This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: SOLUTIONS 1990 Physics C Distribution
of points
Mech l.
(a) 4 points
For an expression of Newton's Law, Fnet = ma 1 point
Fnet = *kv 1 point For calculating the correct acceleration corresponding to the
initial velocity: mag = —kv0
kvb .
a0 = —;; 1 pelnt
For the correct direction, indicated by a negative sign or
the words "to the left" 1 point
(b) 6 points
For the recognition that a = 3% 1 point
QB — E3 1 oint
dt m p
93 =—5 dt:
v m
Upon integration,
£n v = — Et + 2n C
m u = Ce'kt/m
v = v0 at t = 0 gives 0 = no
u = voe'kt/m 1 point L! ”o 0 z —l)0 For showing v = v0 at t = O 1 point
For a curve reasonably representative of exponential decrease 1 point For the curve asymptotically approaching zero as t
approaches infinity 1 point 1990 Physics C Distribution of points Mech 2.
(a) 4 points For use of conservation of energy 1 point 1 mgH = 2mu02 For correct expression for potential energy 1 point For correct expression for kinetic energy 1 point ‘hz . H = —EE 1 paint Alternate solution (Alternate points) a = —g sin 9 (1 point) v2 — uoz = 2ad (or equivalent relevant kinematics) (1 point) (d is the distance traveled up the incline) d = H/sin 6 (1 point) v2 — yo? = —2(g sin ”(H/sin a) = 2gH v = O at highest point Ibz . H = '2? (l p01nt) (b) 5 points For relating kinetic energy K, potential energy U, and work done
by frictional force Wf: K = Wf + U Wf = Ffd Ff = p mg cos 6 d = h/sin 0 Therefore, Wf = (pmg cos €)(h/sin 0) %mv02 = (pmg cos 9)(h/sin 0) + mgh — mgh(p cot 0 + 1) U02 H h = 2g<u cot 0 + l) = p cot 0 + 1 F‘ P‘ P‘ P‘ point
point
point point point 1990 Physics C Distribution
of points (c) 4 points u = $5 = voe'kt/m 1 point Upon integration, x = v0[— %]e'kt/m + C
muo
x = 0 at t = 0 gives C = _2— nm
_E9 [l — e'kt/"J 1 point
(These points were awarded if a correct integration was performed using an incorrect expression for u) X For a reasonable concave—down curve 1 point
For X = 0 at t = O and approaching an asymptote as t approaches infinity 1 point (d) 1 point
For t —~ m, e'kC/m —* 0
mo X = T 1 point 1990 Physics C (C) Distribution of points
Alternate solution (Alternate points)
F = mg sin 9 + pmg cos 0 (one point for each term) (2 points)
a  —g(sin 9 + p cos 6) (1 point)
I)2 — v02 = Zad
d = h/sin 9 (1 point)
—v02 = —2g(sin 0 + y cos 0) h/sin 9
1’02 sin a H sin 0 H .
h _ 2g(sin H + p cos 9) = sin 0 + p cos 9 ~ 1 + p cot 6 (l p01nt)
4 points
For including both translational and rotational kinetic energy
in an equation for conservation of energy:
Ktrans + Krot = méh' 1 point
1 2 .
Krot 51w 1 p01nt
I = 121R2 1 point
w = 3
R
l 2 l 2 El 2 .
2mv0 + 2(mR )[ R = mgh
%mv02 + %mv02 = mgh' 1 point
2
h' = 1—)”— = 2H
g
Alternate solution (Alternate points)
For an expression relating torque to angular acceleration: ET = Ia (1 point)
Taking the torque about the point of contact with the incline:
(*mg)(R)(sin 9)  1'0: 1’ = I + me? = 2an2 (1 point)
a = E (1 point) (This point was awarded only if some indication was
present that the point of contact was used as the
reference point.) 1990 Physics C Distribution of points
. 28
—ng Sin 6 = ZmR E
a = ~§ :ln 9
v2 — v02 = Zad
. h' I
—v02 = —g Sln 6 sin 0 = —gh
2
h' = 39— = 2H (1 point)
g
(d) 2 points
Rotational kinetic energy does not change. Therefore,
1 n .
Emvoz = mgh l paint
ll U0 2 '
h — 2g ~ H l pelnt
Full credit was awarded for merely saying that the
answer is the same as in part (a).
Mech 3.
(a) 2 points
F = kAX 1 point
= ﬁg = (8 kg)(9.8 m/sz) = .
Ax R (1,000 N/m) 0.078 m g: 0.08 m l p01nt
(b) 3 points
For any one of the following equations:
1 k k m .
f = E; E or w — m or T — 2w]; 1 p01nt For correct substitution in any of the above: _ 1_ 1,000 N/m _ 1,000 N/m _ 8 kg .
f ‘ 2n 8 kg 95 w ‘ . 8 kg 95 T " 2“ 1,000 N/m 1 p01nt
1 1 .8 s’ or w = 11 s'1 1 point ll f (The third point was not awarded for T = 0.56 s) 1990 Physics C Distribution of points
(c) 2 points
If the spring pulls the 5—kg block downward such that a > g,
the 5kg block moves faster than the 3kg block can fall,
and they will lose contact. Therefore
amax = g = 9.8 m/s2 2 points
Mathematically:
For 3kg block: 3g ~ Fc  3a, where Fc is the contact force
Fc  0 when the blocks lose contact, so
amax " g
(d) 3 points
Maximum acceleration occurs at the extremes of motion,
when x = A, the amplitude.
amax — 5% or wZA 1 point
For using amax obtained in part (c) 1 point
A .. "Hmax a (8 kg)(9.8 m/sz) or amax .3 9.8 rugs?
k (1,000 N/m) —— wz (11.2 s' )
A = 0.078 m (or answer consistent with part (c)) 1 point
(e) 4 points
For use of conservation of energy, Kmax = Umax 1 point
1 2 .
Kmax = Emu,“ax l paint
1 2 .
Umax = EkA l p01nt
kAZ
2 = __._
Umax m
= ’(l,000 N/m)(0.078 m)2
”max 8 kg
umax  0.87 m/s 1 point
Alternate Solution (Alternate points)
%mx . . . .
vmax = wA or w , for Simple harmonic motion (2 p01nts)
2
um = (11.2 s")(0.078 m) 93 W (1 point)
”max = 0.87 m/s 2; 0.88 m/s (1 point)
1 point was awarded for correct units in at least three answers 1 point 1990 Physics C Distribution
of points E & M 1.
(a) 5 points For an expression of Gauss’s Law 1 point
q
fE‘dA = em:
6('1
4 3 , .
Volume of a sphere = Ewr I paint
Q3113 Qr3
<lent: = 4 = R3 1 point
—7I'R3
3
(One of these last two points was awarded for an erroneous
calculation which indicated a recognition that the enclosed
change depends on r)
Surface area of a sphere = Anrz 1 point
3
2 a, Qr
E4nr 60R
Qr ' .
E W 1 p01nt
(b) 2 points
qenc = Q 1 point
Ednrz = Q_
60
E = _—Q——2 1 point
Aﬁéor
(Second point was not awarded if R was used instead of r)
(c) 2 points
2R < r < SR is inside the conductor, therefore:
2 points E = 0 (Some students continued to treat this shaded area as an
insulator. For an attempt to solve part (c) in this
manner, evidenced by the use of r3 — R3, 1 point was awarded) 1990 Physics c Distribution of points
(d) 4 points
Since E a 0 for r inside the conductor, total charge
enclosed by such a sphere must be zero.
For an indication that the total charge on the inside
surface, 41: has magnitude Q 1 point
For indicating that this charge is negative 1 point
QI .
a = Area I paint
For consistent use of formula for area 1 point
a = ___7‘Q = __Z‘Q
4w(2R) l6nR
(If student indicates that q = O, 1 point is awarded
for indicating a = 0)
(e) 2 points
For r > 3K, E depends on total free charge enclosed,
i.e., Q +3 q
Since qenc = 0 inside the conductor, the net charge go on the
outside surface is:
go =Q+ q lpoint
For consistent use of formula for area 1 point
_ Q + _ Q +,
" 41mg)? ‘ 361ng
E & M 2.
(a) 1 point
For a correct arrow (T) or a statement indicating the
correct direction 1 point
(b) 3 points
Speed as particles enter region III equals speed at which
they travel through region II in a straight line
Felec  QE 1 point
Fmag = QuB 1 point
QvB  QE v = E/B 1 point 1990 Physics C (e) (d) (e) 3 points Region III is, in effect, a mass spectrometer, in which
equating centripetal force to magnetic force allows the
determination of mass mvz Fcen _ R 2 m 3 QBR
U QBZR m E 4 points Distribution
of points The accelerating potential brings the particle to the speed at which it moves through region II. Energy imparted by accelerating potential = QV lz
K — va
_ l 2
QV 2mu
V = mu2/2Q H
r—\
0
m m
N
t_iE
r~—\
WIN
N N :0 RE
V=2—
2 points
U2
a acen = ﬁ
= l E 2
R B
E2
‘3:le Alternate solution = QBE
m [5] ea E2
EEZ m
H (Alternate (l (1 point point point point point point point point point Points) point) point) 1990 Physics C Distribution of points
(f) 2 points
t _ distance
v
= ZWR 2 = £5 1 point
v v
E
 «R B
«RB .
t a _E_ 1 p01nt
(For erroneous solutions, one point could be earned if
n appeared in some sensible fashion that indicated an
attempt to utilize the circular path)
E & M 3.
(a) 2 points
For arrows or words indicating a clockwise direction 2 points
(b) 3 points
Fgravity ’3 Mg 1 point
Fmag  I£B 1 point
g as 
I 2B 1 pelnt
(c) 2 points
6 = IR 1 point
= MgR .
5 28 l p01nt
(d) 2 points
find = d¢/dt where is the magnetic flux 1 point
find a Bﬁv 1 point
(e) 3 points
I’ = Enet/R 1 point
= (E — find)/R 1 point
= ﬁg — 2&3 1 point Bi R (Two of the three points were awarded for calculating the
induced current and not the net current) 1990 Physics C Distribution of points
(f) 3 points
Once the loop reachs terminal velocity, the net force on it
is zero:
(M — Am)g — 1’28 1 point
ﬁg _ B£v
[M —R ]M
2 2
a Mg  B 2 U 1 point
R
2 2
Am = B ﬂ v 1 point ...
View
Full Document
 Spring '09
 Park
 Physics

Click to edit the document details