yindeemark (rry82) – Homework 9 – chelikowsky – (59005)
1
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001
10.0 points
An 83 kg boat that is 7
.
2 m in length is
initially 8
.
3 m From the pier. A 57 kg child
stands at the end oF the boat closest to the
pier. The child then notices a turtle on a rock
at the Far end oF the boat and proceeds to
walk to the Far end oF the boat to observe the
turtle.
8
.
3 m
7
.
2 m
How Far is the child From the pier when she
reaches the Far end oF the boat? Assume there
is no Friction between boat and water.
Correct answer: 12
.
5686 m.
Explanation:
Let
D
= distance oF the boat From the pier
,
= 8
.
3 m
,
L
= length oF the boat
,
= 7
.
2 m
,
M
= mass oF the boat
,
= 83 kg
,
and
m
= mass oF the child
,
and
= 57 kg
,
X
= change in position oF the boat
.
ℓ
= fnal distance oF child From pier
.
d
= fnal distance oF boat From pier
.
D
L
X
d
ℓ
X
We will use the pier as the origin oF the
x
coordinate.
Initially, the center oF mass oF the boatchild
system is
x
cm
=
p
D
+
L
2
P
M
+
D m
M
+
m
±inally, the center oF mass oF the boatchild
system is
x
cm
=
p
D
+
L
2
− X
P
M
+ (
D
+
L
− X
)
m
M
+
m
,
where
X
is the change in position oF the center
oF mass oF the boat. Since the center oF mass
oF the system does not move, we can equate
the above two expressions For
x
cm
p
D
+
L
2
P
M
+
D m
M
+
m
=
p
D
+
L
2
− X
P
M
+ (
D
+
L
− X
)
m
M
+
m
and, solving For
X
, we have
p
D
+
L
2
P
M
+
D m
=
p
D
+
L
2
− X
P
M
+ (
D
+
L
− X
)
m
0 =
−X
M
+ (
L
− X
)
m
X
(
M
+
m
) =
L m
X
=
m
m
+
M
L
=
(57 kg)
(57 kg) + (83 kg)
×
(7
.
2 m)
= 2
.
93143 m
.
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View Full Documentyindeemark (rry82) – Homework 9 – chelikowsky – (59005)
2
The fnal distance
ℓ
oF the child From the
pier is
ℓ
=
D
+
L
− X
= (8
.
3 m) + (7
.
2 m)
−
(2
.
93143 m)
=
12
.
5686 m
.
The fnal distance
d
oF the near end oF the
boat to the pier is
d
=
D
− X
= (8
.
3 m)
−
(2
.
93143 m)
= 5
.
36857 m
.
002
10.0 points
On level ground, a shell is fred From a cannon
to hit an enemy bunker situated a distance
R
away From the cannon. At the highest point oF
the trajectory, the shell accidentally explodes
into two equal parts. The explosion occurs
in such a way that the two pieces have no
velocity in the vertical direction immediately
aFter the explosion. (
i.e.
the internal Forces
From the explosion act only in the horizontal
direction.) One part oF the shell is Found to
hit the ground at
R
2
.
At what distance
d
c
From the cannon would
the second Fragment Fall?
1.
d
c
= 3
R
2.
d
c
=
1
2
R
3.
d
c
=
3
2
R
correct
4.
d
c
=
√
3
2
R
5.
d
c
=
2
3
R
6.
d
c
=
√
2
R
7.
d
c
=
1
3
R
8.
d
c
=
R
9.
d
c
= 2
R
Explanation:
Basic Concepts:
Conservation oF mo
mentum
Center oF Mass:
vr
cm
=
∑
i
m
i
i
∑
i
m
i
Solution:
R
R/2
3R/2
Because the Fragments have the same zero
component oF vertical velocity as the unex
ploded shell, the Fall time oF each Fragment
aFter creation is the same as what the Fall
time From peak altitude oF the original shell
would have been, had it not exploded. Also,
this Fall time is equal to the rise time oF the
original shell to its peak altitude. Let
V
fx
be the
x
component oF velocity oF the Faster
Fragment and
V
shx
be the
x
component oF the
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 Fall '08
 Turner
 Kinetic Energy, Momentum, Work, Correct Answer, kg, Vshx

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