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Oslund (hno56) – Homework 7 – sai – (58015)
1
This printout should have 30 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
Three particles are located in a coordinate
system as shown in the fgure below.
The acceleration oF gravity is 9
.
8m
/
s
2
.
The mass 7 kg is located at the origin, the
mass 5 kg is located on the
x
coordinate, and
the mass 7 kg is located on the
y
coordinate.
x
y
7 kg
7 kg
5 kg
7m
±ind the distance oF the center oF gravity
From the origin.
Correct answer: 3
.
32913 m.
Explanation:
The center oF gravity oF the
x
and
y
compo
nents From the origin respectively are
x
c
=
m
0
0+
m
1
x
1
m
0
+
m
1
+
m
2
=
(5 kg) (8 m)
(7 kg) + (5 kg) + (7 kg)
=2
.
10526 m
and
y
c
=
m
0
m
2
y
2
m
0
+
m
1
+
m
2
=
(7 kg) (7 m)
(7 kg) + (5 kg) + (7 kg)
.
57895 m
.
Then the distance oF the center oF gravity
From the origin is
d
=
±
x
2
c
+
y
2
c
=
±
(2
.
10526 m)
2
+ (2
.
57895 m)
2
=3
.
32913 m
.
002
10.0 points
Giselle is travelling on her bicycle at a speed
oF 15 mph when a car passes her. ±rom her
Frame oF reFerence, she estimates that the car
is going 45 mph toward her and 45 mph away
From her.
What is the speed oF the car From the
Frame oF reFerence oF someone standing on
the ground?
Correct answer: 60 mph.
Explanation:
Since Griselle is traveling in the same direc
tion as the car, it is moving more slowly in his
reFerence Frame than the ground’s by 15 mph.
The speed oF the car in the reFerence Frame oF
the ground is 45 mph.
003
(part 1 oF 4) 10.0 points
A 3 kg block is traveling in the negative
x
direction at 4
.
3m
/
s, and a 7
.
5 kg block is
traveling in the positive
x
direction at 3 m
/
s.
±ind the speed oF the center oF mass.
Correct answer: 0
.
914286 m
/
s.
Explanation:
Let :
m
1
= 3 kg
,
m
2
=7
.
5 kg
,
v
1
=(

4
.
/
s) ˆ
ı,
and
v
2
= (3 m
/
s) ˆ
ı.
±or the center oF mass,
²
i
m
i
$v
i
=
M $v
cm
=
m
1
1
+
m
2
2
m
1
+
m
2
)
cm
cm
=
m
1
1
+
m
2
2
m
1
+
m
2
=
(3 kg) (

4
.
/
s) ˆ
ı
(3 kg) + (7
.
5 kg)
+
(7
.
5 kg) (3 m
/
s) ˆ
ı
(3 kg) + (7
.
5 kg)
=
(0
.
914286 m
/
s) ˆ
ı
.
004
(part 2 oF 4) 10.0 points
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View Full DocumentOslund (hno56) – Homework 7 – sai – (58015)
2
Find the velocity of each block in the center
ofmass reference frame.
1.
±
$u
1
=(

4
.
3m
/
s) ˆ
ı
and
2
= (4
.
/
s) ˆ
ı
2.
±
1

/
s) ˆ
ı
and
2
= (3 m
/
s) ˆ
ı
3.
±
1

5
.
21429 m
/
s) ˆ
ı
and
2
= (2
.
08571 m
/
s) ˆ
ı
correct
4.
±
1
= (2
.
08571 m
/
s) ˆ
ı
and
2
= (2
.
08571 m
/
s) ˆ
ı
5.
±
1
= (4
.
/
s) ˆ
ı
and
2

/
s) ˆ
ı
6.
±
1
= (2
.
08571 m
/
s) ˆ
ı
and
2

5
.
21429 m
/
s) ˆ
ı
7.
±
1

5
.
21429 m
/
s) ˆ
ı
and
2

5
.
21429 m
/
s) ˆ
ı
8.
±
1
= (3 m
/
s) ˆ
ı
and
2

4
.
/
s) ˆ
ı
Explanation:
In the center of mass reference frame, the
velocity of the 3 kg block is
1
=
$v
1

cm

4
.
/
s) ˆ
ı

(0
.
914286 m
/
s) ˆ
ı

5
.
21429 m
/
s) ˆ
ı,
and of the 7
.
5 kg block,
2
=
2

cm
= (3 m
/
s) ˆ
ı

(0
.
914286 m
/
s) ˆ
ı
=
(2
.
08571 m
/
s) ˆ
ı
.
005
(part 3 of 4) 10.0 points
Find the velocity of each block after the colli
sion in the centerofmass reference frame.
Assume:
The collision is head on.
1.
±
$
u
±
1
= (5
.
21429 m
/
s) ˆ
ı
and
$
u
±
2

2
.
08571 m
/
s) ˆ
ı
correct
2.
±
$
u
±
1
= (5
.
21429 m
/
s) ˆ
ı
and
$
u
±
2
= (5
.
21429 m
/
s) ˆ
ı
3.
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 Spring '08
 Kaplunovsky
 Physics, Work

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