pandey (sp34566) – Test 3 Review – swinney – (57305)
1
This
printout
should
have
31
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0points
A mass of 0.2 kg is attached to a massless
spring, which provides the only force acting
on the mass as it executes simple harmonic os
cillation with an angular frequency of 2 rad/s.
If
x
(0) =

0
.
5 m and
v
x
(0) =

1
.
0 m/s
for this oscillator, what is
A
?
The general
solution for a simple harmonic oscillation is
x
(
t
) =
A
cos(
ω t
+
φ
)
.
1.
2.0 m
2.
1.70 m
3.
0.25 m
4.
√
2 m
5.
1.25 m
6.
Zero
7.
1.5 m
8.
1.0 m
9.
0.707 m
correct
10.
0.5 m
Explanation:
The easiest solution involves conservation
of energy, noting that
k
=
m ω
2
.
E
=
1
2
k A
2
=
1
2
m ω
2
A
2
and
E
=
1
2
m v
x
(0)
2
+
1
2
m ω
2
x
(0)
2
.
Equating these,
A
=
radicalBigg
x
(0)
2
+
parenleftbigg
v
x
(0)
ω
parenrightbigg
2
=
radicalBigg
(

0
.
5 m)
2
+
parenleftbigg

1
.
0 m
/
s
2 s
−
1
parenrightbigg
2
= 0
.
707 m
.
002
10.0points
A harmonic oscillator consists of a mass
M
on a horizontal frictionless surface attached
to a spring of negligible mass and with force
constant
k
.
The oscillator has its maximum
kinetic energy (KE) at
x
= 0, and its KE goes
to zero at
x
=
A
.
When it is at
x
=
A
4
, how does its KE
compare to its maximum value;
i.e.
, what is
K
(
x
=
A/
4)
K
max
?
1.
K
K
max
=
7
9
2.
K
K
max
=
15
16
correct
3.
K
K
max
=
1
2
4.
K
K
max
=
5
16
5.
K
K
max
=
7
16
6.
K
K
max
=
1
4
7.
K
K
max
=
13
16
8.
K
K
max
=
5
8
9.
K
K
max
=
3
4
10.
K
K
max
=
1
8
Explanation:
E
=
K
+
U
s
=
1
2
k A
2
=
K
max
is conserved as the oscillator moves. At
x
=
A
4
,
U
s
=
1
2
k
parenleftbigg
A
4
parenrightbigg
2
=
E
16
,
so
pandey (sp34566) – Test 3 Review – swinney – (57305)
2
K
=
E

E
16
=
15
16
E
K
K
max
=
15
16
.
003
10.0points
A student performs a ballistic pendulum
experiment using an apparatus similar to that
shown in the figure.
Initially the bullet is fired at the block while
the block is at rest (at its lowest swing point).
After the bullet hits the block, the block rises
to its highest position, see dashed block in the
figure, and continues swinging back and forth.
The following data is obtained:
the maximum height the pendulum rises is
5 cm,
at the maximum height the pendulum sub
tends an angle of 48
.
2
◦
,
the mass of the bullet is 93 g, and
the mass of the pendulum bob is 793 g.
The final velocity is
v
f
=
radicalbig
2
g h .
Using
conservation of momentum
m
1
v
i
= (
m
1
+
m
2
)
v
f
,
so
v
i
=
bracketleftbigg
(
m
1
) + (
m
2
)
m
1
bracketrightbigg
v
f
=
bracketleftbigg
(
m
1
) + (
m
2
)
m
1
bracketrightbigg
radicalbig
2
g h
=
bracketleftbigg
(93 g) + (793 g)
(93 g)
bracketrightbigg
×
radicalBig
2 (9
.
8 m
/
s
2
) (0
.
05 m)
=
9
.
43113 m
/
s
.
004(part1of4)10.0points
A 8 kg block slides down a frictionless incline
making an angle of 25
◦
with the horizontal.