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Unformatted text preview: 1983 Physics B Solullons (c) 5 points
The frequency of the light is _c_ 3x IO8 __
f—I—SX l0_7—0.6>< lO'SHz The. energy of a photon of thi hf= (4 x 10‘”) (0.6 x10”) the work function, so no phot ; ~
or If one sketched the graph in part (b), one can observe
that the frequency is less than the threshold frequency
of 0.75 x l0” Hz, where the graph intersects the xaxis. ns will be emitted. 1983 Physics C Solutions Mechanics Mech. l. (a) 9 points
i. The velocity vector is tangent to the path.
Since v, = constant, a,r = 0, and the
acceleration vector has only a positive
ycomponent as shown below. v “’1 0 x
ii. The ycomponent of velocity is v). = g;
 42 = 4X 3
By the cham rule, d! dx dt Sincey =lx2, %=x dx
and v}, =x7! = Cx iii. The acceleration is given by ay = so a), = C% = C 2 Alternate solution for ii and iii: ii. Sincey =%x2 and x = CI, y =%C2t2 42 2
Thusvy dt Ct
Since Ct=x,vy=Cx duv’ d iii. ay = —~ = ——(C2t) = C2 Distribution
of points 2 points 3 points 1 point 2 points 1 point
1 point
I point 1 point 1 point
I point
(Alternate points)
(I point)
. (2 points)
(1 point) (2 points) 1983 Physics C Solutions
Distribution of points
(b) 6 points
i. The speed is given by v = . /v,‘2 + vyz' 1 point
. d d dx .
By the chain rule, 0, = a); = i a? = xvx  l pomt
I 2
Sov=‘/vx2(l+x2 = C (1+x2)=C
l + x2 I point
ii. Again the velocity vector is tangent to the path. I point
Since the speed is constant, there is no component
of5’along the path, so E’is centripetal, perpendicular
to 3215 shown. 2 points
y
5’ v
S
0 X
Mech. 2. (a) 4 points
Support Force
T
T "'1"
mlg
On the block, the forces are weight = ng I point
and tension T I point
0n the disk, there are tension T 1 point'
and the pair: mg and supporting force 1 point
(b) i. and ii. 8 points
"On the disk, torque = F = Ia: = %m1R2az 1 point
with F=T~Rand lpoint
acceleration of the block = a = Ra, so a: = a/R 1 point
For the block, Em = mza 1 point
with Fnet = mg  T 1 point
Solving simultaneously gives 1 point
_ Zng d 1 ‘
a — ——2m2 + m! an pomt
T = mm”; 1 point 2m2+ml 1983 Physics C Solutions (b) iii. 3 points By deﬁnition. angular momentum L 4— [w
Angular velocity u; = 1! Since I= émle and z=%.  l 2 _ 2 , _ l
L—ZmlR R t—zmlaRt
Substituting the expression for a from part (b) gives
L __ mlngt — 2m2 + ml Mech. 3. (a) ll points ii. iii. Energy is conserved as the particle slides down
the sphere. so K + U = constant. where gravitational potential energy U = mgh.
If h is measured from the center of this sphere.
0+ng=K+ngcosO so that K = ng (1 cos 0). The centripetal acceleration is
v (1‘. = K
From part i:
émr2 = ng(l  cos 6)
.2 .
so at. =5 =2g(1—c050) '\ R
The only force with a ___‘ tangential component ""4 \\ mg .m u
is the gravitational I / \\
force mg. As the diagram I /
shows. its tangential I/W’ component is mg sin 0 me By Newton's second law, may = mg sin 9. so
a1 = g sin 0 (b) 4 points The particle leaves the sphere when the normal force N has decreased to zero. .
At that point the radial component of the weight provides the centripetal acceleration, so
mg cos 6 = mac
Therefore g cos 0 = ac = 2g(l  cos 0) and cos 0=§ _ 2
or 9— arc cos 3 Distribution
afpoims l point
1 point 1 point 2 points
I point I point 1 point 1 point 1 point 1 point 1 point 2 points 1 point I point 1983 Physics C Solutions Electricity and Magnetism E&M l. (a) (b) E&M 2. (a) 6 points
Gauss‘s Law states that (D = 5E . d3 = qgeniiosedl
0 Apply _it to a sphere of radius r (a < r < b).
Since I; is uniform and directed outward,
§E  dA = E  (area of sphere) = E  4an The enclosed charge is Q.
Therefore E = —Q— 4n£0r3 5 points The potential V0 equals the potential difference
between the spheres, which may be expressed as fed:
bdr Therefore V0 = Tge—J F5
0 (1 Evaluating the integral gives _ Q _l 1’
Vo4neol: r] 4 points .Substitution of the expression for V0 from part (b) gives C=_Q_=ﬂ“_
_Q_ ba ba o 47:60 ab 4 points Initially there is no potential drop across the capacitor, so
8 = [QR
and 8 = 10(10’6)(2 x 106) = 20 volts Distribution
of points 2 points 2 points
1 point 1 point 1 point 1 point 2 points I point 2 points 2 points 2 points
2 points 1983 Physics C Solutions Distribution
of points
(b) 5 points
Current 1' and charge Q are related by
. dQ . .
l: —d— or Q = 51d: lpomt
t
(1)
so that Q = fioe 6 dt 1 point
Evaluation of the antiderivative gives
(1)
Q = — 6i0e " + C 2 points Since Q = 0 when t = O, the constant of
integration C must equal ﬁg, and
" I Q = 6io<l — 6(7)) 1 point
(c) 3 points Q Qmax I 0 Q=0att=0 [point As t—+oo, Q —>Q,m 1 point Exponential shape 1 pomt
(d) 3 points For a charging or discharging capacitor, the time ' constant is RC. . I paint Here, the numerical value of the time constant is 6 seconds, so 2 points 1983 Physics c Solutions E&M 3. (a) 6 points
The ﬁeld at M is down. ’‘ / ’ \ . .
/ \ The path of integration
/ \ is a Circle around I.
_ \ Agipere’s Law states that
I \ §B . ai = no]
l 18) M Applied to the circular
\ <—R path, it gives
\ / B  2m = no!
\ / _ H01
\\ // 50 that B  m
\.___ z
. if
(b) 9 pomts
i. 3 points
At point 0, the B ﬁeld is zero
because the ﬁeld caused by the right current cancels the
ﬁeld caused by the left current. ii. 6 points Bresultant I®<._a—>.<——Ol a——>®I Each current contributes a ﬁeld vector of magnitude 3=__i101_
Zak/aZIryz The vector sum of the two is directed to the right.
Bresultant = 2 ' Bx =2Bc059 = 2  B  <——y—), giving the ﬁnal result
‘ /a2 + y2 poly , directed to the right. Bresultant = W Distribution
of points I point
1 point 2 points l point
1 point 2 points 1 point 1 point 1 point
1 point
1 point I point , I 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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