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Unformatted text preview: 1993 Ihysics 6 Solutions Distribution
of Points Mech 1.
(a) 3 points Using the expression for the energy stored in a spring: U%kx2 lpoint
Substituting: l E 2
U 2 [400 “J (0.5 m) 1 point
U  50 J 1 point (b) 5 points For recognition of conservation of energy or workenergy theorem 1 point
Kinetic energy of block C: K  % mcvcz 1 point
work done (or energy dissipated by) friction: Wf  and 1 point
KU—Wf % mcvcz  U — nmcgd 1 point Solving for no: vc _. /;—2 (v  pmcgd) Substituting:
2
ye  /Z—kg (50 J  (0.4)(4 kg)(10 m/s2)(o.s no)
vc  4.58 m/s 1 point (Full credit also awarded for correct alternate solution
computing anet dx, where Fnet  kx — pFn, to find the
kinetic energy, and then computing the speed.) 1993 mysics C Solutions Mech 1. (c) (d) (Alternate Solution) UNITS: Distribution
of Points (continued)
3 points For any statement of conservation of momentum 1 point mcuc  (mg + mD)uf 1 point Solving for vf:
of  mcvc/(mc + up)
Substituting: Uf  (a kg)(4.58 m/s)/(a kg + 2 kg) 1 point Uf  3.05 m/s 3 points
I
The blobks come to rest when all their kinetic energy
has been dissipated, i.e. AKE  Work done by
frictional force 1 point
%(mc + mD)vf2  p(mc + mD)gd 1 pOint
Solving for d: d  “fa/2M; Substituting: d  (3.05 m/s)2/(2)(0.4)(1o m/sz)
d  1.16 m 1 point (Alternate Points) 2F _ ma (1 point)
Zr #(mc + mu»; 2 2
a  57 (me + mD)  pg  (0.4)(10 m/s )  4 m/s
U2  uoz + Zad (or other appropriate kinematic equations) (1 point)
vzvu2 o—(3 05M)?
d—'28—— 2(_4 m/s) V1.16m P013”
For correct units on all answers 1 point 1993 Physics 0 Solutions Distribution
of Points Mach 2. (a) 3 points F
kv °
"'3
1 point for F0 correctly drawn and labeled 1 point
1 point for ku correctly drawn and labeled 1 point
1 point for N and mg correctly drawn and labeled 1 point
(b) 3 points V
Fnet  ma 1 point
But Fnet  Fb — kv. therefore: 1 point
F0 " ICU  ma
Solving for a:
a  (F0 — kv)/m 1 point
(c) 5 points
a  g! 1 ' t
dt pain
Using the equation from part b:
dv (Fb  kv)
(1) dc  m
Rearranging and integrating:
dv 1
(2)  I E dt 1 point
Changing variables by letting u  F0 — kv, du  k dv:
1 du 1
__E I .5 _ I E d: K 1 point 1n(Fo — kv) — 1nC   g c, where C is a constant v  i [F0  Cekt/m ] A 1993 Ehysics C Solutions Distribution of Points
Mech 2. (continued)
(e) 2 points
Differentiating the expression for velocity from part (c)
to get the expression for acceleration:
dv F0 _ kt/m
 dt  k [(0 (k/m) e
F
m
a
9m
0 I
For correct initial value Fb/m 1 point
Fbr correct shape of curve 1 point (Alternate Solution)
From part (b):
a  i (F0 — kv) Substituting for u from part (c): Eh k Pb kt/m
“I‘m—(1‘9 >.
F0 a _ __ ekt/m
m 1993 Physics 6 Solutions Distribution of Points Mech 3.
(a) 4 points 1  IF 1 point 27  O (or rcw  rccw) 1 point Summing torques about the right end of the rod: Fa! — Mg[§]  0, where Fa is force exerted by axis 1 point Fag—a Fa is directed upward 1 point (Alternate Solution for last two points) (Alternate Points) Sum térques about any other axis and also use 2F  0. For example, summing torques about left end of rod: Mg[§]  Ft!  0, where Ft is force exerted by thread Ft _ Hg ' (1 point) 2F  0, so Ft + F5  Mg  0 FaMgFtMg—I‘g Frga F8 is directed upward (1 point)
(b) 2 points Using Zr  Ia and calculating the torque about the 1 point axis end of the rod: Mg %  M g: a Solving for a: a  g g 1 point 1993 Physics C Solutions Distribution
of Points Mach 3 . (continued) (c) 2 points Using the relation between translational and angular
acceleration: a  at 1 point Substituting r  g and a from previous part: hloﬁ £
2 p
I Nu ton»
on
O 1 point it
I (d) 3 points Using Newton's Second Law: 2F — "a 1 point
but 2F  Mg  Fr 1 point
so Mg  Fr  Ma Substituting 8  %g and solving for Fr:
1
Fr  3 M8 1 point (e) 6 points Using conservation of energy the increase in kinetic energy 1 point
of rotation Kroc is equal to the decrease in potential
energy AU AKrot: ' — Iwz 1 point ' Mg — Sin 6 1 point l Iw2  Mg g sin 6 2
Solving for w: o)  a? sin 9
Substituting I  mZ/a 3 .
w ' is 51“ e 1 point 1993 Physics 6 Solutions Distribution of Points
Mech 3. (continued)
(e) (continued)
(Alternate Solution) (Alternate points
Work done by gravitional force Wg equals the increase in (1 point)
kinetic energy of rotation Krot
Wg  I Mg it cos 9 d9 (1 point)
 m sin 9
2
1 2 .
AKrot  E Iw (1 pomt)
1 2 l
2 It.)  Mg 2 sin 9 Substituting I  1422/3 and solving for w: w  2% sin 9 (1 point) 1993 Physics 0 Solutions Distribution o£ Points
E 6: M 1.
(a) 2 points
If;
1 point for each correct vector 2 points
(If both vectors are reversed from correct directions, then
partial credit of 1 point awarded)
(b) i. 4 points
Gauss's Law:
§ EdA  Qencl/eo (or 4nernC1) 1 point
For 1' > R, using a Gaussian surface that is a cylinder
of radius r and length 1:
§ EdA  E(21rr£) 1 point
Qencl  p(1rR2£) 1 point
E(2me)  p(1rR22)/eo
2 2
E  FR [or 27”ch ] 1 point
_ Zeor r
(b) ii. 2 points
For 1' < K, using a similar Gaussian surface as above:
E(21rr£)  p(1rr2£)/eo 1 point
E  P—r— (or 21rkpr) 1 point 260 1993 thsics c Solutions Distribution of Points
E & M 1. (continued)
to) 3 points
‘_°’}
1 point for first correct vector 1 point
2 points for second correct vector 2 points
(If vectors are reversed from correct directions or if
circular lines of force with counterclockwise arrows
shown, then partial credit of 1 point awarded.)
(d) 4 points
Ampere's Law:
§ Bdl  p0 Iencl, where Iencl is the current enclosed by the 1 point
closed loop of integration (i.e. the current density
times the area)
For r < R, integrating over a loop of radius I:
§ Bod!  B(2wr) 1 point
Iencl  [;%Z] «r2  I g: 1 point 8(2wr)  #01 g: "OI r I‘OJr
B  —;— ﬁg [ or 2 J 1 point 1993 Physics c Solutions Distribution
v of Points E & M 2. (a) i. 2 points
Using the expression for the flux of a uniform field: §  BOA 1 point
Substituting:
¢  abBo 1 point (a) 11. 1 point Using the expression relating the emf and flux: d9
5"3
Both the field and area are constant, so E  zero 1 POint (a) iii. 1 point Since there is no emf there is no current in the loop,
and thus no magnetic force exerted on the loop. 1 point (b) 2 points 2 points When at  t/2, B  Bo cos «/2  zero, i.e. the field has been
decreasing, and is about to change direction. The induced
current will be in a direction to oppose this change, i.e. clockwise. (c) i. 4 points
Calculating the flux:
Q  abBo cos wt 1 point
Calculating the emf: .s  — 99 (negative sign not required) 1 point dt
 awao sin wt Using Ohm’s Law:
I  E/R 1 point R sin wt 1 point I 1993 Physics C Solutions E & M 2. (continued) (a) 11. 3 points A w. Fbr a graph showing a period of 2:
For a graph consistent with answer to (c)i. Fbr coriect orientation of current at n/Z consistent with answer to (b) (i.e., graph positive at n/2 if clockwise current
in (b); graph negative at u/2 if counterclockwise current
in (b)). (c) iii. 2 points The maximum value of the current occurs when sin wt  1. @030
Imax  R Fbr indicating the coefficient of the sin (or cos) term in (c)i. Fbr the correct answer (i.e. the coefficient in (c)i. must be
correct) Distribution
of Points 1 point
1 point 1 point 1 point 1 point ____________________________________________________________________________________ 1993 Physics C Solutions E & M 3. (a) (b) (d) 2 points The force due to the magnetic field provides the centripetal force that causes the ion to move in the semicircle. F  qv x B, so by the righthand rule the magnetic field
must point into the page (or in the z direction). For field being perpendicular to the page
For direction into the page or in the z direction
1 point Between the plates, the electric field must exert a force
opposite to that of the magnetic field. The magnetic force is to the right, and Felec  qE,
so the electric field should point toward the left. Therefore, plate K should have a positive polarity with
respect to plate L. 2 points Using the relation between the electric field and
potential difference for parallel plates: E  V/d Substituting: E  (1500 V)/(0.012 m) E  1.25 x 105 V/m 4 points For a particle to pass between the plates undeflected,
the forces due to the electric and magnetic fields FE and
F3 respectively must be equal in magnitude and in opposite
directions. FEqE EB  qu Therefore, qE  qu Solving for v:
u  E/B — (1.25 x 105 V/m)/(0.20 T)
u  6.25 x 105 m/s DiStribution
of Pbints 1 point 1 point 1 point 1 point 1 point 1 point
1 point 1 point 1 point 1993 Physics c Solutions E & M 3. (continued) (e) 3 points The centripetal force Fc is equal to the force qu due to the magnetic'field. 2
am
Fc'za— 2
an
R qu Solving for m: m  qBR/v Substituting: m  (1.6 x 10'19 C)(0.20 T)(O.SO m)/(6.25 x 105 m/s) 1!:  2.56 x 1026 kg (f) 2 points Substituting 2q into the force equation from part (e)
and solving for the new radius R': _ .22
2qB Substituting the expression for m from part (e): RI .__ti_ 3% _
R ZqB u 12/2 R’  0.25 m UNITS: Additional 1 point awarded if all units are correct Distribution
of Points 1 point 1 point 1 point 1 point 1 point 1 point ...
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This note was uploaded on 01/22/2012 for the course PHYS 2425 taught by Professor Park during the Spring '09 term at CyFair College.
 Spring '09
 Park
 Physics

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