This preview shows pages 1–3. Sign up to view the full content.
Overton, Mays – Homework 16 – Due: Mar 2 2005, 4:00 am – Inst: Turner
1
This printout should have 13 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
The initial kinetic energy imparted to a 0
.
3 kg
bullet is 1955 J.
The acceleration oF gravity is 9
.
81 m
/
s
2
.
Neglecting air resistance, fnd the range oF
this projectile when it is fred at an angle such
that the range equals the maximum height
attained.
Correct answer: 623
.
727 m.
Explanation:
Let :
m
= 0
.
3 kg
,
K
= 1955 J
,
and
g
= 9
.
81 m
/
s
2
.
Assume that the fring height is negligible and
that the bullet lands at the same elevation
From which it was fred. The range
R
is given
by
R
=
v
2
0
g
sin2
θ
±rom kinematics,
x
= (
v
0
cos
θ
)
t
y
= (
v
0
sin
θ
)
t

1
2
g t
2
R
=
h
max
,
so
v
2
0
g
sin2
θ
=
(
v
0
sin
θ
)
2
2
g
v
2
0
g
(2 sin
θ
cos
θ
) =
(
v
0
sin
θ
)
2
2
g
tan
θ
= 4
θ
= 76
◦
.
The kinetic energy is
K
=
1
2
mv
2
0
v
2
0
=
2
K
m
,
so
R
=
v
2
0
g
sin2
θ
=
2
K
mg
sin2
θ
=
2(1955 J)
(0
.
3 kg)(9
.
81 m
/
s
2
)
sin152
◦
=
623
.
727 m
.
002
(part 1 oF 1) 10 points
Two atoms oF equal mass interact with a po
tential energy
U
(
r
) =
A
r
8

B
r
4
A >
0
, B >
0
At what separation is the Force zero?
1.
4
r
A
B
2.
4
r
2
A
B
correct
3.
4
r
B
A
4.
4
r
B
2
A
Explanation:
U
(
r
) =
A
r
8

B
r
4
F
=

dU
dr
=

d
dr
(
Ar

8
) +
d
dr
(
Br

4
)
= +8
Ar

9

4
B r

5
F
= 0
thereFore
8
A
r
9
=
4
B
r
5
2
A
B
=
r
4
r
=
4
r
2
A
B
003
(part 1 oF 1) 10 points
AsoFtballpitcherrotatesa0
.
29kg ballaround
a vertical circular path oF radius 0
.
599 m be
Fore releasing it. The pitcher exerts a 34
.
3 N
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentOverton, Mays – Homework 16 – Due: Mar 2 2005, 4:00 am – Inst: Turner
2
force directed parallel to the motion of the
ball around the complete circular path. The
speed of the ball at the top of the circle is
12
.
7 m
/
s.
The acceleration of gravity is 9
.
8 m
/
s
2
.
If the ball is released at the bottom of the
circle, what is its speed upon release?
Correct answer: 25
.
0982 m
/
s.
Explanation:
Let
U
= 0 at the bottom of the circle.
Given :
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Physics, Work

Click to edit the document details