Wong, Lemuel – Homework 6 – Due: Mar 2 2007, 11:00 pm – Inst: Matt Mackie
1
This
printout
should
have
16
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
NOTE: HW is due 11pm central time,
which is midnight our time.
001
(part 1 of 1) 10 points
Two alpha particles (helium nuclei), each con
sisting of two protons and two neutrons, have
an electrical potential energy of 6
×
10

19
J
.
Given:
k
e
= 8
.
98755
×
10
9
N m
2
/
C
2
, q
p
=
1
.
6021
×
10

19
C
,
and
g
= 9
.
8 m
/
s
2
.
What is the distance between these parti
cles at this time?
Correct answer: 1
.
5379
×
10

9
m.
Explanation:
Let:
U
electric
= 6
×
10

19
J
,
k
e
= 8
.
98755
×
10
9
N m
2
/
C
2
,
q
p
= 1
.
6021
×
10

19
C
,
q
α
= 2
q
p
= 3
.
2042
×
10

19
C
,
and
q
n
= 0 C
.
q
1
=
q
2
= 2
q
p
+ 2
q
n
= 2 (1
.
6021
×
10

19
C) + 2 (0 C)
= 3
.
2042
×
10

19
C
U
electric
=
k
e
q
1
q
2
r
r
=
k
e
q
1
q
2
U
electric
=
k
e
q
2
1
U
electric
= (8
.
99
×
10
9
N
·
m
2
/
C
2
)
×
(3
.
2042
×
10

19
C)
2
6
×
10

19
J
=
1
.
5379
×
10

9
m
.
keywords:
002
(part 1 of 2) 10 points
An electric field does 11 J of work on a
0
.
0009 C charge.
What is the voltage change?
Correct answer: 12222
.
2 V.
Explanation:
Let :
W
= 11 J
and
q
= 0
.
0009 C
.
Work is
W
=
qV
V
=
W
q
=
11 J
0
.
0009 C
=
12222
.
2 V
.
003
(part 2 of 2) 10 points
The same electric field does 22 J of work on a
0
.
0018 C charge.
What is the voltage change?
Correct answer: 12222
.
2 V.
Explanation:
Let :
W
= 22 J
and
q
= 0
.
0018 C
.
The voltage change is
V
=
W
q
=
22 J
0
.
0018 C
=
12222
.
2 V
.
keywords:
004
(part 1 of 1) 10 points
Two insulating spheres having radii 0
.
12 cm
and 0
.
32 cm, masses 0
.
11 kg and 0
.
41 kg, and
charges

4
μ
C and 2
μ
C are released from rest
when their centers are separated by 1
.
2 m
.
How fast is the smaller sphere moving when
they collide?
Correct answer: 15
.
2795 m
/
s.
Explanation:
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Wong, Lemuel – Homework 6 – Due: Mar 2 2007, 11:00 pm – Inst: Matt Mackie
2
Let :
k
e
= 8
.
99
×
10
9
N
·
m
2
/
C
2
,
r
1
= 0
.
12 cm = 0
.
0012 m
,
r
2
= 0
.
32 cm = 0
.
0032 m
,
m
1
= 0
.
11 kg
,
m
2
= 0
.
41 kg
,
q
1
=

4
μ
C =

4
×
10

6
C
,
q
2
= 2
μ
C = 2
×
10

6
C
,
and
d
= 1
.
2 m
.
By conservation of momentum
0 =
m
1
v
1

m
2
v
2
v
2
=
m
1
v
1
m
2
By conservation of energy
(
K
+
U
)
i
= (
K
+
U
)
f
0 +
k
e
q
1
q
2
d
=
1
2
m
1
v
2
1
+
1
2
m
2
v
2
2
+
k
e
q
1
q
2
r
1
+
r
2
k
e
q
1
q
2
d
=
1
2
m
1
v
2
1
+
1
2
m
2
m
1
v
1
m
2
¶
2
+
k
e
q
1
q
2
r
1
+
r
2

k
e
q
1
q
2
1
r
1
+
r
2

1
d
¶
=
m
1
v
2
1
2
m
2
(
m
1
+
m
2
)
.
So
v
2
1
=

2
m
2
k
e
q
1
q
2
m
1
(
m
1
+
m
2
)
1
r
1
+
r
2

1
d
¶
=

2 (0
.
41 kg)
(
8
.
99
×
10
9
N
·
m
2
/
C
2
)
(0
.
11 kg) (0
.
11 kg + 0
.
41 kg)
×
(

4
×
10

6
C) (2
×
10

6
C)
×
1
0
.
0012 m + 0
.
0032 m

1
1
.
2 m
¶
= 233
.
464 m
2
/
s
2
.
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 Spring '08
 Turner
 Physics, Conservation Of Energy, Energy, Potential Energy, Work, Correct Answer, WONG, Matt Mackie

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