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Chapter 9 – Student Solutions Manual
15. We need to find the coordinates of the point where the shell explodes and the velocity
of the fragment that does not fall straight down. The coordinate origin is at the firing
point, the +
x
axis is rightward, and the +
y
direction is upward. The
y
component of the
velocity is given by
v
=
v
0
y
–
gt
and this is zero at time
t
=
v
0
y
/
g
= (
v
0
/
g
) sin
θ
0
, where
v
0
is the initial speed and
0
is the firing angle. The coordinates of the highest point on the
trajectory are
()
2
2
0
00
0
2
20 m/s
cos
sin
cos
sin 60 cos60
17.7 m
9.8 m/s
x
v
xv
tv
t
g
θθ
==
=
=
°
°
=
and
yv
t g
t
v
g
y
=−
=
=
=
°
0
2
0
2
2
0
2
2
1
2
1
2
1
2
20
98
60
153
sin
.
.
m/s
m.
2
b
g
Since no horizontal forces act, the horizontal component of the momentum is conserved.
Since one fragment has a velocity of zero after the explosion, the momentum of the other
equals the momentum of the shell before the explosion. At the highest point the velocity
of the shell is
v
0
cos
0
, in the positive
x
direction. Let
M
be the mass of the shell and let
V
0
be the velocity of the fragment. Then
Mv
0
cos
0
=
MV
0
/2, since the mass of the
fragment is
M
/2. This means
Vv
00 0
2
2 20
60
20
=
°
cos
cos
m/s.
b
g
This information is used in the form of initial conditions for a projectile motion problem
to determine where the fragment lands. Resetting our clock, we now analyze a projectile
launched horizontally at time
t
= 0 with a speed of 20 m/s from a location having
coordinates
x
0
= 17.7 m,
y
0
= 15.3 m. Its
y
coordinate is given by
yy g
t
0
1
2
2
, and
when it lands this is zero. The time of landing is
ty
=
2
0
/
g
and the
x
coordinate of the
landing point is
xxV
txV
y
g
=+
=
+
=
0
2
17 7
2153
53
.
.
.
m
20 m / s
m
2
b
g
b
g
23. The initial direction of motion is in the +x direction. The magnitude of the average
force
F
avg
is given by
3
2
32.4 N s
1.20 10 N
2.70 10 s
avg
J
F
t
−
⋅
=
×
Δ×
The force is in the negative direction. Using the linear momentumimpulse theorem
stated in Eq. 931, we have
–
F
avg
Δ
t
=
mv
f
–
mv
i
.
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View Full Documentwhere
m
is the mass,
v
i
the initial velocity, and
v
f
the final velocity of the ball. Thus,
()
( ) ( )( )
3
avg
0.40kg 14m s
1200 N
27 10 s
67m s.
0.40kg
i
f
mv
F
t
v
m
−
−×
−Δ
==
=
−
(a) The final speed of the ball is

f
v
=
67 m/s.
(b) The negative sign indicates that the velocity is in the –
x
direction, which is opposite to
the initial direction of travel.
(c) From the above, the average magnitude of the force is
.
3
avg
1.20 10 N
F
=×
(d) The direction of the impulse on the ball is –
x
, same as the applied force.
35.
(a) We take the force to be in the positive direction, at least for earlier times. Then the
impulse is
33
3
3.0
10
69
00
3.0
62
93
0
(6.0 10 )
(2.0 10 )
11
23
9.0N s.
JF
d
t
t
t
tt
−−
−
××
×
2
d
t
⎡
⎤
×
−
×
⎣
⎦
⎡⎤
−
×
⎢⎥
⎣⎦
=⋅
∫∫
(b) Since
J = F
avg
Δ
t
, we find
F
J
t
avg
3
9.0 N s
3.0
10 s
3.0
10 N.
Δ
=
⋅
×
−
3
(c) To find the time at which the maximum force occurs, we set the derivative of
F
with
respect to time equal to zero – and solve for
t
. The result is
t
= 1.5
×
10
–3
s. At that time
the force is
F
max
6.0 10
10
2.0 10
10
4.5 10 N.
×
×
=
×
ch
15
2
..
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 Spring '08
 drty
 mechanics

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