homework 21 – YOO, HEE – Due: Mar 20 2008, 4:00 am
1
Question 1, chap 10, sect 1.
part 1 of 1
10 points
What velocity must a car with a mass of
1110 kg have in order to have the same mo
mentum as a 2300 kg pickup truck traveling
at 26 m
/
s to the east?
Correct answer: 53
.
8739 m
/
s (tolerance
±
1
%).
Explanation:
Let :
m
1
= 1110 kg
,
m
2
= 2300 kg
,
and
v
2
= 26 m
/
s to the east
.
vectorp
=
m
1
vectorv
1
=
m
2
vectorv
2
v
1
=
m
2
v
2
m
1
=
(2300 kg) (26 m
/
s)
1110 kg
=
53
.
8739 m
/
s
to the east.
Question 2, chap 10, sect 1.
part 1 of 1
10 points
A(n) 89 kg fisherman jumps from a dock
into a 136
.
2 kg rowboat at rest on the east
side of the dock.
Note:
Assume the boat’s interaction with
the water is frictionless.
If the velocity of the fisherman is 5
.
45 m
/
s
to the east as he leaves the dock, what is the
final speed of the fisherman and the boat?
Correct answer: 2
.
15386 m
/
s (tolerance
±
1
%).
Explanation:
Let east be positive:
Let :
m
1
= 89 kg
,
m
2
= 136
.
2 kg
,
and
v
i,
1
= 5
.
45 m
/
s
.
The boat and fisherman have the same final
speed, and
v
i,
2
= 0 m/s, so
m
1
vectorv
i,
1
+
m
2
vectorv
i,
2
= (
m
1
+
m
2
)
vectorv
f
m
1
vectorv
i,
1
= (
m
1
+
m
2
)
vectorv
f
v
f
=
m
1
v
i
m
1
+
m
2
=
(89 kg) (5
.
45 m
/
s)
89 kg + 136
.
2 kg
=
2
.
15386 m
/
s
,
which is 2
.
15386 m
/
s to the east.
Question 3, chap 10, sect 1.
part 1 of 1
10 points
A(n) 82 kg student holding a 1
.
5 kg physics
textbook stands in the center of a frozen pond
of radius 25 m.
He is unable to walk to
the other side because there is no friction
between his shoes and the ice.
To overcome
this difficulty, he uses recoil:
he throws the
textbook with a horizontal velocity 5
.
8 m
/
s
towards the northern shore of the pond..
How long does it take him to reach the
southern shore?
Correct answer: 235
.
632 s (tolerance
±
1 %).
Explanation:
The recoil works by conservation of mo
mentum:
Since the ice is frictionless, there
are no
external
horizontal force, hence the net
horizontal momentum is conserved,
vector
P
net
= const
.
Initially, there is no motion and hence
vector
P
net
=
0. Consequently, after the student throws the
book, their net momentum is still zero,
P
net
=
M
s
vectorv
s
+
M
b
vectorv
b
=
vector
0
.
Hence, given the book’s velocity, we can find
the student’s velocity as
vectorv
s
=
−
M
b
M
s
vectorv
b
.
Thus, he moves South (in the opposite direc
tion to the book) with speed
v
s
=
M
b
M
s
×
v
b
=
1
.
5 kg
82 kg
×
5
.
8 m
/
s
= 0
.
106098 m
/
s
.
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homework 21 – YOO, HEE – Due: Mar 20 2008, 4:00 am
2
The distance from the pond’s center to the
shore is the radius
R
= 25 m.
So the time
it takes the student to slide through this dis
tance is
t
=
R
v
s
=
25 m
0
.
106098 m
/
s
= 235
.
632 s
.
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 Spring '08
 Turner
 Physics, Energy, Kinetic Energy, Mass, Momentum, Work

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