schopper (as55864) – Ch2HW1 – Antoniewicz – (56465)
1
This
printout
should
have
29
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0points
A ball moves in the direction of the arrow
labeled
c
in the following diagram. The ball is
struck by a stick that briefly exerts a force on
the ball in the direction of the arrow labeled
e
.
Which arrow best describes the direction
of Δ
vectorp
, the change in the ball’s momentum?
a
b
c
d
e
f
g
h
1.
d
2.
b
3.
h
4.
e
correct
5.
c
6.
a
7.
g
8.
f
Explanation:
Recall the definition of impulse:
Impulse =
vector
F
net
Δ
t
= Δ
vectorp.
Therefore, whatever direction the net force
points in, that will be the direction of the
change in the ball’s momentum. We are told
that the force is in the direction of
e
, so that
is the correct answer.
002(part1of3)4.0points
A hockey puck is sliding along the ice with
nearly constant momentum
vectorp
i
=
(
10
,
0
,
8
)
kg
·
m
/
s
when it is suddenly struck by a hockey stick
with a force
vector
F
=
(
0
,
0
,
1400
)
N
that lasts for only 8 milliseconds (0
.
008 s).
What is the final momentum
vectorp
f
of the puck?
Begin by finding just the
x
component,
p
f,x
.
Answer in kg
·
m
/
s.
Correct answer: 10 kg
·
m
/
s.
Explanation:
This is a straightforward application of the
momentum principle. Let the system be the
puck, and assume the only significant interac
tion on the system is with the hockey stick.
Δ
vectorp
=
vector
F
net
,
sys
Δ
t
vectorp
f
−
vectorp
i
=
vector
F
net
,
sys
Δ
t
vectorp
f
=
vectorp
i
+
vector
F
net
,
sys
Δ
t
vectorp
f
=
(
10
,
0
,
8
)
kg
·
m
/
s
+
(
0
,
0
,
1400
)
N
×
(0
.
008 s)
vectorp
f
≈ (
10
,
0
,
19
.
2
)
kg
·
m
/
s
.
This gives us the answers to all three parts.
The
x
component of the final momentum is
p
f,x
=
10 kg
·
m
/
s
.
003(part2of3)3.0points
Find
p
f,y
. Answer in kg
·
m
/
s.
Correct answer: 0 kg
·
m
/
s.
Explanation:
See the explanation to part 1.
004(part3of3)3.0points
Find
p
f,z
. Answer in kg
·
m
/
s.
Correct answer: 19
.
2 kg
·
m
/
s.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
schopper (as55864) – Ch2HW1 – Antoniewicz – (56465)
2
Explanation:
See the explanation to part 1.
005(part1of9)2.0points
You were driving a car with velocity
vectorv
i
=
(
27
,
0
,
13
)
m
/
s
.
You quickly turned and braked, and your ve
locity became
vectorv
f
=
(
9
,
0
,
19
)
m
/
s
.
The mass of the car was 1000 kg.
What
are the components of the (vector) change in
momentum Δ
vectorp
during this maneuver? Start
with the
x
component, Δ
p
x
, and give your
answer in kg
·
m
/
s. Pay attention to signs.
Correct answer:
−
18000 kg
·
m
/
s.
Explanation:
Let’s go ahead and do the vector algebra for
all three components. We’re given the initial
and final speeds, which are nonrelativistic,
and the mass of the car.
Let the system
consist of the car. We have
Δ
vectorp
sys
=
vectorp
f
−
vectorp
i
=
m
car
vectorv
f
−
m
car
vectorv
i
=
m
car
(
vectorv
f
−
vectorv
i
)
= (1000 kg) (
(
9
,
0
,
19
)
m
/
s
−(
27
,
0
,
13
)
m
/
s)
=
(−
18000
,
0
,
6000
)
kg
·
m
/
s
006(part2of9)1.0points
Find Δ
p
y
. Answer in kg
·
m
/
s.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Turner
 Force, Correct Answer, kg, m/s, Fnet

Click to edit the document details