kaplan (hmk378) – Homework 2 – Weathers – (17104)
1
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printout
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have
10
questions.
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before answering.
001
10.0 points
A bullet is fired straight up from a gun with a
muzzle velocity of 288 m
/
s.
The acceleration of gravity is 9
.
8 m
/
s
2
.
Neglecting air resistance, what will be its
displacement after 1
.
1 s?
Correct answer: 310
.
871 m.
Explanation:
Displacement is defined by
s
=
s
o
+
v
o
t
+
1
2
a t
2
.
The initial velocity of the bullet is
v
o
=
v
and its initial displacement is
s
o
= 0
.
The
gravitational acceleration is negative, so the
displacement is
s
=
s
o
+
v
o
t

1
2
g t
2
=
v t

1
2
g t
2
= (288 m
/
s) (1
.
1 s)

1
2
g
(1
.
1 s)
2
= 310
.
871 m
.
002
(part 1 of 2) 10.0 points
An electron in the cathode ray tube of a tele
vision set enters a region where it accelerates
uniformly from a speed of 34400 m
/
s to a
speed of 4
.
93
×
10
6
m
/
s in a distance of 2
.
9 cm.
What is its acceleration?
Correct answer: 4
.
1903
×
10
14
m
/
s
2
.
Explanation:
Basic Concepts:
Constant acceleration
The kinematics equations
v
f
=
v
o
+
at
v
2
f
=
v
2
o
+ 2
as
Solution:
Taking the direction of motion to
be along the
x
axis, the second equation yields
a
=
v
2
f

v
2
o
2
x
=
(
4
.
93
×
10
6
m
/
s
)
2

(34400 m
/
s)
2
2(2
.
9 cm)
·
100 cm
1 m
= 4
.
1903
×
10
14
m
/
s
2
003
(part 2 of 2) 10.0 points
For what length of time is the electron in this
region where it accelerates?
Correct answer: 1
.
16832
×
10

8
s.
Explanation:
The first equation yields
t
=
v
f

v
o
a
=
4
.
93
×
10
6
m
/
s

34400 m
/
s
4
.
1903
×
10
14
m
/
s
2
= 1
.
16832
×
10

8
s
Although the acceleration is very large in this
example, it occurs over a very short time
interval and is a typical value for charged
particles in accelerators.
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 Fall '08
 Weathers
 mechanics, Acceleration, Work, Correct Answer, m/s

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