A 600-g ball A is moving with a velocity of magnitude 6 m/s when it is
hit as shown by a 1-kg ball B which has a velocity of magnitude 4 m/s.
Knowing that the coefficient of restitution is 0.8 and assuming no
friction, determine the velocity of each ball after impact.
SOLUTION
Before
After
v A  6 m/s
(v A ) n  (6)(cos 40)  4.596 m/s
(v A )t  6(sin 40°)  3.857 m/s
vB  (vB )n  4 m/s
(vB )t  0
t-direction:
Total momentum conserved:
m A(v A )t  mB (vB )t  mA (vB )t  mB (vB )t
(0.6 kg)( 3.857 m/s)  0  (0.6 kg)(vA )t  (1 kg)(vB )t
2.314 m/s  0.6 (vA )t  (vB )t
(1)
m A (vA )t  mA (vA )t 3.857  (v A )t
(vA )t  3.857 m/s
(2)
Ball A alone:
Momentum conserved:
Replacing (vA )t in (2) in Eq. (1)
2.314  (0.6)( 3.857)  (vB )t
2.314  2.314  (vB )t
(vB )t  0
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PROBLEM 13.166 (Continued)
n-direction:
Relative velocities:
[(v A ) n  (vB )n ]e  (vB )n  (v A )n
[(4.596)  (4)](0.8)  (vB ) n  (vA )n
6.877  (vB ) n  (vA )n
(3)
Total momentum conserved:
mA (v A )n  mB (vB )n  mA (vA )n  mB (vB )n
(0.6 kg)(4.596 m/s)  (1 kg)(  4 m/s)  (1 kg)(vB ) n  (0.6 kg)(vA )n
1.2424  (vB )n  0.6(vA )n
(4)
Solving Eqs. (4) and (3) simultaneously,
(vA ) n  5.075 m/s
(vB ) n  1.802 m/s
Velocity of A:
tan  
|(v A )t |
|(v A ) n |
3.857
5.075
  37.2

  40  77.2
2
vA  (3.857)  (5.075) 2
 6.37 m/s
Velocity of B:
vA  6.37 m/s
77.2 
vB  1.802 m/s
40 
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237
PROBLEM 13.167
Two identical hockey pucks are moving on a hockey rink at the same speed of
3 m/s and in perpendicular directions when they strike each other as shown.
Assuming a coefficient of restitution e  0.9, determine the magnitude and direction
of the velocity of each puck after impact.
SOLUTION
Use principle of impulse-momentum:
mv1  Imp12  mv 2
t-direction for puck A:
 mv A sin 20  0  m(v A )t
(vA )t  v A sin 20  3sin 20  1.0261 m/s
t-direction for puck B:
 mvB cos 20  0  m(vB )t
(vB )t  vB cos 20  3cos 20  2.8191 m/s
n-direction for both pucks:
mvA cos 20  mvB sin 20  m(vA )n  m(vB )n
(vA )n  (vB )n  vA cos 20  vB sin 20
 3cos 20  3sin 20
(1)
e  0.9
Coefficient of restitution:
(vB )n  (vA )n  e[(vA )n  (vB )n ]
 0.9[3cos 20  ( 3) sin 20]
Solving Eqs. (1) and (2) simultaneously,
(vA )n  0.8338 m/s
(vB )n  2.6268 m/s
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(2)
PROBLEM 13.167 (C
(Continued)
Summary:
( vA )n  0.8338 m/s
20
( vA )t  1.0261 m/s
70
( vB )n  2.6268 m/s
20
( vB )t  2.8191 m/s
70
v A  (0.8338) 2  (1.0261) 2  1.322 m/s
tan  
1.0261
0.8338
  50.9
  20  70.9
vA  1.322 m/s
70.9 
vB  3.85 m/s
27.0 
vB  (2.6268) 2  (2.8191) 2  3.85 m/s
tan  
2.8191
  47.0
2.6268
  20  27.0
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239
PROBLEM 13.168
The coefficient of restitution is 0.9 between the two 60-mmdiameter billiard balls A and B. Ball A is moving in the
direction shown with a velocity of 1 m/s when it strikes ball B,
which is at rest. Knowing that after impact B is moving in the
xdirection, determine (a) the angle  , (b) the velocity of B
after impact.
SOLUTION
(a) Since vB is in the x-direction and (assuming no friction), the
common tangent between A and B at impact must be parallel to the
y-axis
tan  
Thus
250
150  D
  tan 1
250
 70.20
150  60
  70.2 
(b) Conservation of momentum in x(n)direction
mv A cos  m  vB n  m  vA n  mvB
1 cos  70.20   0   vA n  vB
0.3387   vA n   vB 
(1)
Relative velocities in the n direction
e  0.9
 vA cos   vB n  e  vB   vA n
 0.3387  0  0.9   vB   vA n
(2)
(1)  (2)
2vB  0.3387 1.9
vB  0.322 m/s 
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PROBLEM 13.169
A boy located at Point A halfway between the center O of a
semicircular wall and the wall itself throws a ball at the wall in a
direction forming an angle of 45 with OA.. Knowing that after
hitting the wall the ball rebounds in a direction parallel to OA,
determine the coefficient of restitution between the ball
ba and the
wall.
SOLUTION
Law of sines:
sin 
R
2

sin135
R
  20.705
  45  20.705
 24.295
Conservation of momentum for ball in t-direction:
direction:
 v sin   v  sin 
Coefficient of restitution in n:
Dividing,
v (cos  )e  v  cos 
tan 
 tan 
e
e
tan 20.705
tan 24.295
e  0.837 
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241
PROBLEM 13.170
The Mars Pathfinder spacecraft used large airbags to cushion
its impact with the planet’s surface when landing. Assuming
the spacecraft had an impact velocity of 18 m/s at an angle of
45° with respect to the horizontal, the coefficient of restitution
is 0.85 and neglecting friction, determine (a)
( the height of the
first bounce, (b) the length of the first bounce. (Acceleration of
gravity on the Mars  3.73 m/s2.)
SOLUTION
Use impulse-momentum
momentum principle.
mv1  Imp12  mv 2
The horizontal direction (x to the right) is the tangential direction and the vertical direction ( y upward) is the
normal direction.
Horizontal components:
mv0 sin 45  0  mvx
vx  v0 sin 45.
v x  12.728 m/s
Vertical components, using coefficient of restitution e  0.85
vy  0  e[0  (v0 cos 45)]
vy  (0.85)(18cos 45)
v y  10.819 m/s
The motion during the first bounce is projectile motion.
Vertical motion:
Horizontal motion:
1 2
gt
2
v y  (v y )0  gt
y  (v y ) 0 t 
x  vx t
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PROBLEM 13.170 (Continued)
(a)
Height of first bounce:
v y  0:
0  (v y )0  gt
(v y )0
10.819 m/s
 2.901 s
g
3.73 m/s 2
1
y  (10.819)(2.901)  (3.73)(2.901)2
2
t

y  15.69 m 
(b)
Length of first bounce:
y  0:
10.819t 
t  5.801 s
x  (12.728)(5.801)
1
(3.73) t 2  0
2
x  73.8 m 
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243
PROBLEM 13.171
A girl throws a ball at an inclined wall from a height
of 1.2 m, hitting the wall at A with a horizontal
velocity v 0 of magnitude 15 m/s. Knowing that the
coefficient of restitution between the ball and the wall
is 0.9 and neglecting friction, determine the distance
d from the foot of the wall to the Point B where the
ball will hit the ground after bouncing off the wall.
SOLutiOn
Momentum in t direction is conserved.
mv sin 30∞ = mvt¢
(15)(sin 30∞) = v ¢t
vt¢ = 7.5 m/s
Coefficient of restitution in n-direction.
( v cos 30∞)e = vn¢
(15)(cos 30∞)(0.9) = vn¢
vn¢ = 11.69 m/s
Writing v ¢ in terms of x and y components
( v x¢ )0 = vn¢ cos 30∞ - vt¢ sin 30∞
( v x¢ )0 = (11.69)(cos 30∞) - (7.5)(sin 30∞) = 6.374 m/s
( v y¢ )0 = vn¢ sin 30∞ + vt¢ cos 30∞
( v y¢ )0 = (11.69)(sin 30∞) + (7.5)(cos 30∞) = 12.340 m/s
Motion of a projectile.
(origin at 0)
y = y0 + ( v y¢ )0 t -
( gt 2 )
2
y = 1.2 + (12.340 m/s)t - (9.81 m/s2 )
t2
2
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244
PROBLEM 13.171 (continued)
Time to reach Point B
( y B = 0)
Ê 9.81ˆ 2
t
0 = 1.2 + 12.340t B - Á
Ë 2 ˜¯ B
t B = 2.610 s
x = x0 + ( v x¢ )0 t
x = 0 + 6.374t
x B = (6.374)(t B )
= (6.374 m/s)(2.610 s)
x B = 16.63 m
d = x B - 1.2 cot 60
= 15.94 m
d = 15.94 m b
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245
PROBLEM 13.172
Rockfalls can cause major damage to roads and infrastructure. To design
mitigation bridges and barriers, engineers use the coefficient of restitution
to model the behavior of the rocks. The rock A falls a distance of 20 m
before striking an incline with slope  = 40o. Knowing that the coefficient
of restitution between A and the incline is 0.2, determine the velocity of
the rock after the impact.
SOLUTION
Given:
Sketch:
y  20 m
e  0.2
 =40
Apply Work-Energy to determine speed at impact:
T1  Vg1  Ve1  U1NC
3  T2  Vg 2  Ve2
where: Vg1  mgy and T2 
1 2
mvB
2
vB  40 g m/s
Components in n and t directions:
 v B n  vB cos 
 v B t  vB sin 
Conservation of momentum in the tt-direction:
n-components
components using coefficient of restitution:
 v'B t   v B t  vB sin 
e   v B n  0  0   v'B n
 v'B n  e  v B n
Speed of rock after impact:
v'B 
 v'B t   v'B n
2
2
 e 2  v B  n   v B t
2
2
 vB e 2 cos2   sin 2   13.09 m/s
Direction after impact:
tan  
 v'B n e vB cos 

   13.4
vB sin 
 v'B t
v'B  13.09 m/s
26.6 
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PROBLEM 13.173
From experimental tests, smaller boulders tend to have a greater
coefficient of restitution than larger boulders. The rock A falls a distance
of 20 meters before striking an incline with slope  = 45o. Knowing that
h=
= 30 m and d= 20 m, determine if a boulder will land on the road or
beyond the road for a coefficient of restitution of (a) e= 0.2, (b) e= 0.1.
SOLUTION
Given:
y  20 m
h  30 m
Sketch:
d  20 m
 =45
Apply Work-Energy
Energy to determine speed of the rock at impact:
T1  Vg1  Ve1  U1NC
3  T2  Vg 2  Ve2
where: Vg1  mgy and T2 
1 2
mvB
2
vB  40 g  19.809 m/s
Components in n and t directions:
 v B n  vB cos
 v B t  vB sin 
Conservation of momentum in the t-direction:
direction:
 v'B t   v B t  vB sin 
n-components
components using coefficient of restitution:
e  v B n  0  0   v'B n


 v'B n  e  v B n
Speed of rock after impact:
v'B 
 v'B t   v'B n
2
2
 e2  v B  n   v B t
2
2
 vB e2 cos2   sin 2 
 v'B n e vB cos 

   tan 1 e
v'
vB sin 
 B t
Direction after impact:
tan  
x and y components of velocity after impact:
 v'B  x  v 'B cos    
 v'B  y  v 'B sin    
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PROBLEM 13.173 (Continued)
Projectile Motion in y-dir:
Projectile Motion in x-dir:
(a) For e=0.2:
1 2
gt
2
1
0  30  (vB ) y t  gt 2
2
y  y0  (vB ) y t 
x  x0  (vB ) x t
(1)
(2)
v'B  14.285 m/s
 =11.31
 v'B  x  11.885 m/s
 v'B  y  7.924 m/s
Solve (1) for t:
t  1.7939 s or t  3.4094 s
Solve (2) for x:
x  21.32 m > d
(b) For e=0.1:
Rock will land beyond the road
v'B  14.077 m/s
 =5.711
 v'B  x  10.894 m/s
 v'B  y  8.914 m/s
Solve (1) for t:
t  1.7261 s or t  3.5434 s
Solve (2) for x:
x  18.81 m < d
Rock will land on the road
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PROBLEM 13.174
Two cars of the same mass run head-on
into each other at C. After the collision,
the cars skid with their brakes locked
and come to a stop in the positions
shown in the lower part of the figure.
Knowing that the speed of car A just
before impact was 8 km/h and that
the coefficient of kinetic friction
between the pavement and the tires
of both cars is 0.30, determine (a) the
speed of car B just before impact, (b)
the effective coefficient of restitution
between the two cars.
SOLutiOn
(a)
At C:
Conservation of total momentum.
mA = mB = m
20
m/s
9
+ m v + m v = m v¢ + m v¢
A A
B B
A A
B B
20
- + v B = v A¢ + v B¢
9
8 km/h =
(1)
Work and energy.
Car A (after impact):
1
mA ( v A¢ )2
2
T2 = 0
U1- 2 = - F f ( 4)
T1 =
U1- 2 = -m k mA g ( 4 m)
T1 + U1- 2 = T2
1
mA ( v A¢ )2 - m k mA g ( 4) = 0
2
( v A¢ )2 = (2)( 4 m)(0.3)(9.81 m/s2 )
= 23.544 m/s2
v A¢ = 4.8522 m/s
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249
PROBLEM 13.174 (continued)
Car B (after impact):
1
mB ( v B¢ )2
2
T2 = 0
U1- 2 = -m k mB g (1)
T1 =
T1 + U1- 2 = T2
1
mB ( v B¢ )2 - m k mB g (1) = 0
2
v B¢ 2 = (2)(1 m)(0.3)(9.81 m/s2 )
( v B¢ )2 = 5.886 m2 /s2
v B¢ = 2.4261 m/s
From (1)
20
+ v A¢ 2 + v B¢
9
= 2.222 + 5.886 + 2.4261
= 10.5343 m/s
=37.924 km/h
vB =
v B = 37.9 km/h b
(b)
Relative velocities.
+ (-v
¢ - v A¢
A - vB ) e = vB
( -2.222 - 10.5343) e = 2.4261 - 4.8522
e = 0.19019
e = 0.1902 b
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250
PROBLEM 13.175
A 1-kg
kg block B is moving with a velocity v0 of magnitude v0  2 m/s as
it hits the 0.5
0.5-kg sphere A, which is at rest and hanging from a cord
attached at O. Knowing that k  0.6 between the block and the
horizontal surface and e  0.8 between the block and the sphere,
determine after impact (a) the maximum height h reached by the sphere,
(b)) the distance x traveled by the block.
SOLUTION
Velocities just after impact
Total momentum in the horizontal direction is conserved:
m Av A  mB vB  m Av A  mB vB
0  (1 kg)(2 m/s)  (0.5 kg)(vA )  (1 kg)(vB )
4  vA  2vB
(1)
Relative velocities:
(v A  vB ) e  (vB  v A )
(0  2)(0.8)  vB  vA
1.6  vB  vA
(2)
Solving Eqs. (1) and (2) simultaneously:
vB  0.8 m/s
vA  2.4 m/s
(a)
Conservation of energy:
1
m A v12 V1  0
2
1
T1  m A (2.4 m/s) 2  2.88 mA
2
T1 
T2  0
V2  mA gh
T1  V1  T2  V2
2.88 m A  0  0  m A (9.81)h
h  0.294 m 
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PROBLEM 13.175 (Continued)
(b)
Work and energy:
1
1
mB v12  mB (0.8 m/s) 2  0.32mB
T2  0
2
2
U1 2   F f x   k Nx    x mB gx  (0.6)(mB )(9.81) x
T1 
U1 2  5.886mB x
T1  U1 2  T2 :
0.32mB  5.886mB x  0
x  0.0544 m
x  54.4 mm 
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PROBLEM 13.176
A 90-g ball thrown with a horizontal
velocity v0 strikes a 720-g plate
attached to a vertical wall at a height
of 900 mm above the ground. It is
observed that after rebounding, the
ball hits the ground at a distance of
480 mm from the wall when the
plate is rigidly attached to the wall
(Fig. 1), and at a distance of 220 mm
when a foam-rubber mat is placed
between the plate and the wall (Fig.
2). Determine (a) the coefficient of
restitution e between the ball and the
plate, (b) the initial velocity v0 of the
ball.
SOLutiOn
(a)
Figure (1), ball alone
v0 e = ( v B¢ )1
Relative velocities.
Projective motion. t = time for the ball to hit the ground.
0.480 m = v0 e t
(1)
Figure (2), ball and plate
Relative velocities.
+
( v B - vP )e = vP¢ + ( v B¢ )2
v B = v0
vP = 0
v0 e = vP¢ + ( v B¢ )2
(2)
Conservation of momentum.
+
mB v B + mP vP = mB ( - v B¢ )2 + mP ( vP¢ )
(0.09 kg)( v0 ) + 0 = (0.09 kg)( - v B¢ )2 + (0.720 kg)( vP¢ )
vo = -( v B¢ )2 + 8vP¢
(3)
Solving (2) and (3) simultaneously for (1) B/ 2
( v B¢ )2 = v0
(8 e - 1)
9
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253
PROBLEM 13.176 (continued)
Projectile motion.
0.220 m = v0
(8e - 1)
t
9
(4)
Dividing Eq. (4) by Eq. (3),
0.220 8e - 1
=
0.480
9e
4.125e = 8e - 1
e = 0.258 b
(b)
From Figure (1)
Projectile motion.
1 2
gt
2
(9.81) 2
0.900 =
t ,
2
h=
1.80 = 9.81t 2
(5)
Equation (1):
0.480 = v0 et
t=
(0.480)
(0.258) v0
=
1.860
v0
Substituting for t in Eq. (5),
1.800 = (9.81)
(1.860)2
v02
v02 = 18.855
v0 = 4.34 m/s b
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254
PROBLEM 13.177
After having been pushed by an airline employee, an empty
40-kg luggage carrier A hits with a velocity of 5 m/s an
identical carrier B containing a 15-kg
kg suitcase equipped with
rollers. The impact causes the suitcase to roll into the left wall
of carrier B.. Knowing that the coefficient of restitution between
the two carriers is 0.80 and
d that the coefficient of restitution
between the suitcase and the wall of carrier B is 0.30, determine
(a) the velocity of carrier B after the suitcase hits its wall for
the first time, (b)) the total energy lost in that impact.
SOLUTION
(a)
Impact between A and B:
Total momentum conserved:
mAvA  mB vB  mAvA  mB vB
mA  mB  40 kg
5 m/s  0  vA  vB
(1)
Relative velocities:
(vA  vB )eAB  vB  vA
(5  0)(0.80)  vB  vA
(2)
Adding Eqs. (1) and (2)
(5 m/s)(1  0.80)  2vB
vB  4.5 m/s
Impact between B and C(after A hits B
B)
Total momentum conserved:
mB vB  mC vC  mB vB  mC vC
(40 kg)(4.5 m/s)  0  (40 kg) vB  (15 kg) vC
4.5  vB  0.375 vC
(3)
Relative velocities:
(vB  vC ) eBC  vC  vB
(4.5  0)(0.30)  vC  vB
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(4)
PROBLEM 13.177 (Continued)
Adding Eqs. (4) and (3)
(4.5)(1  0.3)  (1.375)vC
vC  4.2545 m/s
vB  4.5  0.375(4.2545) vB  2.90 m/s
(b)
vB  2.90 m/s 
TL  (TB  TC )  (TB  TC )
TB 
1
 40 
mB (vB ) 2  
kg  (4.5 m/s) 2  405 J
2
2


TC  0
TC 
TB 
1
 40 
mB (vB ) 2  
kg  (2.90) 2  168.72 J
2
2


1
 15

mC (vC ) 2  
kg  (4.2545 m/s)2  135.76 J
2
2


 TL  (405  0)  (168.72  135.76)  100.5 J
TL  100.5 J 
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256
PROBLEM 13.178
Blocks A and B each weigh 400-g and block
C has a mass of 1.2 kg. The coefficient of
friction between the blocks and the plane is
m k = 0.30. Initially, block A is moving at a
speed v0 = 5 m/s and blocks B and C are at
rest (Fig. 1). After A strikes B and B strikes C,
all three blocks come to a stop in the positions
shown (Fig. 2). Determine (a) the coefficients
of restitution between A and B and between B
and C, (b) the displacement x of block C.
SOLutiOn
(a)
Work and energy.
Velocity of A just before impact with B.
T1 =
( )
1
1
mA v02 T2 = mA v 2A
2
2
2
U1- 2 = - m kWA (0.3 m)
T1 + U1- 2 = T2
( )
1
1
mA v02 - m k mA g (0.3) = mA v 2A
2
2
2
(v ) = v - 2m g(0.3) = (5) - (2)(0.3)(9.81)(0.3)
(v ) = 23.2342 m /s , (v ) = 4.8202 m/s
2
A
2
A
2
2
0
2
k
2
2
A 2
2
Velocity of A after impact with B.
( v A¢ )2
( )
1
mA v 2A T3 = 0
2
2
U 2-3 = - m k W A (0.075)
T2 =
T2 + U 2-3 = T3 ,
( )
1
mA v 2A - ( m k )mA g (0.075) = 0
2
2
(v ) = 2(0.3)(9.81 m/s )(0.075m) = 0.44145 m /s
2 1
A
2
2
2
2
( v A¢ )2 = 0.6644 m/s
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257
PROBLEM 13.178 (continued)
Conservation of momentum as A hits B.
( v A )2 = 4.8202 m/s
( v A¢ )2 = 0.6644 m/s
+
mA ( v A )2 + mB v B = mB ( v A¢ )2 + mB v B¢ mA = mB
4.8202 + 0 = 0.6644 + v B¢ v B¢ = 4.1558 m/s
Relative velocities (A and B).
+
[( v A )2 - v B ]e AB = v B¢ - ( v A¢ )2
( 4.8202 - 0)e AB = 4.1558 - 0.6644
e AB = 0.724 b
Work and energy.
Velocity of B just before impact with C.
1
1
mB ( v B¢ )22 = mB ( 4.1558)2
2
2
1
T4 = mB ( v B¢ )24
2
( v B¢ )2 = 4.1558 m/s U 2- 4 = - m k mB g (0.3 m)
T2 =
1
1
T2 + U 2- 4 = T4 fi mB ( 4.1558)2 - (0.3)( mB )(9.81)(0.3) = mB ( v B¢ )24
2
2
( v B¢ )4 = 3.9376 m/s
F f = m k WB
Conservation of momentum as B hits C.
mB = 0.4 kg
mC = 1.2 kg
( v B¢ )4 = 3.9376 m/s
+
mB ( v B¢ )4 + mC vC = mB ( v B¢¢)4 + mC vC¢
0.4(3.9376) + 0 = 0.4( v B¢¢)4 + 1.2vvC¢
3.9376 = ( v B¢¢ )4 + 3vC¢
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258
PROBLEM 13.178 (continued)
Velocity of B after B hits C , ( v B¢¢)4 = 0.
(Compare Figures (1) and (2).)
Thus, vC¢ = 1.3125 m/s
Relative velocities (B and C).
(( v B¢ ) 4 - vC )eBC = vC¢ - ( v B¢¢)4
(3.9376 - 0)eBC = 1.3125 - 0
eBC = 0.288 b
(b)
Work and energy (Block C).
1
mC ( vC¢ )2 T5 = 0 U 4 -5 = - m kWC ( x )
2
1
T4 + U 4 -5 = T5
mC (1.3125)2 - (0.3) mC (9.81)( x ) = 0
2
T4 =
x=
(1.3125)2
= 0.29267 m
2 (9.81)(0.3)
x = 293 mm b
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259
PROBLEM 13.179
A 5 kg sphere is dropped from a height y=2
=2 m to test newly designed spring
floors used in gymnastics. The mass of the floor section is 10 kg, and the
effective stiffness of the floor is k= 120 kN/m. Knowing that the coefficient of
restitution between the ball and the platform is 0.6, determine (a) the height h
reached by the sphere after rebound, (b) the maximum force in the springs.
SOLUTION
Given:
mA  5 kg
Sketch before/after impact:
mB  10 kg
y2m
e  0.6
k  120 kN/m
Apply Work-Energy
Energy to determine speed of the sphere just before impact:
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
where: Vg1  mA gy and T2 
1
mAvA2
2
vA  2 gy  6.264 m/s 
Conservation of momentum in the yy-direction:
mAvA  mB vB  mAvA  mB vB
Dividing by mA with ( m B /m A  2) :
- 6.264  vA  2vB
Coefficient of restitution:
vB  vA  e v A  vB

(1)

(2)
 6.264e
Solving Eqs. (1) and (2) simultaneously yields:
vA  0.4176 m/s
vB  3.341 m/s
(a) Apply Work-Energy
Energy to sphere A to determine height it rises:
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
h
where: T1 
1
mAv '2A and Vg2  mA gh
2
v '2A
m/s
2g
h  8.89 mm 
(b) Let  0 be the initial compression of the spring and hB be the additional compression of the spring after
impact. In the initial equilibrium state:
Fy  0: k 0  WB  0 or k 0  mB g
Apply Work-Energy
Energy to determine maximum deflection of the springs (when the plate reaches its
lowest point and the speed is zero):
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PROBLEM 13.179 (Continued)
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
1
1
mB v '2B , Vg1  mB g 0 , Ve1  k 02
2
2
1
2
Vg2  mB g  0  hB  , Ve2  k  0  hB 
2
where: T1 
1
1
1
1
mB v '2B  mB g 0  k 02   mB g  0  hB   k  02  k  0 hB  khB2
2
2
2
2
1
1
Simplify:
mB v '2B   mB ghB  k  0 hB  khB2
2
2
1
1
Substitute k  0  m B g :
mB v '2B   m B ghB  mb ghB  khB2
2
2
Put in known values:
Solving for hB:
Maximum force in spring
120000hB2  111.616  0
hB  0.0305 m
F  k  0  hB 
 k 0  khB
 mB g  khB
Fmax  3758 N 
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261
PROBLEM 13.180
A 5 kg sphere is dropped from a height y=3
=3 m to test a new spring floor used in
gymnastics. The mass of the floor section B is 12 kg, and the sphere bounces
ba
back upwards a distance of 44 mm. Knowing that the maximum deflection of the
floor section is 33 mm from its equilibrium position, determine (a) the
coefficient of restitution between the sphere and the floor, (b) the effective
spring constant k of the floor section.
SOLUTION
Given:
mA  5 kg
Sketch before/after impact:
mB  12 kg
y 3m
h  44 mm
smax  33 mm
Apply Work-Energy
Energy to determine speed of the sphere just before impact:
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
where: Vg1  mA gy and T2 
1
mAvA2
2
vA  2 gy  7.672 m/s  or v A  7.672 m/s
Apply Work Energy to determine the speed of the sphere just after impact:
T2  Vg1  Ve1  U 2NC
3  T3  Vg 3  Ve3
where: T2 
1
mAv '2A and Vg3  mA gh
2
v ' A  2 gh  0.9291 m/s 
(a) Conservation of momentum in the yy-direction: mAvA  mB vB  mAvA  mB vB
Put in known values and solve:
vB 
mA
 vA  vA 
mB
 3.584 m/s
Coefficient of restitution:

vB  vA  e v A  vB
e

vB  vA
vA
Put in known values and solve:
e  0.588 
(b) Let  0 be the initial compression of the spring and smax be the additional compression of the spring
m g
after impact. In the initial equilibrium state:
Fy  0: k 0  WB  0 or  0  B
k
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262
PROBLEM 13.180 (Continued)
Apply Work-Energy to determine the stiffness of the springs (when the plate reaches its lowest point
and the speed is zero). Take the equilibrium position as the datum.
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
where: T1 
1
1
mB v '2B , Ve1  k 02
2
2
Vg2  mB g  smax  , Ve2 
Substitute  0 
mB g
:
k
1
2
k  smax 
2
2
1
1 m g
1
2
mB v '2B  k  B   mB gsmax  k  smax 
2
2  k 
2
mB v '2B k   mB g   2mB g  smax  k  k 2  smax 
2
2
k 2  smax    mB v '2B  2 mB gsmax  k   mB g   0
2
2
Put in known values and solve for k:
k  148.7 kN/m or k  85.5 N/m
k  148.7 kN/m 
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263
PROBLEM 13.181
The three blocks shown are identical. Blocks B and C are at
rest when block B is hit by block A, which is moving with
a velocity v A of 1 m/s. After the impact, which is assumed
to be perfectly plastic (e = 0), the velocity of blocks A and B
decreases due to friction, while block C picks up speed, until
all three blocks are moving with the same velocity v. Knowing
that the coefficient of kinetic friction between all surfaces is
m k = 0.20, determine (a) the time required for the three blocks
to reach the same velocity, (b) the total distance traveled by
each block during that time.
SOLutiOn
(a)
Impact between A and B; conservation of momentum
mv A + mv B + mvC = mv A¢ + mv B¢ + mvC¢
1 + 0 = v A¢ + v B¢ + 0
(1)
Relative velocities (e = 0)
( v A - v B )e = v B¢ - v A¢
0 = v B¢ - v A¢
v A¢ = v B¢
From (1) v ¢A = v B¢ = 0.5 m/s
v = Final (common) velocities
Block C impulse and momentum
+
mc gvc + F f t = mc v
F f = m k mc g
mc gvc + m k mc gt = mc v
0 + 0.2 gt = v
v = (0.2) gt
(2)
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794
264
PROBLEM 13.181 (continued)
Block A and B impulse and momentum
2m(0.5) - 4(0.2)mgt = 2mv
1 - 0.8 gt = 2v
(3)
Substituting v from Eq. (2) into Eq. (3)
1 - 0.8 gt = 0.4 gt
t=
(b)
1 m/s
(1.2)(9.81 m/s2 )
t = 0.08495 s b
Work and energy.
From Eq. (2)
v = (0.2)(9.81)(0.08495) = 0.16667 m/s
Block C:
T1 = 0
T2 =
1
1
m( v )2 = m(0.16667)2
2
2
U1- 2 = F f xC = m kWxC = 0.2WxC
T1 + U1- 2 = T2
xC =
0 + (0.2)( mg ) xC =
1 2
mv
2
(0.16667)2
= 7.0794 ¥ 10 -3 m
2(0.2)(9.81)
xC = 7.08 mm b
Blocks A and B:
T1 =
1
(2m) (0.5)2 = 0.25m
2
T2 =
1
(2m) (0.16667)2 = 0.027779m
2
U1- 2 = -4 m kWgx A = -0.8mgx A = -7.848mx A
T1 + U1- 2 = T2
≠ 0.25m - 7.848mx A = 0.027779m
x A = 0.02832 m
x A = 28.3 mm b
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795
265
PROBLEM 13.182
Block A is released from rest and slides down the
frictionless surface of B until it hits a bumper on
the right end of B. Block A has a mass of 10 kg
and object B has a mass of 30 kg and B can roll
freely on the ground. Determine the velocities of A
and B immediately after impact when (a) e  0,
(b) e  0.7.
SOLUTION
Let the x-direction be positive to the right and the y-direction vertically upward.
Let (vA ) x , (vA ) y , (vA ) x and (vB ) y be velocity components just before impact and (vA ) x ,(vA )y ,(vB ) x , and
(vB ) y those just after impact. By inspection,
(vA ) y  (vB ) y  (vA ) y  (vB ) y  0
Conservation of momentum for x-direction:
While block is sliding down:
0  0  mA (vA ) x  mB (vB ) x
(vB ) x   (vA ) x
(1)
Impact:
0  0  mA (vA ) x  mB (vB ) x
(vB ) x   (vA ) x
(2)
  mA /mB
where
Conservation of energy during frictionless sliding:
Initial potential energies: mA gh for A,
0 for B.
Potential energy just before impact:
V1  0
Initial kinetic energy:
T0  0 (rest)
Kinetic energy just before impact:
1
1
T1  mAvA2  mB vB2
2
2
T0  V0  T1  V1
1
1
1
m Av A2  mB vB2  (mA  mB  2 )vA2
2
2
2
1
 m A (1   ) v A2
2
m A gh 
v A2  (v A ) 2x 
2 gh
1 
vA 
2 gh
1 
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(3)
PROBLEM 13.182 (Continued)
Velocities just before impact:
2 gh
1 
vA 
vB  
2 gh
1 
Analysis of impact. Use Eq. (2) together with coefficient of restitution.
(vB ) x  (vA ) x  e[(vA ) x  (vB ) x ]
  (vA ) x  (vA ) x  e[(vA ) x   (v A ) x ]
(vA ) x  e(vA ) x
Data:
(4)
mA  10 kg
mB  30 kg
h  0.2 m
g  9.81 m/s 2

From Eq. (3),
(a)
e  0:
10 kg
 0.33333
30 kg
(2)(9.81)(0.2)
1.33333
 1.71552 m/s
vA 
(vA ) x  0
(vB ) x  0
vA  0 
vB  0 
(b)
e  0.7:
(v A ) x  (0.7)(1.71552)
 1.20086 m/s
(vB ) x  (0.33333)(1.20086)
 0.40029 m/s
vA  1.201 m/s

vB  0.400 m/s

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267
PROBLEM 13.183
A 340-g ball B is hanging from an inextensible cord attached to a
support C. A 170-g ball A strikes B with a velocity v 0 of magnitude
1.5 m/s at an angle of 60 with the vertical. Assuming perfectly
elastic impact ( e  1 ) and no friction, determine the height h reached
by ball B.
SOLUTION
Ball A alone
Momentum in t-direction conserved
mA  v A t  mA  vA t
 v A t  0   vA t
 vA n  vA
Thus
60
60
Total momentum in the x-direction is conserved.
mAv A sin 60  mB  vB  x  mA  vA  sin 60  mB vB
v A  v0  1.5 m/s
 vB  x  0
0.17 1.5 sin 60  0    0.17  vA  sin 60   0.34  vB
0.2208   0.1472 v A  0.34 vB
(1)
Relative velocity in the n-direction
v A   vB   e  vB cos 30  vA ;
n

 1.5  01  0.866vB  vA
Solving Equations(1) and (2) simultaneously
v B  0.9446 m/s, v A  0.6820 m/s
Conservation of energy ball B
1
2
mB  vB 
2
1 WB
T1 
 3.0232 2
2 g
T1 
T2  0
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(2)
PROBLEM 13.183 (Continued)
V1  0
T1  V1  T2  V2 ;
h
V2  W B h
1 WB
 0.9446 2  0  WBh;
2 g
 0.9446 2  0.0455 m
 2  9.81
h  45.5 mm 
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269
PROBLEM 13.184
An 8-kg cylinder C is released from rest in the position shown and drops
onto a 5-kg platform A which is at rest and is supported by an
inextensible cord attached to a 5-kg counterweight B. Knowing that the
coefficient of restitution between cylinder C and platform A is 0.8,
determine (a) the velocities of C and A immediately after the first impact,
(b) the impulse of the force exerted on platform A by the cord during the
first impact.
SOLUTION
Conservation of energy before impact
(
)
T + V = 0 + 8 kg 9.81 m/s 2 ( 0.15 m ) =
1
(8 kg ) v02
2
v0 = 1.7155 m/s
Cylinder C:
Platform A:
Counterweight B:
Restitution:


0 + ∫ F1 dt − ∫ F2 dt = 5 v′A 
 4 unknowns
0 + ∫ F2 dt = 5 v′A


v′A − vC′ = 0.8 v0

8v0 − ∫ F1 dt = vC′
Simultaneous solution 4 Equations and 4 unknowns
(a ) v0′ = 0, v′A = 1.372 m/s W
(b ) ∫ F2 dt = 6.86 N ⋅s W
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PRoBlEM 13.185
Ball B is hanging from an inextensible cord. An identical ball A is released
from rest when it is just touching the cord and drops through the vertical distance
hA = 200 mm before striking ball B. Assuming e = 0.9 and no friction, determine
the resulting maximum vertical displacement hB of ball B.
Solution
Ball A falls
0
0

T1 + V1 = T2 +
V2
(Put datum at 2)
h = 0.2 m
1
mgh = mv 2A
2
v A = 2 gh = 2(9.81)(0.2) = 1.9809 m/s
Impact
q = sin -1
r
= 30∞
2r
Impulse-Momentum
v B¢ , v At
¢ , v An
¢
Unknowns:
x-dir
0 + 0 = mB v B¢ + mA v An
¢ sin 30∞ + mA v At
¢ cos 30∞
(1)
Noting that mA = mB and dividing by mA
v B¢ + v An
¢ sin 30∞ + v At
¢ cos 30∞ = 0
(1)
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271
PRoBlEM 13.185 (continued)
Ball A alone:
Momentum in t-direction.
-mA v A sin 30∞ + 0 = mA v At
v At
¢ = - v A sin 30∞ = -1.9809 sin 30∞ = -0.99045 m/s
(2)
Coefficient of restitution.
v Bn
¢ - v An
¢ = e( v An - v Bn )
v B¢ sin 30∞ - v An
¢ = 0.9( v A cos 30∞ - 0)
(3)
With known value for vAt, Eqs. (1) and (3) become
v B¢ + v An
¢ sin 30∞ = 0.99045 cos 30∞
v B¢ sin 30∞ - v An
¢ = (0.9)(1.9809) cos 30∞
Solving the two equations simultaneously,
v B¢ = 1.3038 m/s
v An
¢ = -0.8921 m/s
After the impact, ball B swings upward. Using B as a free body
T ¢ + V ¢ = TB + VB
1
mB ( v B¢ )2 ,
2
V ¢ = 0,
TB = 0
T¢ =
where
VB = mB ghB
and
1
mB ( v B¢ )2 = mB ghB
2
hB =
1 ( v B¢ )2
2 g
1 (1.3038)2
2 (9.81)
= 0.08664 m
=
hB = 86.6 mm b
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272
PROBLEM 13.186
A 70 g ball B dropped from a height h0  1.5 m reaches a
height h2  0.25 m after bouncing twice from identical 210-g
210
plates. Plate A rests directly on hard ground, while plate C rests
on a foam-rubber
rubber mat. Determine (a)
(
the coefficient of
restitution between the ball and the plates, (b)
( the height h1 of
the ball’s first bounce.
SOLUTION
(a)
Plate on hard ground(first rebound):
Conservation of energy:
1
1
1
mB v 2y  mB v02  mB gh0  mB vx2
2
2
2
v0  2 gh0
Relative velocities., n-direction
direction:
v0 e  v1
v1  e 2 gh0
  vBx
vBx
t-direction
Plate on foam rubber support at C.
Conservation of energy:
Points  and :
V1  V3  0
1
1
1
1
 ) 2  mB v12  mB (v3 )2B  mB (vBx
 )2
mB (vBx
2
2
2
2
(v3 ) B  e 2 gh0
Conservation of momentum
momentum:
At:
mB (v3 ) B  mP vP  mB (v3 ) B  mP vP
mP 210

3
mB
70
 e 2 gh0  (v3 ) B  3vP
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(1)
PROBLEM 13.186 (Continued)
Relative velocities:
[(v3 ) B  (vP )]e  vP  (v3 ) B
e2 2 gh0  0  vP  (v3 ) B
(2)
Multiplying (2) by 3 and adding to (1)
4(v3 ) B  2 gh0 (3e2  e)
Conservation of energy at ,
Thus,
(v3 ) B  2 gh2
4 2 gh2  2 gh0 (3e 2  e)
3e 2  e  4
h2
0.25
4
 1.63299
h0
1.5
3e2  e  1.633  0
(b)
e  0.923 
Points  and :
Conservation of energy.
1
1
1
 )2  mB v12  mB (vBx
 )2 ;
mB (vBx
2
2
2
1 2
e (2 gh0 )  gh1
2
h1  e2 h0  (0.923)2 (1.5)
h1  1.278 m 
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PROBLEM 13.187
A 2-kg
kg sphere moving to the right with a velocity of 5 m/s strikes at
A the surface of a 9-kg
kg quarter cylinder which is initially at rest and
in contact with a spring of constant 20 kN/m. The spring is held by
cables so that it is initially compressed 50 mm. Neglecting friction
and knowing that the coefficient of restitution is 0.6, determine
(a)) the velocity of the sphere immediately after impact, (b)
( the
maximum compressive force in the spring.
SOLUTION
Momentum:
mv1n  0  mv1n  Mv2 (0.7071)
(2 kg)(5 m/s)(0.7071)  2 kg v1n  9 kg v2 (0.7071)
Restitution:
v2 (0.7071)  v1n  0.6 v1n  0.6(5)(0.7071)
Solve for
 16 
v2    m/s,
 11 
 17 
v1n     (0.7071)
 11 
  1.092801 m/s
v1n

(b) Conservation of energy – cylinder + spring:
1 2 1
1
kx0  M (v2 )2  kx22
2
2
2
2
20, 000
1
20, 000 2
 16 
(0.05)2  (9)   
x2  34.52
2
2  11 
2
x2  0.05875 m,
F  kx2  20,000
N
(0.0587 m) = 1175 N 
m
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PROBLEM 13.188
A 2-kg sphere A strikes the frictionless inclined surface of a 6-kg wedge
B at a 90° angle with a velocity of magnitude 4 m/s. The wedge can roll
freely on the ground and is initially at rest. Knowing that the coefficient
of restitution between the wedge and the sphere is 0.50 and that the
inclined surface of the wedge forms an angle θ = 40° with the
horizontal, determine (a) the velocities of the sphere and of the wedge
immediately after impact, (b) the energy lost due to the impact.
SOLUTION
(a) Momentum of the sphere A alone is conserved in the t-direction.
mA (v A )t = mA (v′A )t
(v′A )t = 0
(vA )t = 0
(v′A )n = v′A
50°
Total momentum is conserved in the x-direction.
mAv A cos 50° + mB vB = mA ( −v′A ) cos 50° + mBv′B
vB = 0
v A = 4 m/s
2 ( 4 ) cos 50° + 0 = 2 ( −v′A ) cos50° + 6v′B
5.1423 = −1.2855v′A + 6v′B
(1)
Relative velocities in the n-direction
(vA − vB ) e = (v′B cos 50° + v′A ); vB = 0, vA = 4 m/s
4 ( 0.5 ) = 0.6428v′B + v′A; 2 = 0.6428v′B + v′A
(2)
Solving Equation (1) and Equation (2) simultaneously
v′A = 1.2736 m/s; v′B = 1.1299 m/s
v′A = 1.274 m/s
v′B = 1.130 m/s
(b)
T lost =
=
50° W
W
1
1
2
2
mAv A2 − mA (v′A ) + mB (v′B ) 

2
2
1
2
2
2 kg )( 4 m/s ) − ( 2 kg )(1.274 m/s )
(


2
2
− (6 kg )(1.130 m/s )  = 10.546 J

Tlost = 10.55 J W
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PROBLEM 13.189
When the rope is at an angle of   30 the 1-kg sphere A has
a speed v0  0.6 m/s. The coefficient of restitution between A
and the 2-kg wedge B is 0.8 and the length of rope l  0.9 m
The spring constant has a value of 1500 N/m and   20.
Determine, (a) the velocities of A and B immediately after the
impact (b)) the maximum deflection of the spring assuming A
does not strike B again before this point.
SOLUTION
Masses:
m A  1 kg
mB  2 kg
Analysis of sphere A as it swings down
down:
  30, h0  l (1  cos  )  (0.9)(1  cos30)  0.12058 m
Initial state:
V0  m A gh0  (1)(9.81)(0.12058)  1.1829 N  m
T0 
1 2 1
mv0  (1)(0.6)2  0.180 N  m
2
2
  0, h1  0, V1  0
Just before impact:
T1 
Conservation of energy:
1
1
mAv A2  (1)v A2  0.5v 2A
2
2
T0  V0  T1  V1
0.180  1.1829  0.5 v A2  0
v 2A  2.7257 m 2 /s2
v A  1.6510 m/s
Analysis of the impact: Use conservation of momentum together with the coefficient of restitution. e  0.8.
Note that the ball rebounds horizontally and that an impulse  Tdt is applied by the rope. Also, an impulse
 Ndt is applied to B through its supports.
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PROBLEM 13.189 (C
(Continued)
Both A and B:
Momentum in x-direction:
m A (v A ) x  0  mA (v A ) x  mB (vB ) x
(1)(1.6510)  (1)(vA ) x  (2)(vB ) x
(1)
(vA )n  (vA ) x cos 
Coefficient of restitution:
(vB ) n  0, (vA )n  (vA ) x cos  , (vB ) x cos 30
(vB ) n  (vA )n  e[(vA )n  (vB )n ]
(vB ) x cos   (vA ) x cos   e[(vA ) x cos  ]
Dividing by cos  and applying e  0.8 gives
(vB ) x  (vA ) x  (0.8)(1.6510)
(2)
Solving Eqs. (1) and (2) simultaneously,
(vA ) x  0.33020 m/s
(vB ) x  0.99059 m/s
(a)
Velocities immediately after impact.
(b)
Maximum deflection of wedge B.
Use conservation of energy:
vA  0.330 m/s

vB  0.991 m/s

TB1  VB1  TB 2  VB 2
1
mB vB2
2
VB1  0
TB1 
TB 2  0
VB 2 
1
k (x)2
2
The maximum deflection will occur when the block comes to rest (ie, no kinetic energy)
1
1
mB vB2  k ( x)2
2
2
( x) 2 
mB vB2 (2)(0.99059 m/s) 2

k
1500 N/m)
( x )  0.0362 m
 x  36.2 mm 
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PROBLEM 13.190
Skid marks on a drag race track indicate that the rear (drive) wheels of a
car skid for the first 18 m and roll with slipping impending during the
remaining 382 m. The front wheels of the car are just off the ground for
the first 18 m, and for the remainder of the race 75 percent of the weight
of the car is on the rear wheels. Knowing that the speed of the car is
58 km/h at the end of the first 18 m and that the coefficient of kinetic
friction is 80 percent of the coefficient of static friction, determine the
speed of the car at the end of the 400-m track. Ignore air resistance and
rolling resistance.
SOLUTION
First 18 m: Since all the cars’ weight is on the rear wheels which skid,
the force on the car is
F = µk N = ( µk )W
 1 hr 
v18 = (58 km/h )(1000 m/km ) 

 3600 s 
= 16.1 m/s
T1 = 0
T2 =
1
1 W 
W
2
2
=   (16.1 m/s ) = (129.6 )
mv18
2
2 g 
g
U1− 2 = ( F )(18 m ) = µk (W )(18 m )
T1 + U1− 2 = T2
W 
0 + 18µkW = (129.6 )  
g
129.6
µk =
= 0.73395
18
( )(9.81)
For 400 m: Force moving the car is for the first 18 m,
F1 = ( µk )(W ) = (0.73395 )W
For the remaining 382 m, with 75% of weight on rear drive wheels and
impending sliding,
F2 = ( µs )( 0.75 )W
µs = µk (0.80 ) = (0.73395 )(0.80 ) = 0.91744
F2 = (0.91744 )(0.75 )(W ) = 0.68808
T1 = 0
T2 =
1 W 
2
  (v400 )
2 g 
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PROBLEM 13.190 CONTINUED
U1− 2 = F1 (18 m ) + F2 (382 m )
= (0.73395 )(W )(18 m ) + (0.68808 )(W )(328 m )
= 13.21W + 262.8W = 276.01W
T1 + U1− 2 = T2
0 + 276.01W =
(
1 W 
2
  (v400 )
2 g 
)
2
v 400
= ( 2 g ) 276.01 = ( 2 ) 9.81 m/s 2 ( 276.01)
2
v400
= 5415.3
v400 = 73.6 m/s
v400 = 265 km/h W
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PROBLEM 13.191
A 60-kg pellet shot vertically from a spring-loaded pistol on the surface of the earth rises to a height of 90 m.
The same pellet shot from the same pistol on the surface of the moon rises to a height of 570 m Determine the
energy dissipated by aerodynamic drag when the pellet is shot on the surface of the earth. (The acceleration of
gravity on the surface of the moon is 0.165 times that on the surface of the earth.)
SOLutiOn
Since the pellet is shot from the same pistol, the initial velocity v0 is the same on the moon and on the earth
Work and energy.
1
Earth:
T1 = mv02
2
U1- 2 = - mg E (90 m) - E L
( E L = loss of energy due to drag))
T1 - 90 mg E - E L = 0
T1 =
Moon:
1 2
mv0
2
T2 = 0
U1- 2 = - mg M (570)
(1)
T2 = 0
T1 - 570 mg M = 0
Subtracting (1) from (2)
(2)
-570mg M + 90mg E + E L = 0
g M = 0.165 g E
m = 0.06 kg
E L = 570(0.06) g M - 90(0.06) g E
= 570(0.06)(0.165) g E - 90(0.06) g E
= 9.81 ¥ 0.06(570 ¥ 0.165 - 90)
= 2.3838 J
E L = 2.38 J b
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281
PROBLEM 13.192
A satellite describes an elliptic orbit about a planet of mass M. The
minimum and maximum values of the distance r from the satellite to
the center of the planet are, respectively, r0 and r1 . Use the
principles of conservation of energy and conservation of angular
momentum to derive the relation
1 1 2GM
GM
  2
r0 r1
h
where h is the angular momentum per unit mass of the satellite and
G is the constant of gravitation.
SOLUTION
Angular momentum:
h  r0 ,
v0  r1 v1
b  r0 v0  r1v1
v0 
Conservation of energy:
h
r0
v1 
h
r1
(1)
1 2
mv0
2
GMm
VA  
r0
TA 
1 2
mv1
2
GMm
VB  
r1
TB 
TA  VA  TB  VB
1 2 GMm 1 2 GMm
 mv1 
mv0 
2
r0
2
r1
1 1
r  r 
v02  v12  2GM     2GM  1 0 
 r0 r1 
 r1r0 
Substituting for v0 and v1 from Eq. (1)
1
r  r 
1
h 2  2  2   2GM  1 0 
 r1r0 
 r0 r1 
 r2  r2 
r  r 
h2
h 2  1 2 20   2 2 (r1  r0 )(r1  r0 )  2GM  1 0 
 r1r0 
 r1 r0  r1 r0
1 1
h2     2GM
 r0 r1 
 1 1  2GM
   2
h
 r0 r1 
Q.E.D. 
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PROBLEM 13.193
A 60-g steel sphere attached to a 200-mm cord can swing
about Point O in a vertical plane. It is subjected to its own
weight and to a force F exerted by a small magnet embedded
in the ground. The magnitude of that force expressed in
newtons is F  3000/r 2 where r is the distance from the
magnet to the sphere expressed in millimeters. Knowing that
the sphere is released from rest at A, determine its speed as it
passes through Point B.
SOLUTION
Mass and weight:
m  0.060 kg
W  mg  (0.060)(9.81)  0.5886 N
Vg  Wh
Gravitational potential energy:
where h is the elevation above level at B.
Potential energy of magnetic force:
3000
dV

( F , in newtons, r in mm)
2
dr
r
r 3000
3000

Vm  
N  mm
 r2
r
F

Use conservation of energy:
T1  V1  T2  V2
Position 1: (Rest at A.)
v1  0
T1  0
h1  100 mm
(Vg )1  (0.5886 N)(100 mm)  58.86 N  mm
2
From the figure, AD  2002  1002 (mm2 )
MD  100  12  112 mm
2
r12  AD  MD
2
 2002  1002  1122
 42544 mm 2
r1  206.26 mm
(Vr )1  
3000
 14.545 N  mm
r1
V1  58.86  14.545  44.3015 N  mm  44.315 103 N  m
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283
PROBLEM 13.193 (Continued)
Position 2. (Sphere at Point B.)
1 2 1
mv2  (0.060)v22  0.030 v22
2
2
(Vg ) 2  0
(since h2  0)
T2 
r2  MB  12 mm
(See figure.)
3000
 250 N  mm  250  103 N  mm
12
T1  V1  T2  V2
(Vm ) 2  
0  44.315 103  0.030v22  250 103
v22  9.8105 m2 /s2
v2  3.13 m/s 
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PROBLEM 13.194
A 22.7 kg sphere A of radius 11.4 cm moving with a velocity of magnitude
v0 = 1.8 m/s strikes a 2 kg sphere B of radius 5 cm which is hanging from
an inextensible cord and is initially at rest. Knowing that sphere B swings
to a maximum height h = .23 m, determine the coefficient of restitution
between the two spheres.
SOLUTION
Angle of impulse force from geometry of A and B
.151 
= 23°
 .164 
θ = cos −1 
Total momentum conserved
Ball A:
Ball B:
(1)
Restitution
e=
e=
v′B cosθ − ( v′A )x cosθ + ( v′A ) y sin θ
v A cosθ
v′B − (v′A )x + (v′A ) y tan θ
vA
=
v A = v0 = 1.8 m/s
( )
v′B − (v′A )x + ( v′A )y .064
.151
2
continued
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285
PROBLEM 13.194 CONTINUED
A:
mAv A sin θ = m A (v′A )x (sin θ ) + mA (v′A ) y ( cosθ )
.064 
.064 
+ (v ′A ) y
1.8
= (v ′A ) x 
 .151 
 .151 
v A tan θ = (v′A ) x tan θ + (v′A ) y ;
.76 = .42(v ′A ) x + (v ′A ) y
A + B : m Av A = mA (v′A )x + mBv′B ;
(2)
(22.7)(1.8 m/s) = (22.7)(v ′A ) x + (2)vB′
(3)
g’s cancel
From equation (1)
From equation (3)
(
)
vB′ = 2 9.81 m/s2 (.23 m ) = 2.12 m/s
(22.7)(1.8) = 22.7(v A′ ) x + 2(2.12)
(v′A ) x = 1.6 m/s
From equation (2)
.76 = .42(1.6) + (v ′A ) y
(v′A ) y = .088 m/s
.064 
2.12 − 1.6 + .088
 .151 
= .309
e=
1.8
e = .309 W
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286
PROBLEM 13.195
A 300-g block is released from rest after a spring of constant
k  600 N/m has been compressed 160 mm. Determine the force
exerted by the loop ABCD on the block as the block passes through
(a) Point A, (b) Point B, (c) Point C. Assume no friction.
SOLUTION
Conservation of energy to determine speeds at locations A, B, and C.
Mass: m  0.300 kg
Initial compression in spring: x1  0.160 m
Place datum for gravitational potential energy at position 1.
Position 1: v1  0
V1 
T1 
1 2
mv1  0
2
1 2 1
kx1  (600 N/m)(0.160 m)2  7.68 J
2
2
Position 2: T2 
1 2 1
mvA  (0.3)vA2  0.15vA2
2
2
V2  mgh2  (0.3 kg)(9.81 m/s 2 )(0.800 m)  2.3544 J
T1  V1  T2  V2 : 0  7.68  0.15v 2A  2.3544
v 2A  35.504 m 2 /s 2
Position 3:
T3 
1 2 1
mvB  (0.3)vB2  0.15vB2
2
2
V3  mgh3  (0.3 kg)(9.81 m/s 2 )(1.600 m)  4.7088 J
T1  V1  T3  V3 : 0  7.68  0.15vB2  4.7088
vB2  19.808 m 2 /s 2
Position 4: T2 
1 2 1
mvC  (0.3)vC2  0.15vC2
2
2
V4  mgh4  (0.3 kg)(9.81 m/s)(0.800 m)  2.3544 J
T1  V1  T4  V4 : 0  7.68  0.15vC2  2.3544
vC2  35.504 m 2 /s 2
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287
PROBLEM 13.195 (Continued)
(a)
Newton’s second law at A:
an 
v 2A 35.504 m 2 /s 2

 44.38 m/s 2

0.800 m
an  44.38 m/s2
F  man : N A  man
N A  (0.3 kg)(44.38 m/s2 )
(b)
N A  13.31 N

Newton’s second law at B:
an 
vB2 19.808 m 2 /s 2

 24.76 m/s 2

0.800 m
an  24.76 m/s2
F  man : N B  mg  man
N B  m(an  g )  (0.3 kg)(24.76 m/s2  9.81 m/s2 )
(c)
N B  4.49 N 
Newton’s second law at C:
an 
vC2 35.504 m 2 /s 2

 44.38 m/s 2

0.800 m
an  44.38 m/s2
F  man : NC  man
NC  (0.3 kg)(44.38 m/s2 )
NC  13.31 N
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
PROBLEM 13.196
A soccer kicking machine has a 5 lb “simulated foot” attached to A
“kicking” attachment goes on the front of a wheelchair, allowing athletes
with mobility impairments to play soccer. The athletes load up the spring
shown through a ratchet mechanism that ppulls
ulls the 2 kg “foot” back to the
position 1. They then release the “foot” to impact the 0.45 kg soccer ball,
which is rolling towards the “foot” with a speed of 2 m/s at an angle
= 30º as shown. The impact occurs with a coefficient of restitution e=
0.75
75 when the foot is at position 2, where the spring is unstretched.
Knowing that the effective friction coefficient during rolling is k = 0.1,
determine (a) the necessary spring coefficient to make the ball roll 30
meters, (b) the direction the ball will travel after it is kicked.
SOLUTION
m A  2 kg, mB  0.45 kg, vB  2 m/s, e  0.75, k =0.1, s 2  0
Given:
 vB t   vB t
Conservation of momentum of the ball in the tt-direction:
 vB sin 30
 1 m/s
Apply Work-Energy
Energy to the ball after impact to when it comes to rest:
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
1
T2  mB v '2B
2
NC
U12  k mB g  30
where:
1
mB v '2B  k mB g  30  =0
2
v 'B  7.672 m/s
Therefore:
Find  v 'B n using:
v 'B 
 v ' B  n   v ' B t
2
2
 v 'B n  v '2B   v 'B t2
 7.607 m/s 
Conservation of momentum of the systems in the nn-direction Impulse momentum diagrams for ball and
foot:
mA  vA n  mB  vB n  mA  vA n  mB  vB n
 vA n 
1
mA  vA   mB  vB   mB  vB  
n
n
n
mA 
(1)
Coefficient of Restitution:
 v'B n   vA n  e  vA n   vB n 
(2)
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289
PROBLEM 13.196 (Continued)
Sub (1) into (2) and solve for  vA n :
1
mA  vA   mB  vB   mB  vB    e  vA    vB  
n
n
n
n
n

mA 
 v'B n 
 vA n =
Sub in known quantities:  vA n =
 mA  mB  v'B n  mB  vB n  emA  vB n
mA  e  1
 2  0.45 7.607   0.45  1.732   0.75  2 
 1.732 
2  0.75  1
 4.805 m/s
(a) Apply Work-Energy
Energy to determine speed of foot before impact:
“Foot in two positions:
T1  Vg1  Ve1  U1NC
 2  T2  Vg 2  Ve2
where:
1
2
1

Ve1  2  ks12  , and T2  mA  vA n
2
2


s1  0.42  0.32  0.3
 0.2 m
mA  v A n
2
Therefore:
k

2s12
2  4.805 
2  0.2 
2
2
k  577.3 N/m 
(b) Direction of the ball:
tan  
 v 'B  n
 v 'B t

7.607
1
  82.5 
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PROBLEM 13.197
A 300
300-g collar A is released from rest, slides down a frictionless rod,
and strikes a 900
900-g collar B which is at rest and supported by a spring
of constant 500 N/m. Knowing that the coefficient of restitution
between the two collars is 0.9, determine (a)) the maximum distance
collar A moves up the rod after impact, (b)) the maximum distance
collar B moves down the rod after impact.
SOLUTION
After impact
Velocity of A just before impact, v0
v0 
2 gh 

2(9.81 m/s 2 )(1.2 m) sin 30
2(9.81)(1.2)(0.5)  3.431 m/s
Conservation of momentum
mAv0  mBvB  mAvA : 0.3v0  0.9vB  0.3vA (1)
Restitution
(vA  vB )  e(v0  0)  0.9v0 (2)
Substituting for vB from (2) in (1)
0.3v0  0.9(0.9v0  v A )  0.3vA
1.2v A  0.51v0
vA  1.4582 m/s, vB  1.6297 m/s
(a)
A moves up the distance d where:
1
mAv A2  mA gd sin 30;
2
1
(1.4582 m/s)2  (9.81 m/s2 )d (0.5)
2
d A  0.21675 m  217 mm 
(b)
Static deflection x0 , B moves down
Conservation of energy (1) to (2)
Position (1) – spring deflected, x0
k x0  mB g sin 30
T1  V1  T2  V2: T1 
1
mBvB2 , T2  0
2
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291
PROBLEM 13.197 (Continued)
V1  Ve  Vg 
x  dB
V2  Ve  Vg  0 0
1 2
kx0  mB gd B sin 30
2
kxdx 
1
k d B2  2d B x0  x02
2


1 2
1
1
kx0  mgd B sin 30  mBvB2  k d B2  2d B x0  x02  0  0
2
2
2

 kd B2  mBvB2 ;
500d B2  0.9(1.6297)2

d B  0.0691 m
d B  69.1 mm 
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292
PROBLEM 13.198
Blocks A and B are connected by a cord which passes over pulleys
and through a collar C.. The system is released from rest when x
 1.7 m. As block A rises, it strikes collar C with perfectly plastic
impact (e  0). After impact, the two blocks and the collar keep
moving until they come to a stop and reverse their motion. As A and
C move down, C hits the ledge and blocks A and B keep moving
until they come to another stop. Determine ((a)) the velocity of the
blocks and collar immediately after A hits C, (b)) the distance the
blocks and collar move after the impact before coming to a stop,
(c)) the vvalue of x at the end of one compete cycle.
SOLUTION
(a)
Velocity of A just before it hits C:
Conservations of energy:
Datum at:
Position:
(v A )1  (vB )1  0
T1  0
v1  0
Position:
1
1
T2  mA (v A )2  mB vB2
2
2
v A  vB (kinematics)
1
11
(5  6)v A2  v A2
2
2
V2  m A g (1.7)  mB g (1.7)
T2 
 (5  6)( g )(1.7)
V2  1.7 g
T1  V1  T2  V2
00
11 2
v A  1.7 g
2
 3.4 
v A2  
 (9.81)
 11 
 3.032 m 2 /s 2
v A  1.741 m/s
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293
PROBLEM 13.198 (Continued)
Velocity of A and C after A hits C:
vA  vC (plastic impact)
Impulse-momentum A and C:
mAvA  T t  (mA  mC ) vA
(5)(1.741)  T t  8vA
(1)
vB  vA ; vB  vA (cord remains taut)
B alone:
mB v A  T  t  mB vA
(6)(1.741)  T t  6vA
Adding Equations (1) and (2),
(2)
11(1.741)  14vA
vA  1.3679 m/s
vA  vB  vC  1.368 m/s 
(b)
Distance A and C move before stopping
stopping:
Conservations of energy
energy:
Datum at :
Position :
1
(mA  mB  mC )(v A )
2
 14 
T2    (1.3681) 2
 2
T2  13.103 J
T2 
V2  0
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294
PROBLEM 13.198 (C
(Continued)
Position :
T3  0
V3  (m A  mC ) gd  mB gd
V3  (8  6) gd  2 gd
T2  V2  T3  V3
13.103  0  0  2gd
d  (13.103)/(2)(9.81)  0.6679 m
(c)
d  0.668 m 
As the system returns to position  after stopping in position , energy is conserved, and the
velocities of A, B, and C before the collar at C is removed are the same as they were in Part (a)
( above
with the directions reversed. Thus, vA  vC  vB  1.3679 m/s. After the collar C is removed, the
velocities of A and B remain the same since there is no impulsive force acting on either.
Conversation of energy:
Datum at :
1
(m A  mB )(vA ) 2
2
1
T2  (5  6)(1.3679)2
2
T2  10.291 J
T2 
V2  0
T4  0 V4  mB gx  mA gx
V4  (6  5) gx
T2  V2  T4  V4
10.291  0  (1)(9.81) x
x  1.049 m 
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295
PROBLEM 13.199
A 2-kg ball B is traveling horizontally at
10 m/s when it strikes 2-kg ball A. Ball A is
initially at rest and is attached to a spring with
constant 100 N/m and an unstretched length
of 1.2 m. Knowing the coefficient of
restitution between A and B is 0.8 and friction
between all surfaces is negligible, determine
the normal force between A and the ground
when it is at the bottom of the hill.
SOLUTION
Ball B impacts on ball A. Use the principle of impulse and momentum.
mv1  Imp12  mv 2
v0  10 m/s
Velocity components:
(v0 ) x  v0
(v0 )n  v0 cos 40 (v0 )t  v0 sin 40
(v A ) x  v A
(v A )n  vA cos 40
(vB ) x  (vB )n cos 40  (vB )t sin 40
Impulse-momentum for ball B alone.
t-direction:
mB (v0 )t  mB (vB )t
(vB )t  (v0 )t  10sin 40  6.4279 m/s
(1)
Impulse-momentum for balls A and B.
x-direction
mB v0  0  mAvA  mB (vB ) x  mB (vB )t
(2)(10)  0  2vA  2[(vB )n cos 40  6.4279sin 40]
2vA  2(vB )n cos 40  11.7365
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(1)
PROBLEM 13.199 (C
(Continued)
Coefficient of restitution.
(e 0.8)
(vB )n  (vA )n  e[0  (v0 )n ]
(vB )n  vA cos 40  (0.8)(10)
(0.8)(10)cos
cos 40
(2)
Solving Eqs. (1) and (2) simultaneously,
vA  6.6566 m/s
(vB )n  1.0291 m/s
As ball A moves from the impact location to the lowest point on the path, the spring compresses and the
elevation decreases. Since friction is negligible, energy is conserved.
T1  V1  T2  V2
1
1
mAvA2  (Ve )1  (Vg )1  mAv22  (Ve ) 2  (Vg ) 2
2
2
Position 1: (Just after impact.)
1
1
mA v A2  (2)(6.6566)2  44.3101 J
2
2
(Ve )1  0 (The spring is unstretched.)
T1 
(Vg )1  0 (Datum)
Position 2: (Lowest point on path.)
1
1
T2  mAv22  (2)v22  v22
2
2
For the spring,
x2  l2  l0  0.4 m  1.2 m  0.8 m
Fe  kx2  (100)(0.8)  80 N
1
1
(V2 )e  kx22  (100)(0.8) 2  32 J
2
2
h2  0.4 m
Elevation above datum:
(V2 ) g  mA g h2  (2)(9.81)(0.4)  7.848
Conservation of energy:
44.310  0  0  v22  32  7.848
v22  20.158 m2 /s2
v2  4.489 m/s
Normal acceleration at lowest point on path:
an 
v22 20.158

 28.798 m/s 2

0.7
an  28.8 m/s2
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297
PROBLEM 13.199 (Continued)
Apply Newton’s second law to the ball.
F  man : N  mg  Fe  man
N  mg  Fe  man
 (2)(9.81)  80  (2)(28.798)
N  157.2 N 
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298
PROBLEM 13.200
A 2-kg block A is pushed up against a spring compressing it a
distance x. The block is then released from rest and slides down
the 20 incline until it strikes a 1-kg sphere B which is
suspended from a 1 m inextensible rope. The spring constant
k  800 N/m, the coefficient of friction
riction between A and the
ground is 0.2, the distance A slides from the unstretched length
of the spring d  1.5 m and the coefficient of restitution
between A and B is 0.8. Knowing the tension in the rope is 20 N
when   30 , determine initial compression x of the spring.
spri
SOLUTION
Data:
mA  2 kg, mB  1 kg, k  800 N/m, d  1.5 m, T  20 N
k  0.2, e  0.8,   20,   30, l  1.0 m
Block slides down the incline:
Fy  0
N  mA g cos   0
N  mA g cos
 (2)(9.81) cos 20
 18.4368 N
F f  k N  (0.2)(18.4368)
 3.6874 N
Use work and energy. Datum for Vg is the impact point near B.
1
1
T1  0, (V1 )e  k x 2  (800) x 2  400 x 2 J
2
2
(V1 ) g  mA gh1  mA g ( x  d )sin   (2)(9.81)( x  1.5)sin 20  6.7104( x  1.5) J
U12  Ff ( x  d )  (3.6874)( x  1.5) J
1
1
T2  mAvA2  (1)(vA2 )  1.000 vA2
2
2
V2  0
T1  V1  U12  T2  V2 : 0  400x 2  6.7104( x  1.5)  (3.6874)( x  1.5)  1.000 v 2A  0
400 x2  3.0231x  4.5346  v2A
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299
(1)
PROBLEM 13.200 (Continued)
Impact:: Conservation of momentum.
Both A and B,, horizontal components
:
mAv A cos  0  mAvA cos   mB vB
2v A cos 20  2vA cos 20  vB
(2)
(vB )n  (vA )n  e[(vA )n  (vB )n ]
Relative velocities:
vB cos   vA  e[v A  0]
vB cos 20  vA  (0.8)v A
(3)
Solving Eqs. (2) and (3) simultaneously,
3.6 cos 20
vA
2 cos 2 20  1
 1.2230v A
vB 
(4)
Sphere B rises:: Use conservation of energy.
1
mB (vB )2 V1  0
2
1
T2  mB v22
V2  mB gh2  mB gl (1  cos  )
2
T1 
1
1
mB (vB ) 2  0  mB v22  mB g (1  cos)
2
2
v22  (1.2230v A )2  2 gl (1  cos  )
T1  V1  T2  V2 :
Tension in the rope:
  1.00 m
an 
v22
 (1.2230vA )2  2 gl (1  cos  )

Fn  mB an :
T  mB g cos   mB an
T  mB (an  g cos  )
T  mB ((1.2230vA )2  2 gl (1  cos  )  g cos  )
20  (1.0)((1.2230vA )2  2(9.81)(1)(1  cos30)  9.81co
9.81cos30)
Solving (5)
Sub (6) into (1)
(5)
vA  3.0739 m/s
(6)
400 x 2  3.0231x  4.9141  0
x  0.107 m 
Solve for x:
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300
PROBLEM 13.201*
The 1-kg ball at A is suspended by an inextensible cord and given an initial
horizontal velocity of v0. If l = 600 mm, xB = 90 mm and yB = 120 mm
determine the initial velocity v so that the ball will enter in the basket. Hint:
use a computer to solve the resulting set of equations.
B
B
SOLUTION
v1 = v0
Let position 1 be at A.
Let position 2 be the point described by the angle θ where the path of the ball changes from circular to
parabolic. At position 2 the tension Q in the cord is zero.
Relationship between v2 and θ based on Q = 0. Draw the free body diagram.
ΣF = 0: Q + mg sin θ = man =
With Q = 0,
v22 = g A sin θ
mv22
A
or v2 = g A sin θ
(1)
Relationship among v0 , v2 , and θ based on conservation of energy.
T1 + V1 = T2 + V2
1 2
1
mv0 − mg A = mv22 + mg A sin θ
2
2
v02 = v22 + 2 g A(1 + sin θ )
(2)
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301
PROBLEM 13.201* (Continued)
x and y coordinates at position 2:
x2 = A cos θ
(3)
y2 = A sin θ
(4)
Let t2 be the time when the ball is in position 2.
Motion on the parabolic path. The horizontal motion is
x = −v2 sin θ
x = x2 − (v2 sin θ )(t − t2 )
x = xB
At Point B,
(tB − t2 ) =
and t = tB .
(5)
From Eq. (5),
A cos θ − xB
vθ sin θ
(6)
y = v2 cosθ − g (t − t2 )
Vertical motion:
y = y2 + (v2 cos θ )(t − t2 ) −
1
g (t − t2 ) 2
2
At Point B,
yB = A sin θ + (v2 cos θ )(t B − t2 ) −
A = 0.6 m, xB = 0.09 m,
Data:
1
g (t B − t 2 ) 2
2
(7)
yB = 0.12 m, g = 9.81 m/s2
With the numerical data,
v2 = 2.426sin θ
Eq. (1) becomes
Eq. (6) becomes
t B − t2 =
(1)′
0.6cos θ − 0.09
v2 sin θ
(6)′
yB = 0.6sin θ + (v2 cos θ )(tB − t2 ) − 4.905(t B − t2 )2
Eq. (7) becomes
(7)′
Method of solution. From a trial value of θ, calculate v2 from Eq. (1)′, t B − t2 from Eq. (6)′, and yB
from Eq. (7)′. Repeat until yB = 0.12 m as required.
B
Try θ = 30°.
v2 = 5.886sin 30° = 1.7155 m/s
0.6cos30° − 0.09
= 0.5009s
1.7155sin 30°
yB = 0.6sin 30° + (1.7155cos30°)(0.5009) − (4.905)(0.5009)2
t B − t2 =
= −0.1865 m
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302
PROBLEM 13.201* (Continued)
Try θ = 45°.
v2 = 5.886sin 45° = 2.040 m/s
0.6cos 45° − 0.09
= 0.2317 s
2.040sin 45°
yB = 0.6sin 45° + (2.040cos 45°)(0.2317) − 4.905(0.2317)2
t B − t2 =
= 0.4952 m
Try θ = 37.5°.
v2 = 5.886sin 37.5° = 1.8929 m/s
t B − t2 =
0.6cos37.5° − 0.09
= 0.335 s
1.8929 sin 37.5°
yB = 0.6sin 37.5° + (1.8929cos37.5°)(0.335) − (4.905)(0.335) 2
= 0.3179 m
By repeated trials,
θ = 33.55° ⇒ yB = 0.1198 ≈ 0.120 m
Substituting θ = 33.55° and y2 = 1.8036 m/s in Eg(2)
v02 = (1.8036) 2 + (2)(9.81)(0.6)(1 + sin 33.55°)
= 21.531 m 2 /s 2
v0 = 4.64 m/s
W
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303
PROBLEM 13.CQ1
Block A is traveling with a speed v0 on a smooth surface when the
surface suddenly becomes rough with a coefficient of friction of 
causing the block to stop after a distance d. If block A were traveling
twice as fast, that is, at a speed 2v0, how far will it travel on the rough
surface before stopping?
(a) d/2
(b) d
(c)
2d
(d) 2d
(e) 4d
SOLUTION
1 2
mv ). The work done
2
by friction to stop the block is U  Fd . This work must equal the amount of kinetic energy before the block
hits the rough patch. Since the kinetic energy is 4x as much it will take 4x the distance to stop.
If the block were traveling twice as fast, its kinetic energy would be 4x as much ( T 
Answer: (e)
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304
PROBLEM 13.CQ2
Two small balls A and B with masses 2m and m respectively are released
from rest at a height h above the ground. Neglecting air resistance, which of
the following statements are true when the two balls hit the ground?
(a)
The kinetic energy of A is the same as the kinetic energy of B.
(b)
The kinetic energy of A is half the kinetic energy of B.
(c)
The kinetic energy of A is twice the kinetic energy of B.
(d)
The kinetic energy of A is four times the kinetic energy of B.
SOLUTION
Since each ball accelerates due to gravity at the same rate, they have the same speed when they hit the ground.
1
Since kinetic energy, T  mv2 , and ball A has twice the mass as ball B, then A will have twice the kinetic
2
energy. Answer: (c)
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305
PROBLEM 13.CQ3
Block A is released from rest and slides down the frictionless
ramp to the loop. The maximum height h of the loop is the same
as the initial height of the block. Will A make it completely
around the loop without losing contact with the track?
(a) Yes
(b) No
(c) need more information
SOLUTION
Answer: (b) In order for A to not maintain contact with the track, the normal force must remain greater than
zero, which requires a non-zero speed at the top of the loop.
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306
PROBLEM 13.CQ4
A large insect impacts the front windshield of a sports car traveling down a road. Which of the following
statements is true during the collision?
(a) The car exerts a greater force on the insect than the insect exerts on the car.
(b) The insect exerts a greater force on the car than the car exerts on the insect.
(c) The car exerts a force on the insect, but the insect does not exert a force on the car.
(d) The car exerts the same force on the insect as the insect exerts on the car.
(e) Neither exerts a force on the other; the insect gets smashed simply because it gets in the way of the car.
SOLUTION
Newton’s third law states that when a body exerts a force on a second body, the second body exerts a force
equal in magnitude and opposite in direction on the first body. Therefore the answer is: (d)
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307
PROBLEM 13.CQ5
The expected damages associated with two types of perfectly plastic collisions are to be
compared. In the first case, two identical cars traveling at the same speed impact each
other head on. In the second case, the car impacts a massive concrete wall. In which case
would you expect the car to be more damaged?
(a) Case 1
(b) Case 2
(c) The same damage in each case
SOLUTION
In both cases the car will come to a complete stop, so the applied impulse will be the same. With the same
impulse, the same damage should occur. Answer: (c)
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308
PROBLEM 13.CQ6
A 5 kg ball A strikes a 1 kg ball B that is initially at rest. Is it
possible that after the impact A is not moving and B has a speed
of 5v?
(a) Yes
(b) No
Explain your answer.
SOLUTION
Answer: (b) No.
Conservation of momentum is satisfied, but the coefficient of restitution equation is not. The coefficient of
restitution must be less than 1.
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309
PROBLEM 13.F1
The initial velocity of the block in position A is 9 m /s. The coefficient
of kinetic friction between the block and the plane is k  0.30. Draw
impulse-momentum diagrams that could be used to determine the time
it takes for the block to reach B with zero velocity, if  20°.
SOLUTION
Answer:
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310
PROBLEM 13.F2
A 20-N collar which can slide on a frictionless vertical rod is
acted upon by a force P which varies in magnitude as shown.
Knowing that the collar is initially at rest, draw impulsemomentum diagrams that could be used to determine its
velocity at t 3 s.
SOLUTION
Answer:
Where

t2  3
t1  0
Pdt is the area under the curve.
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311
PROBLEM 13.F3
The 15-kg suitcase A has been
propped up against one end of a
40-kg luggage carrier B and is
prevented from sliding down by
other luggage. When the luggage
is unloaded and the last heavy
trunk is removed from the carrier,
the suitcase is free to slide down,
causing the 40-kg carrier to move
to the left with a velocity vB of
magnitude 0.8 m/s. Neglecting
friction, draw impulse-momentum
diagrams that could be used to
determine (a) the velocity of A as
it rolls on the carrier and (b) the
velocity of the carrier after the
suitcase hits the right side of the
carrier without bouncing back.
SOLUTION
Answer:
(a)
(b)
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312
PROBLEM 13.F4
Car A was traveling west at a speed of 15 m/s and car B was traveling north at an
unknown speed when they slammed into each other at an intersection. Upon
investigation it was found that after the crash the two cars got stuck and skidded
off at an angle of 50° north of east. Knowing the masses of A and B are mA and mB
respectively, draw impulse-momentum diagrams that could be used to determine
the velocity of B before impact.
SOLUTION
Answer:
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313
PROBLEM 13.F5
Two identical spheres A and B,each of mass m, are attached to an inextensible
inelastic cord of length L and are resting at a distance a from each other on a
frictionless horizontal surface. Sphere B is given a velocity v0 in a direction
perpendicular to line AB and moves it without friction until it reaches B' where the
cord becomes taut. Draw impulse-momentum diagrams that could be used to
determine the magnitude of the velocity of each sphere immediately after the cord
has become taut.
SOLUTION
Answer:
Where v A y  vB y since the cord is inextensible.
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314
PROBLEM 13.F6
A sphere with a speed v0 rebounds after striking a frictionless inclined plane as
shown. Draw impulse-momentum diagrams that could be used to find the velocity
of the sphere after the impact.
SOLUTION
Answer:
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315
PROBLEM 13.F7
An 80-Mg railroad engine A
coasting at 6.5 km/h strikes a 20Mg flatcar C carrying a 30-Mg
load B which can slide along the
floor of the car (k 0.25). The
flatcar was at rest with its brakes
released. Instead of A and C
coupling as expected, it is observed
that A rebounds with a speed of 2
km/h after the impact. Draw
impulse-momentum diagrams that
could be used to determine (a) the
coefficient of restitution and the
speed of the flatcar immediately
after impact, and (b) the time it
takes the load to slide to a stop
relative to the car.
SOLUTION
Answer:
(a)
Look at A and C (the friction force between B and C is not impulsive) to find the velocity after impact.
(b)
Consider just B and C to find their final velocity.
Consider just B to find the time.
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316
PROBLEM 13.F8
Two frictionless balls strike each other as shown. The coefficient of restitution
between the balls is e. Draw the impulse-momentum diagrams that could be used
to find the velocities of A and B after the impact.
SOLUTION
Answer:
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PROBLEM 13.F9
A 10-kg ball A moving horizontally at 12 m/s strikes a 10-kg block B. The
coefficient of restitution of the impact is 0.4 and the coefficient of kinetic
friction between the block and the inclined surface is 0.5. Draw impulsemomentum diagrams that could be used to determine the speeds of A and B
after the impact.
SOLUTION
Answer:
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PROBLEM 13.F10
Block A of mass mA strikes ball B of mass mB with a speed of vA as shown.
Draw impulse-momentum diagrams that could be used to determine the speeds
of A and B after the impact and the impulse during the impact.
SOLUTION
Answer:
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