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Unformatted text preview: 1985 Physics 0 Solutions Mechanics Distribution
of points
Mech. I. (a) 5 points
"or = (50 m/s) 5i" 37. = (50 "1/5) (60) = 30 m/s 1 point
y =yo + toy! + %at2 3' point
0 = (35 m) + (30 m/s)! — g (10 m/sz);2 2 points i 1 point for y = O. I point for correct substitutions on
right—hand side) For corretion solution. I = 7 s 1 point
Full credit was given for alternate equivalent solutions. (b) 3 points R = 11),: I point
R = (50 ms) (cos 37‘) (7 5) (Le, for correct substitution) 1 point
For correct solution. R = 280 m I point (c) 5 points 1:4 = PM = 40 ms 1 point
1'5 = l'o = 50 m S 1130111!
For [(2 For any mention of energy I point For conserving energy I point I ~. I s
; mrc' = ; mro' + mgh — 2
PC _ V, r0 + :gh = V (50 ms)‘ + 2H0 m s)(35 m) = v/3200 m‘,’sz rc = 56.6 m 's 1 point
Full credit was given for alternate equivalent solutions (e.g.. if time of travel from point B to point C is known from part (a). kinematic equations can be used) (d) 2 points
For any use of momentum or center of mass 1 point
.HIO kg) = (30 m) (6 kg)
x = 18 m I point
Mech. 2. (a) 5 points
f1 = “r‘v l POint
f, = mg sin 0 or w sin 0 I point
.‘V = mg cos 0 or w cos 0 I point
_ L — m2  
u, — N  mg cos 9  tan 9 2 pomts
Alternate solution using F = ma: /.4/Iernale points) '
.V  mg cos 0 = 0 (1 point)
If, — mg sin 0 = ma (I point)
f, = u,.V (I point)
u}mgcosBmgsin9=ma
=mgsinlﬂwnaz a a
11, mg cos 6 tan 0 + g cos 0 (l pomt)
u = 0. therefore it, = tan 0 (1 point) 57 1985_Physics C Solutions Distribution
of points
(b) 5 points
AE = AK + AU
The initial and ﬁnal points are at the highest point
on the incline and the point of full spring compression.
respectively.
AK = 0 I point
AL' = til/gm"), + AUSW,g
Ab’gmuy = mgh = mg(d 1 x) sin 9 2 points
Amp“g =  £16K: 2 points AE = mg(d + x) sin 6 — %k.\:2 I point was deducted il'the student did not write down
the ﬁnal expression for AE. 2 points were awarded if the student tried to solve this
part using frictional force and showed
AE =j;(d + .v) or AE = WNW + x) or AE = pkmg cos 6(d + x) [a (c) 5 points ' Wm = A5 =ft(d + x) I point
fk = ilkV I point
titAV“! + x ) = mg sin 0 (d + x) — ékﬁ 2 points 1 point was deducted if the factor of(d + x) was missing
from left hand side. .V = mg cos 0 mg sin 0 (d + x)  ékx2
“k = mgcosBldﬁx)
__ , _ kx2 .
pt  tan 8 2—~——————(d + x)mg c059 . or I pomt
_ kx:
“k _ “i _ 1(d + .r )mg cos 9 Since M < u.. the minus sign in the above expression
is important. If the signs were interchanged in this
expression. I point was deducted. 1985 Physics C Solutions Mach. 3. (a) 4 points T I T2
2m
mg 2mg For indicating tension vectors
For indicating weight vectors For indicating different tensions on each block
For indicating different weights for each block (The last 2 points were also awarded if such indications
were made in the solution of subsequent parts.) (b) 5 points
i. For applying F = ma
T.  mg = ma
2mg  T2 = 2ma
The sign of the acceleration is arbitrary in these
two equations. ll. [1 = (T: "' Ti),
I point for torque = (at
I point for torque = (T;  T.)r (c) 3 points
a
1 = " ,
For using the system of three equations obtained in part (b)
Tl = m(g + a) T;=2m(ga) T;T,=mg3ma 2% = (mg  3ma)r [tr1 =gmr a=g ‘OIN (d) l point
Ti =m<s+a>=mtg +§gi = 34mg
Full credit was awarded for answer that was consistent with
previous work. Distribution
of points I point
I point
I point
1 point I point
I point
I point 2 points I point I point I point I point 1985 Physics C Solutions Distribution
of points
(e) 2 points
N = 7mg + T1 + T; or a clear indication that the normal
force is different from the static weight 1 point
r = 2m — 3 —  m
2 g 93 9 g
N=7mg+%mg+195mg=%§mg lpoint
Electricity and Magnetism
E&M l. (a) 4 going
IE  dA = q to or 4nkq 2 points
E(2n:rL) = gig, or 4nkq I point
= _3__ 2_k9 
E lncorL or rL ! pomt
(b) 5 points
.b—e _.
1/ = 'i E  dE (V = [Edi was acceptable) 2 points
b
q dr .
=   l t
J; 22:501. r P°'"
_ q a .
V— — 2n£0L [In rlb lpomt
= __q_ ‘ ’2 1181 '2 
V MOL 1%) or L I a I pomt
No deduction was taken for reversal of limits of integration.
(c) 3 points
_ 1 ,
Co — V 2 pomts
_ ZmOL L .
C0 ‘ ln (b a) 0' 2!: Who) ‘ P°"“ I point was deducted if the value of C 0 was negative. as
would be the case if the expression contained In (a ,b). (d) 3 points
C = C1 + C; or the capacitors are parallel I point
C=k%Co=§Co lpoint
C2 = gco I point
c = g Co Full credit was awarded for correct ﬁnal answer. regardless
of the method used by the student. 1985 Physics c Solutlone E&M 2. (a) 3 points
V = R or equivalent (may appear here or in part (c))
For correct substitutions. V = 3000 V and R = 5 x 105 9
For correct answer, i, = .006 A (b) 3 points
V7 0 For starting curve at the origin
For showing V2 asymptotic to a positive ﬁnal value
For showing V2 monotonically increasing and exponential (c) 2 points i, = i; or % = 153—202Tv5 or a statement indicating the series nature of the circuit in this situation
For correct answer. 1';  .002 A (d) 2 points
Q =CVand/or VZOOOV
For correct answer. Q  .0l C (e) 2 points
U  % QV or equivalent For correct answer. U  [0 1 Distribution
of points 1 point
I point
I point I point
I point
I point I point I point I point
I point I point
I point 1985 Physics 0 Solutions 3 oints  . . .
(n P ,2 Distribution
of points For nonzero initial value 1 point lpoint l
' For showing 1'; asymptotic to zero
hing the ﬁnal value For showing 1': monotonically approac
and exponential. 1 point
l point was deducted if parts (a) and (c) were both correct
but no work was shown. '
l point was awarded if at least one graph showed an
exponential form and no other points were available.
E&M 3. (a) Spoints
C = — ‘18? (negative sign not required) 2 points
(I) = BA 1 point
«19 , 42 4
d1 d1 ‘
A = m3 I point
8=‘f—g 1tr2=(60'l"s)n(.5m)z = 1511 v=47v lpoint
(b) 4 points _.
For showing E directed to the left at point P on diagram 1 point
8 = Ed l point
= .nr l point
__ § _ _§_ ‘_ 47 v __ .
E—d_2nr_2n(.5m)—‘5v’m lpomt
(c) 3points
F = '1}: I point
I“ = my ' I point
Thus ’1"; = qtB
 19 — i
l.6 x l0 C .5 m l0 T = 8‘8106mgs ‘ point FEE:
m 9.11 x 10‘" kg 1985 Physics C Solutlons (d) 3 points
F = ma
F = (15
Thus mg = (15 19 ,
a=LE=1L6xlo CﬁlSVm2=2.6xlolzm/sl m 9.11x10'3’kg 1 point was awarded for merely stating that a = dv d: Distribution
ofpoints '1 point
1 point I point ...
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 Spring '09
 Park
 Physics

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