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 Document
Unformatted text preview: Mechanics '1 Scoring Guide Solutions Note: For this salution, vectm quantities to the right are deﬁnﬁd as positiva.
(a) 3 points Far a correct exprcasion relating impulse and force
a 1:: 1M:
5 0R FA: F0: any indication that the impulse is the area "under: the curve I mag)(z x 30:“ MM x 10” s) + (2 x 103 NXZZ x 10” s) For the correct answar
I a 12 N w s (b) 3 points Far a caveat bxpressian relating impulxe and momentum
“z Imfmdv OR A}? 1;
F01" a correct axpmssien for the impuise in terms 0f the mass and 5;)ch ofme ball
I m 3719(1)“ w vb.) ! Sinca thE: bail is initially at rest, 1)“ == 0 .
Substituting and solving for v M: 1)“ == 1 /mb 1)“ = (MN ~ s)/(5 kg) For the correct answer
1) M a 2.4 mfg 8O Distribution
of points 1 point 1 point I. point 1 {mint 3; point 1 {mini Physics C m Mechanics Distribution Solutions
0f points (c)
i. 2 points For aither an cxpressien for the impulse cm the cube, with an indication
that the: value for the cube is equal and opposite that of the ball, or
conservation of mumcnmm I m mom“ w v“) and I: «~12N*s
OR mtvci 2"“ mbvbl + mtvaf
Selving for thﬁ 596% cf the, cube afmr the collisicm:
v” m {I + mg!) ,)/mg OR vﬁf : (mava  mbvwﬂmg (711 1 point map12 N . s + (as kg)(26 m/s)}/(0.5 kg) OR m [(05 WNW) m/s)  (Skg)(2.4 m/s)]/(0.5 kg) For a cormct answer 1)an “Mn/s
L 1 paint it I point The spew 0131311135! a‘beve has a positive: value, so the cube is moving
:9 the right. 1 point:
((1) 3 points Fm a ammo: axpression for the kinetic: energy dissipated I paint AKchi+KMK‘
1?}
For 21 came: general exprcssion for kinetic enargy: 1
K 2:: *2” mt)2 ipaim I ’2 ii 2
AK w "5(mcvﬂ + mb‘uM » may“) “W” ghmkgm mi 92 + <5 kg>c234 m M  (0.5 km miSﬂ Fm the: camera: answer 1 pwim AK 2:154} 81 Machanics 1 Scoring Guide Solutions (a) 3 points
Beth objects takt: the same amount of time to reach the ﬂoor. For the kinematic aquation fer the vertica} motion 1 2
37% vﬂté‘ Egt Th3 initial velocity i3 zeroﬁ Solving for t :
t w Jilly/g
a JE(1.2m)/(10m/sﬁ) a 402433 must Fm a cartect axpregsion for hcrizontai motion with no acceleration
x m vxt Applying this mutation to the two objects2: xb :2 (2411113)“).55) z 1.2m xc m (2 m!s)(0.5 3) == Tim Ax m 31,, w xc For the: correct answar
Ax : 02 m 82 Digtribuﬁon
of points 1 paint I paint 1 paint WV. Physics C m Mechanics Mechanics 2 ﬂeeting Guide Solutions Distribution
of points
(a)
i. 3 points
. . d U
The trummum occurs when  m 0.
er
For reaming that one must take the derivative of U 1 point
For getting the dcrivaxive equal m zero 1 point, Evaluating the darivative:
a? U a (  1 Mb (if rﬂ l? r2 =0 a ab w...sz 2
b r9 Solving for re:
.2. r .953.
b w r; 2 2
ramb For {ha comet answer "1 point
r0 m 19 ii. I paint The minimum potential energy is found by evaluating U (rO ). b b
UﬂwUUQ matz; + For the correct answer 1 point Uamla 93 Mechanics 2 Scoring Guide Salutions
(b) 3 points
. . w (W *
Uttmg the rclatmntahtp F r  W, the shape of the graph can be detertmnad
d U
from the gamma} expteﬁsimn for 72; found in part (ah, or qualitatively fmm the slope: of the U (I) graph. Fm a positive fame for F < r0 , smooth and momtonicany decreasing
as r increases, that times not cross the: y axis For a zero force at r m I}; , with the curve passing smhothly through
the): axitt Fat 3 negative force for r > I}; , smooth and monotonic, that approaches
a canstant value (In the abs/cum 0f credit fer the graph, indicating that F a « 7 mccived 1 point.) 94 Distribution
at“ points 1 point 1 point: 1 point Physics C m Mechanics
Solutions Distribution
of points
(0) 4 points
For indicating that energy is conserved 1 point
E m K +1] a constant
For recognizing that the initial kinetic energy is new 1 point
rO ‘ 1 2 = “amt? + (/05)
, r0 .
For evaluating U “f i paint
U .332; t. .2... 23?. .. 22.
:2. "" ‘1 2b + b " 2
Substituting:
.1. “Dz a» 3g.” 2 w E.
2 m “” 2 a "' 2
Solving fer the speed:
1) w J a/ m
1 point For using U€3 m Zn from part (a)ii, and substituting for a
v w «I UQ {’2m (ti) 2 points By conservatien of energy, the partieie wiii again come to rest (instantanenusiy
have new velocity) at a point r1 where its petential energy equate that at r0 / ’2,» r0 Utiri) : UL?) Au 2 paints (Some students interpreted “at rest” to mean permanently at rest. Fuli credit
was awarded for saying that the particle never comes to rest.) (e) 2 points The particle will cseillate, with the end points of the motion
being I'm/2 and my 1 point For indicating that the particie will csciiiate
1 point For indicating, imphcitiy 0r explicitly, that energy is not lost 95 Mechanics '3 Scoring Guide Solutians Distribution
of points
(a) 3 points
For a 00mm expression for centripetal acceleration 1 point;
2
1)
a a: «7 OR at) 2:“
For expressing v a: m in terms 0f given quantities 1 paint
in r 22:
u w W OR a) a 
T T
For the cormct angwer 1 paint,
4ﬁ2ﬂ
a; w W
T 4.4
(b) 4 points
The centripetal force, on star/1. is due to the gravitational force excnad
by star 3 .
Fm: a currect axpression for the oeniripetal force on star A 1 paint
Fa a Mia. OR M‘uf/r.
Fm: a comet expreszaion for the gravitational force:
(1354“?!b v
FG m 2 ! pmm:
r
F01" using r m r“ + rb 1 point
Fm setting F c and F 0 equal 1 point
41: 2Va GM bM a
M: T2 In 2
(r. + r13)
Sawing far Mb:
3, 2
M x 431: rn(r‘ + r93) b GT2 Solutions Distribution of points
(c) 2 points
The same calculatioas as above can be performed with the roles of star A
and star B switched.
For realization of this symmetry 1 point
For tho correct answer I point
415% (r. + rt)2
M m l
‘ GT
(Alternate solution} Qéltemare points)
Writing {ha expression for the center of mass with respect to point P :
 M at“ + M,m r" 0
Mi 4» Mb “
For the simpliﬁed version of this aquarion (1 point)
Ml rat a Mb r?)
r
Ma a M b “2.
rl
Substituting the expression for M b from part (b) to obtain the ﬁnal answer (1 point)
4:53:13 {rh + “)2
M: g x 2
GT
(:1) 3 points
I a 2 mi:2
For indicating that I 0‘ mrz 1 point
For knowing that in this case the constant of proportionality is 1
for each star. 1 point
For the corroct answer 1 point 2 2
I  MJ" + Marl, Mechanics» 3 Scoring Guide Snluﬁons Distribution
of points (6) 3 points For indicating that the tom! angular momentum is the sum of the individual stars’ momenta 1 point
L m L1I + Lb
For a correct general axpmssion for angular momentum 1 paint L w m1) 1' OR 10)
L a My.“ + Mbvbrb OR (My: +Mbr:)£o 2m. 2mb 23':
L m M‘ T r“ + MbTrb OR (Mari +Mbri)?
Fm Wk: 00mm anﬁwer I paint ‘21::
a ?(Miri +Mbr:) ...
View
Full Document
 Spring '09
 Park
 Physics

Click to edit the document details