Solution 3.182
The speed of the plug upon impact is
2g 2(32.2)(2) 11.35 ft/secvh== =
Solution 3.183
2
2
0
+=
t
HMdtH
Solution 3.184
22 22
00
0; 2 ( ) 2 (2 ) (2 ) 0
2244
33
/44
Hmrrmrr
nTT mr mr
ωω
ωω
Δ= − =
=Δ = =
Solution 3.185
0, so const.
== =
MH H
mυ
Solution 3.186
Solution 3.187
0
0
0
00 0
2
0
mg
1mg
2
dH
HMdk
dt
Hutk
===−
=−
Solution 3.188
HH
+=
Solution 3.189
Speeds:
()
2g 2g sin30 g
BB
vh
ρρ
== °=
D
30° ρ
ρ
y
(Note: d = ut)
mg
m
d
Solution 3.190
Conservation of angular momentum:
()
()
()()
() ()
θ
−
−= −
=
=−=
228 23
66
22
22
11 111
188,500 3.439 10 333,000 4.095 10
2 2 5280
50 10 75 10
153,900 ft/sec
153,9
AB
B
B
rB
rr
v
v
vvv −=
22
00 125,700 88,870 ft/sec
Solution 3.191
System angular momentum conserved during impact:
+=
12
00
:
HH
Solution 3.192
2
relea
2
se posit
ion
Solution 3.193
()
11 12 2 2
0.4 14 0.4 9
cos30 4 cos30
32.2 12 32.2 12
9.68rad/sec
TVU TV
ω
ω
−
°= °
=
′
++ = +
Solution 3.194
()
0
40 2 / 60 4.19 rad/s
ωπ
==
[][]
()
()( )( )
5 0.620 3.00 0.524 4.19 2 5 9.81 0.1243
6.850 12.190 5.34 V
U
=×−×+
=− + =
Solution 3.195
2g , 2g
vhv h
′′
==
a
r
Solution 3.196
→+ = +
=
′′
=
11 2 2 11 2 2
21 21
410410
:0.8 20 3
mv mv mv m v
evv
Solution 3.197
System linear momentum:
3(0.7) 4( 0.5) 3 4
vv
+
→+−=
′′
+
4
0.1686 m/s (right)
v
=
Average acceleration for 3-kg body:
0.1914 0.7
vvv
′
Δ−
−−
Solution 3.198
Conservation of linear momentum:
0 0
0
0
Solution 3.200
112 2
2g , 2g , 2g
vhv hv h
== =
Solution 3.201
t-momentum: cos cos (1)
vv
θ
θ
′
=
θ
n
υ‘
Solution 3.202
2g 2 32.2 3
v
H==××
cos75.9
°
2
222
1.132ft
2g 2 32
.2
s
== =
×
Solution 3.203
Spring deflection: 2233
TU T
−
+=
22
2
2
11
0
22
11 1 0
22 2
1
2
BB
mv k
e
mvk
em
vk
δ
δ
δ
′−=
+
−=
+
=
0
H = 3‘
10°
y
s
Solution 3.204
22
2
222
111
AA BB AA
AA
Tmvmvmv
mm
′′
Δ=− − +
Solution 3.205
22
000
3.06 m/s
3.33m/s
tn
vvv
=
=+=
0
0
2
2
y
20°
n
y
t
Solution 3.206
Solution 3.207
I. Drop of m1
2
1
2.80 m/s
=
II. Collision +
←⎯⎯
21 21
12
:0.7 (2)
evv
==
0.896 m/s left
v
′=
III. Subsequent spring compression:
22
11
:0
vvk
δ
′
+= − =
D
d√—
2
d√—
Solution 3.208
Conservation of system linear momentum:
′
6898 (4)
yy
ms
vv
′′
−=−
Solve Eqs. (1)−(4):
()
3
tan 2.92 10 degrees
x
s
v
θ
=
=
′
Solution 3.209
2
d
d
Solution 3.210
()
2g ,
yyy
vhhvev
′
=′+ =
2
S
olve for g
thhh
=′±′+
0
d
Solution 3.211
For each sphere, 0:
t
GΔ=
A 3
12 sin (a)
v
θ
′′
−=−
Restitution:
cos cos
0.4 (d)
BBAA
vv
θθ
′′′′
−
=
y
xA
θυ
A
‘
Solution 3.212
′
== °
=
10 sin30 5 m/
s
vv
′′ ′
−−
==
′
(2) :0.75
xx
xx xx
xx
B
BB
B
A
AA
A
vv vv
evv
v7.58m/s
x
B
′
=
′
5.12 m/s at 77.8
Magnitudes and directions
9.67 m/s at 38.4
AA
BB
v
v
θ
θ
==
′
==°
°
′
()
()
22
1
22
2
1
Initial : 10 6 68 m
2
1
Final : 5.12 9.67 59.8 m
2
68 59.8 100% 12.06%
68 ()
Tm
Tm
n
=+=
=+=
−
==
θx
υ
B
= 6 m/s
Solution 3.213
()
()
=+=
1
22
2
2
1
5.20 2 7.84 75.1m
6)%
8
Tm
Solution 3.214
()
()
()
()
g
2
63
J
3.13m/s
For anvil 0
12.8 10 0.024 29.4 10 0.024
e
e
TV V
V
=
Δ+Δ +Δ =
Δ×
=+
3.13
B
x
AF
o
F = kx = 2.8(10
6
)x N
600 kg
3000 kg
υυ‘
υ
1
ΔV
e
Solution 3.215
BB
1 5.125 16 ft
32.2 32.2 sec
95.9 ft/sec, 26.5 , 1.184 ft/sec
So 95.9 ft/sec at 46.5
AB A
BA
B
vv
v
α
β
′′
==°=−
′==
°
υ
B
‘
tn
Solution 3.216
50 cos , 50 sin
10 10 1
50 cos 5 co
AA
xy
AB
vv
tv
αα
==
== =
2
2
10 2
25cos 5cos
1g
2
B
ABy
yyvt t
αα
=+ −
′
2
2
2
50 cos
1
Use (tan 1) to obtain
cos
α
α
=+
y
Solution 3.217
Coefficient of restitution
applies to velocity components in n-dir.
0.02722, 0.1650 m or
δ
δ
== 165.0 mm
δ
=
10 m/sec
60°
υ‘
Solution 3.218
Impact velocity components at A:
0,2g
xy
Vvv d==−
2
At vertical line CC:
0
00
0
2
(
): g
22
(
c
d
xdv vt
dd
dv dv d
−=
−−