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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:) ...
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 Spring '09
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