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Unformatted text preview: 1986 Physics C
SOLUTION 2 F = 0
2T = (80 kg + 20 kg)(10 m/sz) = 1000 N
T é 500 N
(b) 4 points
Again considering the forces on the person and platform,
2 F = II
2T  1000 N = (100kg)(2 m/sz)
1 pt. 1 pt. 1 pt.
T = 500 N
(c) 5 points
Considering the forces on the person only, 2 F = no
Normal + 600 N  (80 kg)(10 m/szl = (80 kg)(2 m/sl)
v W W \W‘/
1 pt. 1 pt. 1 pt. 1 pt.
Normal = 360 N
(d) 2 points
work ugh
P‘s ETEE = t = (I9)V P = (1000 N)(O.4 m/s) = 400 N Distribution
of points 1 point lJ points 1 point (4 points 1 point points 1 point point 1 point Physics C
Mach 2. (a) 6 points P.E. = H.E. 1 point
th = %Hv= + %Iu= 7 points
Ly. W 1 pt. 1 pt. w = v/R 1 point th = =Hu2 + %[§HR=][§] . 7 .
2. any. «0 éI'ILI. = Wye
_ 10
v — ~7gh
‘ r = luv= = §M h 1 oint
1 ‘trans g 7 9 P
.. , _ 1 . _ 1 2 _ 2 .
11. 'ert — Elw  5H1! — 7149’? 1 pmnt
(b) b polnts
i. v3 = 235 1 point
_ g: = (10/7) gh = g A .
a ' 2: 2(h/sin e) 7 9 5‘" ° 1 9°‘nt
Alternate Solutions
Method 1: (Alternate Points) Summing the forces parallel to the plane, 2 F = ma (1 point)
Hg sin 9  f = Ha
  £1 2 Mg sin 6  gﬁa  Ha sin 9 (1 point) iwj Physics C
mean 2, (b) i. (alternate solutions continued) Method 2:
Summing the torques about the point of contact, 3 P = In (1 point)
_ (Hg sin elk _ g sin 9
a — — ' 7
MR3 + EMR2 ;R
o .1
a = GR .3 = “7’9 sin e (1 point)
ii. P = In 1 point
p = fR 1 point but a = % ’ 1 point
2 2 .
so f  §HR% = §H[gg sin 9)
f = §Hg sin 6 1 point
Alternate Solution (Alternate Points)
2 F = ea (1 point)
Hg sin 6  f  Ha (1 point)
F=Hgsin9H[.$79sin e]
(2 points)
f = §Hg sin 9
(c) 1 point
K“H  agh 1 point
(d) 2 points
The rotational kinetic energy 0‘ the hollow sphere i! 1 point greater than for the solid sphere.
because the moment of inertia is greater. 1 point Mech 3.
(a) 4 points
r =  %% g; u =  I F(x)dx 1 point
A
u = ~1' (kx3) dx 1 point
0
_ k.“ A .
U — 3— 10 1 point
 I’
U = f: 1 point (p) 4 points From conservation of energy, maximum kinetic energy
equals potential energy at point of release % Hyman: = % kA‘ 2 points
Vt
"max: = 1 point
,, k .
vnaI f= A ~ 2, 1 paint
(c) 4 points
From conservation of energy, E‘o‘ = K + U
when R = U , E‘o‘ = U + U = 2U
_ l
U ' Egtot
..b . t
k: = % [kg 1 3 points
A .
x = ¢‘/‘ 1 point
.(11) 3 points
The period of oscillation decreases 1 point
For a spring that obeys Hooke's law, F c x, and the
period is independent of amplitude. For the springsin
this problem, F c x', so the force and hence acceleration
increase at a greater rate with displacement than for the
Hooke's law spring.
2L 2 points
For a Hooke's law spring v”M x A, and
T a A I constant.
Vac:
For the spring in the problem, from part (b),
v.0' I A3, and T c s c —%— Physics C
534” 1 . (a) (b) (c) (d) (e) (f) 7 a points All vectors drawn with correct sense (higher to lower
potential) All vectors drawn perpendicular to equipotential lines For showing vectors at all three points (1 point for
missing vectors) 2 points
Magnitude of electric field is greatest at point T, because equipotential lines are closest together near T. (Note: Some students misinterpreted the question to mean
at which of the points L, N, and V, from part (a), is the
field greatest. For these students, an answer of point U received credit.) 4 points
='AZ
E Ax For correct substitutions
AV = 10 V
.02 m 500 V/m Ax=
E: 2 points V — V = 40 V  5 V = II 3 35 V (Note: deviation of $2 in the answer was acceptable)
3 points H = «W H = (S x 10": C) (40 V  30 V) H = S x 10‘** J 1 point No, 9; Same, 9; same as (a),
9: Work does not depend on path point point point point point point point
point point points point points point Physics C EtM 2.
(a) 4 points‘
V '= IR 1 point
For correctly obtaining 20 n as the resistance of the 1 point
combination of 10 ﬂ and 30 n resistors
RT = 5 9 +—20 ﬂ = 25 Q 1 point
 §§_V a
IT ' 25 a 1 A
IF = ézr = % A 1 point
(D) 2 points
1R = % A’ 9: same as in part (a) 2 points
(c) 4 points
a = CAV 1 point
AV = 10V 2 points
0 = 100 uC 1 point
(d) 4 points f
In the static situation, there is no potential difference
across the inductor; the situation is the same as if the
inductor were a resistanceless wire.
.1. = 1 + 1
R,, 10 Q 30 ﬂ
1 point
RI‘ = 7.5 9
RT. = s + 217.5) a 20 n 1 point
 '5‘ v 5 .
IT  33 a = z A I paint
1 = 31 = l§ A = o 9375 A 1 oint
R 4 T 16 I ' p (e) 1 point The current through each 30 R resistor is glr. Since 2 I I 0 at point A, 3 1 1 5
1L = 1,,  IR  317  317  5[; A] 1L 2 g A  0.624 A 1 point Physics C
Ean 3. (a) 3 points f B 0 d2 = uol, where I = :7
EA]. .. .
B = 2',  points
= unﬁt _ .
B zﬁr 1 point
(b) 1 point
Induced current in loops is counterclockwise 1 point
(c) 7 points
{3 = I B ' d5 2 points
b+a
= u; .
{B I 2', bdr 1 point
a
Ib a+b .
13 = 23;— 1n [ a ] , I = at 1 point
8 = N 9% (Note: sign ignored in grading. Magni 1 point tude of 8 was all that was required.) e = (1) § [ leQi—E ln [ azb ] ] 8 8 E3%2 ln [ ég ] 1 point . E cb a+ . 1 8 E 3‘ E§;§ ln [ a J I paint
(d) 1 point Force on the loop is away from the wire. However, the 1 point point was awarded only if (d) was consistent with (b).
If (h) was incorrect (i.e., clockwise); but (d) was
consistent (*orce toward wire), credit was given. (e) 3 points F I 12x3 ' 1 point
For recognition that F,., I .F.  F... 1 point
rn,‘  1:8,  1t8,+.  12(3,  a...) For correct substitutions for: I, §ron part (c)
B, from part (a) for r  a and r  b + a 1 point
z. using t I length of side of loop I b rm 1"[‘?]][%:1—b] ...
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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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