Version 012 – Midterm 01 – florin – (56930)
1
This
printout
should
have
18
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0points
An 800 kg car skidding due north on a level
frictionless icy road at 100 km
/
h collides with
a 2110 kg car skidding due east at 44
.
8 km
/
h
in such a way that the wreckage sticks to
gether and skid as shown in the picture.
2110 kg
44
.
8 km
/
h
100 km
/
h
800 kg
v
f
θ
N
At what angle East of North do the two
coupled cars skid?
1. 36.8846
2. 61.7675
3. 36.1832
4. 50.8886
5. 37.3848
6. 49.7584
7. 55.7113
8. 46.2873
9. 59.1821
10. 54.4406
Correct answer: 49
.
7584
◦
.
Explanation:
Let :
m
1
= 800 kg
,
v
1
= 100 km
/
h
,
m
2
= 2110 kg
,
and
v
2
= 44
.
8 km
/
h
.
m
2
v
2
p
2
p
1
m
1
v
1
p
f
=
m
f
v
f
θ
N
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 .
Looking at the
x
and
y
components of mo
mentum,
p
fx
=
m
2
v
2
= (2110 kg) (44
.
8 km
/
h)
×
parenleftbigg
10
3
m
km
parenrightbigg parenleftbigg
1 h
3600 s
parenrightbigg
= 26257
.
8 kg m
/
s
,
and
p
fy
=
m
1
v
1
= (800 kg) (100 km
/
h)
×
parenleftbigg
10
3
m
km
parenrightbigg parenleftbigg
1 h
3600 s
parenrightbigg
= 22222
.
2 kg m
/
s
.
Since we are asked to find the angle from the
y
axis instead of the
x
,
tan
θ
=
p
fx
p
fy
=
26257
.
8 kg m
/
s
22222
.
2 kg m
/
s
= 1
.
1816
θ
= arctan
parenleftbigg
26257
.
8 kg m
/
s
22222
.
2 kg m
/
s
parenrightbigg
= 49
.
7584
◦
.
002
10.0points
A hanging wire made of gold with diameter
0
.
09 cm is initially 2 m long and a cubic atomic
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Version 012 – Midterm 01 – florin – (56930)
2
structure. When a 62 kg mass is hung from it,
the wire stretches an amount 1
.
13 cm. A mole
(
N
A
= 6
.
02
×
10
23
atoms) of gold has a mass
of 197 g
/
mol, and its density is 19
.
3 g
/
cm
3
.
Use
g
= 9
.
8 m
/
s
2
.
Based
on
these
experimental
measure
ments, find the approximate value of the effec
tive spring stiffness of one interatomic bond
in gold.
1. 46.7481
2. 54.6406
3. 66.3317
4. 40.2155
5. 51.9423
6. 60.4593
7. 41.8459
8. 43.4275
9. 40.9284
10. 42.5678
Correct answer: 43
.
4275 N
/
m.
Explanation:
Let
M
be the weight of the mass hung from
the wire, let
d
be the diameter of the wire,
and let
l
and
x
be the original length of the
wire and the amount by which it stretches,
respectively. The tension in the wire is equal
to the weight of the hanging mass, since the
wire’s mass is negligible.
Therefore, using the definition of Young’s
modulus, and letting
Y
=
F
A
L
Δ
L
=
4
M g
π d
2
l
x
=
4(62 kg)(9
.
8 m
/
s
2
)
π
(0
.
09 cm)
2
parenleftbigg
2 m
1
.
13 cm
parenrightbigg
= 1
.
69042
×
10
11
N
/
m
2
.
To find the stiffness of an interatomic bond,
we use the expression
k
=
Y a,
where
k
is the stiffness and
a
is the length of a
single interatomic spacing. We find the value
of
a
by first finding the volume of a cubic
lattice of atoms in the gold, and then taking
the cubed root.
Converting the given quantities into the
necessary units, we find that the density and
the molar mass of gold are
ρ
= 19300 kg
/
m
3
M
= 0
.
197 kg
/
mol
,
respectively. Now the volume of a cubic lat
tice of atoms is given by
V
=
(
19300 kg
/
m
3
)
−
1
(0
.
197 kg
/
mol)
parenleftbigg
1 mol
N
A
parenrightbigg
= 1
.
69556
×
10
−
29
m
3
.
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 Fall '08
 LEON
 Mass, Correct Answer, m/s, Iiia

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