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Unformatted text preview: 1994 Physics C Solutions Distribution
of Points Mech l.
(a) 3 points For the correct expression for the energy stored in a spring: 1
U = 5—ka 1 point
For substituting the correct values:
1 U . 3(200 N /m)(O.4 my  1 point For the correct answer (including units): U = 16 J 1 point
(b) 3 points For applying conservation of energy 1 point Kinetic energy of ball/block combination = energy stored in spring For a correct expression for kinetic energy 1 point 1 2 3‘le = U v1  .IZU/M = .J' 2(16J)/(2.l kg) For the correct answer. u1 = 3.9 m/s lpoint
(c) 3 points For applying conservation of momentum 1 point For a correct expression for momentum p 1 point mv = le u  le /m = (2.1 kg)(3.9 m/s)/(0.l kg)
For the correct answer:
v = 81.9m/s lpoint (d) 2 points For indicating that the maximum compression of the spring will be less than 0.4 meters. 1 point For a correct explanation referring to the motion of the blocks or their kinetic energy 1 point Some examples:
The 0.8kg block will begin to move before the spring can compress as much, due to the force that the spring exerts on it .
The kinetic energy of the center of mass of the system in this case is nonzero, so some of the initial energy must remain as kinetic energy. Less energy is available as potential
energy, so the spring compression is less. Distribution of Points Mech 1. (cont)
(e) 4 points Two principles must be applied to calculate the velocity of the 0.8 kg block at the instant the spring is no longer compressed: conservation of momentum and conservation of energy. For indicating conservation of momentum is applicable 1 point le  Mv2 + sza For substituting: (2.1 kg) (3.9 m/s) [or 8.2 kg  m/sz]  (2.1 kg)v2 + (8 kg)v3 1 point For indicating conservation of energy is applicable 1 point 1 z 1 2 1 2 3M1)l  5M1;z +3M2 Us For substituting: 1 1 1 2(2.1 kg) (3.9 m /s)2 [or 16 J] = 3(21 kg)v22 + E(8 kg)v32 1 point The equation 3.9 m/s = 123 — v2 or — (v2  v3),whichcanbeobtained from the two equations above, was accepted as a substitute equation, but only if one of
those equations was also present. ' ——_—————___—______—_—__—__ Mech 2.
(a) 4 points Ktolzl=K nal+K malaria rotational For a correct expression for translational kinetic energy
For a correct expression for rotational kinetic energy 1 2 1 ‘2
Km]  Emu + 31w . . v
For substituting w = —r i 2 i E 2 v2 7 2 + 2(5mr){,2)— = 775(25kg)<10m/s)2 K‘om a In the absence of other credit, one point was awarded for indicating that the total kinetic
energy is the sum of translational and rotational kinetic energies. (*0 .
1. 4pomts For an indication that conservation of energy is applicable
K m, (leaving incline) = K tom (bottom of incline) — U For correct use of U = mgh
Tami): = K ,°Q(bottom of incline) — mgh = 1,7501 — (25 kg)(10m/sz) (3 m)
a 1,000J v 1/170(1,000 J)/(25 kg) For correct substitutions
v  7.56 m /s Full credit could be earned for this part if rotation was omitted both here and in (a). If rotation was included in (a) but omitted here, or if translation was included in (a)
but omitted here, a maximum of 2 points could be earned. 1 1 .
The statementfmu2 a mgh or mgh = é—mvz + 310:)2 was awarded 1 pomt.
The statement vf2 = v02  2g}; was accepted as equivalent to the energy equation
without rotation. Distribution
of Points 1 point
1 point 1 point 1 point 1 point 1 point 1 point
1 point Mech 2. (cont) (b) (cont)
ii. 1 point As it leaves the top of the incline the sphere’s velocity will be directed parallel
to the incline, i.e. at 25° above the horizontal.
For any clear indication that student realized this fact (c) 4 points For a correct kinematic equation for the vertical motion, indicating use of the
component of the velocity . r 1 2 2
y, = y0 +voyt + 3th (or vy = 120, — 2g(y — yo»
For recognizing that y f  y0  —3 m, or otherwise correctly accounting for the difference in height
Substituting the appropriate values: )2, a 0,):0 = 3m, on = 7.56m/s(sin25°),g  —10m/s, and solving the quadratic equation:
12 = 1.16 s,  0.52 s, where the positive solution is the applicable one. (Separate calculations for appropriate intervals could also be done, and added to
obtain!2  1.16 s.) ' For a correct equation for the horizontal distance, indicating use of the component
of the velocity x = vmt  vocos 25°t = (7.56 m/s) cos 25° (1.157 s)
x  7.93 m (d) 2 points
For indicating that the speed of the sphere as it leaves the top of the incline would
be less than in (b). For a correct explanation For example:
All of the potential energy gained is at the loss of the translational kinetic energy. Distribution
of Points 1 point 1 point 1 point 1 point 1 point 1 point 1 point Distribution
of Points Mech 3. (a) 3 points Using the general relationship between potential energy and force:
U  I F dR f
GM,m
U  I R, dR
For correct limits on the integral 1 point For an expression in which the force is proportional to 1 /R2 1 point
For having the remainder of the force expression correct 1 point G M m
1 point was awarded for the expression F = R; in the absence of any other correct work. (b) 2 points The total energy of the satellite at pointA is the sum of the kinetic and potential energies. 1 2 GM em
E A = —2—mvo — a
For a correct expression for the kinetic energy (with v0) 1 point For a correct expression for the potential energy (with a) 1 point (c) 2 points For a correct expression for angular momentum:
LIw ORmvr ORpxr ‘ lpoint For correctly substituting v  v0 and r a a 1 point L  mvoa (d) 3 points For any indication of use of conservation of angular momentum: 1 point For correctly substituting velocities and radii 1 point muBb a mvoa For the correct solution:
voa ” " b 1 point B Mech 3. (cont)
(d) (cont) (Alternate solution)
For any indication of use of conservation of energy
For correctly substituting velocities and radii GM‘m GM‘m mvo fmvB  For the correct answer (i.e. solving the equation for v or v2) v3  [002 + 26M,[ ll” In a circular orbit, the centripetal force is provided by gravity.
For any expression of Newton’s second Law, F  ma (e) 3 points For the correct expression for the centripetal acceleration, a  — mvz GMem For the correct answer, with r a a
GMe ’U (0 2 points The work done is equal to the change in the total energy of the satellite. Since the potential energy remains constant, the change in total energy is equal to
the change in kinetic energy. For any indication that the work equals the change in kinetic energy W  AK  —m
For the correct answer with all substitutions made: W——m (.2  1.02) . GM, 2
_v0 Distribution
of Points (Alternate points)
(1 point)
(1 point) (1 point) 1 point
1 point 1 point 1 point 1 point Distribution
of Points
E & M 1. (a) 4 points For a correct expression for the potential: V: 1 J‘dq 1 Q ._._.. or — 1 point
r r
4m: 0 41:80 Since all of the charge islat the same distance, simply substitute the total charge and the appropriate distance.
For correct expressions for calculating the total charge: Q = M 1 point
2 == 23R lpoint
1 23R A
41teo R
For a correct answer: Va A
V = —— OR Zak?» 
280 1 pomt (b) 2 points The electric ﬁeld from each inﬁnitesimal piece of the ring is canceled by the ﬁeld
from a piece on the opposite side of the ring. '
Therefore, E a 0. 2 points (c) 2 points The total charge on the rod is simply the charge per unit length times the length of the rod. Q  h! For using the correct expression for the length of the rod: Z : 2RB 1 point For the correct answer: Q " 29m 1 point
(d) 2 points Once again, since all of the charge is at the same distance, simply substitute the total
charge and the appropriate distance in the expression for the potential.
For using the answer to (c) for the total charge 1 point 1 20ml 43:20 R
For the correct answer. 6}» .
V = OR 21(9)» 1 pomt
2m: 0 E & M 1. (cont)
(e) 5 points For a correct expression {or the electric field: 1 d l
E = f ~—qz— or £2
43:30 43tso R
Since all the charge is at the same distance, calculating the electric ﬁeld requires
integration over the angle only. Also, the vertical components of the field cancel, so the integration can be limited to the horizontal components. dq  M? d4)
For incltiding the cos 4) term in the horizontal component
9
E 1 AR 4) dd)
= —2— cos
43tso _9 R
1 9» . a
 — sm (1)
41120 R I
1 A . .
= 4“ f (Sine — sm(—0» For a correct expression for the electric field:
A sine 2k x sin 6
OR —— 2.11:8 oR R
For correctly indicating the direction of the field Distribution
of Points 1 point 1 point
1 point 1 point
1 point Distribution
of Points
E & M 2. (a) v
i. lpomt The shuttle end is negative. 1 point
Due to the motion of the wire, the free electrons in it have a velocity to the right.
Using the righthand rule, the force on the electrons due to the magnetic ﬁeld is toward the shuttle.
ii. 2 points Electrons in the wire stop moving when the force due to the electric field created by the
differencein charge between the ends of the wire counteracts the magnetic force. qE = qu E = v8 The emf V( potential difference between the ends of the wire) is the electric ﬁeld multiplied by the length 3 of the wire. V = 128! 1 point  (7,600 m/s) (3.3 x 10“ T) (20 x 103 m)
V  5,016 V 1 point (b) 2 points For using Ohm’s Law: I = V/R lpoint
= (5,016 V)/(10,000 $2) I = 0.5016 A 1 point (C) .
1. 2p01nts For a correct expression for the magnetic force on a current: F  I d! x B 1 point
The current is perpendicular to the ﬁeld, so the magnitude of the force is: F  MB a (0.5016 A) (20 x 103 m) (3.3 x 10'5 T)
F  0.331 N 1 point (If a student used 20 m for the length of the tether instead of 20 x 103 m in both
(a) and (c), only one answer point was lost.) ii. 1 point By the right—hand rule, the direction of the magnetic force is opposite the
shuttle’s velocity. 1 point Distribution of Points
E & M 2. (cont)
(d) 4 points
The change in the shuttle’s energy is equal to the energy dissipated due to the
resistance of the wire.
AU  Pt ' 1 point
P  12R _ ' 1 point
AU=Pm
For correct substitution:
AU  (0.5016 A)2 (10,000 Q) (7 d) (24 h /d) (60 min/h) (60 5 /min) 1 point
AU  1.52 x 10° J lpoint
(Alternate solution) (Alternate points)
The change in energy equals the work done on the shuttle. .
AE  F d (1 point)
F d = th (1 point)
For correct substitution:
AE = (0.33 N)(7,600 m/s)(7 d)(24 hr/d)(60 min/hr)(60 s/ min) (1 point)
AE — 1.5 x 109 J I (1 point)
(e) 2 points
If current was forced to ﬂow the other way, the direction of the magnetic force
on the current would be reversed. This force would do work on the shuttle, and the
resulting gain of energy would cause an increase in the radius of the orbit (it would
“boost”the orbit). ‘
For correctly indicating that the radius would increase 1 point
For a correct explanation of the cause ‘ 1 point
For correct units in three answers and no incorrect units 1 point —————————————————____________—_—______—_ E&M3. (a) .
1. 4pomts For a correct expression for Ampere’s Law:
§B  d! = uoIm
For indicating the correct path of integration, either by drawing a circle on the ﬁgure or using 2m in Ampere’s Law
The enclosed current is the current per unit area for the inner conductor multiplied
by the area inside the distance r. 2
1m  1172 = 1’7
m2 0
Ir2
3(2117)  Ito—2"
a
u Ir
B . ° 2
2m
ii. 2 points
For a < r< b, the enclosed current is the total current in the inner conductor. I m a I
B (237R) = no]
pt I
B .. _°_
23R
(b) lpoint For r >c, the net current enclosed is zero.
Therefore, the ﬁeld is also zero. Distribution
of Points 1 point 1 point 1 point 1 point 1 point 1 point 1 point Distn'bution of Points
E & M 3. (cont)
(c) 4 points
0 a b i c
For a linear increase for O < r < a (or answer consistent with (a)i) 1 point
For a l/r decrease for a < r < b (or answer consistent with (a)ii) 1 point
For any decreasing function for b < r < c 1 point
For a zero function for r > c (or answer consistent with (b)) 1 point
A maximum of three points could be earned if the graph was discontinuous
(‘9 .
1. 2 pomts
For a correct expression for the magnetic force on a moving point charge:
FqvaorFaqu lpoint
Substituting the expression for B from (a)iz
qvp. I
F  ° 1 point
2:1:R
ii. 1 point Cross Section For correctly indicating the direction of the force 1 point Distribution of Points
E & M 3. (cont)
(e) lpoint
The answers to (d) would not change. 1 point Only the current inside the radius r has any effect on the magnetic ﬁeld and
hence the charge. M— ...
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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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