homework 09 – FRENNEA, KYLE – Due: Mar 28 2007, 4:00 am
1
Question 1
part 1 of 1
10 points
Two ice skaters approach each other at right
angles. Skater A has a mass of 72
.
5 kg and
travels in the +
x
direction at 2
.
88 m
/
s. Skater
B has a mass of 62
.
5 kg and is moving in the
+
y
direction at 0
.
982 m
/
s. They collide and
cling together.
Find the final speed of the couple.
Correct answer: 1
.
6121
m
/
s (tolerance
±
1
%).
Explanation:
From conservation of momentum Δ
p
= 0
m
A
v
A
ˆ
ı
+
m
B
v
B
ˆ
= (
m
A
+
m
B
)
v
f
Therefore
v
f
=
radicalbig
(
m
A
v
A
)
2
+ (
m
B
v
B
)
2
m
A
+
m
B
=
radicalbig
(208
.
8 kg m
/
s)
2
+ (61
.
375 kg m
/
s)
2
72
.
5 kg + 62
.
5 kg
= 1
.
6121 m
/
s
Question 2
part 1 of 2
10 points
m
1
1
m
2
before
after
v
v
v
2
v
m
1
m
2
3
4
α
β
An
m
2
= 1
.
6 kg can of soup is thrown
upward with a velocity of
v
2
= 5
.
6 m
/
s.
It
is immediately struck from the side by an
m
1
= 0
.
6 kg rock traveling at
v
1
= 8
.
2 m
/
s.
The rock ricochets off at an angle of
α
= 67
◦
with a velocity of
v
3
= 6
.
4 m
/
s.
What is the angle of the can’s motion after
the collision?
Correct answer: 57
.
7764
◦
(tolerance
±
1 %).
Explanation:
Basic Concepts:
Conservation of Mo
mentum
p
before
=
p
after
.
Solution:
Horizontally
m
1
v
1
+
m
2
(0) =
m
1
v
3
cos
α
+
m
2
v
4
cos
β ,
so that
m
2
v
4
cos
β
=
m
1
v
1

m
1
v
3
cos
α
(1)
= (0
.
6 kg) (8
.
2 m
/
s)

(0
.
6 kg) (6
.
4 m
/
s) cos 67
◦
= 3
.
41959 kg m
/
s
.
Vertically
m
1
(0) +
m
2
v
2
=
m
1
v
3
sin
α
+
m
2
v
4
sin
β ,
so that
m
2
v
4
sin
β
=
m
2
v
2

m
1
v
3
sin
α
(2)
= (1
.
6 kg) (5
.
6 m
/
s)

(0
.
6 kg) (6
.
4 m
/
s) sin67
◦
= 5
.
42526 kg m
/
s
.
Thus
tan
β
=
m
2
v
4
sin
β
m
2
v
4
cos
β
=
(5
.
42526 kg m
/
s)
(3
.
41959 kg m
/
s)
= 1
.
58652
,
and
β
= arctan(1
.
58652)
= 57
.
7764
◦
.
Question 3
part 2 of 2
10 points
With what speed does the can move imme
diately after the collision?
Correct answer: 4
.
00815 m
/
s (tolerance
±
1
%).
Explanation:
Using equation (1) above,
v
4
=
m
1
v
1

m
1
v
3
cos
α
m
2
cos
β
=
(0
.
6 kg) (8
.
2 m
/
s)
(1
.
6 kg) cos57
.
7764
◦

(0
.
6 kg) (6
.
4 m
/
s) cos(67
◦
)
(1
.
6 kg) cos(57
.
7764
◦
)
= 4
.
00815 m
/
s
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homework 09 – FRENNEA, KYLE – Due: Mar 28 2007, 4:00 am
2
or using equation (2) above,
v
4
=
m
2
v
2

m
1
v
3
sin
α
m
2
sin
β
=
(5
.
6 m
/
s)
sin(57
.
7764
◦
)

(0
.
6 kg) (6
.
4 m
/
s) sin(67
◦
)
(1
.
6 kg) sin(57
.
7764
◦
)
= 4
.
00815 m
/
s
.
Question 4
part 1 of 2
10 points
A 2040 kg car skidding due north on a level
frictionless icy road at 219
.
6 km
/
h collides
with a 3651
.
6 kg car skidding due east at
180 km
/
h in such a way that the two cars
stick together.
3651
.
6 kg
180 km
/
h
219
.
6 km
/
h
2040 kg
v
f
θ
N
At what angle (

180
◦
≤
θ
≤
+180
◦
) East
of North do the two coupled cars skid off at?
Correct answer: 55
.
7231
◦
(tolerance
±
1 %).
Explanation:
Let :
m
1
= 2040 kg
,
v
1
= 219
.
6 km
/
h
,
m
2
= 3651
.
6 kg
,
and
v
2
= 180 km
/
h
.
m
2
v
2
p
2
p
1
m
1
v
1
p
f
=
m
f
v
f
θ
N
Basic Concepts:
Momentum Conserva
tion,
K
=
1
2
mv
2
,
vectorp
=
mvectorv .
During the collision, the total momentum
of the two car system will be conserved
vectorp
f
=
vectorp
1
+
vectorp
2
=
m
1
vectorv
1
+
m
2
vectorv
2
p
fx
ˆ
ı
+
p
fy
ˆ
=
m
1
v
1
ˆ
+
m
2
v
2
ˆ
ı .
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 Kinetic Energy, Mass, Work, Moment Of Inertia, Rotation, Correct Answer

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