homework 08 – ALIBHAI, ZAHID – Due: Mar 21 2007, 4:00 am
1
Question 1
part 1 of 1
10 points
Two air blocks with masses 184 g and
184 g are equipped with identical springs
(
k
= 1560 N
/
m)
.
The blocks move toward
each other with identical speeds of 1 m
/
s on a
horizontal air track and collide, compressing
the springs.
184 g
1 m
/
s
1560 N
/
m
184 g
1 m
/
s
1560 N
/
m
Find
the
maximum
compression
of
the
spring attached to the 184 g mass.
Correct answer: 1
.
08604
cm (tolerance
±
1
%).
Explanation:
Let :
k
= 1560 N
/
m
,
m
1
= 184 g
,
m
2
= 184 g
,
and
bardbl
vector
v
1
bardbl
=
bardbl
vector
v
2
bardbl
=
v
= 1 m
/
s
.
Note:
Only the kinetic energy in the center
ofmass system is equivalent to the potential
energy of the compressed springs.
The centerofmass velocity
v
cm
is
v
cm
=
m
1
−
m
2
m
1
+
m
2
v
=
(184 g)
−
(184 g)
(184 g) + (184 g)
(1 m
/
s)
= 0 m
/
s
.
The total kinetic energy of both masses
is equal to the total potential of both com
pressed springs.
1
2
k x
2
1
+
1
2
k x
2
2
=
1
2
[
m
1
+
m
2
] [
v
2
−
v
2
cm
]
=
1
2
[
m
1
+
m
2
]
v
2
radicalBig
x
2
1
+
x
2
2
=
radicalbigg
[
m
1
+
m
2
]
v
2
k
=
braceleftbigg
[(184 g) + (184 g)]
×
(1 m
/
s)
2
(1560 N
/
m) (1000 g
/
kg)
bracerightbigg
1
/
2
×
(100 cm
/
m)
= 1
.
5359 cm
.
Each of the two springs absorb the same
amount of potential energy (onehalf the total
potential energy);
i.e.
,
x
=
x
1
=
x
2
,
we have
radicalBig
x
2
1
+
x
2
2
=
√
2
x ,
so
x
=
radicalbigg
x
2
1
+
x
2
2
2
=
(1
.
5359 cm)
√
2
=
1
.
08604 cm
.
Question 2
part 1 of 1
10 points
An object of mass
m
is moving with speed
v
0
to the right on a horizontal frictionless
surface, as shown, when it explodes into two
pieces. Subsequently, one piece of mass
36
41
m
moves with a speed
v
36
/
41
=
v
0
9
to the left.
m
v
0
before
36
41
m
v
0
9
5
41
m
v
5
/
41
after
What is the speed
bardbl
vectorv
5
/
41
bardbl
of the other piece
of the object?
1.
bardbl
vectorv
5
/
41
bardbl
=
77
5
v
0
.
2.
bardbl
vectorv
5
/
41
bardbl
=
41
36
v
0
.
3.
bardbl
vectorv
5
/
41
bardbl
=
1476
5
v
0
.
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homework 08 – ALIBHAI, ZAHID – Due: Mar 21 2007, 4:00 am
2
4.
bardbl
vectorv
5
/
41
bardbl
=
205
36
v
0
.
5.
bardbl
vectorv
5
/
41
bardbl
= 41
v
0
.
6.
None of these
7.
bardbl
vectorv
5
/
41
bardbl
=
41
5
v
0
.
8.
bardbl
vectorv
5
/
41
bardbl
=
46
36
v
0
.
9.
bardbl
vectorv
5
/
41
bardbl
= 9
v
0
.
correct
10.
bardbl
vectorv
5
/
41
bardbl
= 5
v
0
.
Explanation:
The horizontal component of the momen
tum is conserved, so
0 +
m v
0
=
36
41
m v
36
/
41
+
5
41
m v
5
/
41
0 +
m v
0
=
36
41
m
parenleftBig
−
v
0
9
parenrightBig
+
5
41
m v
5
/
41
m v
0
=
−
36
369
m v
0
+
5
41
m v
5
/
41
5
41
v
5
/
41
=
parenleftbigg
369
369
+
36
369
parenrightbigg
v
0
5
41
v
5
/
41
=
405
369
v
0
v
5
/
41
=
405
369
41
5
v
0
bardbl
vectorv
5
/
41
bardbl
=
9
v
0
.
Question 3
part 1 of 1
10 points
A gadget of mass 20
.
24 kg floats in space
without motion.
Because of some internal
malfunction, the gadget violently breaks up
into 3 fragments flying away from each other.
The first fragment has mass
m
1
= 6
.
45 kg
and speed
v
1
= 4
.
38 m
/
s while the second
fragment has mass
m
2
= 5
.
62 kg and speed
v
2
= 2
.
87 m
/
s. The angle between the veloc
ity vectors
vectorv
1
and
vectorv
2
is
θ
12
= 80
◦
.
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 Spring '08
 Turner
 Energy, Kinetic Energy, Mass, Work, Correct Answer, kg, m/s

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