PROBLEM 13.188 (Continued)
Sphere A: Momentum in t–direction:
sin 0 ( )
AA A At
mv m v
θ
′
+=
2
1
1
2
2
2
1
2
0
0
1()
2
B BB
B
B
B
T mv
V
T
V kx
=
=
=
= ∆
PROBLEM 13.188 (Continued)
The maximum deflection will occur when the block comes to rest (ie, no kinetic energy)
22
11
a speed
0
0.6 m/s.v=
The coefficient of restitution between A
and the 2-kg wedge B is 0.8 and the length of rope
0.9 ml=
The spring constant has a value of 1500 N/m and
20 .
θ
= °
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
PROBLEM 13.189 (Continued)
Both A and B:
Momentum in x-direction:
() 0 () ()
AAx AAx BBx
mv mv mv
′′
+= +
PROBLEM 13.190
A 32,000-lb airplane lands on an aircraft carrier
and is caught by an arresting cable. The cable is
inextensible and is paid out at A and B from
mechanisms located below dock and consisting
of pistons moving in long oil-filled cylinders.
Knowing that the piston-cylinder system
maintains a constant tension of 85 kips in the
cable during the entire landing, determine the
landing speed of the airplane if it travels a
distance
95 ftd=
after being caught by the
cable.
SOLUTION
2
32000 lb 993.79 lb s /ft
W
1
1
PROBLEM 13.191
A 2–oz pellet shot vertically from a spring–loaded pistol on the surface of the earth rises to a height of 300 ft.
The same pellet shot from the same pistol on the surface of the moon rises to a height of 1900 ft. 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.
2
1
L
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,
0
r
and
1
r
. Use the
principles of conservation of energy and conservation of angular
momentum to derive the relation
2
01
112GM
rr h
+=
where h is the angular momentum per unit mass of the satellite and
G is the constant of gravitation.
SOLUTION
Angular momentum:
0 0 11
0 0 11
,h r v rv
b r v rv
hh
= =
= =
01
rr
2
01
rr h
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
2
3000/Fr=
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
1
PROBLEM 13.193 (Continued)
Position 2. (Sphere at Point B.)
2 22
22 2 2
22
11
(0.060) 0.030
22
( ) 0 (since 0)
g
T mv v v
Vh
= = =
= =
2
23.13 m/sv=
PROBLEM 13.194
A 50-lb sphere A of radius 4.5 in. moving with a velocity of magnitude
0
v
= 6 ft/s strikes a 4.6–lb sphere B of radius 2 in. which is hanging from
an inextensible cord and is initially at rest. Knowing that sphere B
swings to a maximum height h = 0.75 ft, determine the coefficient of
restitution between the two spheres.
SOLUTION
Angle of impulse force from geometry of A and B
1
6
cos 22.62
6.5
θ
−
= = °
2
A
PROBLEM 13.194 (Continued)
( ) ( ) ( ) ( )
: sin sin cos
AA AA AA
xy
A mv mv mv
θθ θ
′′
= +
2.5 2.5
tan ()tan ();6 () ()
66
A Ax Ay Ax Ay
vv v v v
θθ
′′ ′ ′
=+=+
( ) ( )
15 2.5 6
AA
xy
vv
′′
= +
(2)
( )
50 50 4.6
: ; (6 ft/s) ( )
AA A A BB Ax B
x
A B mv m v mv v v
g gg
′′ ′ ′
+=+ =+
(3)
g’s cancel
From equation (1)
2
2(32.2 ft/s )(0.75 ft) 6.9498 ft/s
B
v′= =
From equation (3)
(50)(6) 50( ) 4.6(6.9498)
Ax
v′
= +
( ) 5.3606 ft/s
Ax
v′=
From equation (2)
15 2.5(5.3606) 6( )
Ay
v′
= +
( ) 0.2664 ft/s
Ay
v′=
2.5
6.9498 5.3606 0.2664 60.2834
6
e
−+
= =
0.283e=
PROBLEM 13.195
A 300–g block is released from rest after a spring of constant
600 N/mk=
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.
PROBLEM 13.195 (Continued)
(a) Newton’s second law at A:
222
2
35.504 m /s 44.38 m/s
A
v
C
C
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 pulls 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
θ = 30º as shown. The impact occurs with a coefficient of restitution e =
0.75 whe
n the foot is at position 2, where the spring is unstretched.
Knowing that the effective friction coefficient during rolling is
m
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
BA AB
nn nn
PROBLEM 13.196 (Continued)
Sub (1) into (2) and solve for :
( )
An
v
( ) ( ) ( ) ( ) ( ) ( )
1
B AA BB BB A B
n n n n nn
A
v‘ mv mv mv ev v
m
′
− +− =−
PROBLEM 13.197
A 300–g collar A is released from rest, slides down a frictionless rod,
and strikes a 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
PROBLEM 13.197 (Continued)
2
10
1sin 30
2
e g BB
V V V kx m gd=+= + °
( )
0
22
2 00
0
12
2
B
xd
eg B B
V V V kxdx k d d x x
+
′′
=+= = + +
∫
( )
2 22 2
0 00
1 11
sin 30 2 0 0
2 22
B BB B B
kx mgd m v k d d x x+ °+ = + + + +
22 2 2
; 500 0.9 (1.6297) 0.0691 m
B BB B B
kd m v d d∴= = =
69.1 mm
B
d=
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
( 0).e=
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 value of x at the end of one compete cycle.
SOLUTION
PROBLEM 13.198 (Continued)
PROBLEM 13.198 (Continued)
Position :