pels (kcp389) – Homework 7 – hill – (666)
1
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49
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before answering.
001
(part 1 of 4) 10.0 points
Calculate the magnitude of the linear momen
tum for each of the following cases
a) a proton with mass 1
.
67
×
10
−
27
kg mov
ing with a velocity of 6
×
10
6
m
/
s.
Correct answer: 1
.
002
×
10
−
20
kg
·
m
/
s.
Explanation:
Let :
m
= 1
.
67
×
10
−
27
kg
and
v
= 6
×
10
6
m
/
s
.
Momentum is
vectorp
=
mvectorv ,
so
p
=
(
1
.
67
×
10
−
27
kg
) (
6
×
10
6
m
/
s
)
=
1
.
002
×
10
−
20
kg
·
m
/
s
.
002
(part 2 of 4) 10.0 points
b) a 1
.
6 g bullet moving with a speed of
358 m
/
s to the right.
Correct answer: 0
.
5728 kg
·
m
/
s.
Explanation:
Let :
m
= 1
.
6 g
and
v
= 358 m
/
s
.
p
= (1
.
6 g) (358 m
/
s)
·
1 kg
1000 g
=
0
.
5728 kg
·
m
/
s
.
003
(part 3 of 4) 10.0 points
c) a 8 kg sprinter running with a velocity of
13
.
8 m
/
s.
Correct answer: 110
.
4 kg
·
m
/
s.
Explanation:
Let :
m
= 8 kg
and
v
= 13
.
8 m
/
s
.
p
= (8 kg) (13
.
8 m
/
s)
=
110
.
4 kg
·
m
/
s
.
004
(part 4 of 4) 10.0 points
d) Earth (
m
= 5
.
98
×
10
24
kg) moving with
an orbital speed equal to 29700 m
/
s.
Correct answer: 1
.
77606
×
10
29
kg
·
m
/
s.
Explanation:
Let :
m
= 5
.
98
×
10
24
kg
and
v
= 29700 m
/
s
.
p
=
(
5
.
98
×
10
24
kg
)
(29700 m
/
s)
=
1
.
77606
×
10
29
kg
·
m
/
s
.
005
10.0 points
A 2 kg steel ball strikes a wall with a speed
of 14 m
/
s at an angle of 50
.
7
◦
with the normal
to the wall.
It bounces off with the same
speed and angle, as shown in the figure.
x
y
14 m
/
s
2 kg
14 m
/
s
2 kg
50
.
7
◦
50
.
7
◦
If the ball is in contact with the wall for
0
.
105 s, what is the magnitude of the average
force exerted on the ball by the wall?
Correct answer: 337
.
803 N.
Explanation:
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pels (kcp389) – Homework 7 – hill – (666)
2
Let :
M
= 2 kg
,
v
= 14 m
/
s
,
and
θ
= 50
.
7
◦
.
The
y
component of the momentum is un
changed. The
x
component of the momentum
is changed by
Δ
P
x
=
−
2
M v
cos
θ .
Therefore, using impulse formula,
F
=
Δ
P
Δ
t
=
−
2
M v
cos
θ
Δ
t
=
−
2 (2 kg) (14 m
/
s) cos 50
.
7
◦
0
.
105 s
bardbl
vector
F
bardbl
=
337
.
803 N
.
Note:
The direction of the force is in negative
x
direction, as indicated by the minus sign.
006
(part 1 of 3) 10.0 points
The force of magnitude
F
x
acting in the
x
direction on a 2
.
6 kg particle varies in time as
shown in the figure.
0
1
2
3
4
5
0
1
2
3
4
F
x
(N)
t
(s)
Find the impulse of the force.
Correct answer: 9 N
·
s.
Explanation:
The impulse is the area under the
F
versus
t
graph;
i.e
, the sum of the area of the rectangle
plus two triangles, so
Impulse = 2
bracketleftbigg
(3 N) (2 s)
2
bracketrightbigg
+ (3 N) (1 s)
= 9 N
·
s
007
(part 2 of 3) 10.0 points
Find the final velocity of the particle if it is
initially at rest.
Correct answer: 3
.
46154 m
/
s.
Explanation:
Given :
m
= 2
.
6 kg
and
v
i
= 0 m
/
s
.
Impulse = Δ
p
F
Δ
t
=
m
(
v
f
−
v
i
)
m v
f
=
F
Δ
t
+
m v
i
v
f
=
F
Δ
t
m
+
v
i
=
9 N
·
s
2
.
6 kg
+ 0 m
/
s
= 3
.
46154 m
/
s
008
(part 3 of 3) 10.0 points
Find the final velocity of the particle if it
is initially moving along the
x
axis with a
velocity of
−
3 m
/
s.
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 Fall '08
 Guzman
 Physics, Kinetic Energy, Mass, Momentum, Work, Velocity, Correct Answer, Wnc

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