Patel (pnp223) – homework 21 – Turner – (92160)
1
This
printout
should
have
11
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
(part 1 of 2) 10.0 points
An attack helicopter is equipped with a 20
mm cannon that fires 135 g shells in the
forward direction with a muzzle speed of
324 m
/
s.
The fully loaded helicopter has a
mass of 1770 kg. A burst of 89
.
8 shells is fired
in a 5
.
33 s interval.
What is the resulting average force on the
helicopter?
Correct answer: 736
.
933 N.
Explanation:
The impulse imparted to the shells equals
the change in momentum:
F
av
Δ
t
= Δ
mv
The mass change is
Δ
m
=
n m
= (89
.
8 shells) (135 g) = 12
.
123 kg
,
so the average force is
F
=
v
Δ
m
t
=
(324 m
/
s) (12
.
123 kg)
5
.
33 s
=
736
.
933 N
.
Since the velocity of the shells is much greater
than the velocity of the helicopter, there is no
need to use relative velocity.
002
(part 2 of 2) 10.0 points
By what amount is its forward speed re
duced?
Correct answer: 2
.
21913 m
/
s.
Explanation:
From conservation of momentum Δ
p
= 0,
so
Δ
v
=
Δ
m v
M
=
(12
.
123 kg) (324 m
/
s)
1770 kg
=
2
.
21913 m
/
s
,
003
10.0 points
A(n) 2
.
9 kg object moving at a speed of
4
.
3 m
/
s strikes a(n) 0
.
98 kg object initially
at rest.
Immediately after the collision, the
2
.
9 kg object has a velocity of 1
.
2 m
/
s directed
29
◦
from its initial line of motion.
What is the speed of the 0
.
98 kg object
immediately after the collision?
Correct answer: 9
.
77155 m
/
s.
Explanation:
Let :
m
1
= 2
.
9 kg
,
v
1
i
= 4
.
3 m
/
s
,
v
1
f
= 1
.
2 m
/
s
,
m
2
= 0
.
98 kg
,
and
θ
= 29
◦
.
p
1
i
p
1
f
p
2
f
29
◦
Momentum is conserved, therefore
vectorp
1
i
=
vectorp
1
f
+
vectorp
2
f
p
2
2
f
=
p
2
1
i
+
p
2
1
f
−
2
p
1
i
p
1
f
cos
θ
= (12
.
47 kg m
/
s)
2
+ (3
.
48 kg m
/
s)
2
−
2 (12
.
47 kg m
/
s) (3
.
48 kg m
/
s) cos 29
◦
= 91
.
702 kg
2
m
2
/
s
2
,
and
p
2
f
=
radicalBig
(91
.
702 kg
2
m
2
/
s
2
)
= 9
.
57612 kg m
/
s
,
so
p
2
f
=
m
2
v
2
f
.
Solving for
v
2
f
, we have
v
2
f
=
p
2
f
m
2
=
(9
.
57612 kg m
/
s)
(0
.
98 kg)
=
9
.
77155 m
/
s
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Patel (pnp223) – homework 21 – Turner – (92160)
2
004
(part 1 of 2) 10.0 points
A 1480 kg car skidding due north on a level
frictionless icy road at 265
.
95 km
/
h collides
with a 2234
.
8 kg car skidding due east at
197 km
/
h in such a way that the two cars
stick together.
2234
.
8 kg
197 km
/
h
265
.
95 km
/
h
1480 kg
v
f
θ
N
At what angle (
−
180
◦
≤
θ
≤
+180
◦
) East
of North do the two coupled cars skid off at?
Correct answer: 48
.
202
◦
.
Explanation:
Let :
m
1
= 1480 kg
,
v
1
= 265
.
95 km
/
h
,
m
2
= 2234
.
8 kg
,
and
v
2
= 197 km
/
h
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 Turner
 Physics, Kinetic Energy, Momentum, Work, kg, m/s, patel

Click to edit the document details