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30 January 2013
AP Physics Test Ch 14,22,23
Name:
1)
A mass
period
n? hanging on a spring
r
with a spring constant k executes simple harmonic motion with a
tr the same mass is huns
oscilration
a.
b.
e.
LJa =
increases by a factor of 2.
decreases by a factor of 2.
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d
rroma .orfn
ot
decreases by a ractor ot
is not affected.
or
E;quency
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2) Any body moving with simple harmonic motion is being acted on by a force that is
constant.
proportional to a sine or cosine function of the displacement.
proportional to the inverse square of the displacement.
directly proportional to the displacement.
proportional to the square of the displacement.
3)
r^r.t&-
'€ * Lt * &, ,
The force constant of a massless spring is 25.0 N/m. A mass of 0.45 kg is oscillating in simple
harmonic motion at the end of the spring with an amplitude of 0.32 m. The maximum speed of
massis
t[e
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''t' t--n
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rrl
,' ,y'.:Y;; t/* y'24! "){ *t **.)L
,/., "&z
u" 10 m/s.
ty'
U,-"<r_.a
tZ"
.}
,Hg
+,/\ tt.t /,.
-, .l{
4) The equation of a body in simple narmoniJ-m k;Tr:/O
-/ = 8.0 cos(2}r +
{)
a
where y is in centimeters and f is in seconds. The frequency of the oscillations is
a.
b.
El
LD =-2-'; J- ";
nl10 Hz.
nl4 Hz.
,6)rcnu=.
'*4
e.
{- #.
8 Hz.
20H2.
G
/..F:
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-,@.""
5) The period of a simple pendulum of length 9.8 m and mass 5.0 kg is approximately
g
d
0.1 s
0.16 s
2.0 s
5.0 s
6.3 s
e/
---r'-
{-
?{,
{
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l-
Page
1
AP Physics Test Ch
6)
14,22,23
30 January 2013
The energy of a simple harmonic oscillator could be doubled by increasing the amplitude by a factor
of
i
a. ot, or IfJ)
b. 1.0
€,to'r,
e.
J/
*/
*d4='
*iu{4'
G*&w*
= *-*t{&. * zb: ,{e'
4:* zn- -='? '4-* =,lLft
4.0
7) A body moving in simple harmonic motion has maximum
a.
b.
c.
@
e.
8)
maximum velocity.
maximum kinetic energy.
minimum Potentialcn::Py
minimum kinetic energy'
zero displacement.
e L flz**"-a*.* **
,ut
r4* Fnp'**n**.
/
fJ**'"
7
For a simple pendulum to have a period of 1 second it must hbve a length of
a.
tff)
*-61
1.0 m
o.zs m
e.a m
d.4,om
e.
g)
acceleration when it has
* * ax
/r4-
a/
4
2.7
m
-7'*
n
z-.ff
.
' *";L
f,7 yp,E{u
= -**+'e'l
A pendulum is oscillating with a total mechanical energy Ee. When the pendulum is at its maximum
displacement, the kinetic energy K and the potential energy U are
a.
@c.
d.
e.
K =Tz Es; U =Y2E6
x= o; IJ=Eo
K=EoiU=Eo
K=Eo;U--0
K=EoiU=YzEo
10) A meter stick is used as a physical pendulum, with the axis of suspension at one end of the stick. lf
the moment of inertia of the stick about the axis is (113)ML" where M and L are the mass and length
of the meter stick, respectively, how does the simple harmonic period I of the physical pendulum
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1) lf the charge on one electron is -1 .6 x 10-1s C (Coulomb), and an object is determined to have a
negative charge of '160 nC, you can conclude that the object has an excess of
a.
b.
c.
Gr
e.
10' electrons
electrons
electrons
1o12 electrons
10'o electrons
1010
/ac"4 *t/(/6xtD
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*-9,
1011
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30 January 2013
AP Physics Test Ch 14,22,23
12) Three charges +q, +Q, and -Q are placed at the corners of an
equilateral triangle as shown. The net force on charge +q due to the
other two charges is
a.
b.
l,Y--
vertically up.
't
./''
;:
'.'
fa,!rr
vertically down.
zeto.
d.
horizontal to the left.
o.,
-a
*Qo
d.i horizontal to the right.
13) Three charges are located at 100-m intervals along a horizontal line: a charge of -3.0 C on the left,
+2.0 C in the middle, and +1.0 C on the right. What is the resultant force on the 1.0 C charge due
the other two
--',
(a;11
charges
r !_
@
e
tisjt
v
I 19:I l? ll?
{ t.t xlooNtotheteft
r*/'
;. z.s)t0'Ni;inJrignt ? 2 /2,\t?
- ,'"
d. 2.s x io6 N to the teft rzr. -=t
6
**
-r.
to ,r
r
= Li r_?l
-*l
'-a/
lD' I
'
ei
\
tr+ ,=%-/r)( *
'e-,'
I
i 7d*
14) Charges Qr = -Q and Qz = +4q are placed as shown. The electric field E is zero at one, and only
one, of the five positions indicated by the numbered dots. Which is dot is it?
4b
Gt
b.2
Qt----
c.3
d.4
e.5
8s
_..-
.\-**--**J2 3 4
E)4?
2
L----s
F --.*
E*7€;
?
_.4
offi
15) A proton is moving horizontaily north in an erectric fietd that
force on the oroton is
triffir];fr
erectric
t zeto.
upward.
/6r/
uc. downward.
d.
e.
to the west.
to the east
l*20 gm--->F-t[ *----*'
16) The electric field at point A is zero. What is charge Qr?
."tt &* le + t€) ".
'"fi +zz 1tc QPt
-t
gJ e /-* "o#
b -321tc
c. The field cannot be z6ro at A for any value of Q7.
d. +16 pC
e. -16 pC
+Lq I
-
.a-
C
(rO.t*)
C-<t @/
I
Q1
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AP Physics Test Ch 14,22,23
30 January 2013
17) ln the figure, the direction of the electric field at a point equidistant
from two equally charged bodies A and B is indicated by i vector.
The direction of the vector indicates that
1B) A surface is so constructed that, at all points on the surface,
can be said that
D
!
/-^} '-
a. both A and B are positive.
b. both A and B are negative.
A is positive and B is negative.
e
Y. B is positive and A is negative.
e. B is negative and A is neutral.
:'
\3/--
r\
r
--i>
t--------@
I
I
-I
the E vector points inward. Therefore,
g. the surface encloses a net positive charge.
surface encloses a net negative charge.
@tne
Y the surface encloses no net
"iarg..on the surface is necessarily
d. the surface vector AA at all points
parallel to the electric
field vector E.
the surface vector AA at all points on the surface is necessarily perpendicular
to the
electric field vector ,E .
e.
19) The figure shows a surface enclosing the charges 2q and _q. The
net flux through the surface surrounding the two chaiges is '
Q/eo
2qleo
-qleo
zero
None of these is correct.
20) A rod of infinite tength has a charge per unit length of l.
1=q71,
Gauss's Law makes it easy to determine that the
electric field strength at a perpendicular distance r
from the rod is, in terms of k=(4xes)-1,
a. k2/r,
b. k)./r
h);rYl'
zero.
L-.e:t
21) consider a solid sphere of charge of radius R and uniform
charge density p throughout the sphere.
Using Gauss's Law we can compute the electricfield E everywliere
in sjaEe as i'function of the
radial distance r from the center of the sphere. which
of the'sketches most accuratety represents the
behavior of the field both inside and outside the sphere?
a.
b.2
c.3
1
@o
e.5
E
AP Physics Test Ch 14,22,23
30 January 2013
22\ The electric field at the surface of a conductor
a.
b.
c.
d.
is parallel to the surface,
depends only on the total charge of the conductor.
depends only on the surface area of the conductor.
depends only on the curvature of the surface of the conductor.
depends on the surface area of the conductor, on the curvature of the conductor, and
on the charge on the conductor.
23) An electrically neutral rectangular conducting slab is placed in a
uniform electric field as indicated in the diagram to the right. Which of
the sketches best illustrates the situation when the system has come
to equilibrium following the insertion of the slab in the field? [The
vectors represent the presence of an electric field, and the oositive and
negative signs (+,-) represent charges.l
a.
{D
\--{
-+ tl
-->
--
I
z
?
d.
4
e.
q
e
--------s.
+1.;E
--z<-P
-----{..---*
-H
+l->
----l-l)
--1F-t>
-fr>
€
tlll
-l- +l-+
-l-l--->
---1+
ff
(1)
24) A spherical shell of radius 9.0 cm carries a uniform surface charge density o = 9.0 nC/m2. The
electric field at r = 8.9 cm is approximately
a.
b.
/e.
( o)
=.
1.0 kN/C
0.13 kN/c
0.32 kN/C
zero
o.6s kN/c
25) Two concentric conducting spherical shells of radii r, and 12 ( r, . r, ) contain charge densities o1 end
o2, respectively. For r > 12 (outside the larger sphere) the electric field is
a.
c.
ko, f r'
k(o, + or)lr'
d.
4tzkorr| f 12
b.
--*
j '
:-. Ii 4";!-to-
ko, f 12
O*(o,r,'+orr|)f
/-3' r, *r/
L,l
qrl :.
,'
P'-
F/
7 7r
: 1rL
r-
'Qr*-4
/f"t
t
\
d4r
)=
=
-'-f
€;
-G-
J- )n (o*' V' & F'r G€:
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