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Unformatted text preview: 1 998
The College Board
Advanced Placement Examination PHYSICS C
SECTION II TABLE OF INFORMATION FOR 1998 CONSTANTS AND CONVERSION FACTORS UNITS PREFIXES
l uniﬁed atomic mass unit. lu = 1.66 X 10—27 kg Name MID‘21 F—aqp—r Pmﬁx 34m
= 93t MeV/cz meter in 109 giga G
Proton mass. mp = 157 x 10'27 kg kilogram kg 106 mega M
Neutron mass, : ’27 3 .
mﬂ 1.67 x 10 kg second S 10 kilo 1:
El t mass. _ 31
“C m“ mE  9.11 x 10 kg A 10.2 cemi c
Magnitude of the electron charge, 8 = 1.60 X 1019C amp“:
3 . .
Avogadro's number, No x 602 X 1013 moi1 kelvin K 10 “1111‘ m
. _6 _
Universal as constant, = . 10 micro u
g R 8.31 J/(rnol K) mole m0]
Boltzmann's constant, k5 = 133 x 1043 J/K 109 mm H
. s hertz Hz
Speed “ﬁght. 6 = 3.0% 10 m/s 1012 pm p
Planck‘s constant, _ ~34 . newton N
h ‘ 6'63 x 10 15 J 5 mm Pa VALUES OF TRIGONOMETRIC FUNCTIONS
= 4.]4x 10— eV ' s P ‘ FOR COMMON ANGLES
he = 1.99x10‘25 J ' m joule J
= 1.24><103 eV  rim watt w
Vacuum permittivity, 50 = 335 x 10"” CQ/N . m2 “Dummb C
Coulomb's law constant, k : 1/4n50 = 90X 109 N . mq/C? VD“ V
Vacuum permeability, p0 = 4It x 107 (T . m) /A ohm Q 35
Magnetic constant, kt : “0/4” = {077(1 . m) /A henry H
Universal gravitational constant, G = 6.67 X 10” [If/kg . 52 farad F @2 Via/2
ACCBlemliotia due]:1 to gravity tesla T
att 6 art 'ssu ace, _ 2
g ' 9'3 m” degree 4/5 3/5
1 atmosphere pressure, 1 atm = 1.0 x 10’ N/m2 Celsius °c .
5
= 1.0x 10 Pa clean0m 19 volt 6V
lelectron volt, iev = Léox 10 J
l angstrom. HR = Ix io’“’m The following conventions are used in this examination.
1. Unless otherwise stated, the frame of reference of any problem is assumed to be inertial,
11. The direction of any electric current is the direction of ﬂow of positive charge (conventional current). 111. For any isolated electric charge, the electric potential is deﬁned as zero at an inﬁnite distance from the charge. Copyright © 1998 by College Entrance Examination Board and Educational Testing Service.
All rights reserved. ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 1998 MECHANICS ELECTRICITY AND MAGNETISM 1; = no + at a = acceleration F: —— ‘1qu A = area
1 F = force 47”” r B = magnetic field
5 = so + not + E at2 f 2 frequency E = E C = capacitance
h = height :1 d = distance
32 = 302 + gag; _ so) I = rotational inertia E = electric field
J: impulse ﬁEdA =62 5=emf
2F = FM! 2 ma K = kinetic energy 0 F = force
d k = spring constant _ d_V I = current
F = a]; f = length E _ '— dr L = inductance
L = angular momentum E = length
J = I F d? = AP m = mass V: L 23 n = number of loops of wire
.. N = normal force 47‘“) r per unit length
p — mv P _ _ 1 qu .—
. power U = qV _ — — P — power
an'c S W p = momentum 47*" r Q = charge
a . r = distance q = point charge
W _ IF ds 3 = displacement C = g R = resistance
1 2 T = eriod : = distance
K_§mr’i t=ti3me C=K€0A t=time
dW ‘ U = potential energy d U = potential or stored energy
= E? v = velocity or speed Cp = 2 Cl. V = electric potential
W = work 1 v = velocity or speed
Aug = mgh x = displacement . I 1 1 p = resistivity
2 u = coeff1c1ent of friction E = 6 rpm = magnetic flux
1: 2 s , r . .
ac = — = (,3 , 9 = angle x = dielectric constant
r 't = torque
1' = r x F to = angular speed (1 = angular acceleration ADVANCED PLACEMENT PHYSICS C EQUATIONS FOR 1998 GEOMETRY AND TRIGONOMETRY Rectangle A = area A = bh C = circumference
Triangle V = volume 1 S = surface area A _ E bk [7 = base
Circle h = height A = m2 E = length C = 2m w = width
Parallelepiped r = radius V = Ewh
Cylinder 2 V = m f S = 21:14? + 21tr2
Sphere _E 3
V—3n:
S=41tr Right Triangle CALCULUS df_ﬂ.ﬂ nix—dot dx d x_ x
a(e)—e d 1
E(lnx)—:t i (sin x)  cos x
dx _ % (cos x) = —sin x J'xﬂdx= 1 11+] 1: ,natl
n+1
Iexdx=ex
Idi=1n lxl
x Icosxdx=sinx Isinxdx=—cosx 1 998 PHYSICS C SECTION II, MECHANICS Time—~45 minutes
3 Questions
Directions: Answer all three questions. The suggested time is about 15 minutes for answering each of the questions, which are worth 15 points each. The parts within a question may not have equal weight. Show all your work in the pink booklet in the spaces
provided after each part, NOT in this green insert. Air Track Mech. 1. Two gliders move freely on an air track with negligible friction, as shown above. Glider A has a mass of 0.90 kg and
glider B has a mass of 0.60 kg. Initially, glider A moves toward glider B, which is at rest. A spring of negligible
mass is attached to the right side of glider A. Strobe photography is used to record successive positions of glider A at
0.10 5 intervals over a total time of 2.00 3, during which time it collides with glider B. The following diagram represents the data for the motion of glider A. Positions of glider A at the end of each 010 s
interval are indicated by the symbol A against a metric ruler. The total elapsed time I after each 0.50 s is also indi
cated. 1.10 1.20
A A A A A A A A A A A A A A A A A A A A A
0.00s 0.505 . 1.003 1.505 2.005 r (a) Determine the average speed of glider A for the following time intervals. i. 0.105t00.305
ii. 0.905t01.108 iii. 1.70 s to 1.90 s GO ON TO THE NEXT PAGE Copyright © 1998 by College Entrance Examination Board and Educational Testing Service. All rights reserved. 1998 PHYSICS C—MECHANICS (b) On the axes below, sketch a graph. consistent with the data above, of the speed of glider A as a function
of time t for the 2.00 s interval. ) \II
S S
{t {x .t
I f. G
.A
P
. T.
3 VA
0 .m 0 F.
I 0 r. O N
l1l—II1I1I1II1IJIJII1I1IIII1JII1 u p lltdll_ll1l1llqllulJlJII.II1IIII_I1J'I1 .
__ _ __ _ __ _ _. . __ _ 7. s . _ __ _ __ _ __ _ __ _ _ 0. EL
_Lt_ __I.‘ _ __ _“_ __ e __ ___ __ __ _#_I__t_ u”
. _ ._.IHII_ _IJ._.I_I4l—ui_ ﬁJI... ..nL II._.I_I4 _IJlﬂJlﬁth 1. ._ut_ .._. T
_ _ __ _ ._ _ _, _ __ _ _ L _ _ __ _ __ _ ._ _ _, ﬂ _ nu
II_LI_I____I_tF.I___LI_ m II_LI_I____I_F_I__1_._
q..._._ln__tJ._._._.ﬁl_ﬂ_._. m e ﬁ_._._IJﬁJ._._n—_Il_ﬂl_._. T
__ __ .. __ __ __ __ . f .m ._ F. __ _. __ __ ._ _ Mu
IIPLIFtrLIPLIeLILIrLIrLIk s t IIhLIpirLlrLlhtrplrLlrle nu
_ __ ._ _ __ __ . __ __ m f _ __ ._ _ __ __ . __ __
_ _ __ _ __ _ __ _ __ _ _ a o _ _ __ _ __ . __ _ ._ _ _ nu
t+LI+ITLTLI+IT+ITL1+LI+ m n InTLI+ITLITLI+IT+ITLIr11+ nu
. _ __ _ __ _ __ _ _. h _ D. .m . _ __ _ ._ __ _ __ _ __
d _ _ﬂ __ _ ._ __ . __ _ mw W a _ __ __ _ a. _ __ __ _ _ m
ItrLILIrLIrletT+IrLIrLI+ . t n ItrIT+IrLIrLt+ITLIrLIrLI+ .
_ __ __ __ __ __ .ﬂ. _ l .1 nm . __ __ __ __ .. __ _n 1
_ __ __ _ __ __ _ _. __ m _ __ __ __ _ __ __ _ __
II14I4I1414t411414+4l4 t a t+J41414+I14I14ﬂJI4
.. __ __ __ __ ___ __ ﬂ S _ ___ n. __ ._ __ ___
II_LI_I__I._1_IF_I__I__ _ Y a I__I_IF_ _LI_iF_I__I_LI_
.._474 uIJﬁ_Aﬂi_ ﬁJlﬂ .1 B ._.l_._.__I1_._._n—_1_«._._.
_ __ __ _ __ __ _ __ __ w . __ __ _ __ __ . __ __
IIrLtFIFLIrLielrptrLlrLt» .m r IIhLiFIrLIrLlhIFFlrLIrLtk
__ __ __ __ _ﬂ __ __ _ d R _. __ __ __ __ _. __ _
_ __ _. _ __ __ _ __ __ e H _ _. __ _ ,_ __ _ __ __
IIrLlpIrLIrLtklrplrLlrLrp m g IlrLtFIFLLrLIhIFFtrLIrLIh
__ _ __ _ _. _ __ _ __ _ m f. .. __ _ __ _ __ _ __ __
. __ _ __ __ __ __ ._ _ 0 .1 0 __ __ __ ._ __ __ __ _ n
II+LI+ITLIr1+IT+tTLIrL1+ 0 .d I+IT+ITL1TLI+T+ITLITLI+ ﬂ
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II+LI+IT4ITLI+T+IT¢I+LI+ e w. II+LI+ITLITLt+IT+tTLI+L1+
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IIﬁJt4114thal14t141J4 g t IlﬁJt414I144114141144
_ __ __ . __ _. _ __ __ d d _ __ u. _ _. __ . __ __
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_ _ .w h . n. __ _ ,_ __ . __ _—
_ _ e W __ __ _. __ __ __ ___
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i, m h tIrIFFIFLtr4IrITLI L 5
. C _ _ _ _ _ _ _ _ _ .
0 m a __ __ __ __ ._ n
1 k II+LI+1TLITLI+IT+I
ﬂ 5 __ ._ __ __ __
k M _ ___ __ __ __
w 0 IIﬁIW4I1JI1414I14I
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w m _ ._ __l__I_t__
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C
( 1998 PHYSICS C—MECHANICS A graph of the total kinetic energy K for the twoglider system over the 2.00 s interval has the following shape.
K0 is the total kinetic energy of the system at time I = 0. KO) 0 0.50 1.00 1.50 2.00 (d) i. Is the collision elastic? Justify your answer. ii. Brieﬂy explain why there is a minimum in the kinetic energy curve at t = 1.00 s. GO ON T0 THE NEXT PAGE Mech. 2. 1998 PHYSICS C—MECHANICS A space shuttle astronaut in a circular orbit around the Earth has an assembly consisting of two small dense spheres,
each of mass m, whose centers are connected by a rigid rod of length l2 and negligible mass. The astronaut also has a device that will launch a small lump of clay of mass m at speed 110 . Express your answers in terms of m, U0 , l2,
and fundamental constants. I——9~——I
m.—.m
it
C
m (3) Initially, the assembly is “floating” freely at rest relative to the cabin, and the astronaut launches the clay
lump so that it perpendicularly strikes and sticks to the midpoint of the rod, as shown above. i. Determine the total kinetic energy of the system (assembly and clay lump) after the collision. ii. Determine the change in kinetic energy as a result of the collision. I—— 9 ————
m.——.m
Too
0 (b) The assembly is brought to rest, the clay lump removed, and the experiment is repeated as shown above, with
the clay lump striking perpendicular to the rod but this time sticking to one of the spheres of the assembly. i. Determine the distance from the left end of the rod to the center of mass of the system (assembly and clay .
lump) immediately after the collision. (Assume that the radii of the spheres and clay lump are much smaller
than the separation of the spheres.) ii. On the figure above, indicate the direction of the motion of the center of mass immediately after the
collision. iii. Determine the speed of the center of mass immediately after the collision. iv. Determine the angular speed of the system (assembly and clay lump) immediately after the collision. v. Determine the change in kinetic energy as a result of the collision. GO ON TO THE NEXT PAGE Mech. 3. 1998 PHYSICS C—MECHANICS Block 1 of mass m] is placed on block 2 of mass m2, which is then placed on a table. A string connecting block 2
to a hanging mass M passes over a pulley attached to one end of the table, as shown above. The mass and friction of the pulley are negligible. The coefﬁcients of friction between blocks 1 and 2 and between block 2 and the tabletop
are nonzero and are given in the following table. Coefﬁcient Between Coefﬁcient Between Blocks 1 and 2 Block 2 and the Tabletop
Static ,uxz
Kinetic ,un Express your answers in terms of the masses, coefficients of friction, and g, the acceleration due to gravity. (51) Suppose that the value of M is small enough that the blocks remain at rest when released. For each of the
following forces, determine the magnitude of the force and draw a vector on the block provided to indicate
the direction of the force if it is nonzero. i. The normal force N] exerted on block 1 by block 2 ii. The friction force ﬁ exerted on block 1 by block 2 1998 PHYSICS 'C—MECHANICS iii. The force Texerted on block 2 by the string iv. The normal force N2 exerted on block 2 by the tabletop v. The friction force f2 exerted on block 2 by the tabletop (b) Determine the largest value of M for which the blocks can remain at rest. (c) Now suppose that M is large enough that the hanging block descends when the blocks are released. Assume
that blocks l and 2 are moving as a unit (no slippage). Determine the magnitude 0 of their acceleration. (cl) Now suppose that M is large enough that as the hanging block descends, block 1 is slipping on block 2.
Determine each of the following. i. The magnitude a} of the acceleration of block 1 ii. The magnitude (:2 of the acceleration of block 2 STOP END OF SECTION II, MECHANICS 1 998 PHYSICS C
SECTION II, ELECTRICITY AND MAGNETISM
Time—45 minutes 3 Questions Answer all three questions. The suggested time is about 15 minutes for answering each of the questions. which are worth 15 points each. The parts within a question may not have equal weight. Show all your work in the pink booklet in the spaces
provided after each part, NOT in this green insert. E&M. 1. Note: Figure not drawn to scale. The small sphere A in the diagram above has a charge of 120 ttC. The large sphere BI is a thin shell of
nonconducting material with a net charge that is uniformly distributed over its surface. Sphere B, has a mass
of 0.025 kg, a radius of 0.05 m, and is suspended from an uncharged, nonconducting thread. Sphere BI is in equilibrium when the thread makes an angle 6 = 20° with the vertical. The centers of the spheres are at the same
vertical height and are a horizontal distance of 1.5 m apart, as shown. (a) Calculate the charge on sphere Bl.
(b) Suppose that sphere BI is replaced by a second suspended sphere 82 that has the same mass, radius,
and charge, but that is conducting. Equilibrium is again established when sphere A is 1.5 m from sphere 32 and their centers are at the same vertical height. State whether the equilibrium angle 62 will be less
than, equal to, or greater than 20°. Justify your answer. GO ON TO THE NEXT PAGE 1998 PHYSICS C—E 8: M The sphere 32 is now replaced by a very long, horizontal. nonconducting tube, as shown in the top view
below. The tube is hollow with thin walls of radius R = 0.20 m and a uniform positive charge per unit length of]. = +0.10 [AC/m. / Tube (radius R) Top View (c) Use Gauss‘s law to show that the electric field at a perpendicular distance r from the tube is given 1.8 x 103
l" by the expression E = N/C, where r>R and r is in meters. ((1) The small sphere A with charge 120 [JC is now brought into the vicinity of the tube and is held at a
distance of r = 1.5 In from the center of the tube. Calculate the repulsive force that the tube exerts on the sphere. (e) Calculate the work done against the electrostatic repulsion to move sphere A toward the tube from a
distance r = 1.5 m to a distance r = 0.3 m from the tube. GO ON TO THE NEXT PAGE E&M.2. 1998 PHYSICS C—E & M R1=IOQ S 3
Al
e=2ov C=15uF L=2.0H
R2=200 In the circuit shown above, the switch S is initially in the open position shown, and the capacitor is uncharged. A voltmeter (not shown) is used to measure the correct potential difference across resistor R1 . (a) (b) (C) (d) (6) (f) On the circuit diagram above, draw the voltmeter with the proper connections for correctly measuring
the potential difference across resistor R1. At time t = 0, the switch is moved to position A. Determine the voltmeter reading for the time
immediately after t = 0. After a long time, a measurement of potential difference across R1 is again taken. Determine for this later
time each of the following. i. The voltmeter reading
ii. The charge on the capacitor At a still later time t = T, the switch S is moved to position 3. Determine the voltmeter reading for the
time immediately after t = T. A long time after t = T, the current in R1 reaches a constant final value If.
i. Determine If. ii. Determine the ﬁnal energy stored in the inductor, Write, but do not solve, a differential equation for the current in resistor R! as a function of time t after
the switch is moved to position B. GO ON TO THE NEXT PAGE 1998 PHYSICS C—E 8: M E & M. 3. A conducting bar of mass m is placed on two long conducting rails a distance it apart. The rails are inclined at
an angle 9 with respect to the horizontal, as shown above, and the bar is able to slide on the rails with negligible friction. The bar and rails are in a uniform and constant magnetic field of magnitude B oriented perpendicular to
the incline. A resistor of resistance R connects the upper ends of the rails and completes the circuit as shown. The bar is released from rest at the top of the incline. Express your answers to parts (a) through (d) in terms of m, 9, 6,
B, R, and g. (3) Determine the current in the circuit when the bar has reached a constant ﬁnal speed.
(h) Determine the constant final speed of the bar. (c) Determine the rate at which energy is being dissipated in the circuit when the bar has reached its
constant final speed. ' (d) Express the speed of the bar as a function of time I from the time it is released at t = 0.
(6) Suppose that the experiment is performed again, this time with a second identical resistor connecting the rails at the bottom of the incline. Will this affect the ﬁnal speed attained by the bar, and if so, how? Justify
your answer. STOP END OF SECTION II, ELECTRICITY AND MAGNETISM ...
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This note was uploaded on 02/19/2011 for the course PHYS 102 taught by Professor Allen during the Spring '11 term at Northwestern IA.
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