brown (twb493) – extra
credit 02 – turner – (57340)
1
This
printout
should
have
10
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0points
Two rocks are tied together with a string of
negligible mass, and thrown into the air. At a
particular instant, rock 1, which has a mass of
0
.
1 kg, is moving with velocity
v
1
=
(
7
,
−
5
,
4
)
m/s, and rock 2, which has a mass of 0
.
25 kg,
is moving with velocity
v
2
=
(−
7
,
4
,
4
)
m/s.
What is the magnitude of the momentum
of the system of the two rocks?
Correct answer: 1
.
82003 kg
·
m
/
s.
Explanation:
Let :
m
1
= 0
.
1 kg
,
m
2
= 0
.
25 kg
,
v
1
x
= 7 m
/
s
,
v
1
y
=
−
5 m
/
s
,
v
1
z
= 4 m
/
s
,
v
2
x
=
−
7 m
/
s
,
v
2
y
= 4 m
/
s
,
and
v
2
z
= 4 m
/
s
.
The total momentum of the system of the two
rocks is
vectorp
total
=
vectorp
1
+
vectorp
2
,
so
The magnitude of the total momentum is

vectorp
total

=

vectorp
1
+
vectorp
2

=
{
(
p
1
x
+
p
2
x
)
2
+ (
p
1
y
+
p
2
y
)
2
+(
p
1
z
+
p
2
z
)
2
}
1
/
2
=
{
(
m
1
v
1
x
+
m
2
v
2
x
)
2
+ (
m
1
v
1
y
+
m
2
v
2
y
)
2
+(
m
1
v
1
z
+
m
2
v
2
z
)
2
}
1
/
2
=
braceleftBig
[(0
.
1 kg)(7 m
/
s)
+(0
.
25 kg)(
−
7 m
/
s)]
2
+ [(0
.
1 kg)(
−
5 m
/
s)
+(0
.
25 kg)(4 m
/
s)]
2
+ [(0
.
1 kg)(4 m
/
s)
+(0
.
25 kg)(4 m
/
s)]
2
bracerightBig
1
/
2
= 1
.
82003 kg
·
m
/
s
.
002
10.0points
A
charged
particle
is
acted
upon
by
a
force
of
magnitude
120N
over
a
time
t.
Given
that
the
initial
momentum
is
(15
,
10
,
0)
kg.m/s
and that the final momen
tum is (20
,
20
,
−
10)
kg.m/s
, find
t
and the
force
F
(in Newtons).
1.
0.125s, (80
,
40
,
40)
N
2.
0.25s, (20
,
40
,
−
40)
N
3.
0.125s, (
−
80
,
40
,
80)
N
4.
0.125s, (
−
40
,
−
80
,
80)
N
5.
0.25s, (20
,
−
40
,
40)
N
6.
0.125s, (40
,
80
,
−
80)
N
correct
7.
0.125s, (40
,
80
,
80)
N
Explanation:
Given that
vector
p
i
= (15
,
10
,
0)
kg.m/s
vector
p
f
= (20
,
20
,
−
10)
kg.m/s
vector

F

= 120
N
We use the formula for the magnitude of
the impulse
=

vector
p
f
−
vector
p
i

=

vector
F

t
t
=
(

vector
p
f
−
vector
p
i

)

vector
F

= 0
.
125
s
vector
F
=
(
vector
p
f
−
vector
p
i
)
t
= (40
,
80
,
−
80)
N
003
10.0points
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brown (twb493) – extra
credit 02 – turner – (57340)
2
Consider four vectors
vector
F
1
,
vector
F
2
,
vector
F
3
, and
vector
F
4
, with
magnitudes are
F
1
= 47 N,
F
2
= 22 N,
F
3
=
18 N, and
F
4
= 46 N, and
θ
1
= 120
◦
,
θ
2
=
−
150
◦
,
θ
3
= 24
◦
, and
θ
4
=
−
72
◦
, measured
from the positive
x
axis with the counter
clockwise angular direction as positive.
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 Spring '08
 Turner
 Physics, General Relativity, Correct Answer, Standard gravity

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